category
stringclasses 107
values | title
stringlengths 15
179
| question_link
stringlengths 59
147
| question_body
stringlengths 53
33.8k
| answer_html
stringlengths 0
28.8k
| __index_level_0__
int64 0
1.58k
|
|---|---|---|---|---|---|
special relativity
|
Doppler redshift in special relativity
|
https://physics.stackexchange.com/questions/68132/doppler-redshift-in-special-relativity
|
<p>I came across this exercise in Elementary General Relativity by Alan MacDonald:</p>
<p>A source of light pulses moves with speed v directly away from an observer at rest in an inertial frame. Let $ \Delta t_e $ be the time between the emission of pulses, and $ \Delta t_o $ be the time between their reception at the observer. Show that $ \Delta t_o = \Delta t_e + v\Delta t_e $.</p>
<p>Based on my understanding of special relativity, the space-time interval between two events as measured from two inertial frames of reference should be the same. Therefore,
$$ \Delta t_e^2 = \Delta t_o^2 - \Delta x^2 $$
$$ \implies \Delta t_e^2 = \Delta t_o^2 - v^2\Delta t_o^2 $$
$$ \implies \Delta t_o = (1 - v^2)^{-1/2}\Delta t_e $$</p>
<p>which is not the same relation. What is wrong with my reasoning?</p>
|
<p>Your answer is right assuming $\Delta t_e$ is the interval between emission <em>as measured by the emitting source itself</em>. The given answer is right assuming $\Delta t_e$ is the time between emission <em>as measured by the observer</em>. It seems as though this problem is aiming at a lower level than your current understanding of relativity; you put too much thought into it.</p>
| 400
|
special relativity
|
Relativistic corrections to classical physics formulae
|
https://physics.stackexchange.com/questions/71943/relativistic-corrections-to-classical-physics-formulae
|
<p>How are classical formulas in physics (such as p = mv, or kinetic energy, or maxwell distribution of speeds) treated with the appropriate relativistic correction/modification? Is it done by using the Lorentz transformation equations? Could anyone give me a few examples of relativistic corrections to classical formulae? Thanks.</p>
|
<p>Alfred Centauri's wiki link certainly gives you a nice list of such things, but that entry doesn't exactly show you how such quantities reduce to their non-relativistic counterparts. </p>
<p>In particular, in the context of classical mechanics, one can generically show that relativistic expressions reduce to their non-relativistic counterparts in the limit of velocities $v$ that are small compared to the speed of light $c$.</p>
<p>The key to seeing this for many such quantities is to notice that the parameter $\gamma$ which appears everywhere in relativistic expressions is close to $1$ when $v\ll c$. Specifically, note that $\gamma$ is defined in terms of the ratio $\beta = v/c$ as follows:
\begin{align}
\gamma = \left(1-\beta^2\right)^{-1/2}
\end{align}
When $\beta$ is small, namely when we are doing physics for velocities $v$ much smaller than $c$, the following Taylor expansion of $\gamma$ about $\beta = 0$ becomes relevant;
\begin{align}
\gamma = 1 + \frac{1}{2}\beta^2+ O(\beta^4)
\end{align}
In particular, the first two terms are a very good approximation at low speeds. Here's an example:</p>
<p><strong>Kinetic Energy.</strong></p>
<p>The relativistic definition of the kinetic energy of a particle of mass $m>0$ is it's total energy minus its rest energy;
\begin{align}
K = E-mc^2
\end{align}
where the total energy is defined as
\begin{align}
E = \gamma mc^2
\end{align}
Combining these facts, we see that the kinetic energy is given by
\begin{align}
K = (\gamma - 1)mc^2 = \frac{1}{2}\beta^2 mc^2 + O(\beta^4) = \frac{1}{2}mv^2 + O(\beta^4)
\end{align}
where I used the Taylor expansion of $\gamma$ about $\beta = 0$ in the first step, and the definition of $\beta$ in the second step. In other words, to lowest non-vanishing order in $\beta$, the relativistic expression for the kinetic energy of a massive particle is the non-relativistic expression.</p>
| 401
|
special relativity
|
Minkowski Metric
|
https://physics.stackexchange.com/questions/76134/minkowski-metric
|
<p>Hi all I'm a little confused </p>
<p>So I have that in special relativity time is included as a coordinate so that in 1 spatial dimension we have 2 space time coordinates. The most basic metric is the Minkowski metric given by $\left[ {\begin{array}{*{20}{c}}
{ - 1}&0 \\
0&1
\end{array}} \right]$ so now if I have a vector $\left[ {\begin{array}{*{20}{c}}
1 \\
1
\end{array}} \right]$ and I use Einstein Summation Convention with the metric I get $-1 + 1 =0$ ... so what does this tell me about the length squared of the vector.. why is the $0$ significant ? </p>
<p>EDIT: I found out that if the length squared = $0$ the interval is then light-like and light-like particles have world-lines confined to the light cone and the square
of the separation of any two points on a light-like world line is zero. Is this correct ?</p>
<p>Help is very much appreciated</p>
<p>Thank you =)</p>
|
<p>We assume that we are working in units where the speed of light $c=1$. Given a vector with time and space components $A=(A^t, A^x)$, the Minkowski square of the vector is
\begin{align}
A^2 = -(A^t)^2+(A^x)^2
\end{align}
We can classify the vector $A$ according to the sign of its square length. In particular, in the given signature (with the minus sign in the time coordinate) we use the term <strong>timelike</strong> to refer to any vector with $A^2<0$, <strong>lightlike</strong> to refer to any vector with $A^2=0$, and <strong>spacelike</strong> to refer to any vector with $A^2>0$.</p>
<p>Therefore, the vector you wrote down is said to be lightlike.</p>
<p>One way of attaching physical significance to these terms is by considering trajectories of particles moving in spacetime. If a massive particle moves in spacetime, then the tangent vector to its trajectory will be timelike, while when a massless particle moves through spacetime, the tangent to its trajectory will be lightlike. In order for the tangent vector to a trajectory to be spacelike, it would have to be moving faster than light.</p>
<p>So the vector you wrote down could be the tangent vector to the path of a massless particle moving through spacetime, like a photon.</p>
| 402
|
special relativity
|
Is it possible to derive the invariant spacetime interval from Einstein's two postulates for SR?
|
https://physics.stackexchange.com/questions/76716/is-it-possible-to-derive-the-invariant-spacetime-interval-from-einsteins-two-po
|
<p>In many textbooks, the interval</p>
<p>$$
I = -(c\Delta t)^2 + (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2
$$</p>
<p>is taken for granted as the same for two events in any reference frame.</p>
<p>Is it possible to derive this just from the two postulates,</p>
<ol>
<li>That the laws of physics are the same in all reference frames</li>
<li>That the speed of light <em>c</em> is constant for all reference frames?</li>
</ol>
| 403
|
|
special relativity
|
Time dilation apeears in the both frame: Where is the problem?
|
https://physics.stackexchange.com/questions/83841/time-dilation-apeears-in-the-both-frame-where-is-the-problem
|
<p>Let we consider time in stationary frmae t, and respect the to stationary frame moving frame time $t^ \prime $ .</p>
<p>According to lorentz transformation,</p>
<p>$$t^\prime= \frac{t-\frac{vx}{c^2}}{\sqrt{1-\frac{v^2}{c^2}}}$$ and according to inverse Lorentz transformation, $$t= \frac{t^\prime +\frac{vx^\prime}{c^2}}{\sqrt{1-\frac{v^2}{c^2}}}$$</p>
<p>If we measure a event we have got,$$ t= t_2 -t_1$$
then, $$t= \frac{t_2^ \prime -t_2^ \prime }{\sqrt{1-\frac{v^2}{c^2}}}$$
$$t= \frac{t_0}{\sqrt{1-\frac{v^2}{c^2}}}$$</p>
<p>Therefore t is greather than $t_0$.</p>
<p>But if we calculate the time dilation in otherway, we have got,
$$t_0 = t_2^\prime -t_1^\prime $$</p>
<p>then, $$t_0= \frac{t_2 -t_2 }{\sqrt{1-\frac{v^2}{c^2}}}$$
$$t_0= \frac{t}{\sqrt{1-\frac{v^2}{c^2}}}$$
Therefore $t_0$ is greather than $t$.</p>
<p>I think I am doing wrong somewhere, because I have proven the stationary time longer and after some time I got the opposite. Can you please tell me that where did I get wrong?</p>
|
<p>For defining time dilation you first need to understand 'proper time interval'.
Proper time(denoted $\Delta t_0$) is time taken in an inertial frame in which two events occur at same position. This proper time interval is dilated in all other frames of reference. </p>
<p>Let two events occur at $(x_1,t_1)$ & $(x_2,t_2)$ in $S$ frame and they occur at $(x'_1,t'_1)$ & $(x'_2,t'_2)$ in $S'$ frame.</p>
<p>The Lorentz transforms for $x$ & $t$ can be written as </p>
<p>$x'=\gamma(x-vt)\\
t'=\gamma(t-\dfrac{vx}{c^2})$ </p>
<p>If $x_1=x_2$, </p>
<p>then using Lorentz transforms one can write, </p>
<p>$t'_2-t'_1=\Delta t'=\gamma(\Delta t)$</p>
<p>Which means the time interval will appear to be dilated in $S'$ frame. </p>
<p>Where as if you use inverse Lorentz transforms to go from $S'$ to $S$ then, </p>
<p>Notice $x'_2 \neq x'_1$ </p>
<p>Therefore,</p>
<p>$\Delta t=t_2-t_1 \neq \gamma \Delta t'$ because time interval $\Delta t'$ is not proper.</p>
<p>In this answer I have used $\gamma=\dfrac{1}{\sqrt{1-\frac{v^2}{c^2}}}$</p>
| 404
|
special relativity
|
Special Relativity and current in wire
|
https://physics.stackexchange.com/questions/83908/special-relativity-and-current-in-wire
|
<p>If I am a stationary observer and the electrons are moving relative to me,then shouldn't its density increase according to special relativity and thereby create an altogether negative net charge.</p>
|
<p>Increase in density doesn't mean that it creates negative net charge. The charge is preserved. </p>
| 405
|
special relativity
|
Speed of Light and Special Relativity
|
https://physics.stackexchange.com/questions/91344/speed-of-light-and-special-relativity
|
<p>If light is travelling vertically upwards and I am travelling horizontally and perpendicular to light, the velocity of light in the vertical direction will get reduced. How does this happen? How does motion in x-direction cause a reduction in speed of light in y-direction ?</p>
<p>Also, in Lorentz Transformation, suppose two observers measure a rod, can one of the observers see that the other person who is moving w.r.t. him, measure the rod to be less than his own measurement ? In other words, can the metre sticks of a moving observer get bigger than the observer who views the moving observer and sees himself to be at rest ?</p>
|
<p>You're asking so many questions that the only way to answer satisfactorily would be to completely rejustify special relativity. I suggest you take a look at a book like <a href="http://rads.stackoverflow.com/amzn/click/0393097935" rel="nofollow">Special Relativity</a> which does such a thing!</p>
<p>I'll try to answer your first question:</p>
<blockquote>
<p>How does this happen?</p>
</blockquote>
<p>If a photon is at position $(0,0,ct)$ at time $t$, that is, it has a spacetime event at time t of $e=(ct,0,0,ct)$, one has to apply the Lorentz Transform, which would yield $e'=(ct',x',y',z')=(\gamma c t,-\beta\gamma ct,0,ct)$. So $ct=ct'/\gamma$, and therefore:</p>
<p>$$e'=(c t',-\beta c t',0,ct'/\gamma)$$</p>
<p>Now, if we want to find the speed of the photon in this frame, that will be the magnitude of the spatial portion of the event $e'$, differentiated by $t'$ (and in this case, since everything is linear, that has the same effect as dividing by $t'$):</p>
<p>$$\|(-\beta c,0,c/\gamma)\|=\sqrt{c^2(\beta^2+(1-\beta^2))}=c$$</p>
<p>It moves at the speed of light, still. The reduction in the y-direction, in this frame of reference, is a consequence of the Lorentz transformation, which can be viewed as a consequence of the constancy of the speed of light. Of course you can notice that if the motion in the y direction <em>didn't</em> decrease, motion in the $x$ and $z$ directions would have to stay the same for the speed of light to stay constant, but this doesn't make sense, because it implies the beam of light is being dragged along with your reference frame.</p>
<blockquote>
<p>Also, in Lorentz Transformation, suppose two observers measure a rod,
can one of the observers see that the other person who is moving
w.r.t. him, measure the rod to be less than his own measurement ? In
other words, can the metre sticks of a moving observer get bigger than
the observer who views the moving observer and sees himself to be at
rest ?</p>
</blockquote>
<p>I'll limit my reply to this. Try to work it out from the definition of a Lorentz transformation. The answer to "can one of the observers see that the other person who is moving w.r.t. him, measure the rod to be less than his own measurement" is <em>yes</em>. In your next sentence, if you mean to imply that some observer measures a meter stick at longer than a meter, the answer is <em>no</em>. A meter stick at rest is a meter, and a meter stick flying by (along its axis) in either direction is slightly shorter than a meter, by a factor of $1/\gamma$.</p>
| 406
|
special relativity
|
How is length contraction reconciled with other objects occupying space?
|
https://physics.stackexchange.com/questions/91084/how-is-length-contraction-reconciled-with-other-objects-occupying-space
|
<p>Say I have a ball at 0.999999% the speed of light going past the Sun toward Earth. Now from the ball's reference frame, the distance between Earth and Sun is the same length as the ball's diameter. Why is the ball occupying the entire space between the Earth and Sun? What happened if a comet was between the Earth and Sun in the ball's path? Would the comet be inside the ball?!</p>
|
<p>You should check out the barn paradox! It's about the same thing.</p>
<p>The problem is that there's an extra effect in relativity you haven't accounted for: observers don't agree on the order of events. For example, in the earth frame, we may have the ordering</p>
<ol>
<li>Back of ball at sun</li>
<li>Something passes between sun and Earth</li>
<li>Front of ball at Earth</li>
</ol>
<p>In the ball's frame of reference, 1 and 3 happen at the same time. So it appears to mean that 2, which is between 1 and 3, must also happen at that time. But there's actually no such guarantee! The ordering may be</p>
<p>2'. Something passes between earth and sun</p>
<p>1'/3'. Ball occupies space between earth and sun</p>
<p>which resolves the paradox.</p>
| 407
|
special relativity
|
Synchronisation of clocks
|
https://physics.stackexchange.com/questions/93463/synchronisation-of-clocks
|
<p>How can two clocks be synchronised with each other at some instant without being at the same place and same time $?$ considering that simultaneity is a relative concept .</p>
|
<p>Here's the standard way in flat spacetime. Let's say you want to produce a synchronized pair of clocks that are a spatial distance $d$ away from one another, then perform the following steps:</p>
<ol>
<li><p>Construct two identical clocks such that they start ticking when they receive a special light signal. Call the clocks clock $1$ and clock $2$.</p></li>
<li><p>Before you engage either clock with the light pulse, set clock $1$ to time $0$ and clock $2$ to time $d/c$ where $c$ is the speed of light in vacuum.</p></li>
<li><p>Still before you engage either clock, move clock $2$ a spatial distance $d$ away from clock $1$.</p></li>
<li><p>Send out the special light signal from right next two clock $1$ toward clock $2$ so that it immediately starts clock $1$. </p></li>
</ol>
<p>The signal will reach clock $2$ at precisely the time $d/c$ later, so once clock $2$ is engaged, it will be synchronized with clock $1$ for all later times.</p>
| 408
|
special relativity
|
Moving Clocks Time Problem
|
https://physics.stackexchange.com/questions/93675/moving-clocks-time-problem
|
<p><a href="http://www.pitt.edu/~jdnorton/teaching/HPS_0410/chapters/Special_relativity_rel_sim/index.html" rel="nofollow">http://www.pitt.edu/~jdnorton/teaching/HPS_0410/chapters/Special_relativity_rel_sim/index.html</a></p>
<p>Talking about the situation of clocks shown on this page. Clocks A&B. Now suppose clock B is moving towards A with constant velocity. A sees B coming towards it, and figures out that the clock B is running slow, and by symmetry B says the same. Now, B & A meet at a space-time and actually compare readings. Now they will know one of the clocks ran slower. How is this possible</p>
| 409
|
|
special relativity
|
Relativity and speed of light again - two opposite light sources
|
https://physics.stackexchange.com/questions/94492/relativity-and-speed-of-light-again-two-opposite-light-sources
|
<p>Two light sources emit light at the same moment but in opposite directions. At what speed the distance between two light fronts is increasing? <strong>c</strong> or <strong>c</strong> * 2?</p>
<p>Note, that there is only one coordinate system here - a system, where these two light sources are placed and they don't move.</p>
|
<p>I think you are asking: If I turn on a lightbulb, I can imagine a sphere of light spreading radially outward from the bulb at the speed of light. How fast is the diameter of the sphere increasing in the lightbulb's frame? The answer is 2*c.</p>
| 410
|
special relativity
|
Inner product of four-vectors in special relativity
|
https://physics.stackexchange.com/questions/32762/inner-product-of-four-vectors-in-special-relativity
|
<p>Reference) "Feynman lectures on Physics Vol.3 , p.7-4 ."</p>
<p>With four vectors $x_{\mu} = (t,x,y,z)\ , \ p_{\mu} = (E,p_{x},p_{y},p_{z})$</p>
<p>the inner product of these two four vectors is scalar invariant and equals to $Et - \overrightarrow{p} \overrightarrow{x}$ . Alright.</p>
<p>But I cannot understand why $p_{\mu}x_{\mu}$ is just $Et$ in the rest frame as Feynman writes in the book above.</p>
|
<p>The value of $xp=x^\mu p_\mu=\eta^{\mu\nu}x_\mu p_\nu$ (with the sum over repreted indices) is invariant, i.e. if $A$ and $A'$ denote two frames, then the number given by the product $xy$ is the same as $x'y'$. </p>
<p>In one frame $A$, the expression $xp$ might evaluate to $Et-\vec x \vec p$, in another frame $A'$ where $x_0'\equiv t',\vec p' = 0$ (the rest frame) you'll have $x'p'=E't'-\vec x' \vec p'=E't'$. So the product can look like something which only contain energy and time this way.</p>
<p>By the invariance discussed in the first sentece, i.e. $xp=x'p'$, you have $Et-\vec x \vec p=E't'$. Not more, not less. In general $E't'\ne Et$, so probably it's not written explicitly that the energy and time variable on the other page really are these quantities in another frame.</p>
| 411
|
special relativity
|
Airplane example of special relativity
|
https://physics.stackexchange.com/questions/43258/airplane-example-of-special-relativity
|
<p>I'm struggling with an introductory example of special relativity. We haven't done the math yet so I would like an explanation based only on the fact that the speed of light is the same in every inertial frame.</p>
<p>An airplane travels east with a certain speed. There is a "clock" at both ends of the airplane. If there is a flash in the middle of the airplane and we're in the airplane's inertial frame, both clocks will register the flash at the same time, say 3.</p>
<p>But if we are an observer standing on the ground, in our inertial frame the light will reach the clock at the end of the airplane faster than the clock in the front. So the clock in the back may reads 3 when the clock in the front only reads 1, for example.</p>
<p>Question: Let's say the airplane has a speed v. The light's speed as it travels in the inertial frame of the observer towards the back will c+v, and the speed as it travels towards the front will be c-v. Let's say the distance from the middle to the back and front respectively is L. Then the time it takes for the light to travel from the middle to the back is L/(c+v) and to the front L/(c-v).</p>
<p>Shouldn't these be the same? If not, why can we say that the clocks will read 3 when the light hits them, no matter what frame is used? This is the argument the book uses to say that the clock must be less than 3, i.e. 1, in the front when the light hits the clock in the back in the observer's inertial frame.</p>
|
<p>You started by sentences that indicated that you realize that the speed of light is $c$ regardless of the speed of the source and the observer. However, you quickly wrote another sentence that indicated that the conclusion about knowledge in the previous sentence was an illusion:</p>
<blockquote>
<p>The light's speed as it travels in the inertial frame of the observer towards the back will $c+v$, and the speed as it travels towards the front will be $c-v$.</p>
</blockquote>
<p>No. This is not how it works according to special theory of relativity. The speed of light is always $c$, it is never $c+v$ or $c-v$. Because the light from the middle of the aircraft gets to the rear end of the aircraft earlier (because the rear end is moving towards the light while the front end tries to escape it and the light has the same speed in both directions), it just proves another fact of relativity, namely the "relativity of simultaneity". Two events that are simultaneous according to the aircraft's inertial system, $t_1=t_2$, are not simultaneous according to the inertial system attached to the Earth's surface: $t'_1\neq t'_2$.</p>
<p>Objects are contracted in the direction of the motion. The front-rear length of an object moving by speed $v$ is
$$ L = L_0 \cdot \sqrt{1-v^2/c^2}$$
where $L$ is the length of the object in its own rest frame. The formula to add velocities that replaces $u+v$ is
$$V_{\rm total} = \frac{u+v}{1+uv/c^2}$$
See <a href="https://physics.stackexchange.com/questions/23625/how-to-deduce-the-theorem-of-addition-of-velocities">two derivations</a> of this formula.</p>
<p>Check that with $u=c$, you get $V_{\rm total}=v$: the speed of light "added" to anything else in the relativistic way is still the speed of light. All these things may be derived from a more general <a href="http://en.wikipedia.org/wiki/Lorentz_transformation#Boost_in_the_x-direction" rel="nofollow noreferrer">Lorentz transformation</a> that maps the coordinates $(t,x,y,z)$ from one inertial system to another.</p>
| 412
|
special relativity
|
Special Relativity
|
https://physics.stackexchange.com/questions/45308/special-relativity
|
<p>Could someone explain to me how <a href="http://en.wikipedia.org/wiki/Special_relativity" rel="nofollow">special relativity</a> works?</p>
<p>I know there are thousands of sources and databases of knowledge out there, but I find it difficult to understand, even after reading up on those sources.</p>
<p>(Note: if you're an admin to close my question down, would you please be so nice as to point out something to help me with this question?)</p>
|
<p>I think I could give some intuitive look on SR. It is not very hard to understand the basic overview of SR. There are only <a href="http://en.wikipedia.org/wiki/Postulates_of_special_relativity" rel="nofollow noreferrer">two postulates</a> and not more that that..! But, there are many sites which provide a bit <em>wrong infos</em>. And, <em>Whooo</em> - found it... Your question is just <a href="https://physics.stackexchange.com/questions/31/lay-explanation-of-the-theory-of-relativity">a duplicate</a>.</p>
<p><strong>First postulate:</strong> "The laws of physics hold good (are the same) in all <em>inertial</em> frames of reference". </p>
<p>First of all, SR declares that all motions are relative. Mass, length, space & time, etc. are not independent but they are all <em>inter-dependent</em> according to Einstein's view which discarded <em>absolute space</em> and for now, it is treated <em>obsolete</em>.</p>
<p>Say for example (the <em>most basic</em> one) - You and your friend are traveling in parallel but exactly the opposite direction at speeds $v_1$ and $v_2$. Relative to you (you are a rest frame <em>now</em>), your friend would be past you at $v_1+v_2$. Both of you would experience some effects (mentioned below) and would also measure different distances, time, etc. The physical laws would be the same because you guys are in an uniform motion. If both were racing each other, one would measure the other's velocity as $v_1$ ~ $v_2$. And, another thing I'd like to note - If both travel near $c$, you'd have to take Lorentz factor $\gamma$ into account. So, you'd have $\Delta l,\Delta t, \Delta m$ and even relativistic acceleration. But, these are noticeable only to the worst cases (like above $0.5 c$).</p>
<p><strong>Second postulate:</strong> "The speed of light ($c$) is the same in all <em>inertial</em> frames of reference".</p>
<p>Wherever you both go, you guys will measure the speed of light to be the same value $c$ because you guys are still in inertial frame. This postulate is perhaps given a greater priority because it specifically says that information could not be transferred at velocities above $c$.</p>
<hr>
<p>Thus, SR concluded some new facts like <a href="http://en.wikipedia.org/wiki/Time_dilation" rel="nofollow noreferrer">slowing <em>time</em></a>, <a href="http://en.wikipedia.org/wiki/Length_contraction" rel="nofollow noreferrer">shorting length</a>, <a href="http://en.wikipedia.org/wiki/Relativistic_mass" rel="nofollow noreferrer">apparent mass</a>, Could <a href="http://en.wikipedia.org/wiki/Faster_than_light" rel="nofollow noreferrer">Faster than light</a> be possible?, <a href="http://en.wikipedia.org/wiki/Twin_paradox" rel="nofollow noreferrer">got stuck by a twin paradox</a>, the possibilities for <a href="http://en.wikipedia.org/wiki/Tachyons" rel="nofollow noreferrer">faster than light</a>, etc., etc... A great thing to note is - all these effects could be experienced by you <em>only</em> if you travel comparatively near $c$. Oh... And the most important ones - <strong><a href="http://en.wikipedia.org/wiki/Mass_energy_equivalence" rel="nofollow noreferrer">Mass-energy equivalence</a>, <a href="http://en.wikipedia.org/wiki/Minkowski_spacetime" rel="nofollow noreferrer">Space-time</a> and <a href="http://en.wikipedia.org/wiki/Lorentz_transformation" rel="nofollow noreferrer">Lorentz transformation</a></strong>.</p>
<hr>
<p>For more info, please refer the duplicate one. A best reference would be a <strong>Simulation</strong>. Once you have some <a href="http://www.astronomy.ohio-state.edu/~pogge/Ast162/Unit5/sr.html" rel="nofollow noreferrer">basic understanding on SR</a>, please see <a href="http://www.anu.edu.au/Physics/new/rtrDownload.php?rtrVersion=RTR_v1.6.0.msi" rel="nofollow noreferrer">Real Time Relativity</a>. It is totally amazing. Of course, I found it (some time ago) in Lubos' blog.</p>
<p><strong>Note:</strong> Gravity also affects objects. Actually, the effect of gravity on objects were generalized by Einstein to expand SR to <a href="http://en.wikipedia.org/wiki/General_relativity" rel="nofollow noreferrer"><strong>GR</strong></a> (took some years though) </p>
| 413
|
special relativity
|
what is use of relativistic action?
|
https://physics.stackexchange.com/questions/52122/what-is-use-of-relativistic-action
|
<p>this is relativistic action:
$$S=\int_C \mathcal {L}dt$$
where the $\mathcal{L}$ is $-m_oc^2\gamma^{-1}$
what is use of relativistic action!?</p>
|
<p>The use of an action is do derive all the dynamical equations of a theory from the least-action principle, $\delta S=0$ (action is minimized along the right path). Quantum mechanically, the use of an action is to derive the transition amplitudes from an initial state to the final state by summing over histories weighted by $\exp(iS/\hbar)$ in a quantum theory.</p>
<p>Similarly. The use of a relativistic action is do derive all the dynamical equations of a relativistic theory from the least-action principle, $\delta S=0$ (action is minimized along the right path). Quantum mechanically, the use of a relativistic action is to derive the transition amplitudes from an initial state to the final state by summing over histories weighted by $\exp(iS/\hbar)$ in a relativistic quantum theory.</p>
<p>Yes, I used the copy-and-paste and added the adjective "relativistic" to the second paragraph. ;-)</p>
| 414
|
special relativity
|
Why does 'proper length' exist as a notion?
|
https://physics.stackexchange.com/questions/52912/why-does-proper-length-exist-as-a-notion
|
<p>$$\text{proper time}= \tau= \sqrt{dt^2-d\mathbf{s}^2}$$</p>
<p>$$\text{proper length}= L= \sqrt{-dt^2+d\mathbf{s}^2}$$</p>
<p>What tangible benefit is brought about by calling $i \tau$ 'proper length' (applying when $\Im(L)=0$ (the spacetime intervals are spacelike))?</p>
<p>Could one extend the notion of the interval to simulatneusly cover both proper time and distance? Call this $\text{Interval}=dt^2-d\mathbf{s}^2$, <em>which can be imaginary</em>. Is it impossible to merge proper time and length because they are fundamentally different things, as proper time is a explicitly measurable quantity, but proper distance must be determined implicitly?</p>
<p>My problem is that I don't understand the rationale behind having <a href="http://en.wikipedia.org/wiki/Proper_length" rel="nofollow">proper length</a>, rather than a cheap mathematical trick to make the interval real when $ dt^2-d\mathbf{s}^2<0$.</p>
|
<p>This isn't complex analysis. There is no reason to complexify anything. These notions do not both exist for a single pair of events. Either two events are timelike separated from one another, and there is a proper time between them that all observers agree on, or the two events are spacelike separated, and there is a proper distance between them that all observers agree on.</p>
<p>This is just a fundamental result of having a metric isn't positive definite. The invariant interval between the two events tells us whether they are spacelike or timelike separated, sure, and that's why proper time and proper distance are closely related, but they are not the same thing, and I think a complexification just confuses things.</p>
<p>Edit: the proper distance between two events is the distance that would be measured between them in the frame in which they are simultaneous. This is also called proper length because it measures the length of an object in its own rest frame.</p>
| 415
|
special relativity
|
What are the uses of proper length as a parameter?
|
https://physics.stackexchange.com/questions/55007/what-are-the-uses-of-proper-length-as-a-parameter
|
<p>Proper time is used to parameterize the world line of a moving particle in a way which is Lorentz invariant, which is elegant and powerful. Since space and time are usually treated on the same footing, I'd expect proper length to be a powerful parameter of some sort, yet I've never come across it being used in this way.</p>
<p>What are the uses of Proper length as a parameter?</p>
|
<p>Proper distance and proper time are both really the calculation of the metric disitance between two events. In general relativity the infinitesimal metric is:</p>
<p>$ds^2 = g_{\mu\nu}dx^\mu dx^\nu$</p>
<p>In special relativity (with units where $c=1$ ), space-time is flat so the Cartesian Minkowski metric between two points is:</p>
<p>$\Delta s^2 = \Delta t^2 -\Delta x^2 -\Delta y^2 -\Delta z^2 $</p>
<p>In general, when $c=1$, the proper time, $\Delta \tau$, and the proper distance, $\Delta L$, are related to the metric by:</p>
<p>$\Delta L^2 = - \Delta \tau^2 = - \Delta s^2$</p>
<p>So for two points that have a time-like separation, such as two points on the worldline of a massive particle the proper time will be a real positive number and is just this metric rewritten as follows:</p>
<p>$\Delta \tau = \sqrt{ \Delta t^2 -(\Delta x^2 +\Delta y^2 +\Delta z^2)/c^2 }$</p>
<p>Where I have put the speed of light in explicitly to make it clear that the proper time has the dimensions of time. Note that all observers will agree on the value of the proper time between two events on the worldline of a massive particle.</p>
<p>Now for two points that are at a space like separation this proper time would become imaginary so it is customary to rewrite it as a real positive proper distance as follows:</p>
<p>$\Delta L = \sqrt{\Delta x^2 +\Delta y^2 +\Delta z^2 - \Delta t^2 c^2 }$</p>
<p>Where, again, I have put the speed of light in explicitly to make it clear that the proper distance has the dimensions of spatial distance not time. if tachyons existed, this would be the proper distance that all observers would agree is the proper distance between two events on the world line of a tachyon. Even without tachyons, the proper distance between two space-like separated events is the spatial distance between those events in a reference frame where they are measured to be simultaneous (for space-like separated events there is always a frame where they will be simultaneous - this the relativity of simultaneity).</p>
<p>Note that for massless particles, such as photons, $\Delta \tau = \Delta L = 0$.</p>
<p>So there really isn't any fundamental distinction between proper time and proper distance - they are both different ways of representing the same metric which happens to give you a real number for events on either time-like or space-like event separations respectively.</p>
| 416
|
special relativity
|
Does an accelerating spaceship move backwards due to length contraction?
|
https://physics.stackexchange.com/questions/62008/does-an-accelerating-spaceship-move-backwards-due-to-length-contraction
|
<p>Let's assume I have a spaceship in front of me let's say at 1000000km distance. Now let's assume I have also a stationary wall just behind the spaceship at 999999km. Initially the spaceship's speed is 0.</p>
<p>Now let's accelerate the spaceship rapidly to relativistic speeds.</p>
<p>Due to the length contraction the spaceship and it's distance from me contracts. </p>
<p>Now I see a paradoxical situation: I and the wall is in the same frame of reference. From the wall's point of view the spaceship come closer to the wall momentarily and zips away. From my point of view the spaceship can come closer to me than the wall (due to length contraction rate ), effectively smacking into the wall during the acceleration.</p>
<p>How to resolve this apparent paradox?</p>
<p>(or where is the origin of the contraction?)</p>
|
<p>The paradox of your situation lies in which frame of reference you're using. In reality, there is no paradox. It seems as though you have a decent grasp on relativistic effects, so I'll spare you the gory details. However, you problem is that in your and the wall's frame, the distance between you and the spaceship won't change.</p>
<p>Relativistic length contraction only affects things that are travelling at relativistic speeds in your frame. So in your frame, it is the spaceship that will contract, you will see it as being shorter, but the location of the ship won't change initially (assuming instantaneous acceleration). The same is true from the wall's perspective. What I believe you were thinking of is that in the spaceship's frame of reference, the distance between you and it will contract.</p>
<p>This doesn't present a problem though, because while the spaceship sees the distance between it and the wall shrink, the ratio of lengths between it to the wall and it to you remains the same. So no, it does not move backwards.</p>
| 417
|
special relativity
|
Faraday tensor, antisymmetric electromagnetic tensor
|
https://physics.stackexchange.com/questions/62309/faraday-tensor-antisymmetric-electromagnetic-tensor
|
<p>I want to write $F^{\mu \nu}F_{\mu \nu}$ in terms of $F_{\mu \nu}F^{\mu \nu}$. How to do it?</p>
|
<p>The two expressions you wrote down are the same because for each pair $(\mu,\nu)$, the quantities $F^{\mu\nu}$ and $F_{\mu\nu}$ are real numbers and can be commuted past one another.</p>
| 418
|
special relativity
|
Doppler shift of radio signals to an accelerating observer
|
https://physics.stackexchange.com/questions/62851/doppler-shift-of-radio-signals-to-an-accelerating-observer
|
<p>Suppose a man leaves from Earth to a star which is 1000 light years away. He accelerates to a velocity such that the entire trip lasts a year, from the reference frame of the rocket.</p>
<p>Now lets pretend the person in the rocket wants to have a transmission of the radio to him.</p>
<p>Due to time dilation, when a day passes on earth only a few seconds pass on the rocket ship, so from the travellers frame of reference, as he accelerates the frequency of transmissions goes up. </p>
<p>As he arrives at his destination the frequency of transmissions should go down.</p>
<p>Is this correct?</p>
|
<p>You are <em>not quite</em> correct (see edits). Except I wouldn't say that "in reality it takes just slightly over 1000 years" - the rocket frame is no less real than the Earth's frame. As far as the actual numbers go: at one gee acceleration it takes about a year in the rocket frame to accelerate, and a year again to deccelerate, so two rocket frame years altogether. You can achieve a one year trip at a higher acceleration but your passenger might feel squished. :)</p>
<p>Edit: Oops, read it wrong. From the rocket frame he receives <strong>fewer</strong> messages per second, not more, since he's moving <em>away</em> from the Earth. I'll put up a spacetime diagram to illustrate this later. Lagerbaer gets it right in his answer.</p>
<p>Edit2: Okay. Here it is:</p>
<p><img src="https://i.sstatic.net/JteFg.png" alt="enter image description here"></p>
<p>I made it a shorter 2 ly trip than your 1000 ly one just so we can see what's happening on the plot. Nothing essential changes because of this, things are just easier to see. The blue curve is the rocket accelerating away from the Earth at $x=0$ until it reaches the halfway point and begins deccelerating. The total proper time for the rocket is 1 year, but about 2.4 yr elapses on the Earth.</p>
<p>The red dashed lines are regular messages sent from Earth at 0.05 yr intervals. Notice that in the middle of the journey the ship receives very few messages. Here are the arrival times:</p>
<p>$$
\begin{array}{cc}
\text{Earth time signal sent} & \text{Rocket time signal received} \\
0. & 0 \\
0.05 & 0.0574144 \\
0.1 & 0.137814 \\
0.15 & 0.273035 \\
0.2 & 0.623843 \\
0.25 & 0.815773 \\
0.3 & 0.912568 \\
0.35 & 0.977809 \\
\end{array}
$$</p>
<p>Note that between the fourth and fifth messages 0.35 yr elapses in the rocket frame, compared to the 0.05 yr between them in the Earth frame!</p>
<p>Spacetime diagrams like this are the only way to get intuition about relativity. Learn to love them. :)</p>
<hr>
<p>Edit3: With a minor tweak of the code I can run the numbers for your design journey - 1000 lyr in a proper time of one year. The plot is unreadable, but the stats are:</p>
<p>Acceleration: $20\ \mathrm{g}$</p>
<p>Earth time: $1000.1\ \mathrm{yr}$</p>
<p>Max speed: $0.999999995\ c$</p>
<p>$$
\begin{array}{cc}
\text{Earth time signal sent} & \text{Rocket time signal received} \\
0. & 0 \\
0.01 & 0.0111428 \\
0.02 & 0.0254581 \\
0.03 & 0.0455048 \\
0.04 & 0.079222 \\
0.05 & 0.229245 \\
0.06 & 0.915324 \\
0.07 & 0.95177 \\
0.08 & 0.972725 \\
0.09 & 0.987495 \\
0.1 & 0.99891 \\
\end{array}
$$</p>
| 419
|
special relativity
|
Speed of Light and Information
|
https://physics.stackexchange.com/questions/64419/speed-of-light-and-information
|
<p>Einstein in his thought experiment(?) for the constancy of speed of light in vacuum in all frames reasoned , that if speed of light of vacuum isn't constant than you'll be able to perceive effect before action .</p>
<p>But then this must happen in all mediums , but it doesn't ,that means if you live in a medium with high optical density , you can perceive effect before action ? How can this be justified in a medium but not in vacuum ?</p>
<p>And supposedly if we weren't able to see and sound was our only source to get information , then shouldn't speed of sound too be a cosmic speed limit ?</p>
|
<blockquote>
<p>Einstein in his thought experiment(?) for the constancy of speed of light in vacuum in all frames reasoned , that if speed of light of vacuum isn't constant than you'll be able to perceive effect before action .</p>
</blockquote>
<p>Hm, as far as I know his reason for saying that the speed of light is constant is the following:</p>
<p>The speed of light in a vacuum is something that comes out mathematically from Maxwell's equations.</p>
<p>Einstein assumed that physics is beautiful and got rid of the privileged (aether) frame, and then took the following situation:</p>
<p>Alice and Bob are in identical boxes floating around in space, with Bob's box moving at $v$ relative to Alice, with no external (or cross-) influences. Both of their boxes contain world-class all-purpose physics laboratories (in technical terms: a reference frame). If the privileged frame existed, then <em>even though all other experiments show that the boxes are subjected to the same physical conditions</em>, any experiment made to measure the speed of light (this can be easily done by tossing a bunch of electrons around) would give different results. However, this doesn't sound quite right, since $c$ is something that pops out of Maxwell's equations--it's a fundamental constant (unlike the speed of sound, which depends on the circumstances leading to the sound). If the privileged frame held true, then Maxwell's equations would get ugly for moving frames.</p>
<p>Also, the "ugly" aether-versions of Maxwell's equations didn't hold up to experiments like the Michelson-Morley experiment.</p>
<p>Note that this does not apply for light in a medium. When light travels in a medium, Alice and Bob can detect a difference in their situations as they can measure the relative speed of the medium (which differs by $v$ from their frames).</p>
<p>The time dilation bit came after this postulate as far as I can tell, and depends on it. The reasoning in your first sentence("you'll be able to perceive effect before action .") can't be made unless you assume that simultaneity is frame dependent, and for that you need the framework of special relativity, and for that you need the speed of light to be constant.</p>
<hr>
<p>Side note: In any medium, the maximum speed of any readable communication of information is $c$. Particles going faster than light in a medium exist (for example, cooling moderator rods from a nuclear power plant kept in water emit faster-than-light-in-water neutrons, which emit <a href="https://en.wikipedia.org/wiki/Cherenkov_radiation" rel="nofollow">Cherenkov radiation</a> (a sort of "sonic boom for light")</p>
| 420
|
special relativity
|
A video conference between earth and a space shuttle
|
https://physics.stackexchange.com/questions/8539/a-video-conference-between-earth-and-a-space-shuttle
|
<p>I have just started looking into special relativity and I have come up with an intriguing <em>gedanke</em>, as Einstein himself called such theoretical thought experiments.</p>
<p>Imagine a space shuttle traveling through space at a constant velocity close to $c$. As the shuttle passes earth, a previously set-up camera starts broadcasting from earth to the shuttle. Since radio waves travel at the speed of light, the shuttle is receiving a constant transmission feed, assuming the camera is broadcasting 24/7.</p>
<p>Now, from what I have understood of special relativity so far, time will flow slower for the astronaut than for the earthlings. Hence, assuming $v=0.8c$, the astronaut will after 30 years have received a video transmission 50 years long!</p>
<p>Is my reasoning correct, that even though the transmission is live, the astronaut would actually be watching things that happened many years ago, while still receiving the "live" feed, which would be stored/buffered in the shuttles memory, thus making it possible for the astronaut to fast-forward the clip to see what happened more than 30 years after passing the earth?</p>
<p>My second question is, what happens when we consider the space shuttle to be at rest and the earth to be moving instead? If that would imply that it has been 50 years from the astronauts point of view, while only 30 years have passed on earth, then the astronaut would run out of video material after the first 30 years of watching the broadcast. Then what?</p>
<p>I hope it makes sense. Thank you!</p>
|
<p>The astronaut will not have received 50 years worth of transmissions at the time you specify. </p>
<p>Let's start by stating precisely what time dilation means in this instance. Consider two "events": </p>
<ol>
<li>The people on Earth have a party to celebrate 50 years of radio broadcasting.</li>
<li>The astronaut on board the ship has a party to celebrate 30 years of travel.</li>
</ol>
<p>In a reference frame in which the Earth is at rest, those two events are simultaneous. That's what we mean when we say that the astronaut's clocks run slow.</p>
<p>But the astronaut won't receive the 50th year of radio broadcasting until much later -- to be precise, 200 more years of Earth-time will elapse before he receives this signal. (Let's check this: in 200 years at $0.8c$, he will travel 160 light-years. Add that to the 40 light-years he's already gone, and you find that he's 200 light-years from Earth at the time. The radio signal will just be reaching that distance at that time.)</p>
<p>Of course, because of time dilation, that's only 120 additional years of astronaut time. Still, it means the astronaut can't see into the future.</p>
<p>To say it another way, at the moment in question (when the astronaut throws his 30-year party), the radio signal he is receiving is one that left Earth merely 10 years after he passed Earth. In Earth's reference frame, that radio signal traveled a distance of 40 light-years (since that's how far away the astronaut is at that moment), and took 40 years to do it (50 years minus 10 years).</p>
| 421
|
special relativity
|
Physical interpretation of equation for relativistic aberration
|
https://physics.stackexchange.com/questions/9507/physical-interpretation-of-equation-for-relativistic-aberration
|
<p>I'm working in a book on relativity. The author states that if $u$ and $u'$ are a velocity referred to two inertial frames with relative velocity $v$ confined to the $x$ axis, then the quantities $l$, $m$, $n$ defined by </p>
<p>$$
(l, m, n) = \frac{1}{|u|}(u_x, u_y, u_z)
$$</p>
<p>and</p>
<p>$$
(l', m', n') = \frac{1}{|u'|}(u'_x, u'_y, u'_z)
$$</p>
<p>are related by </p>
<p>$$
(l', m', n') = \frac{1}{D}(l - \frac{v}{u}, m\gamma ^{-1}, n\gamma ^{-1})
$$</p>
<p>and that this can be considered a relativistic aberration formula.
The author gives the following definition for $D$, copied verbatim. </p>
<p>$$
D = \frac{u'}{u}\left( 1 - \frac{u_xv}{c^2}\right) = \left[1 - 2l\frac{v}{u} + \frac{v^2}{u^2} - (1 - l^2)\frac{v^2}{c^2} \right]^{\frac{1}{2}}
$$</p>
<p>My question is why the author finessess the expression into the third/final form. I was able to get it but it seems like a pain. Why is that better than the second expression? It seems more difficult to calculate and not at all clear or meaningful.</p>
<hr>
<p>Also, in case it's not clear, $\gamma =1/ \sqrt{1 - \frac{v^2}{c^2}}$ and $|u| = |(u_x, u_y, u_z)| = \sqrt{u_x^2 + u_y^2 + u_z^2}$ </p>
|
<p>The last form avoids any usage of $u'$ as well as $u_x$ which may be considered derived or non-covariant quantities, respectively, so the last form may be superior in many contexts. Moreover, it makes it much easier to imagine how far the result is from one - because it resembles a Taylor expansion of a sort.</p>
<p>But even if one didn't write the comments above, why would it be a problem to write a result in yet another mathematically equivalent way?</p>
| 422
|
special relativity
|
How is energy conserved when a moving charge has false ideas about positions of other charges
|
https://physics.stackexchange.com/questions/11075/how-is-energy-conserved-when-a-moving-charge-has-false-ideas-about-positions-of
|
<p>An electron is shot towards a target that is negatively charged. While the electron is traveling, the target makes an abrupt move towards the electron. While the information that the target moved is traveling from the target to the electron, the electron behaves like an electron that is moving towards a target that is in the original position.</p>
<p>How can energy be conserved when an electron that is moving towards a nearby charge behaves like it was moving towards a far away charge? Seems we end up with electron being at 2 meters distance from the target, while the electron had enough energy to travel to at most 4 meters distance from the target.</p>
<p>It also seems to me that "moving the target requires energy" is not a solution to this problem.</p>
|
<p>Moving the target requires energy.</p>
<p>Suppose the target is an infinite plane with constant charge density $\sigma$. It will not radiate when you move it because the electric field is constant everywhere. Suppose the test charge $q$ is small enough that its radiation is negligible.</p>
<p>The electric field of the plane is $2\pi\sigma$ in the direction perpendicular to the plane.</p>
<p>The charge begins a distance $d$ from the plane. The potential energy in the system is $-2\pi\sigma q d$ (define to be zero when $d=0$). The force on the charge is $2\pi\sigma q$ and by Newton's third law there is an equal and opposite force on the plane. (We are assuming there is no radiation, so momentum of the charge carriers is conserved, and Newton's third law holds.)</p>
<p>The charge also has some kinetic energy $T$.</p>
<p>We move the plane towards the charge a distance $\Delta d$. This takes energy because the force of the charge on the plane does negative work. The energy required is the force multiplied by the distance, or $2\pi\sigma q \Delta d$.</p>
<p>The new distance of the charge from the plane is $d - \Delta d$, so the new potential energy is $-2\pi\sigma q (d-\Delta d)$. The potential energy has increased by $2\pi\sigma q \Delta d$, exactly the amount of work that had to be put into the system to move the plane. The charge still has kinetic energy $T$, so energy is conserved in that the change in energy of the system is equal to the energy that was used to move the plane.</p>
| 423
|
special relativity
|
Non-interchangeability of time-like intervals
|
https://physics.stackexchange.com/questions/12265/non-interchangeability-of-time-like-intervals
|
<p>I am reading Landau's Volume 2 of the course of theoretical physics. I have a doubt after reading the first few pages of it which I explain below.</p>
<p>Landau first defines intervals and on pages 5 and 6 shows that two events having time like interval between them can never occur simultaneously in any reference system. Then he goes on to construct a 2D space-time graph (for visualization) with an event O occurring at (0,0,0,0). Then he considers any event which occurs in future in that frame and is time-like w.r.t. origin and says on page no. 7, </p>
<blockquote>
<p>But two events which are separated by a time-like interval cannot
occur simultaneously in any reference system. Consequently, it is
impossible to find a reference frame in which any of the events in
region aOc occurred "before" the event O, i.e. at time t<0.</p>
</blockquote>
<p>The argument above just proves that because interval square should be positive, i.e. the events can't be simultaneous. But, if I replace the difference in time in the original frame with its negative in my proposed frame and let the space distance between them to be same in both frames, then I get an in my proposed frame an interval which is time like but in it the order of events is changed. Am I making some gross error or Landau has missed some argument?</p>
|
<p>If you believe that (a) timelike separated events cannot be simultaneous in any reference frame, <em>and</em> (b) the set of inertial frames is (in some appropriate sense) a continuous set, then L&L's conclusion follows. After all, if there were two frames in which the order of two timelike separated events differed, then by continuously transforming one frame into another, you could find one in which they were simultaneous.</p>
<p>But without some such additional assumption, you're right that the conclusion doesn't logically follow. There are coordinate systems that preserve the spacetime interval but flip the direction of time, such as the substitution $t\to -t$ that you mention. As BebopbutUnsteady observes in a comment to Karsus Ren's answer, we often use the term "orthochronous Lorentz transformation" to refer to a transformation that preserves the direction of time. The full group of Lorentz transformations (i.e., of all transformations that preserves the interval) includes both orthochronous and non-orthochronous components, which are not connected to each other. Physically, we usually only consider the orthochronous ones. </p>
<p>You do have to be careful with the terminology: sometimes people use "Lorentz transformation" to mean just the orthochronous ones; sometimes it's the full group.</p>
<p>By the way, pretty much the same thing applies to spatial reflections: is $x\to -x$ (leaving $y,z,t$ unchanged) a Lorentz transformation? After all, it preserves the spacetime interval. Often we refer to non-reflecting Lorentz transformations as "proper." So when people are being careful with their terminology they often refer to "proper orthochronous Lorentz transformations."</p>
| 424
|
special relativity
|
Looking for specific Relativity example
|
https://physics.stackexchange.com/questions/14958/looking-for-specific-relativity-example
|
<p>Many years ago (in the '70s I think) I read an explanation of the meaninglessness of simultaneity at large distances. The example had to do with two people walking along a sidewalk in opposite directions, and an alien race on a planet millions of light-years away planning an invasion of the Solar System. The example showed that in one walker's reference frame the invasion fleet had departed, but in the other reference frame the fleet had not. </p>
<p>At the time, the explanation made perfect sense, but I have forgotten the details and have never run across this example again.</p>
<p>Does anybody know where this was, or have the text of the explanation?</p>
|
<p>I think you are talking about the <a href="http://en.wikipedia.org/wiki/Rietdijk%E2%80%93Putnam_argument" rel="nofollow">Rietdjik-Putnam argument</a>.</p>
| 425
|
special relativity
|
Relative Synchronicity
|
https://physics.stackexchange.com/questions/15906/relative-synchronicity
|
<p>Einstein said that the synchronization of two clocks is dependant on the velocity of the observer. But I feel a conceptual contradiction can be made:</p>
<p>There are two observers A and B. Observer 'A' faces direction X, and will be labeled "stationary." Another observer B faces direction X and is travelling rapidly in that direction as well on a collision course with 'A'. Both observers are holding two clocks; One in each hand, with hands held perpendicular to direction X. Both observers hit a "synchronization" button on the clocks before colliding.</p>
<p>My expectation in this case is that when the moving observer B halts to greet observer A - both observers will agree the B-pair clocks are synchronized and the A-pair clocks are synchronized (though all four clocks are not necessarily synchronized with eachother, and is not at issue in this question).</p>
<p>Bottom line: what was considered "synchronicity" by the speedy individual is accepted by the stationary individual. This seems to contradict special relativity.</p>
<p>Now if my scenario were modified slightly, then I believe special relativity would apply. If observer A were to hold the A-pair clocks with one held out before herself and the other held out behind herself (instead of the original left and right perpendicular angles), then finally I would expect there to be disagreement between the two observers when they stop to meet eachother.</p>
<p>All this suggests that position is an unaccounted for primary component of relativity, but from what little I know of SR, it doesn't hardly factor in at all. Can someone explain what I'm missing?</p>
|
<p>I don't think you're missing anything. If the two clocks are not separated in the dimension of the velocity of the systems (say this is the x dimension), then they are synchronized with respect to both systems. Try thinking about light clocks held this way. In both systems, light takes the same time to go from B to his left-hand clock back to B as it would for light to go from B to the right-hand clock and back to B. Thus, the clocks are synchronized in both systems.
Summary: This does not contradict the two axioms of special relativity.</p>
| 426
|
special relativity
|
why evaluate at lambda = 0
|
https://physics.stackexchange.com/questions/16432/why-evaluate-at-lambda-0
|
<p>I am trying to understand Herbert Goldstein's introduction to 4-vectors. He describes a 1-D curve in spacetime $ P_(\lambda) $ then he says a 4 vector is defined as the tangent vector to this curve $$ v = \biggr ( \frac {dP} {d\lambda}\biggr)_{\lambda =0} $$ </p>
<p>why is $ \lambda $=0? what does that have to do with anything? I have been staring at this for like 20 minutes I still don't understand what he is talking about... it's giving me problems because i need to understand this part later because it is relevant to how tensors transform
also he says $ \lambda $ is a measure of a length along the curve... i don't really follow that point either... i though $ \lambda $ could be any parameter like proper time etc.</p>
<p>any help on this??</p>
|
<blockquote>
<p>he says a 4 vector is defined as the tangent vector to this curve</p>
</blockquote>
<p>That is <em>not</em> true in general. A four-vector is not always defined as the tangent vector to a curve. In the book they are computing a tangent vector to a curve in 3+1D spacetime; the tangent vector is just <em>one example</em> of a four-vector.</p>
<p>In particular, the formula given tells you how to compute the tangent vector at a specific point $\mathcal{A}$. Since the curve runs from $\mathcal{A}$ to $\mathcal{B}$ and is parametrized by $\lambda \in [0,1]$, $\lambda = 0$ is the value which corresponds to the point $\mathcal{A}$. So if you're going to define the tangent vector at $\mathcal{A}$, you need to set $\lambda = 0$.</p>
| 427
|
special relativity
|
Lorentz Transformation via Geometry
|
https://physics.stackexchange.com/questions/16845/lorentz-transformation-via-geometry
|
<p>Today, I was tutoring and explained the space-time. I explained how one can convert North-South into West-East by rotating, and how you can convert time into space with velocity.</p>
<p>Below the Energy-Momentum stuff the book had some problems. One was the following:</p>
<blockquote>
<p>Given an event at (1 Lightsecond, 2 s), give it's coordinates in a frame of reference that is moving with 0.6 of speed of light to it.</p>
</blockquote>
<p>This the diagram I came up with, in blue there is the event. (Not really to scale.)</p>
<p><a href="http://wstaw.org/m/2011/11/11/m8.png" rel="noreferrer">http://wstaw.org/m/2011/11/11/m8.png</a></p>
<p>Then I projected the point onto the orange axes, and I get smaller values on each. This makes sense to me, as a moving observer would see thing smaller and slower, resulting in less values in either measurement.</p>
<p>But how do I get the orange lines? I cannot see how I get this angle. If I got them, I could tell you the speed of the reference frame, but the speed is given here …</p>
<p>And did I do the projection of the speed right?</p>
|
<p>I'm going to take $c=1$. The top orange line, which is the $t'$ axis in your drawing, is drawn with a slope of $1/0.6$; this follows from the definition of velocity. The bottom orange line, which is the $x'$ axis, has a slope of 0.6.</p>
<p>What's a little harder is to get the scale on the orange axes. The scale follows from the fact that area is preserved by Lorentz transformations. The velocity of 0.6 turns out to correspond to a Lorentz transformation in which a square is distorted into a parallelogram with its long axis stretched by a factor of 2, and its short axis contracted by $1/2$.</p>
<p>If you want to see this approached developed in more detail, see ch. 7 of <a href="http://www.lightandmatter.com/area1sn.html" rel="nofollow">this online book</a>. (I'm the author.)</p>
<p>There are a couple of recent commercial textbooks that use similar geometric approaches: </p>
<ul>
<li><p>Mermin, It's About Time: Understanding Einstein's Relativity</p></li>
<li><p>Takeuchi, An Illustrated Guide to Relativity</p></li>
</ul>
| 428
|
special relativity
|
A possible absolute reference system
|
https://physics.stackexchange.com/questions/17055/a-possible-absolute-reference-system
|
<p>What about considering the microwave background radiation (2.7K if I remember well) as a reference system with some absolute character? Please explains if this question make sense and possible answers.
Thank you.</p>
|
<p>No more absolute than using the position of distant Quasars</p>
| 429
|
special relativity
|
What would an observer see if he/she flew toward a clock at relativistic speeds?
|
https://physics.stackexchange.com/questions/19370/what-would-an-observer-see-if-he-she-flew-toward-a-clock-at-relativistic-speeds
|
<p>If an observer approaches a clock at a significant fraction of the speed of light, would they see the clock's hands moving at a faster or slower than usual rate?</p>
<p>I figure there are two competing effects at play - time dilation and diminishing distance.</p>
|
<p>You can see this as an example of the relativistic Doppler shift (for equations, see eg: <a href="http://en.wikipedia.org/wiki/Relativistic_Doppler_effect">http://en.wikipedia.org/wiki/Relativistic_Doppler_effect</a> ).</p>
<p>The hands of the clock are moving with some angular frequency and you are moving with a velocity v towards the clock. It follows that the frequency you are seeing will be higher, thus the clocks hands will move faster.</p>
<p>Conceptually this makes sense. Suppose a picture of the clock is emitted each second. Since you're moving towards the clock, you will pick up one of those pictures more often than once per second, thus making the clock seem to run faster.</p>
| 430
|
special relativity
|
Einstein's Famous Thought Experiment Contradiction
|
https://physics.stackexchange.com/questions/20532/einsteins-famous-thought-experiment-contradiction
|
<p>Putting Special Relativity into the General Relativity category as is current practices submerges important aspects of Einstein's 1905 paper, which I recently read in a 1952 Dover paperback (The Principle of Relativity). That paper is totally unlike the modern presentations in texts. I noticed a marked discrepancy of the analysis in that first paper with what was actually described as happening. This has to do with the famous thought experiment. I modified that experiment and still found a discrepancy that I would like explained.</p>
<p>For the thought experiment, Einstein calculates a round trip time for a pulse of light from inception to reflection back to the inception point as measured in a moving frame. The round trip time is given as $T = (2D/c)/\sqrt {1 - {{(v/c)}^2}} ,$ where D is the distance from the inception point to a mirror moving in the same frame. Einstein’s hypothesis is that the on-board time is dilated to the same extent as the length is contracted, such that the on-board RT time as measured is independent of whether the frame is moving or not, hence, giving rise to the notion that a moving frame cannot know it is moving without an outside reference.</p>
<p>To the contrary, any symmetrical effect, such as the dilation and contraction, no matter what their forms, cancel in the above equation. This leaves an RT time discrepancy that is clearly measurable by the on board system, which knows what the speed of light is and what the distance D is. This then contradicts the hypothesis that the moving observer cannot know or measure that they are moving. Furthermore, they can tell how fast they are moving.</p>
<p>Let us modify the thought experiment. I put a clock at both the inception and the reflection points, both synchronized, and perform the same experiment. I measure when the pulse is reflected. All dilation and contraction effects are the same for all on-board systems and paths. I launch the pulse, measure the outbound transit time and the reflected transit time. I then compare the results. The two times are asymmetrical. Their combined time is the total time calculated above. Hence, the on-board system knows how fast it is moving.</p>
<p>If one further modifies the thought experiment into a radar experiment with external retro-reflectors on an aircraft, one finds the same asymmetries and the same time differences comparing an approaching aircraft with a receding aircraft. In this case, only the Lorentz contraction would be occurring, but the asymmetry is still present.</p>
<p>How is it possible to reconcile the above conclusions with the modern notions that the moving platform cannot know it is moving? I'd prefer an answer using the same math and logic Einstein used in his first paper.</p>
|
<p>You are confused because you are not taking the relativity of simultaneity into account in your analysis. When you have two mirrors moving to the right with velocity v, you are thinking that it takes light a long time to go to the right, but a very short time to go to the left, because in one direction, the mirror is going along with the light, and in the other direction, the mirror is going toward the light.</p>
<p>You conclude that the fellow who is riding along with the two mirrors can't see the same time for the forward leg of the light circuit between the two mirrors as for the backward leg. This conclusion is wrong, because the fellow riding along with the two mirrors has a failure of simultaneity--- the notion of "right now" is altered, so that the "right now" moment at a distance x along the direction of motion is further into the future by vx.</p>
<p>This asymmetrical simultaneity failure means that as the light is going to the right, it is going into the future of the moving observer slower than when it is going to the left, so that from the point of view o the moving observer, both the forward motion and the backward motion of the light go an equal number of time steps into the future, relative to the notion of simultaneity appropriate for the moving observer.</p>
<p>This is explained with a diagram in this answer: <a href="https://physics.stackexchange.com/questions/12435/einsteins-postulates-minkowski-space-in-laymans-terms/13621#13621">Einstein's postulates $\leftrightarrow$ Minkowski space for a Layman</a></p>
<p>If you understand why the line of simultaneous events for the moving observer slopes up as viewed by the stationary observer, you will immediately understand why the forward and backward motions are symmetric for the moving observer. This is explained in the first and second diagram of the linked answer, which explicitly use right-moving and left-moving light to establish the slope of the simultaneity line.</p>
| 431
|
special relativity
|
Is radiation pressure constant in this experiment, and does force of a spring change?
|
https://physics.stackexchange.com/questions/21419/is-radiation-pressure-constant-in-this-experiment-and-does-force-of-a-spring-ch
|
<p>Let's say a light clock consists of two parallel mirrors, some photons bouncing between the mirrors, and a spring that pulls the mirrors together with the same force that the photons push them apart.</p>
<p>Now we start accelerating the mirrors and the spring, in the direction parallel to the mirrors. Whenever a photon falls off from the end of the light clock, we put a similar photon at the front part of the light clock.</p>
<p>Now my question is: does the distance between the mirrors change?</p>
<p>(Actually I'm interested whether there are any changes in the forces in the spring)</p>
|
<p>The way I understand your description, the motion of the light and the spring tension are perpendicular to the direction of acceleration. Consider the problem in the (instantaneous) rest frame of the mirrors and spring.</p>
<p>Let's use coordinates where x is the direction of the photon's motion (in the unmoving rest frame), and y is in the direction of acceleration (and motion) of the mirrors.
The relevant equations are spring force $F = kx$, photon momentum in the mirror (x) direction = $P_x = \hbar k_x$, photon rate $1/T$ where $T$ is the time between photons, and the photon force = change in momentum per unit time = $2\hbar k_x / T$. The requirement for no change to the mirror is that $kx = 2\hbar k_x/T$.</p>
<p>Clearly the spring is just a spring in the moving rest frame and so there's no change to the spring force $kx$.</p>
<p>As for the photons, their momentum perpendicular to the direction of velocity (or acceleration) is unchanged, so there's no change to the momentum any single photon imparts to the mirror $2\hbar k_x$. Uh, note that since the photon reverses its direction, it imparts twice its momentum to the mirror. Another way of saying the same thing is to note that stationary and moving observers agree on the number of photon wavelengths $\lambda_x$ between the two mirrors. They also agree on the distance between the two mirrors. Therefore they agree on the wave number $k_x = 2\pi/\lambda_x$ for the photon's momentum in the x direction.</p>
<p>So what's left is the rate at which the photons impact the mirror $1/T$. This rate does not depend on the acceleration (relativity problems rarely do). Instead it depends on the velocity of the mirror.</p>
<p>As the mirror velocity increases, the distance traveled by the photons in the moving frame increases. Thus the time between photons increases and the photon force decreases.</p>
<hr>
<p>Of course in the unmoving rest frame the photon rate is unchanged, it is only in the moving frame that the photon rate changes. This effect is called time dilation. Consequently, the non moving observer, on noting the force on the mirror, must conclude that the spring constant $k$ has changed due to the motion of the spring. For a discussion of this interesting effect, see: <a href="http://www.mathpages.com/home/kmath068/kmath068.htm" rel="nofollow">http://www.mathpages.com/home/kmath068/kmath068.htm</a></p>
| 432
|
special relativity
|
Pushing with a lorentz contracting stick
|
https://physics.stackexchange.com/questions/21919/pushing-with-a-lorentz-contracting-stick
|
<p>If I use a stick to push and accelerate an object, my hand pushes one end of the stick distance $x$, while the other end of the stick pushes the object distance $y$.</p>
<p>Distance $y$ is smaller than distance $x$, because of Lorentz contraction of the stick. </p>
<p>My hand does work $Fx$.</p>
<p>Work $Fy$ is done on the object.</p>
<p>Energy $F \cdot(\text{Lorentz contraction of the stick})$ seems to disappear.</p>
<p>So I'm asking, what happens to the "missing" energy?</p>
<p>EDIT: In this thought experiment pushing causes the object and the stick to accelerate, which causes the stick to Lorentz-contract. In extreme case the length of the stick becomes zero, which means my hand moved a distance of the stick's length kind of unnecessarily. Shorter stick saves energy. </p>
<p>EDIT2: I noticed that "lost" energy approaches zero, when force approaches zero. This suggests the energy loss is linked to deformation of the stick.</p>
<p>EDIT3: This very simple problem may be very difficult to understand, so I ask this way: A good push rod is rigid. Relativity says rigid push rods don't exist. So what kind of energy goes into a push rod, that is as rigid as relativity allows, when we use the push rod, using moderate force, and the speed that the push rod is accelerated to, is relativistic?</p>
|
<p>$y=x$</p>
<p>For a constant pushing velocity, lorentz contraction is constant. It's just a smaller, rigid rod, solve classically.</p>
<p>V2:</p>
<p>The missing energy went into accelerating the stick, of course. I'm not sure if you even are allowed to use an accelerating situation in SR.</p>
| 433
|
special relativity
|
Factor 2 and equations for the weak gravitational field
|
https://physics.stackexchange.com/questions/24188/factor-2-and-equations-for-the-weak-gravitational-field
|
<p>By using some axioms people derives equation for Lorentz force and, then, Maxwell's equations from the Coulomb's law and Lorentz transformations. When I used analogical methodology for Newton's law of universal gravitation, I derived some equations like gravitomagnetic equations deriving from the GR equations in the weak field limit. But I lost factor 2, which multiplies "magnetic" component. It's binded with representation of gravitation field as the tensor. Can you explain me, how to, using methodology described above and representative of gravitation field by tensor, get factor 2?</p>
|
<p>What's going on here is that the relativistic force is</p>
<p>$$ {d^2x^\mu \over d\tau^2} = - \Gamma^\mu_{\alpha\beta} \dot{x}^\alpha\dot{x}^\beta$$</p>
<p>Where the Christoffel symbol is</p>
<p>$$ \Gamma^\mu_{\alpha\beta} = -{1\over 2}g^{\mu\nu}(g_{\alpha\nu,\beta} + g_{\beta\nu,\alpha} - g_{\alpha\beta,\nu} ) $$</p>
<p>The factor of 2 in your derivation almost cetainly the half in front of the right hand side. The metric tensor is the square of the length of a vector, and the preservation of length in the parallel transport is what requires the factor of 1/2, the weak field metric tensor is <em>twice</em> the Newtonian potential.</p>
<p>The weak field limit has $g_{\mu\nu}= \eta_{\mu\nu} + h_{\mu\nu}$, and the weak field gravitational force field $\Gamma$ is</p>
<p>$$ \eta_{\mu\nu} \Gamma^\mu_{\alpha\beta} = -{1\over 2} h_{\alpha\nu,\beta} -{1\over 2} h_{\beta\nu,\alpha} + {1\over 2} h_{\alpha\beta,\nu} $$</p>
<p>You want to expand this noncovariantly. In noncovaraint equations (like the Lorentz force law) you write the derivative of the velocity and the velocity using time, not proper time. To do this, separate out the 0 index, and write the force law (using ${d\over d\tau}=\gamma{d\over dt}$ as:</p>
<p>$$ \gamma {d \over dt }(\gamma v^i) = - \Gamma^i_{00} + 2\gamma \Gamma^i_{0j}v^j + \gamma^2 \Gamma^i_{jk} v^j v^k $$</p>
<p>Where $\gamma={1\over \sqrt{1-v^2}}$ (there's a reason nobody writes weak field GR in nonrelativistic notation). The result in terms of the weak field metric is:</p>
<p>$$ {d\over dt}{\gamma v_i} = {1\over 2\gamma} h_{00,i} -{1\over \gamma} h_{0i,0} - {1\over 2} v^j (h_{i0,j} + h_{j0,i} - h_{ij,0}) - {\gamma\over 2} v^j v^k (h_{ji,k} + h_{ki,j} - h_{jk,i}) $$</p>
<p>To get the Newtonian limit, set v to zero, and the time derivative of h to be small (in Newtonian mechanics, everything is changing slowly). Then the leading term leads to to identify the Newtonian potential as half the 00 component of h.</p>
| 434
|
special relativity
|
What are the increasingly sophisticated ways to perform a Lorentz transformation?
|
https://physics.stackexchange.com/questions/29387/what-are-the-increasingly-sophisticated-ways-to-perform-a-lorentz-transformation
|
<p>Since Einstein first derived the Lorentz transformations, their generalisation and execution has changed over the century. So starting with those first derived by Einstein: </p>
<p>What are the main, increasingly sophisticated ways, of carrying out a Lorentz transformation today? What are their advantages/disadvantages over the preceding methods?</p>
|
<p>"Since Einstein first derived the Lorentz transformations...", that's priceless. </p>
<p>The Lorentz transformations, motivated as specific linear transformations of in a vector space (i.e. matrices) have not and will never change. What changes is notation. If by "...execution has changed over the century" you mean how to implement these efficiently, then I guess that's a comp. science question.</p>
<p>Regarding "...and generalizations...", as I said, the fundamental abstract transformation group is the same. Its finite <a href="http://en.wikipedia.org/wiki/Representations_of_the_Lorentz_Group" rel="nofollow">representations</a> (how to transform this and that object, and which objects are there, which can be transformed) are fairly easy to grasp.</p>
| 435
|
special relativity
|
The temperature in space
|
https://physics.stackexchange.com/questions/29418/the-temperature-in-space
|
<ol>
<li><p>We know that cosmic microwave background temperature is about 2.7K. But what temperature we will measure in space using a simple Kelvin thermometer in the shadow? Can it be lower than 2.7K?</p></li>
<li><p>Suppose a space ship is flying in our solar system with a speed of 0.99c relative to earth. Will it measure a different temperature? Much higher? And if so, will the intense heat cause the ship itself to warm up?</p></li>
</ol>
|
<p>Re question 1: It's relatively simple to calculate temperatures in space because since there is no air, there is no heat conduction or convection. A body in space can only absorb heat by absorbing radiation, and can only lose heat by radiating.</p>
<p>The CMB behaves like a black body with a temperature 2.725K, so any body in equilibrium with it will also be at this temperature. If there are no other sources of heat your thermometer will have the same temperature as the CMB. However it's hard to achieve this. You could argue that you could shade your thermometer from the Sun, but whatever you use as a sunshade will eventually heat up and start radiating, and it will then heat your thermometer. The only way you'd get your thermometer down to 2.725K would be to put it in interstellar space or possibly even intergalactic space.</p>
<p>Re question 2: Travelling fast will indeed blue shift the CMB and raise it's temperature, and that will heat your ship. In principle if you travel fast enough the blue shifted CMB would vaporise your spaceship.</p>
<p>This actually happens to the Earth (well, not the vaporising bit!). The CMB is hotter in the direction of travel of the Earth. See <a href="http://apod.nasa.gov/apod/ap010128.html">http://apod.nasa.gov/apod/ap010128.html</a> for details.</p>
| 436
|
special relativity
|
Computing inverse Lorentz transformations in matrix form
|
https://physics.stackexchange.com/questions/408382/computing-inverse-lorentz-transformations-in-matrix-form
|
<p>I have a question about finding inverse Lorentz transformations explicitly, in matrix form:</p>
<p>Suppose I have a Lorentz transformation $\Lambda^\mu_{\;\nu}$, with matrix representation $\underline{\underline{\Lambda}}$. The inverse Lorentz transformation $(\Lambda^{-1})^\mu_{\;\nu}=\Lambda_\nu^{\;\mu}$ can be found as $\Lambda_\nu^{\;\mu}=g_{\nu\rho}g^{\mu\sigma}\Lambda^\rho_{\;\sigma}$.</p>
<p>My question concerns how I should write the latter in matrix form, in order to compute it explicitly. If I write $g_{\mu\rho}\Lambda^\rho_{\;\sigma}g^{\nu\sigma}$ then, to me, that indicates the matrix expression is $\underline{\underline{\mathbb{g}}}\underline{\underline{\Lambda}}\underline{\underline{\mathbb{g}}}$. To me, this seems like the correct order as the indices of the neighbouring objects "match".</p>
<p>But if I look in a textbook, the inverse transformation is listed as $g^{\nu\sigma}\Lambda^\rho_{\;\sigma}g_{\mu\rho}$ instead, which gives $\underline{\underline{\mathbb{g}}}\underline{\underline{\Lambda}}^\mathrm{T}\underline{\underline{\mathbb{g}}}$. Which form is correct, and why?</p>
| 437
|
|
special relativity
|
Time dilation formula question
|
https://physics.stackexchange.com/questions/408604/time-dilation-formula-question
|
<p>I understand that the formula for time dilation is given as</p>
<p>$$T = T_0\gamma = \frac{T_0}{\sqrt{1-v^2/c^2}}$$</p>
<p>Where T is moving with velocity <em>v</em> seen from $T_0$. Though, this is when an event is occuring, since the value for $T$ would in any case of $v>0$ be greater than $T_0$, meaning that the event takes longer time for $T$. However, in a situation like the twin paradox, if we ignore acceleration and all that, $T_0$ should be greater than $T$, since the time would tick slower for $T$. So what is the exact formula for such a situation, how is it derived, and is it true that abovementioned formula is only true for an event happening?</p>
|
<p>In your case when the twin on the spaceship travels with constant speed, the twins will see each other age at the same rate. Both of them could say that the other one is traveling relative to him, so the other one should age slower.
But they age at the same rate in this case. This is because speed is simmetrically relative.</p>
<p>The aging difference comes when the traveling twin gets to the point of return.
There, he has to accelerate and decelerate. That is the way for him to return.
At that point, because acceleration is absolute as per GR, the traveling twin will age slower. </p>
<p>The reason for that is that as per GR, the object (his spaceship) is accelerating when turning, and that effect is the same as a gravitational field, and in a stronger (stronger then earth) gravitational field, the traveling twin will slow down in the time dimension, he will age slower (compared to the twin on earth).</p>
<p>Why that is happening is because the four speed vector's magnitude has to stay c. If the traveling twin accelerates, his spatial vectors will change, and because the magnitude of the four speed vector has to stay c, the time component will decrease to compensate. So the spaceship and the traveling twin will move in the time dimension slower at the point of return, and so he will age slower then the twin on earth.</p>
| 438
|
special relativity
|
Paradox of the twins in time dilation
|
https://physics.stackexchange.com/questions/411161/paradox-of-the-twins-in-time-dilation
|
<p>In the twin paradox from what I understand both observer see each other's time dilated so they always believe that the other frame is younger. Finally because the space ship frame has to make many accelerations it results that he will be younger. However, what if the frame in the space ship doesn't have to make any accelerations to come back to earth and could simply just come back like if the space was a looped or that there was a wormhole , so in this case who would be younger ?</p>
|
<p>The traveling twin will be younger, because the paths are not symmetric.</p>
<p>When considering paths in these spaces, you have to consider homotopy classes. Two paths are in the same homotopy class if they can be continuous deformed into one and other.</p>
<p>It the standard twin paradox, the traveling twin's path can be deformed into the stationary twin's path (and to a point); however, because the traveling twin experiences acceleration, his elapsed proper time is shorter.</p>
<p>It can be shown among call curves between 2 points is a given homotopy class, only one corresponds to an inertial observer, and it is that path the experiences the most proper time.</p>
<p>In your question, one of the paths travels around a cylindrical (or toroidal) dimension. Each path can be characterized by a winding index, which is a topological invariant. It cannot be changed by a change of coordinates, reference frame, or deformation. The observer with winding number 0 experiences the most proper time.</p>
<p>According to the topologists: "The spatial topology thus imposes privileged frames among the class of all inertial frames, and even if the principle of relativity remains valid locally, it is no longer valid at the global scale. This is a sign that the theory of relativity is not a global theory of space-time."</p>
<p>Details can be found in this (very tractable for non topologists) paper: <a href="https://arxiv.org/pdf/0910.5847.pdf" rel="nofollow noreferrer">https://arxiv.org/pdf/0910.5847.pdf</a> </p>
| 439
|
special relativity
|
Proof that an observer cannot be in two places at once
|
https://physics.stackexchange.com/questions/422405/proof-that-an-observer-cannot-be-in-two-places-at-once
|
<p>If you went at light speed, you would literally and instantly teleport from the sun to the Earth. Any slower would create the perception of time. In other words, we appear frozen to the light (or observer going at c).</p>
<p>Still, the light can never appear frozen to us. Its speed is always the same, so that if we left the sun at 0.999c, the light that left the sun and appears outside our spaceship’s window going in our direction is also going at c. </p>
<p>That sounds like it’s breaking physics, but since everything’s relative, it doesn’t. If you can be in two places at once - that is, observe yourself on the spaceship and on earth, wouldn't it break relativity? </p>
|
<p>No object with rest mass can travel at c. Everything is relative to c, the speed of light in vacuum (when measured locally).</p>
<p>For an observer on Earth, it takes 8 minutes for a photon to reach Earth. That is why an observer on Earth (observer who has rest mass) experiences time.</p>
<p>A photon does not have a reference frame. You cannot say from the view of the photon. </p>
<p>What you could say, is that you could try to look at it the way around, that everything is relative to the speed of light. You have to slow down from speed c, and to do that, you need to gain rest mass, that is how you can experience time.</p>
<p>For light, the emission at the sun and the absorption on Earth happens instantly. The photon does not experience time as we do, the speed of the photon in the time dimension is 0.</p>
<p>To imagine how it is to go that fast in space, you could chech what a neutrino experiences when traveling from the sun to Earth. From the neutrino's frame, time passes really slow (compared to clocks on Earth). The neutrino will see and experience this travel as it only takes very little time on the neutrino's own clock as per SR.</p>
<p>For an observer on Earth, the neutrino still seems to arrive in a little more then 8 minutes.</p>
<p>What you are referring to, that the photon is at the two places at the same time, is called a lightlike worldline. For the photon, the place of emission in the sun and the place of absorption in spacetime is zero distance.</p>
<p>For the neutrino, the same distance in spacetime is not zero distance. it is called a timelike worldline.</p>
<p>The answer to your question is that at the same time at different places means that the neutrino cannot be at different places at the same time because its worldline is timelike. In spacetime, the photon's worldline is lightlike, and for the photon the spacetime distance is zero between the emission and absorption. You cannot say that the neutrino is at the same time at different places. You could though view it as if the photon was experiencing time so that it would travel a zero distance (in spacetime) from emission to absorption.</p>
| 440
|
special relativity
|
what is the behavior of an electronic oscillator at relativistic speeds?
|
https://physics.stackexchange.com/questions/433230/what-is-the-behavior-of-an-electronic-oscillator-at-relativistic-speeds
|
<p>Imagine an LC circuit in which the capacitor consists of two metal discs of area <span class="math-container">$A$</span> and spacing <span class="math-container">$D_0$</span> which is connected to a solenoid coil which for convenience has length <span class="math-container">$D_0$</span>, cross-sectional area <span class="math-container">$A$</span> and contains <span class="math-container">$N$</span> turns of wire. The capacitance of the capacitor at rest is designated <span class="math-container">$C_0$</span> and the inductance of the coil is <span class="math-container">$L_0$</span> where</p>
<p><span class="math-container">$$C_0 = \frac{e_0A}{D_0}$$</span>
and
<span class="math-container">$$L_0 = \frac{u_0N^2A}{D_0}$$</span></p>
<p>Where <span class="math-container">$e_0$</span> is the permittivity of free space and <span class="math-container">$u_0$</span> is the permeability of free space. </p>
<p>We provide a means of momentarily asserting a current through the coil or a voltage on the capacitor in order to kick it into oscillation. The circuit contains no resistance and so its oscillations do not die away. We also provide an ammeter in series with the circuit that has a needle that swings back and forth to allow nearby observers to measure the period of the oscillations. The circuit’s resonant frequency at rest is </p>
<p><span class="math-container">$$f_0 =\frac{1}{\sqrt{L_0C_0}}$$</span></p>
<p>We orient the capacitor and the inductor so their axies of symmetry are pointing in one direction, check the circuit’s resonant frequency by watching the ammeter needle, and then accelerate the circuit in that direction. As the circuit’s speed <span class="math-container">$v$</span> becomes relativistic, both the capacitor and the inductor experience foreshortening, as seen by a stationary observer nearby: the gap <span class="math-container">$D_0$</span> between the plates appears to have shrunk to <span class="math-container">$D(v)$</span>, and the length <span class="math-container">$D_0$</span> of the coil also appears to have shrunk to <span class="math-container">$D(v)$</span> where we use the length contraction factor <span class="math-container">$\gamma = \sqrt{1-\frac{v^2}{c^2}}$</span> to calculate <span class="math-container">$D(v)$</span>:</p>
<p><span class="math-container">$$D(v) = D_0\sqrt{1-\frac{v^2}{c^2}}$$</span></p>
<p>That observer then calculates the capacitance <span class="math-container">$C(v)$</span> of the foreshortened capacitor and the inductance <span class="math-container">$L(v)$</span> of the foreshortened coil as</p>
<p><span class="math-container">$$C(v) =\frac{e_0A}{D(v)}$$</span>
which is greater than <span class="math-container">$C_0$</span>, and</p>
<p><span class="math-container">$$L(v) = \frac{u_0N^2A}{D(v)}$$</span>
which is greater than <span class="math-container">$L_0$</span>. </p>
<p>Because the foreshortened capacitor’s capacitance went up and the foreshortened coil’s inductance went up, we would expect the resonant frequency to go down to some value <span class="math-container">$f(v)$</span> where </p>
<p><span class="math-container">$$f(v) = \frac{1}{\sqrt{L(v)C(v)}}$$</span></p>
<p>Substituting in the values for <span class="math-container">$L(v)$</span> and <span class="math-container">$C(v)$</span>, we get</p>
<p><span class="math-container">$$f(v) =f_0 \sqrt{1-\frac{v^2}{c^2}} $$</span></p>
<p>So we used the length contraction factor <span class="math-container">$\sqrt{1-\frac{v^2}{c^2}}$</span> to shrink the width of the capacitor gap and the length of the coil, and then obtained an equation for a shrinkage in the resonant frequency of the circuit. </p>
<p>Now if we had instead applied the gamma factor of relativistic time dilation directly to the time between peaks of the moving circuit’s oscillation, we would get exactly the same reduction in the resonant frequency of the circuit as we did by relativistically foreshortening the physical dimensions of the oscillator’s components. </p>
<p>Is this seeming coincidence worth understanding more deeply, or is it simply a trivial result due to some circularity in my reasoning? </p>
|
<p>Interesting question, unfortunately circuit theory is non relativistic. Circuits that satisfy the assumptions of circuit theory in their rest frame do not satisfy them in other frames. For example, one of the assumptions of circuit theory is that all of the components of a circuit have no net charge (Nilsson and Riedel, Electric Circuits). But a current density in one frame is a charge density in another frame, so some components gain net charge (Panofsky and Phillips, Classical Electricity and Magnetism). Also, the geometry of a circuit does not matter in circuit theory, but it does matter in relativity. </p>
<p>So, to answer the question, I would say that it is probably not merely a coincidence, but I am not sure that there is much that can be done to pursue it more deeply. The circuit theory is fundamentally a non relativistic approximation, so it would require quite a bit of theoretical work to get it to the point where it could be proven that it is more than a coincidence </p>
| 441
|
special relativity
|
Car-garage paradox with just one door
|
https://physics.stackexchange.com/questions/436396/car-garage-paradox-with-just-one-door
|
<p>Special relativity implies the possibility of some apparently paradoxical situations, which can ususally be made sense of if one applies the theory rigorously. One of these is the car-garage paradox: a car speeds towards a garage which, at rest, is slightly shorter than the car. From the reference frame of the garage, the car appears shorter, so that it will fit into the garage (if the speed is high enough, which we assume it is). From the reference frame of the car, the garage appears shorter, so that the car will not fit. I have seen some very nice solutions, like <a href="https://www.quora.com/What-is-the-solution-to-the-ladder-paradox-in-special-relativity" rel="nofollow noreferrer">this one</a>, where the garage has a front door and a back door and the caveat is that their opening and closing times depend on the reference frame, so that in the end the car gets through in both cases, although in one case it is longer than the garage.</p>
<p>But what happens if there is only a front door? Of course the car will eventually crash, so that there are some non-inertial computations involved, if one wants to do things rigorously. But in any case, if the garage door closes immediately (let it be almost as fast as light and very small) after the rear of the car has passed it, then in the garage's frame the door will close and then the car will crash, whereas in the car's frame the door will not be able to close, and since after the car crashes (decelerates to <span class="math-container">$0$</span>) it is still longer than the garage (neglect shortening due to the accident), the door will not close at all, ever. But this is impossible, since the fact that the door is closed or open for all future times should not depend on the reference frame.</p>
<p>Is there a way to make sense of this applying the theory of special relativity and without waving hands and just saying "this situation is not physically possible"? I know it isn't, but one can certainly modify it appropriately so that it is (e.g. transforming the car into a particle and the door into a sensor,...).</p>
|
<blockquote>
<p>it is still longer than the garage (neglect shortening due to the accident)</p>
</blockquote>
<p>You cannot neglect the shortening. The rear of the car is moving forward. The front of the car is having a collision.</p>
<p>The crucial point is that information about this collision cannot reach the back of the car faster than the speed of light. So the rear of the car will always continue forward (at the same speed) until information about the collision reaches it. </p>
<p>The situation is set up so that the rear of the car will not get information about the collision until after it has entered the garage. Only at that point could it be slowed. </p>
<blockquote>
<p>this relativistic shortening of the car is not the same as the one we observe in car accidents, is it?</p>
</blockquote>
<p>No, not exactly. The "shortening" is really just a different point of view. But it does mean that everyday concepts of rigid objects break down at relativistic speeds. </p>
<p>It's not that the collision is necessary to shorten the car. It's that whether the collision happens or not (maybe you have a choice of whether there's an open door), the rear of the car is going to enter the garage at a particular time in the garage frame. There's nothing you can do to the front of the car to prevent it. </p>
<blockquote>
<p>What if it is not a car but a bar of a very hard material, and at the end of the garage there is a very hard elastic wall that can make the bar decelerate in a negligible length. So at the end the bar fits inside the garage, it has not broken, it has not bent. But it is shorter. Have the particles composing it got closer together?</p>
</blockquote>
<p>Relativity tells us that the concept of a perfectly rigid material must be impossible. If you push on one end of a material, the other end will not react until after sufficient time for a signal has reached it. That signal must be no faster than the speed of light. </p>
<p>It doesn't tell us exactly what the behavior of the material is. It could compress, it could fail/explode. Real materials tend to either flow or shatter. But if you suppose that you can stop the front of the object, it will necessarily compress (in some manner) because the rear of the object will not stop simultaneously.</p>
| 442
|
special relativity
|
Rest mass of electron-positron would be the same as their energy in annihilation?
|
https://physics.stackexchange.com/questions/442709/rest-mass-of-electron-positron-would-be-the-same-as-their-energy-in-annihilation
|
<p>If the combined mass of an electron and a positron was approximately <span class="math-container">$1$</span> MeV/<span class="math-container">$c^2$</span>, then would the total energy of their annihilation be equal to <span class="math-container">$9 \times 10^{16}$</span> MeV? Why do we not multiply the rest mass by <span class="math-container">$c^2$</span> to get the energy that would be produced?</p>
<p>Please explain in layman terms.</p>
|
<p>We do multiply the mass by <span class="math-container">$c^2$</span> to get the energy. The mass is roughly <span class="math-container">$1$</span> MeV/<span class="math-container">$c^2$</span> and when we multiply this by <span class="math-container">$c^2$</span> we get <span class="math-container">$1~c^2$</span> MeV/<span class="math-container">$c^2$</span> = <span class="math-container">$1$</span> MeV.</p>
| 443
|
special relativity
|
Length Contraction Scenario
|
https://physics.stackexchange.com/questions/460049/length-contraction-scenario
|
<p>Suppose a space ship is traveling from star A to star B at some significant fraction of the speed of light. In the frame of the ship, the distance A to B is less than the distance in A and B's rest frame. Is it possible for the ship to quickly increase its speed so that in its frame the ship is then closer to A compared to when it was going slower? That is, in the time interval that it is accelerating it moves further away from A due to its speed but the effect of increased length contraction has it end up closer? That sounds like an odd situation to be in.</p>
|
<p>This is an interesting puzzle and I don't think I agree with the previous answer.</p>
<p>Suppose the ship is stationary, halfway between A and B, which are <span class="math-container">$L$</span> apart. In a very short (negligible) time it accelerates towards B at high speed. A and B are now <span class="math-container">$L/\gamma$</span> apart. The ship is still halfway between A and B (playing the 'negligible time' card). So A was <span class="math-container">$L/2$</span> away and is now <span class="math-container">$L/2\gamma$</span> away. A gets closer. </p>
<p>If you want, you can avoid the 'negligible time' argument and the fact that acceleration is off-limits for special relativity by hypothesising that another ship travelling at constant high speed just happens to pass the stationary one, and the two pilots compare notes and data as they pass. </p>
<p>Odd situation - yes. But not inconsistent. </p>
| 444
|
special relativity
|
Question regarding Lorentz Transformation formula
|
https://physics.stackexchange.com/questions/466211/question-regarding-lorentz-transformation-formula
|
<p>So the Lorentz Transformation formation equation are
<span class="math-container">$$x′=\gamma(x−vt),$$</span></p>
<ol>
<li>Does <span class="math-container">$x′$</span> and t represent time and position at one event(one instance) or do they represent two events- meaning is <span class="math-container">$x'$</span> actually <span class="math-container">$x_2-x_1$</span></li>
<li>I am rather confused on when to use the Lorentz equation for solving problems? Can't I just use the time dilation and length contraction formulas depending on the problem's constraints?</li>
</ol>
|
<p>The equation you've given is not the full Lorentz transformation. There is also a transformation giving t'.</p>
<blockquote>
<p>Does x' and t represent time and position at one event(one instance) or do they represent two events- meaning is x′ actually x2−x1</p>
</blockquote>
<p>It works for either purpose.</p>
<blockquote>
<p>Can't I just use the time dilation and length contraction formulas depending on the problem's constraints?</p>
</blockquote>
<p>No. If length contraction and time dilation were all there was, then it would be equivalent to simply changing the units used for time and distance. The Lorentz transformation also says that different observers don't agree on whether events happened in the same place (also true for Galilean relativity) and that they don't agree on whether events happened at the same time.</p>
| 445
|
special relativity
|
3 variations of the Twin Paradox
|
https://physics.stackexchange.com/questions/467099/3-variations-of-the-twin-paradox
|
<p>Here I state and try to answer three variations of the twin paradox</p>
<p><strong>1) "Classical" problem, no acceleration, no turn around</strong></p>
<p>Consider the case where there's a stationary planet, and a moving spaceship moving at close to the speed of light, starting at the left going right at constant velocity. Now imagine at some instant, two beings spawn on earth and in the spaceship. They are both aged 0 at that point. Now as the spaceship continues, both frames remain inertial. Which twin is older at any given point?</p>
<p>According to one of the answers here (the 4th answer down to be exact): <a href="https://physics.stackexchange.com/questions/2554/how-is-the-classical-twin-paradox-resolved">How is the classical twin paradox resolved?</a>, there is no solution to this paradox. Either A or B will be older depending on who you ask (person A or B or anyone else).</p>
<blockquote>
<p>Is this problem reasonable and is the "solution" correct?</p>
</blockquote>
<p><strong>2) Normal problem, no acceleration, yes turn around</strong></p>
<p>Everyone knows this one; the paradox is resolved because the person traveling changes direction. <a href="http://www.physicsmatt.com/blog/2017/1/18/the-twin-paradox-in-special-and-general-relativity" rel="nofollow noreferrer">http://www.physicsmatt.com/blog/2017/1/18/the-twin-paradox-in-special-and-general-relativity</a>. Been there, done that.</p>
<p><strong>3) Weird problem, no acceleration, no turn around, curved universe?</strong></p>
<p>Now I was thinking, "what if I combine the two?". Imagine the earth is like an asteroids game, or it's a sphere/torus. Going in a certain direction for long enough means that I eventually end up where I started. Consider the case I stated in (1), where two beings aged 0 appear on the earth and on the spaceship. The spaceship proceeds to go at a constant velocity to the right, never changing direction, never under any gravitational field, but due to the nature of the universe they are in, the spaceship twin ends up reaching the earth again. <em><strong>Now at this point, who is older?</strong></em> The solution for (2) won't work because there's no turning point, right?
In the link I provided:</p>
<blockquote>
<p>You do this by just putting a sheet above the one you started with (or to the side, if moving in that direction). Nothing wrong with that, and no edges to worry about crossing.</p>
<p>The trick is though, when the traveler gets to this new point, you yell "surprise" and identify the new point with the point in the original sheet using the symmetry of the torus.</p>
<p>So the poor sap thinks they're getting away from their twin, only for you to change the nice new spot they picked out for themselves to the place they started from. It's a mathematical sleight-of-hand, but it is the easiest way to see exactly who is younger in the Twin Paradox on the torus.</p>
</blockquote>
<p>And my question that I don't feel like the other question sufficiently addresses:</p>
<blockquote>
<p>A specification question:
<strong>I don't quite see how <em>this particular</em> "mathematical sleight of hand" makes it so that the moving twin is aging less. Sorry if I'm being slow, but could anyone explain that part more thoroughly? I still don't see how even in an <em>inertial constant-velocity frames</em> one twin is aging slower.</strong></p>
</blockquote>
<p>Feel free to tell me that this problem is nonsensical and stupid--I'm a complete noob so there is a high chance that I have no idea what I'm talking about (please tell me if that's the case)</p>
|
<p>The question is not stupid. So in the normal twin paradox, the traveling twin's path includes acceleration, and hence can be distinguished from the "stationary" twin.</p>
<p>In the toroidal version, both twins are inertial, but the traveling twin's path cannot be continuously deformed to a point. As explained in <a href="https://arxiv.org/pdf/0910.5847.pdf" rel="nofollow noreferrer">https://arxiv.org/pdf/0910.5847.pdf</a>, this introduces <em>"new kind of asymmetry
between the spacetime paths joining two events, an asymmetry which is not due to acceleration but to the multiply connected topology.... all the inertial frames are not equivalent, and the topology introduces a preferred class of inertial frames."</em></p>
<p>The authors go onto explain that, <em>"The loss of equivalence between inertial frames is due to the fact that a multiply connected spatial topology globally breaks the Poincaré group."</em></p>
| 446
|
special relativity
|
Is length contraction and time dilation symmetrical between non-accelerating reference frames?
|
https://physics.stackexchange.com/questions/477472/is-length-contraction-and-time-dilation-symmetrical-between-non-accelerating-ref
|
<p>You awaken on a deep space observing station. You do not know its acceleration history but right now there is no gravity and you are billions of light years from the nearest other molecules. A light-year-long series of high speed, high resolution video cameras stretches out left to right as viewed from your observation deck. Each camera has a big LED clock attached to it, which is kept in sync with the clock in your observation post, and which appears in the foreground of every picture. Also there is a yardstick next to each camera so passing observers can measure relative speed.</p>
<p>Your friend awakens aboard a 40m-long spaceship that’s adrift in deep space. He has no memory. After awhile his ship happens to drift in a line exactly parallel to your cameras. His ship has a big LED clock on the side that your cameras can see. His ship itself also has a high-speed camera aboard that records pictures of your cameras’ clocks each time it passes one, and in the foreground of its pictures, the clock attached to his ship is also visible.</p>
<p>Neither of you can ever remember having experienced gravity or acceleration, so it is unknown which of you is the one who accelerated to achieve the speed difference. It’s entirely possible you both accelerated the same amount in opposite directions. Who knows.</p>
<p>From your observation deck, after 1.001001001001001 years, your friend’s ship finally passes the last camera. Upon observing the complete set of photos, you see that your clock and his ship’s clock were exactly in sync in the first photo taken when he initially passed your first camera. Due to the elapsed time, you calculate his velocity was 0.999x the speed of light. Due to time dilation, the clock on the exterior of his ship has only elapsed 16 days, 8 hours, 3 minutes from the first picture you took. Also, his ship only appears to be 1.7884 m in length—4.4471% of the original length. </p>
<p>Meanwhile, your friend’s ship’s clock has also taken photos of your cameras’ clocks as they whizzed past. Finally, there was a quick wifi transmission between his clock and your final clock, over which connection, the final photographs were exchanged.</p>
<p><strong>The Questions</strong></p>
<p>To him, how far would your line of clocks appear to stretch? I calculate 0.044515 LY, due to your length contracting relative to him. </p>
<p>Since you are moving 0.999 c relative to him, I calculate that he would only experience 16 days, 8 hours, 3 minutes having passed while your line of cameras zipped past him. Is this correct? If so, why does he experience a different amount of elapsed time compared to you?</p>
<p>If I’m right then the photo of his own ship that he receives from your final camera will show 16 days, 8 hours, 3 minutes as having elapsed on the outside clock of his ship while your camera’s own foreground clock will indicate 1.001001... years as having passed. Is that what would happen, no matter who was the one that accelerated? Or does some acceleration that occurred in the past affect things? If so, why?</p>
<p>Lastly, the final picture you receive from him would show the same elapsed times as the picture he received from you: his ship has 16 days, 8 hours, 3 minutes, and your final clock has 1.001... years in both pictures. Correct?</p>
<p>If I’m correct on the above points, then I am confused, because it seems asymmetrical. </p>
<p>I would have expected that if you saw his clock elapsed by 16 days, 8 hours, 3 minutes while you experienced 1.001... years, then likewise, he would see:</p>
<ul>
<li>your final clock shows 16 days, 8 hours, 3 minutes passed for you</li>
<li>his clock shows a year had passed for him</li>
<li>his own calculation of your velocity, using the shrunken yardsticks next to your cameras as a guide, would be 0.999c</li>
</ul>
<p>However in this scenario, would either of you receive the wifi data exchange? Your final clock would send it after 1.001... years, but to him, your final clock only shows 16 days 8 hours 3 minutes have passed—so how can he be receiving a file that your clock only sends after 1.001... years?</p>
|
<blockquote>
<p>I am confused, because it seems asymmetrical.</p>
</blockquote>
<p>It is asymmetrical because the vehicles are asymmetrical. The length of the 40m craft really has no bearing on the time of the interaction. So for you, the length contraction doesn't matter.</p>
<p>The length of the light-year-long observatory does matter. So the length contraction observed by your friend matters a lot.</p>
<blockquote>
<p>Each camera has a big LED clock attached to it, which is kept in sync with the clock in your observation post</p>
</blockquote>
<p>Let's amend this sentence:</p>
<p>Each camera has a big LED clock attached to it, which is kept in sync <em>in your frame of reference</em> with the clock in your observation post.</p>
<p>With that in mind, some of your problems should disappear. </p>
<blockquote>
<p>Your final clock would send it after 1.001... years, but to him, your final clock only shows 16 days 8 hours 3 minutes have passed</p>
</blockquote>
<p>No, your final clock shows a timestamp of 1.001... years, but that does not represent an elapsed time for him. To him, your last clock is not synced to the others and is set almost one year ahead of the first clock. </p>
<hr>
<p>Addressing later comment:</p>
<blockquote>
<p>I was already assuming your light-year-long row of cameras were in your reference frame.</p>
</blockquote>
<p>I'm not sure what you mean by this. They're in both reference frames. The clocks only show the same time in the observatory's rest frame though. They show different times in other frames.</p>
<blockquote>
<p>What I don’t understand is why both his and your final pictures would show your final clock showing 1.001 years, but his final clock at only ~16 days.</p>
</blockquote>
<p>Because they aren't synchronized in the friend's reference frame. They aren't set to the same zero point. Instead of showing the same time, he would describe the clocks as differing by about a year.</p>
<blockquote>
<p>In that case, for you, his clock is running slow, and for him, your clock is running fast. But each of you is stationary from your own perspective, so shouldn’t you both perceive the others’ clock to be running slower than your own? If not, then why?</p>
</blockquote>
<p>"running slow" can be interpreted in different ways. The clock tick rate and how it's set. </p>
<p>Both of them would observe the other clocks to have a run rate that is slower than their local clocks. What the friend sees is that the end clocks on the observatory are both ticking at a slower rate than his clocks. But the last clock is set to a time about a year in advance of the first clock.</p>
<p>So the friend sees the last clock at the observatory to read "1.001001...years", but interprets this that it's set nearly a year ahead of the first clock. Not that it took a year to get from one side to the other.</p>
| 447
|
special relativity
|
What does Ethan Siegel mean by "The fact that there are no stationary, oscillating in-phase electric and magnetic fields led to Special Relativity."
|
https://physics.stackexchange.com/questions/496197/what-does-ethan-siegel-mean-by-the-fact-that-there-are-no-stationary-oscillati
|
<p>In <a href="https://medium.com/starts-with-a-bang/this-one-anomaly-is-driving-physicists-to-search-for-light-dark-matter-774ec0eb7023" rel="nofollow noreferrer">This One ‘Anomaly’ Is Driving Physicists To Search For Light Dark Matter</a>, what does Ethan Siegel mean by "The fact that there are no stationary, oscillating in-phase electric and magnetic fields led to Special Relativity."</p>
|
<p>In a traveling wave, the magnitude of the magnetic field is proportional to the magnitude of the electric field at each point in space (they can't be equal because they have different units, but they are related by <span class="math-container">$|E| = c|B|$</span>). This means the <em>nodes</em> and <em>antinodes</em> of the two fields are at the same places, and hence the magnetic and electric fields are "in-phase." See <a href="http://www.physics.miami.edu/~zuo/class/fall_05/supplement/Figure32_10.jpg" rel="nofollow noreferrer">this picture</a>. </p>
<p>However, in a standing wave (like between two mirrors in a cavity), the nodes of magnetic field are at the antinodes of the electric field and vice-versa, so they are "out-of-phase." See <a href="http://www.physics.miami.edu/~zuo/class/fall_05/supplement/Figure32_18.jpg" rel="nofollow noreferrer">this picture</a>. </p>
<p><em>My guess</em> is that the author is saying that there is no standing wave where the nodes and antinodes of the two fields are aligned. That is, there is no way to make a traveling wave stationary, even relative to you when you try to catch up to it. </p>
<p>This relates to Einstein's thought experiment that PM 2Ring linked in the comments. </p>
| 448
|
special relativity
|
How can two different inertial frames observe two different events?
|
https://physics.stackexchange.com/questions/509553/how-can-two-different-inertial-frames-observe-two-different-events
|
<p>Consider an apparatus similar to that used in Michelson-Morley experiment kept in a frame (S') moving with a constant speed (v) to the right of an inertial frame (S).</p>
<p>The apparatus consists of a light source, a partially silvered glass (B) in front of the source and two mirrors facing B (at a length L from B) and perpendicular to each other.</p>
<p>Light leaves the source and is reflected as well as transmitted at B. Let <span class="math-container">$T_1$</span> and <span class="math-container">$T_2$</span> be the total times to reach the mirrors from B and return back.</p>
<p>According to Special Relativity, the speed of light (c) is constant in both the frames. Thus, for S' <span class="math-container">$T_1=T_2=2L/c$</span> and for S <span class="math-container">$T_1=\frac{2L/c}{1-v^2/c^2}$</span> and <span class="math-container">$T_2=\frac{2L/c}{\sqrt{1-v^2/c^2}}$</span>.</p>
<p>From this it is evident that the two beams of light reach back B simultaneously in S' but not in S. But how can the two frames observe two different events - light reaching simultaneously at the same point and light not reaching simultaneously?</p>
|
<p>If the arrival of the photons at a specific point (B) is simultaneous from the viewpoint of one inertial observer, it will be simultaneous from the viewpoint of any other inertial observers.</p>
<p><a href="https://i.sstatic.net/G6D1o.gif" rel="nofollow noreferrer"><img src="https://i.sstatic.net/G6D1o.gif" alt="Figure"></a></p>
<p>Your calculations for the upper mirror is correct. That is, the total time <span class="math-container">$T$</span> needed for the photon to return to the splitter, as measured in <span class="math-container">$S$</span>, is:</p>
<p><span class="math-container">$$T=\frac{2L/c}{\sqrt{1-v^2/c^2}}$$</span></p>
<p>However, when calculating the time for the photon emitted towards left, you must be noticed that the assigned mirror escapes from the mirror so that the photon has to travel a longer path than the Lorenz contracted <span class="math-container">$L/\gamma$</span>. Therefore, we have:</p>
<p><span class="math-container">$$L/\gamma+vT_1=cT_1 \rightarrow T_1=\frac{L/\gamma}{c-v}$$</span></p>
<p>As the photon reflects back from the left mirror towards the splitter, you must consider that the splitter approaches the photon at <span class="math-container">$v$</span>. Therefore, the traveled path is smaller than <span class="math-container">$L/\gamma$</span> so that we can write:</p>
<p><span class="math-container">$$L/\gamma-vT_1=cT_1 \rightarrow T_1=\frac{L/\gamma}{c+v}$$</span></p>
<p>The total time is thus calculated to be:</p>
<p><span class="math-container">$$T=T_1+T_2=L/\gamma \left( \frac{1}{c-v}+\frac{1}{c+v} \right)=\frac{2L/\gamma}{c^2-v^2}=\frac{2L/c}{\sqrt{1-v^2/c^2}}$$</span></p>
<p>This time equals to the prior one. That is, both of the photons move parallel and perpendicular to motion direction reach the splitter simultaneously. </p>
| 449
|
special relativity
|
Special Relativity - Perpendicular Boosts Equaling to a Rotation after a Boost
|
https://physics.stackexchange.com/questions/515663/special-relativity-perpendicular-boosts-equaling-to-a-rotation-after-a-boost
|
<p><strong>[Question]</strong>
I recently read that two perpendicular Lorentz boosts equal to a rotation after a boost. Can anyone here show me an example of this happening? Thank you for your time and assistance!</p>
<p><em>Source: None - (not a homework question)</em></p>
|
<p>Here is an explicit example. The matrix rows and columns are in the usual order <span class="math-container">$t,x,y,z$</span>.</p>
<p><span class="math-container">$$
\left(
\begin{array}{cccc}
\frac{2}{\sqrt{3}} & 0 & -\frac{1}{\sqrt{3}} & 0 \\
0 & 1 & 0 & 0 \\
-\frac{1}{\sqrt{3}} & 0 & \frac{2}{\sqrt{3}} & 0 \\
0 & 0 & 0 & 1 \\
\end{array}
\right)
\left(
\begin{array}{cccc}
\frac{2}{\sqrt{3}} & -\frac{1}{\sqrt{3}} & 0 & 0 \\
-\frac{1}{\sqrt{3}} & \frac{2}{\sqrt{3}} & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
\end{array}
\right)
=
\left(
\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & \frac{4 \sqrt{3}}{7} & -\frac{1}{7} & 0 \\
0 & \frac{1}{7} & \frac{4 \sqrt{3}}{7} & 0 \\
0 & 0 & 0 & 1 \\
\end{array}
\right)
\left(
\begin{array}{cccc}
\frac{4}{3} & -\frac{2}{3} & -\frac{1}{\sqrt{3}} & 0 \\
-\frac{2}{3} & \frac{25}{21} & \frac{2}{7 \sqrt{3}} & 0 \\
-\frac{1}{\sqrt{3}} & \frac{2}{7 \sqrt{3}} & \frac{8}{7} & 0 \\
0 & 0 & 0 & 1 \\
\end{array}
\right)
$$</span></p>
<p>I will let you confirm the equality, and that the left-hand-side represents a boost by <span class="math-container">$c/2$</span> along the <span class="math-container">$x$</span>-direction followed by a boost by <span class="math-container">$c/2$</span> along the <span class="math-container">$y$</span>-direction, and that the right-hand-side represents a boost by <span class="math-container">$\sqrt{7}c/4$</span> in the direction <span class="math-container">$(2/\sqrt{7},\sqrt{3/7},0)$</span> followed by rotation around the <span class="math-container">$z$</span>-axis by <span class="math-container">$\cos^{-1}(4\sqrt{3}/7)$</span> or <span class="math-container">$8.21$</span> degrees.</p>
<p>It helps to have the <a href="https://en.wikipedia.org/wiki/Lorentz_transformation#Proper_transformations" rel="nofollow noreferrer">formula for a general boost matrix</a>, which is</p>
<p><span class="math-container">$$
\left(
\begin{array}{cccc}
\gamma & -\gamma \beta n_x & -\gamma \beta n_y & -\gamma \beta n_z \\
-\gamma \beta n_x & 1+(\gamma-1)n_x^2 & (\gamma-1)n_xn_y & (\gamma-1)n_xn_z \\
-\gamma \beta n_y & (\gamma-1)n_yn_x & 1+(\gamma-1)n_y^2 & (\gamma-1)n_yn_z \\
-\gamma \beta n_z & (\gamma-1)n_zn_x & (\gamma-1)n_zn_y & 1+(\gamma-1)n_z^2 \\
\end{array}
\right).
$$</span></p>
<p>To get a <a href="https://en.wikipedia.org/wiki/Wigner_rotation" rel="nofollow noreferrer">Wigner rotation</a>, the two boosts don't have to be perpendicular; they just have to be non-colinear. Their composition can also be expressed as a rotation followed by a boost, rather than a boost followed by a rotation. If you express the composition of the boosts as a rotation followed by boost, the resulting rotation will be the same as before, but the resulting boost will be different. For example,</p>
<p><span class="math-container">$$
\left(
\begin{array}{cccc}
\frac{2}{\sqrt{3}} & 0 & -\frac{1}{\sqrt{3}} & 0 \\
0 & 1 & 0 & 0 \\
-\frac{1}{\sqrt{3}} & 0 & \frac{2}{\sqrt{3}} & 0 \\
0 & 0 & 0 & 1 \\
\end{array}
\right)
\left(
\begin{array}{cccc}
\frac{2}{\sqrt{3}} & -\frac{1}{\sqrt{3}} & 0 & 0 \\
-\frac{1}{\sqrt{3}} & \frac{2}{\sqrt{3}} & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
\end{array}
\right)
=
\left(
\begin{array}{cccc}
\frac{4}{3} & -\frac{1}{\sqrt{3}} & -\frac{2}{3} & 0 \\
-\frac{1}{\sqrt{3}} & \frac{8}{7} & \frac{2}{7 \sqrt{3}} & 0 \\
-\frac{2}{3} & \frac{2}{7 \sqrt{3}} & \frac{25}{21} & 0 \\
0 & 0 & 0 & 1 \\
\end{array}
\right)
\left(
\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & \frac{4 \sqrt{3}}{7} & -\frac{1}{7} & 0 \\
0 & \frac{1}{7} & \frac{4 \sqrt{3}}{7} & 0 \\
0 & 0 & 0 & 1 \\
\end{array}
\right)
$$</span></p>
<p>Now the boost direction is <span class="math-container">$(\sqrt{3/7},2/\sqrt{7},0)$</span>.</p>
<p>If you do the two original boosts in the opposite order, you'll get different results since they don't commute.</p>
<p>ADDENDUM: Wondering how to decompose the product of a general Lorentz transformation into a boost and a rotation? See <a href="https://physics.stackexchange.com/questions/516020/special-relativity-how-to-tell-a-new-boosts-magnitude-direction-and-rotatio/516126#516126">this related question</a>.</p>
| 450
|
special relativity
|
How will the length of a moving rod change for an observer 'seeing' it?
|
https://physics.stackexchange.com/questions/516209/how-will-the-length-of-a-moving-rod-change-for-an-observer-seeing-it
|
<p>I was studying length contraction and it considered the following scenario.</p>
<p>A rod is moving at a velocity <span class="math-container">$v$</span> with respect to a frame <span class="math-container">$S$</span>. A frame <span class="math-container">$S'$</span> observes the rod stationary and thus measures proper length <span class="math-container">$l_0$</span> of the rod. Now, if the observer in <span class="math-container">$S$</span> frame measures the length <span class="math-container">$l$</span> of rod at same time t, then it can be shown that <span class="math-container">$$l_0=\gamma l$$</span></p>
<p>It is further mentioned that if the <span class="math-container">$S$</span> observer <strong>looks</strong> at the moving rod, it won't see it as shorter. It further says that</p>
<blockquote>
<p>If the time that is required for the light from each point on the
rod to reach the observer’s eye is taken into account, the overall effect is that of making
the rod appear as if it is rotated in space.</p>
</blockquote>
<p>I don't know what it means and how to prove it. Also, will a similar situation occur with <em>time dilation</em> ? </p>
<p>I'd be grateful if someone could help me with this. </p>
|
<p>There are two separate effects which should not be confused, one being length contraction and the other being Terrell rotation. </p>
<p>Length contraction is the reduction in the measured length of an object when it is measured in a frame moving relative to it.</p>
<p>Terrell rotation is a change to the apparent length and orientation of an object when viewed by an observer moving relative to it.</p>
<p>The distinction between the two definitions I have given relies upon an understanding of the difference between viewing and measuring an object. You can get a feel for this by imagining the difference in relation to a stationary object. Imaging you have a train stationary alongside a platform. You might measure its length at 200m, say. Its apparent length will depend upon the position from which you view it. If you stand close to one end of the train and look along it, it will appear foreshortened, an effect of perspective.</p>
<p>The Terrell effect is analogous to the foreshortening effect of perspective on a stationary object. </p>
<p>Calculating the effect of terrell rotation is not straightforward.</p>
| 451
|
special relativity
|
How does a photon decide what to hit along it's path?
|
https://physics.stackexchange.com/questions/531552/how-does-a-photon-decide-what-to-hit-along-its-path
|
<p>In general, let us have a light emitting point A and two light absorbing points B and C, such that the three points fall on a straight line and B is somewhere between A and C. For simplicity, let us consider all photons emitting from A to travel along this same line in the direction of the points B and C. Also, let the distance from A to B be AB and the distance from A to C be AC. </p>
<p>For the photon, special relativity states that there should be maximum space contraction along it's direction of motion. That means that, for the photon, A, B and C are all at the same point in space. This results in the photon effectively "reaching" B and C at the same time. How then does the photon "decide" that it must be absorbed by point B and not C, so as to be consistent with what an outside observer would see (point B "casting a shadow" on point C)?</p>
| 452
|
|
special relativity
|
Analogue to special relativity in other physical systems
|
https://physics.stackexchange.com/questions/541139/analogue-to-special-relativity-in-other-physical-systems
|
<p>It's hard to get an intuitive grasp of the Lorentz transformations; I was wondering if the same mathematical formulas – hyperbolic rotation – appear under disguise in other physical systems. Note: I doubt there's a direct <em>mechanical analogy</em>, with a speed limit much lower than <span class="math-container">$c$</span> (though that would be great, obviously – perhaps similar in spirit to the optical fibre analogue of event horizons); I'm more thinking of completely different quantities obeying mathematically-equivalent equations. </p>
|
<p>Here are two related abstracts that might be of interest to you.
(I have not read the papers to comment.)</p>
<ul>
<li><p>Geometrical interpretation of optical absorption<BR>
J. J. Monzón, A. G. Barriuso, L. L. Sánchez-Soto, and J. M. Montesinos-Amilibia<BR>
Phys. Rev. A 84, 023830 – Published 17 August 2011<BR>
<a href="https://doi.org/10.1103/PhysRevA.84.023830" rel="nofollow noreferrer">https://doi.org/10.1103/PhysRevA.84.023830</a><BR><BR>
"We reinterpret the transfer matrix for an absorbing system in very simple geometrical terms. In appropriate variables, the system appears as performing a Lorentz transformation in a (1+3)-dimensional space. Using homogeneous coordinates, we map that action on the unit sphere, which is at the realm of the Klein model of hyperbolic geometry. The effects of absorption appear then as a loxodromic transformation, that is, a rhumb line crossing all the meridians at the same angle."
<BR><BR></p></li>
<li><p>Fresnel formulas as Lorentz transformations<BR>
Juan José Monzón and Luis L. Sánchez-Soto<BR>
Journal of the Optical Society of America A Vol. 17, Issue 8, pp. 1475-1481 (2000) <BR><a href="https://doi.org/10.1364/JOSAA.17.001475" rel="nofollow noreferrer">https://doi.org/10.1364/JOSAA.17.001475</a><BR><BR>
"From a matrix formulation of the boundary conditions we obtain the fundamental invariant for an interface and a remarkably simple factorization of the interface matrix, which enables us to express the Fresnel coefficients in a new and compact form. This factorization allows us to recast the action of an interface between transparent media as a hyperbolic rotation. By exploiting the local isomorphism between SL(2, 𝐶) and the (3+1)-dimensional restricted Lorentz group SO(3, 1), we construct the equivalent Lorentz transformation that describes any interface."</p></li>
</ul>
<p>For different geometric approach to special relativity,
I'll suggest my own approach<BR>
"Relativity on Rotated Graph Paper"
American Journal of Physics 84, 344 (2016);<BR>
<a href="https://doi.org/10.1119/1.4943251" rel="nofollow noreferrer">https://doi.org/10.1119/1.4943251</a><BR>
where the emphasis is on a different geometrical figure: the "light-clock diamond", traced out by the spacetime paths of the light signals in a light-clock.
You can play with a visualization at <a href="https://www.geogebra.org/m/HYD7hB9v" rel="nofollow noreferrer">https://www.geogebra.org/m/HYD7hB9v</a> .</p>
| 453
|
special relativity
|
Simultaneity and special relativity
|
https://physics.stackexchange.com/questions/554156/simultaneity-and-special-relativity
|
<p>Suppose, in inertial reference frame <span class="math-container">$F_1$</span>, observers A and B are at rest, each having torch, and are separated by some distance and we have put machine M at middle of A and B.</p>
<p>Machine M has light bulbs on both sides ,right and left, so that if it catches light from A which is at left ,then machine M glows left light bulb, similar with right bulb.Also, if it senses both reaching at same instant of time then it start to make noise.</p>
<p>Now consider another inertial reference frame <span class="math-container">$F_2$</span> which is moving at constant speed <span class="math-container">$v$</span> with respect to <span class="math-container">$F_1$</span> to the right.</p>
<p>Now ,in frame <span class="math-container">$F_1$</span> , both A and B turn on their torches at same instant of time, say <span class="math-container">$t=0$</span> and both rays reach at M at <span class="math-container">$t=t_1$</span>, and machine M makes noise indicating that those events were "simultaneous" in <span class="math-container">$F_1$</span>.</p>
<p>Now we know these events are not simultaneous in <span class="math-container">$F_2$</span>, in other words ,person seating in <span class="math-container">$F_2$</span> will say ,"I should not hear sound from machine M."
But somehow machine makes noise.(or it doesn't make noise?)</p>
<p>So does this mean according to <span class="math-container">$F_2$</span> ,machine is malfunctioning?</p>
|
<p>The machine M responds to events right there at the machine---the events of light arriving from left, light arriving from right. So the machine is reporting that the <em>light arrival</em> events are simultaneous at M. This is fine; all reference frames will agree that two things happening at the same place and time do indeed happen at the same place and time.</p>
<p>But when we interpret M to be reporting that the emission events are simultaneous, now we have a frame-dependent interpretation. What M is really saying is "well the two light beams reached me simultaneously, so what I can claim is that if the emitters are at distances <span class="math-container">$d_1$</span> and <span class="math-container">$d_2$</span>, then the emission times were <span class="math-container">$d_1/c$</span> and <span class="math-container">$d_2/c$</span> before now. So if <span class="math-container">$d_1=d_2$</span> then the emission events were simultaneous. And if <span class="math-container">$d_1 \ne d_2$</span> then the emission events were not simultaneous."</p>
<p>What happens in your scenario is that in frame <span class="math-container">$F_1$</span> the two distances are equal, whereas in frame <span class="math-container">$F_2$</span> they are not equal.</p>
<p>By the way I would always recommend learning to use spacetime diagrams when learning special relativity. </p>
| 454
|
special relativity
|
Can different observers observe different realities?
|
https://physics.stackexchange.com/questions/569274/can-different-observers-observe-different-realities
|
<p>The question is extention of thought experiment proposed by Einstein of light pulses in a train cart called <a href="https://en.m.wikipedia.org/wiki/Relativity_of_simultaneity#:%7E:text=Einstein%27s%20thought%20experiments-,Einstein%27s%20train,to%20strike%20at%20different%20times." rel="nofollow noreferrer">Einstein's train</a>. Assume that light pulses are emitted from the <strong>end of cart</strong>, instead from the center, such that stationary observer, sees the light pulses at the center crossing each other at the same time. <a href="https://en.m.wikipedia.org/wiki/File:Traincar_Relativity1.svg" rel="nofollow noreferrer">image of this case from real experiment is given here</a>
And now let there a new observer which has a relative motion between cart and himself. So similarly <a href="https://en.m.wikipedia.org/wiki/File:Traincar_Relativity2.svg" rel="nofollow noreferrer">to this case</a>, the new observer will see that two pulses, coming at the center, in two different instants of time.
Now there are two things</p>
<ol>
<li>I can consider two pulses crossing each other at the center as a single event because they have same spacetime coordinates. So events don't change in relativity but here they are apparently. Because now the event is non-existential here in <strong>any other frame</strong> than the stationary one. (Light pulses don't reach at the center at the same time) So where is that, I am getting wrong?</li>
<li>It's consequences can change reality. For example if I place a detector at the center and what it does is, it (making it dramatic) kills the schrodinger's cat (I know the name looks familiar) in the <strong>glass</strong> box placed at the center, if it records that two pulses have reached at it simultaneously. (Since it is placed in the cart at the same spatial point with the center, we can approximately say that detection and killing is simultaneous). But for everyother observer the cat would be alive. Why?</li>
</ol>
|
<p>You're assuming that events that are simultaneous in one frame will be simultaneous in another frame too which is not the case. That is the whole point of the Einstein's train thought experiment. If in the train's frame, the light pulses are emitted at the same time from the ends, then to an observer in the train the light pulses will meet at the center. To an observer outside the train moving with a velocity relative to the train, the light pulses will still meet at the center of the train but the outside observer will say that the light pulses at the ends were not emitted at the same time. To the outside observer, light from one side was emitted earlier than light from the other side so that the two pulses meet at the center of the train.</p>
| 455
|
special relativity
|
Why does Feynman say “That is all there is to the theory of relativity – it just changes Newton’s law by introducing a correction factor to the mass”
|
https://physics.stackexchange.com/questions/594901/why-does-feynman-say-that-is-all-there-is-to-the-theory-of-relativity-it-just
|
<p>In the Feynman Lectures, he lays out how in special relativity, the mass is adjusted by a factor of <span class="math-container">$\sqrt{1 - v^2 / c^2}$</span> and then writes:</p>
<blockquote>
<p>For those who want to learn just enough about it so they can solve problems, that is all there is to the theory of relativity – it just changes Newton’s law by introducing a correction factor to the mass</p>
</blockquote>
<p>Does Feynman mean that you can derive the other phenomena described in special relativity (time dilation, length contraction) from only this correction to the mass? If so, how? If not, what other postulates are necessary?</p>
|
<p>The claim is wrong because Newton's second law does <em>not</em> become <span class="math-container">$F=m \gamma a$</span>.</p>
| 456
|
special relativity
|
Energy of a $n$-particle system in special relativity
|
https://physics.stackexchange.com/questions/621603/energy-of-a-n-particle-system-in-special-relativity
|
<p>Consider an inertial frame <span class="math-container">$S'$</span>. With respect to this frame of reference consider a system of <span class="math-container">$n$</span> particles. The <span class="math-container">$k$</span>-th particle has rest mass <span class="math-container">$m_{0,k}$</span> and it moves with speed <span class="math-container">$u_k$</span>. Can we say that the mass of the system is <span class="math-container">$$\mathbf{m=\sum_{k=1}^{n}m_{0,k}\gamma_{u_{k}}}$$</span> with total energy <span class="math-container">$E=mc^2$</span> ,where <span class="math-container">$\gamma_{u_{k}}$</span> is given by <span class="math-container">$\frac{1}{\sqrt{1-\frac{{u_k}^2}{c^2}}}$</span> ? <br />
And also, how does the relativistic energy expression contain the information about potential energy?</p>
|
<p><span class="math-container">$\sum\gamma_k m_k c^2$</span> is called the "total relativistic energy" <span class="math-container">$E_{rel,tot}$</span>.
<BR>
Although some have called <span class="math-container">$\frac{1}{c^2}E_{rel,tot}=\sum\gamma_k m_k$</span> the "relativistic mass",
because of misconceptions by novices, its use is discouraged.
(Further comments at <a href="https://physics.stackexchange.com/questions/606375/invariant-rest-mass-vs-proper-velocity">Invariant rest mass vs Proper velocity</a> )</p>
<p>The "invariant mass of the system" (or the "rest mass of the system")
<span class="math-container">$m_{sys}$</span> is essentially the magnitude of the total 4-momentum:
<span class="math-container">\begin{align}m_{sys}c^2
&=\sqrt{\tilde P_{tot}c\cdot \tilde P_{tot}c}\\
&=\sqrt{\left(\sum \tilde P_k\right)c\cdot\left(\sum \tilde P_k\right)c}\\
&=\sqrt{\left(\sum E_{rel,k}\right)^2-\left(\sum \vec p_{rel,k}c\right)^2}\\
&=\sqrt{\left(\sum \gamma_k m_k c^2\right)^2-\left(\sum \vec p_{rel,k}c\right)^2}\\
\end{align}</span></p>
| 457
|
special relativity
|
How does Lorentz transforming forwards, then backwards, stay consistent?
|
https://physics.stackexchange.com/questions/635048/how-does-lorentz-transforming-forwards-then-backwards-stay-consistent
|
<p>Let me take you through the logic in my head...</p>
<ul>
<li>In frame S, you have coordinate <span class="math-container">$x$</span></li>
<li>Transform to frame S' with velocity <span class="math-container">$v$</span> so the coordinate is now <span class="math-container">$x' = \gamma x$</span></li>
<li>Now treat the S' frame as if you started there.</li>
<li>Transform to a frame S'' moving at velocity <span class="math-container">$-v$</span>. The coordinate is now <span class="math-container">$x'' = \gamma x'$</span></li>
<li>However, S'' and S are the same frame so <span class="math-container">$x'' = x$</span></li>
<li>So <span class="math-container">$x = \gamma x' = \gamma ^2 x$</span></li>
</ul>
<p><strong>I would like to clarify, I'm using t=0 for everything here</strong></p>
<p>How does this make sense, what am I missing?</p>
|
<p>The Lorentz transformation always transfoms not only coordinates but also time. In fact you can consider it as a sort of a "rotation" in <span class="math-container">$(t,x)$</span> "plane".</p>
<p>Whe you start with an event <span class="math-container">$(0,x)$</span> in the new frame it will have <span class="math-container">$t'=-\gamma \frac{v}{c^2} x$</span>, <span class="math-container">$x'=\gamma x$</span>. You see that even though all events <span class="math-container">$(0,x)$</span> are simultaneous in the initial frame, in the new frame they have different <span class="math-container">$t'$</span>. This is known as a relativity of simultaneity and in my experience most "paradoxes" in special relativity originate from people forgetting about this fact.</p>
<p>Now if you apply the reverse Lorentz transformation you have
<span class="math-container">$$x''=\gamma x'+\gamma v t'=\gamma^2(1-\frac{v^2}{c^2})x=x$$</span>
Similarly you will get <span class="math-container">$t''=0$</span></p>
| 458
|
special relativity
|
How can a twin travelling astronaut paradox be resolved?
|
https://physics.stackexchange.com/questions/657975/how-can-a-twin-travelling-astronaut-paradox-be-resolved
|
<p>This is a variant of the twin paradox. But having each of the twin astronauts take off in opposite directions and returning to meet such that all aspects of acceleration and velocity are the same. Both should observe the other’s clock being slow for the whole trip but when they meet they should have aged equally. How can this be (or where is the error in this setup)?</p>
|
<p>The mistake you are making is that you are forgetting about the relativity of simultaneity. Although each twin's clock will seem to tick slowly compared with clocks stationary in the other twin's reference frame, those other clocks will appear to be out of synch to the twin who is moving relative to them. Over the course of their identical return journeys, the effects of time dilation observed by the twins are exactly cancelled by the effects of the relativity of simultaneity, so the twins will have aged by the same amount when they meet again.</p>
| 459
|
special relativity
|
What is the meaning of "clocks and rods" in special relativity?
|
https://physics.stackexchange.com/questions/662666/what-is-the-meaning-of-clocks-and-rods-in-special-relativity
|
<p>In my introductory text about special relativity, I am told to consider the whole coordinate system as consisting of meter sticks (rods) joined by clocks. I am told to consider a light pulse which is used to synchronize the clocks, with each clock stopping when it receives the pulse and correcting for the time taken for the light to reach it, so that all the clocks are synchronized. If any event occurs, it stops the clock at which the event occurs and makes a mark on the rod.</p>
<p>I don't understand the physical significance of this thought experiment or what it is supposed to show.</p>
|
<p>It looks to me like the <em>way</em> the "clocks and rods" idea has been introduced to you is more complicated than necessary, and this has caused you to have trouble understanding it. It is in fact very simple and down to earth, although it can be <em>used</em> for some subtle purposes.</p>
<p>It's sad to see what a state you are in after reading your "introductory text" which failed in showing how utterly awesome lattices of clocks and rods are. You sound unenthusiastic about relativity, and even a little bit depressed. My heart goes out to you. It's so sad that something as, if may say so, <em>beautiful</em>, as the clocks and rods lattice thought experiment was put across to you in a way that failed to fire your imagination. The reason is probably that the original use of the clocks and rods lattice thought experiment was by Albert Einstein in 1905, and his purpose was not to <em>explain</em> anything, but rather to make his theory as rigorous as possible. And, as happens all too often in physics (and even more so in math) students (even beginners) are talked to as if they were miniature professors, full of skepticism and therefore in need of proof. What follows is based on the idea that you want to understand relativity, not see it proved or made more rigorous than it already is.</p>
<p>First, you need to understand what "synchronized" means here. The use of high tech light pulses and so on can make it seem like this word is being used in some special, different sense in relativity. In fact, the word has its everyday meaning.</p>
<p>To understand this, consider when a screen gang of criminals sychronize their watches, as part of preparing to rob a bank, say. One way this could happen is for the gang leader to visit each of his underlings and adjust the underling's watch (or supervise the underling while he does so) so that it is displaying the same time as his watch (which we could say is the "master watch"), and then warns the underling to use only this watch, and not adjust the time on it, until after the robbery has been completed. Once the gang leader has visited all his underlings, it can be said that they have successfully completed the task known as "synchronizing watches" in every movie.</p>
<p>The clocks in the rods and clocks lattice are synchronized in the exact same sense of the word. There's no need to use light pulses or radar or radio signals to do it. You could carry a master clock with you and adjust each clock in the lattice, one by one, by visiting it and putting the master clock and the lattice clock next to each other and adjusting the time on the lattice clock so that it matches that of the master clock.</p>
<p>The point of the light or radio pulses is that it is a quick and convenient way for a physicist to do it in practice. It has nothing to do with the meaning of "synchronized" or anything to do with the fundamental nature of the thought experiment. These are ordinary clocks, telling ordinary time, synchronized in the ordinary way, which just means that they all display the same time once the synchronizing process is completed.</p>
<p>It does <em>not</em> mean that the sound of the ticks reaches your ears simultaneously. Clocks at the far end will seem to tick just after clocks at the near end. This is just signal delay.</p>
<p>Likewise, if the lattice is big enough or you have a video camera that high speed enough, the finite speed of light means that you will see the second hand of a clock nearby tick over before you see the second hand of a faraway clock tick over. This is just signal delay again. In one case it's due to sound taking time to cross space, and in the other due to light taking (approximately a million times less) time to cross space, and does not change the fact that the clocks are synchronized. In fact, if you <em>did</em> see all the clocks ticking over at exactly the same time, despite some being near and others far, that would show that they were <em>not</em> synchronized.</p>
<p>The way the lattice is used to make measurements is that you have a technician manning each clock. Each clock is attached to the point where six rods meet at right angles. We can call this a node in the lattice. It has a definite location, and the time is displayed there by a clock, and there is exactly one technician there, ready to make measurements. The clock doesn't necessarily "join" (as you put it) the rods to each other. To do that it would have to have zero size, or it would add to length of the rods.</p>
<p>Now imagine a rocket flying through the lattice. When the front of the rocket passes a node in the lattice, the technician at that node notes the time displayed by his clock, and that the front of the rocket passed at that time. A little bit later the back of the rocket passes, and the technician notes the fact, and notes the time that it happened, in the same way. Then he writes a report, seals it in an envelope and mails it to the bookkeeper, who, along with all the envelopes from all the other technicians, opens the envelope, and makes a table containing all the data, and from that is able to work out what the position was of the the front of the rocket at what time, and likewise for the back of the rocket.</p>
<p>There's no advantage to stopping any of the clocks, not even in an automated lattice.</p>
<p>Note that when you wrote in your question,"If any event occurs, it stops the clock at which the event occurs and makes a mark on the rod." you are essentially referring to an automated version of the lattice where when, say, the front of the rocket passes a node it stops the clock belonging to that node, and makes a mark on the rod (it would work equally well if it made a mark on the clock). The mark might be red because the dye on the front of the rocket is red, or blue because the back or the rocket is blue, although in this case, since the clock, once stopped, cannot be used to make a second time measurement, the time that the blue back of the rocket passes can't be recorded by that node). If you are only interested in the front of the rocket, marking the rod adds nothing, because the stopped clock has a fixed and known or measurable position in the lattice. In fact, it's confusing to say that "the rod" (which or the six rods that meet there?) gets a mark. Where the rod? If there is to be a mark, it needs to be at the exact same location as the clock is at, and like I said, why not just mark the clock in that case. If that's what your introductory text says, I suggest you look at some other texts as well.</p>
<p>The bookkeeper can use the data he receives in the envelopes to calculate the length of the rocket, as well as it's velocity, at any time that it was inside the lattice. If the rocket is accelating, that can be calculated, too, with the same when and where precision. All this from those simple reports sent in by those technicians, that only specified what was where and when. They didn't themselves try to measure the speed, direction, or acceleration.</p>
<p>Even more fun is when the front and the back of the rocket each have a clock attached to it. Then the technician can add to his report what each moving clock is displaying as it passes him.</p>
<p>You might be wondering about the time it takes for the light to get from the moving clock to the lattice technician's eye. The idea in this thought experiment is that the measurement of the moving clock's displayed time is done <em>at the node</em>, in effect by a process involving direct contact, something like printing, or brass rubbing (except instantaneos). The clocks are imagined as being points, or of negligible size. The point of this is to avoid having to consider what the effects of signals, especially signal delay might be.</p>
<p>Many of the misunderstandings of relativity are due to misunderstanding the role of signal delay, so it makes sense to take them out of the picture when relativity is being introduced. Using the lattice achieves this by allowing all measurements to be made <em>at</em> each event, with negligible signal delay. The light might travel a meter or two, say, from the front of the rocket to the detection device in a typical lattice, which is near as darn at it nothing. It's negligible. There's no need to completely eliminate signal delay. On the other hand, it's worth noting that it is <em>possible</em> to eliminate all signals from the thought experiment, and this could be helpful for some students. This is done by using <em>direct contact</em>.</p>
<p>There are many ways that you could have the clock(s) on a rocket moving at high speed read by direct contact with a lattice that is at rest. One way would be to have the time on the clock converted to a binary number, and then indicate that number by a row of pins on a flat rod. Pin out means a one and and pin in means a zero. The pins that are out would leave marks on any rod they struck. Or the pins could electrically charged in opposite ways, or magnetized in opposite ways. As the row of pins passes a node the pattern of electric charge or magnetism could be recorded automatically.</p>
<p>You may be thinking that this lattice is a lot like a frame of reference, and you'd be right. It is in effect a thought experiment <em>embodiment</em> of a frame of reference. Just as you can have moving, and even accelerated frames of reference, you can have your lattice do those things, too. There's very little difference in practice between a lattice of clocks and rods (and technicians if it's not automated) and a frame of reference. I guess a frame of reference would be the mathematical abstraction you get from the lattice's behavior.</p>
<p>Anyway, the lattice is the ideal way to introduce relativity, and the beauty of it is that the way it works is entirely Newtonian, and you can teach it and use it as part of teaching Newtonian mechanics. The process of construction and calibration is just common sense.</p>
<p>It's a way to clearly explain, and rigorously define, what a frame of reference is. Frames of reference, relative motion, and even moving and accelerated frames of reference all have meaning in Newtonian physics. So mastering the lattice can be done without touching relativity, but it creates the ideal solid understanding (of Newtonian physics) needed for trouble free assimilation of Einstein's special theory of relativity.</p>
<p>In a nutshell, if you have doubts about the behavior of signals, in effect you do not trust your own eyes. What do you do in that case? Feel your way. And that's exactly what the lattice is for. The way it works is tactile. Events within it are in a sense "felt" (detected by direct contact).</p>
<p>At normal "low" nonrelativistic speeds, that rocket will behave in a normal Newtonian way. It's length won't change nonnegligibly, for example. It's clocks will tick at a constant rate and so on. All this can be verified <em>using</em> the lattice (if you built one, or an equivalent piece of kit).</p>
<p>When you repeat the experiments with relativistic speeds, e.g. half the speed of light, the lattice will show you a couple of interesting things. First, the length of the rocket as measured by the lattice, which means within the frame of reference of the lattice which is at rest in our example would be less. This means that for us, at rest, the rocket <em>really</em> is shorter. It's not some sort of optical illusion, which is a common way relativity is explained, some sort of trick of the light that becomes real for unclear reasons. When the rocket is travelling at half the speed of light, there <em>are</em> a lot of interesting "tricks of the light". For example, the light from the rocket will seem to show the rocket flying diagonally ("rotated") but this isn't what is actually happening, and the lattice measurements would show that it's orientation hasn't changed for anyone in any frame of reference. It's just that the light from the far side the rocket arrives later, making it look rotated, but <em>only</em> look rotated. So as you can see (no pun intended), there's a lot be gained from taking signals out of the picture.</p>
<p>The clocks and rods lattice is one of the most underrated things in physics education. I can't believe how few articles devoted to explaining what a frame of reference make any mention of it. It's absolutely crucial (no pun intended).</p>
<p>A great thing about the lattice is that you can see first what Newtonian physics and relatistic physics have in common, and later what they don't have in common.</p>
<p>The exact same lattice still serves even when rotating and <em>even</em> when spacetime is not flat, for example near a black hole. It just gets a bit bent out of shape, so to speak. So it will still be useful even when you start studing general relativity.</p>
<p>You can also have one 3D lattice moving through another in a thought experiment. In a lab you can a have a "2D lattice", or more likely, a "1D lattice", that is to say, a sheet or more likely a line of clocks and rods moving over another, each one taking readings of its own and the other's clocks and their positions. This helps with visualizing the Lorentz transformation.</p>
<p>The clocks can be candles that burn down slowly as time passes for easy visualization in a thought experiment.</p>
<p>The alert reader will be wondering about the relativistic effects on the master clock when it is transported around the lattice when the clocks are being synchronized. The answer to that is threefold. First, as long as the lattice can fit into a medium sized city, and the master clock is moved about no faster than an automobile can carry it, the time lost by it due to slowing of the master clock due to its journey aroung the lattice will not even be <em>detectable</em>, let alone nonneglibible. Second, the effects of relativity are huge and there's really no need to even synchonize the clocks. You can just use some clocks that are all showing the current time. The use of pulses of light to synchronize the clocks is liable to mislead the student into thinking that superprecision is needed in sychronizing the clocks and that therefore the change in length of the rocket are slight, and hard to detect. Nothing could be further from the truth. The rocket is <em>half as long</em> as it was at rest. That's not a subtle effect. Third, even if the lattice is so big that moving the master clock around it could conceivably cause it to lose a detectable amount of time, you can reduce that time to as little as you want simply by moving it <em>slowly</em> enough. By taking twice as long to visit all the clocks you cut the total time lost by the clock by a factor of about two. So a large distance is no problem if travelled slowly enough.</p>
<p>I hope that helps. Many people learn of the rods and clocks lattice from an awesome book by Lewis Epstein called "Relativity Visualized". One reviewer called it "the gold nugget of the relativity books". If you liked this answer you'll love Relativity Visualized.</p>
<hr />
<p>Here's a screenshot of part of Wikipedia article called "Einstein synchronization" included as evidence that what I wrote above is correct. I used a screen shot because copy pasting didn't work with the symbols. Sorry it's so small. Windows has a screen "magnifier" that may help (shortcut: press the equals sign key while holding down one of the Windows icon keys) or try right clicking in the image and selecting "open image in new tab" from the pop up menu.</p>
<p><a href="https://i.sstatic.net/zEPEp.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/zEPEp.png" alt="Screen shot of part of the Wikipedia article called @Einstein sychronization" /></a></p>
| 460
|
special relativity
|
How does Wigner rotation manifest itself in 3-forces?
|
https://physics.stackexchange.com/questions/687497/how-does-wigner-rotation-manifest-itself-in-3-forces
|
<p>There is no gravity. Two lasers, one in <span class="math-container">$(x, 0, 0)$</span> and the other in <span class="math-container">$(-x, 0, 0)$</span>, are motionless. Both of them are pointing upward (in z direction) toward circumference of a motionless disk. The disk will start moving upward in z direction, if we turn lasers on, due to the force <span class="math-container">$Fs$</span> exerted by the poynting vectors. Since the lasers are assumed to be the same, disk cannot experience torque, or any kind of rotation for that matter.</p>
<p>Now, assume that the disk is bounded to move in z direction only and two lasers are rotating on circumference of a circle placed on xy plane (with radius <span class="math-container">$x$</span> and center of <span class="math-container">$(0, 0, 0)$</span>) with constant speed. If we turn them on, again due to the symmetry of the system, disk starts moving upward without rotation; according to lab observer (S) who is at the rest with respect to disk at <span class="math-container">$t=0$</span>. We want to know the force <span class="math-container">$Fr$</span> exerted by one of the lasers on disk from S point of view. To do this, we can consider an inertial frame <span class="math-container">$S'$</span> which is momentarily at the rest with respect to one of lasers and use Lorentz transformation matrix on 4-force of <span class="math-container">$S'$</span> to get <span class="math-container">$Fr$</span>. According to <span class="math-container">$S'$</span>, poynting vector's force of the laser is <span class="math-container">$Fs$</span>, because after all, it is motionless in this frame. So we have</p>
<p><span class="math-container">$$F' = (0, 0, 0, Fs)^T$$</span>
<span class="math-container">$$F = \Lambda(vx,vy,0) F'$$</span></p>
<p>Where <span class="math-container">$vx$</span> and <span class="math-container">$vy$</span> are velocity of <span class="math-container">$S'$</span> in <span class="math-container">$S$</span> viewpoint and <span class="math-container">$\Lambda(vx,vy,0)$</span> is <em>pure</em> boost matrix given by</p>
<p><span class="math-container">$$ \Lambda(vx, vy, 0) = \left(
\begin{array}{cccc}
\gamma [\{\text{vx},\text{vy},0\}] & -\frac{\text{vx} \gamma [\{\text{vx},\text{vy},0\}]}{c} & -\frac{\text{vy} \gamma [\{\text{vx},\text{vy},0\}]}{c} & 0 \\
-\frac{\text{vx} \gamma [\{\text{vx},\text{vy},0\}]}{c} & \frac{\text{vx}^2 \gamma [\{\text{vx},\text{vy},0\}]+\text{vy}^2}{\text{vx}^2+\text{vy}^2} & \frac{\text{vx} \text{vy} (\gamma [\{\text{vx},\text{vy},0\}]-1)}{\text{vx}^2+\text{vy}^2} & 0 \\
-\frac{\text{vy} \gamma [\{\text{vx},\text{vy},0\}]}{c} & \frac{\text{vx} \text{vy} (\gamma [\{\text{vx},\text{vy},0\}]-1)}{\text{vx}^2+\text{vy}^2} & \frac{\text{vy}^2 \gamma [\{\text{vx},\text{vy},0\}]+\text{vx}^2}{\text{vx}^2+\text{vy}^2} & 0 \\
0 & 0 & 0 & 1 \\
\end{array}
\right)$$</span></p>
<p>So of course <span class="math-container">$F = F'$</span> but since we are interested in <span class="math-container">$Fr$</span> (z component of 3-force) we can simply use the fact that</p>
<p><span class="math-container">$$f = \frac{(0, 0, Fs)^T}{\Lambda_{11}} = (0, 0, Fs/\gamma [\{\text{vx},\text{vy},0\}] )^T$$</span></p>
<p>So <span class="math-container">$Fr = Fs/\gamma [\{\text{vx},\text{vy},0\}]$</span>, which is expected, since due to the rotational Doppler shift (or rather, the change of electric and magnetic field magnitudes) force of lasers decreases. However, the force of both lasers are the same, so net torque is zero.</p>
<p>Assume another observer S'' who moves in x direction with velocity <span class="math-container">$ux$</span> with respect to S. Again, we are interested in z component of 3-force of S''. This time however, since S'' and S' axis are not collinear, we have to consider Wigner rotation. Simply put, since S' and S are connected by <span class="math-container">$\Lambda(vx, vy, 0)$</span> and S'' and S are connected via <span class="math-container">$\Lambda(ux, 0, 0)$</span> we can say that S'' and S' are connected by</p>
<p><span class="math-container">$$F'' = \Lambda'F'=\Lambda(-ux,0,0) \Lambda(vx,vy,0) F'$$</span></p>
<p>which again gives <span class="math-container">$F'' = F'$</span> but for three force we have</p>
<p><span class="math-container">$$f' = \frac{(0, 0, Fs)^T}{\Lambda'_{11}}=\frac{(0, 0, Fs)^T}{\gamma [\{-u,0,0\}] \gamma [\{\text{vx},\text{vy},0\}]-\frac{u \text{vx} \gamma [\{-u,0,0\}] \gamma [\{\text{vx},\text{vy},0\}]}{c^2}}$$</span></p>
<p>Or
<span class="math-container">$$Fr'' = \frac{Fs}{\gamma [\{-u,0,0\}] \gamma [\{\text{vx},\text{vy},0\}]-\frac{u \text{vx} \gamma [\{-u,0,0\}] \gamma [\{\text{vx},\text{vy},0\}]}{c^2}}$$</span></p>
<p>Now, here is the problem. Under appropriate coordinate transformation <span class="math-container">$vx = -rwsin(wt)$</span> and <span class="math-container">$vy = rwcos(wt)$</span> we can immediately see that <span class="math-container">$Fr = \sqrt{1-r^2w^2/c^2}Fs$</span> but</p>
<p><span class="math-container">$$Fr'' = \frac{\sqrt{1-r^2w^2/c^2}\sqrt{1-u^2/c^2} Fs}{(1 + \frac{u \ r \ w \ \sin(wt)}{c^2})}$$</span></p>
<p>Note that all unprimed coordinates belong to S. The last equation makes me uncomfortable, because it is time dependent; meaning, depending on the location of lasers, their force changes. If we attach a few sensors on circumference of the disk for measuring force of lasers in z direction, according to S'' their value should change through time, periodically. While S argues that force is not time dependent, and S'' claim is not true. One might argue that the mechanism of sensors change from S'' point of view, which is true. But this change will not and cannot be time dependent, since S and S'' are connected via <span class="math-container">$\Lambda(u, 0, 0)$</span>. So, where is the problem?</p>
<p>P.S: This question was inspired from an article cited here <a href="https://physics.stackexchange.com/questions/685215/thought-experiment-and-possible-contradiction-between-kinematics-and-dynamics-in">Thought experiment and possible contradiction between kinematics and dynamics in special relativity</a> but I understood neither the article fully, nor the question of the author. I came up with my own version with simpler formalism.</p>
<p>Edit 1: If you are not comfortable with sensors and measuring forces, you can consider this: S can apply constant force <span class="math-container">$Fo = -Fr$</span> to disk, to keep its velocity constant. On the other hand, according to S'' we have <span class="math-container">$Fo''=-Fr/\gamma(u, 0, 0)$</span> which cannot cancel time dependent force of <span class="math-container">$Fr''$</span>.</p>
<p>Edit 2: In the case anyone wondering how <span class="math-container">$\Lambda'$</span> looks like, it is like this:</p>
<p><span class="math-container">$$\left(
\begin{array}{cccc}
\gamma [\{-u,0,0\}] \gamma [\{\text{vx},\text{vy},0\}]-\frac{u \text{vx} \gamma [\{-u,0,0\}] \gamma [\{\text{vx},\text{vy},0\}]}{c^2} & \frac{u \gamma [\{-u,0,0\}] \left(\text{vx}^2 \gamma [\{\text{vx},\text{vy},0\}]+\text{vy}^2\right)}{c \left(\text{vx}^2+\text{vy}^2\right)}-\frac{\text{vx} \gamma [\{-u,0,0\}] \gamma [\{\text{vx},\text{vy},0\}]}{c} & \frac{u \text{vx} \text{vy} \gamma [\{-u,0,0\}] (\gamma [\{\text{vx},\text{vy},0\}]-1)}{c \left(\text{vx}^2+\text{vy}^2\right)}-\frac{\text{vy} \gamma [\{-u,0,0\}] \gamma [\{\text{vx},\text{vy},0\}]}{c} & 0 \\
\frac{u \gamma [\{-u,0,0\}] \gamma [\{\text{vx},\text{vy},0\}]}{c}-\frac{\text{vx} \gamma [\{-u,0,0\}] \gamma [\{\text{vx},\text{vy},0\}]}{c} & \frac{\gamma [\{-u,0,0\}] \left(\text{vx}^2 \gamma [\{\text{vx},\text{vy},0\}]+\text{vy}^2\right)}{\text{vx}^2+\text{vy}^2}-\frac{u \text{vx} \gamma [\{-u,0,0\}] \gamma [\{\text{vx},\text{vy},0\}]}{c^2} & \frac{\text{vx} \text{vy} \gamma [\{-u,0,0\}] (\gamma [\{\text{vx},\text{vy},0\}]-1)}{\text{vx}^2+\text{vy}^2}-\frac{u \text{vy} \gamma [\{-u,0,0\}] \gamma [\{\text{vx},\text{vy},0\}]}{c^2} & 0 \\
-\frac{\text{vy} \gamma [\{\text{vx},\text{vy},0\}]}{c} & \frac{\text{vx} \text{vy} (\gamma [\{\text{vx},\text{vy},0\}]-1)}{\text{vx}^2+\text{vy}^2} & \frac{\text{vy}^2 \gamma [\{\text{vx},\text{vy},0\}]+\text{vx}^2}{\text{vx}^2+\text{vy}^2} & 0 \\
0 & 0 & 0 & 1 \\
\end{array}
\right)$$</span></p>
<p>I omitted this matrix from original post since I thought knowing its full form is useless, but I changed my mind. Of course, none of calculations was done by hand, it was done by Mathmatica (that's where those redundant curly brackets come from) so unless there is a problem with my logic, matrices should be fine. Moreover, I extracted 3-force from 4-force according to this Wikipedia page <a href="https://en.wikipedia.org/wiki/Acceleration_(special_relativity)" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Acceleration_(special_relativity)</a>, knowing that the first element of <span class="math-container">$\Lambda'$</span> represents <span class="math-container">$\gamma''$</span> since if <span class="math-container">$\Lambda'$</span> acts on four velocity of <span class="math-container">$S'$</span>, the coefficient of first element of the four velocity represents <span class="math-container">$\gamma''$</span> which in this case is <span class="math-container">$\Lambda'_{11}$</span>.</p>
| 461
|
|
special relativity
|
Understand result of Lorentz transform
|
https://physics.stackexchange.com/questions/689301/understand-result-of-lorentz-transform
|
<p>Assume an event that happens at P=(ct:2,x:4) in some inertial frame of reference S. Assume a second inertial frame S' in standard configuration and <span class="math-container">$\beta=4/5$</span> (<span class="math-container">$\gamma=5/3$</span>). The Lorentz transform is:</p>
<p><span class="math-container">$$ P'=\frac{5}{3}\begin{pmatrix}
1 & -4/5\\
-4/5 & 1\end{pmatrix}\begin{pmatrix}
2\\
4\end{pmatrix}=\begin{pmatrix}
-2\\
4\end{pmatrix}$$</span>.</p>
<p>That means that the event is perceived in S' before t=0. When S' pass by (ct:0,x:0) it could "say" to S the information about the event. In this way, S knows its future.</p>
<p>An error in my calculations or an apparent paradox ?</p>
|
<p>Information can never travel faster than light. You can draw a light cone eminating from the origin and the point <span class="math-container">$P$</span>, both before and after the transform. Since <span class="math-container">$P$</span> is outside the lightcone of the origin it can't communicate anything with the origin.</p>
<p>In relativity causality means that if a point <span class="math-container">$P$</span> is outside a light cone it is impossible to get it inside the light cone with a Lorentz transformation.</p>
| 462
|
special relativity
|
How does the principle of relativity imply that photon clocks and mechanical clocks experience time dilation the same way?
|
https://physics.stackexchange.com/questions/699949/how-does-the-principle-of-relativity-imply-that-photon-clocks-and-mechanical-clo
|
<p><strong>Context for this question:</strong> There is a famous thought experiment used to explain time dilation that uses two mirrors and a photon to set up a clock. The two mirrors are placed parallel to one another, and a photon is sent travelling perpendicular to the planes of the mirror, bouncing back and forth. Each time the photon hits a mirror, the clock ticks. When it is viewed by an observer travelling at relative velocity perependicular to the direction of the photon, the mirror-clock ticks more slowly due to the apparent zigzagging motion of the light. I started wondering why this is true in general, and not merely a feature of this particular type of clock. I found <a href="https://physics.stackexchange.com/questions/570966/why-is-the-photon-clock-equivalent-to-all-clocks">this earlier question</a> which asked just that. This question is a follow-up to the answer provided by robphy on that post.</p>
<p>Robphy states that all clocks must experience the same phenomenon by invoking the principle of relativity:</p>
<blockquote>
<p>An inertial observer carries both a light clock and a mechanical wristwatch, which agree when all are at rest. If they don't agree when the inertial observer is moving [with nonzero constant velocity] carrying these clocks, then that observer can distinguish being at rest from traveling with nonzero constant velocity.</p>
</blockquote>
<p>I don't understand this answer. Why wouldn't the two clocks agree agree? If the inertial observer is moving with nonzero constant velocity carrying the clocks, wouldn't the situation be identical to the observer being in the rest frame for both clocks? So why would the clocks have different measurements at all?</p>
|
<p>You seem to think that Robphy uses the principle of relativity as a kind of self-evident universal truth (like 1+1=2), which you could come up by yourself if you think hard enough. He then seems to deduce something even more obvious from it (equivalence of light clocks and everyday clocks). But you believe in neither truth and ask us how Robphy came to think that way.</p>
<p>The problem is that the principle of relativity is not something trivial like some axioms of math, but <strong>it is a law of nature that has been repeatedly confirmed by observation</strong>. We could well have lived in a different world, where it were possible to distinguish between absolute rest and motion. One possibility for such a distinction would be if light and everyday clocks went different when moving at various velocities.</p>
<p>But even if all clocks always go synchronously in all systems, that is not a "proof" of the principle of relativity. The principle of relativity tells us that there have never been found any experiments whatsoever (either with clocks or anything else), that allow us to distinguish between absolute rest and movement. Possibly, we have not searched hard enough and in 500 years from now we could find such an experiment, but at the moment that is the state of affairs (and to be sure, broad consensus is that it is pretty unlikely that special relativity will ever be broken). In that sense, finding such clocks or conditions would invalidate the principle of relativity, because the principle of relativity specifically says that there are no such clocks.</p>
<p>So we can't answer your question <a href="https://www.youtube.com/watch?v=36GT2zI8lVA" rel="noreferrer">"why"</a>, we can only confirm that none of us knows any way to distinguish between absolute rest and motion, of which differently going clocks depending on velocity would be one example. In the same sense, I don't know of anybody who can cancel gravitation. That doesn't mean that anti-gravitation doesn't exist, nor does it automatically imply that we just have to search hard enough to find anti-gravitation.</p>
| 463
|
special relativity
|
Relativistic inelastic collisions
|
https://physics.stackexchange.com/questions/731793/relativistic-inelastic-collisions
|
<p>Suppose a clay ball of mass <span class="math-container">$m$</span>, travelling at speed <span class="math-container">$v=\sqrt{3}c/2$</span>, collides with an identical clay ball that is at rest. They undergo a perfectly inelastic collision.</p>
<p>After performing the relevant calculations (using two separate methods), I found that <span class="math-container">$M=\sqrt{6}m$</span> (where <span class="math-container">$M$</span> is the mass of the composite ball) and <span class="math-container">$u=c/\sqrt{3}$</span> (where <span class="math-container">$u$</span> is the speed of the composite ball).</p>
<p>What I don't understand is how the mass of the composite ball can be greater than the sum of the two balls' separate masses, while its speed is also greater. Where does the extra energy come from?</p>
<p>I would have thought that if the mass of composite ball is greater than <span class="math-container">$2m$</span>, its speed will be smaller than the pre-collision speed of just the one clay ball. I've looked at similar questions online (such as a question apparently from the Feynman lectures, where <span class="math-container">$v=4c/5$</span>, rather than <span class="math-container">$\sqrt{3}c/2$</span>), and in each case, if the mass went up then the speed went down and vice versa.</p>
<p>An explanation would be much appreciated (and if I messed up my algebra, pointing that out would also be appreciated).</p>
<p>Note: I didn't use the expression 'rest mass' for any of the masses above because the question didn't either. Not sure if that's relevant/important.</p>
| 464
|
|
special relativity
|
How can I read articles by famous scientists for free?
|
https://physics.stackexchange.com/questions/729882/how-can-i-read-articles-by-famous-scientists-for-free
|
<p>I am high school student. I want to read the articles of famous scientists. These articles must be in their original language and not edited in any way. For example, I want to read Special Relativity, which was published by Einstein in 1905. In the German language and with an exact copy from 1905. Where can i find this article? I found a few sites(gutenberg project, libgen, sci-hub). But I don't know if these sites provide what I want. Can you guide me?</p>
|
<p>Einstein published in German Annalen der Physik, look this up in wiki
<a href="https://en.wikipedia.org/wiki/Annalen_der_Physik" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Annalen_der_Physik</a> and you find most of his german work free to download. for other authors go to there name in wiki and look up the biblographie and the journals they published in .</p>
| 465
|
special relativity
|
Spacetime diagram in Rindler coordinates
|
https://physics.stackexchange.com/questions/440400/spacetime-diagram-in-rindler-coordinates
|
<p>I am currently studying the <a href="https://en.wikipedia.org/wiki/Rindler_coordinates" rel="nofollow noreferrer">Rindler coordinates</a> </p>
<p><span class="math-container">$$T = x \sinh(a t) , \, X = x \cosh(at).$$</span></p>
<p>I am trying to understand the connection between the Rindler coordinates and their <a href="https://upload.wikimedia.org/wikipedia/commons/5/56/Rindler_chart.svg" rel="nofollow noreferrer">Minkowski diagram</a>. </p>
<ul>
<li>Why is <span class="math-container">$t= \pm \infty$</span> in the case of <span class="math-container">$x=0$</span> and why does that correspond to a line with a <span class="math-container">$45^\circ$</span> angle? </li>
<li>What are the straight lines for constant <span class="math-container">$t$</span> and how can I obtain them from the Rindler coordinates? </li>
<li>Why are the lines for constant <span class="math-container">$x$</span> curved?</li>
</ul>
<p>I feel like I am failing to understand this Minkowski diagram at a very basic level and clearing that up will help a lot. I already read the wikipedia page but I couldn't wrap my heard around the connection between the coordinates and the diagram yet.</p>
|
<p>Start with the coordinate system <span class="math-container">$(T,X)$</span>, so that every point in two-dimensional spacetime is labeled by a pair of numbers <span class="math-container">$(T,X)$</span>. Define <span class="math-container">$t$</span> and <span class="math-container">$x$</span> implicitly by
<span class="math-container">$$
T =x\,\sinh(at)
\tag{1}
$$</span>
<span class="math-container">$$
X =x\,\cosh(at).
\tag{2}
$$</span>
To depict the relationship graphically, consider a graph whose vertical and horizontal axes are labeled by <span class="math-container">$T$</span> and <span class="math-container">$X$</span>, respectively.
To the questions, consider the identities
<span class="math-container">$$
\frac{T}{X} = \frac{\sinh(at)}{\cosh(at)}
\tag{3}
$$</span>
<span class="math-container">$$
X^2-T^2=x^2.
\tag{4}
$$</span>
Now, here are the answers to the specific questions, in a different order.</p>
<blockquote>
<p>What are the straight lines for constant <span class="math-container">$t$</span> and how can I obtain them from the Rindler coordinates? </p>
</blockquote>
<p>Equation (3) says that the ratio <span class="math-container">$T/X$</span> is completely determined by <span class="math-container">$t$</span>, regardless of <span class="math-container">$x$</span>. Therefore, the points with a given value of <span class="math-container">$t$</span> are the same as the points with a given ratio <span class="math-container">$T/X$</span>. These are straight lines through the origin in the <span class="math-container">$T,X$</span> plane. The maximum and minimum possible slopes are <span class="math-container">$+1$</span> and <span class="math-container">$-1$</span>, respectively, because these are the maximum and minimum possible values of the right-hand side of equation (3). </p>
<blockquote>
<p>Why are the lines for constant <span class="math-container">$x$</span> curved?</p>
</blockquote>
<p>Equation (4) says that the combination <span class="math-container">$X^2-T^2$</span> is completely determined by <span class="math-container">$x$</span>, regardless of <span class="math-container">$t$</span>. Therefore, the points with a given value of <span class="math-container">$x$</span> are the same as the points with a given value of <span class="math-container">$X^2-T^2$</span>. To visualize this, choose a particular value for <span class="math-container">$x$</span> and write equation (4) like this:
<span class="math-container">$$
X = \pm\sqrt{T^2 + x^2}.
\tag{5}
$$</span>
Since <span class="math-container">$x$</span> is a constant (we chose its value), this equation gives us <span class="math-container">$X$</span> as a function of <span class="math-container">$T$</span>. We get two curves, one for each sign of the square root. For <span class="math-container">$x\neq 0$</span>, each of these curves is a hyperbola. The one on the negative-<span class="math-container">$X$</span> side passes through the point <span class="math-container">$(T,X)=(0,-|x|)$</span>, and the one on the positive-<span class="math-container">$X$</span> side passes through the point <span class="math-container">$(T,X)=(0,|x|)$</span>. To see that the asymptotic slopes of these hyperbolas are <span class="math-container">$\pm 1$</span>, consider equation (5) in the limit <span class="math-container">$T\rightarrow \pm\infty$</span>. In this limit, the term <span class="math-container">$x^2$</span> is negligible, which means that the asymptotes are <span class="math-container">$X=\pm\sqrt{T^2}=\pm|T|$</span>. The smaller the value of <span class="math-container">$x^2$</span>, the more closely the hyperbolas "hug" these asymptotes. In the limiting case <span class="math-container">$x=0$</span>, the hyperbolas become the asymptotes themselves. And this answers the next question...</p>
<blockquote>
<p>Why is <span class="math-container">$t=\pm \infty$</span> in the case of <span class="math-container">$x=0$</span> and why does that correspond to a line with a <span class="math-container">$45^\circ$</span> angle? </p>
</blockquote>
<p>Equation (4) says that if <span class="math-container">$x=0$</span>, then <span class="math-container">$X=\pm T$</span>. These are the two <span class="math-container">$45^\circ$</span> lines through the origin. And if <span class="math-container">$X=\pm T$</span>, then equation (3) says <span class="math-container">$t=\pm \infty$</span>. As described above, this is just a limiting case of the pair of hyperbolas we get for any given non-zero value of <span class="math-container">$x$</span>.</p>
| 466
|
special relativity
|
Symmetrical twin paradox without changing direction
|
https://physics.stackexchange.com/questions/201700/symmetrical-twin-paradox-without-changing-direction
|
<p>If I understood well, in the special relativity </p>
<p>1- A stationary observer sees other moving observer's clock works more slowly than the stationary clock. </p>
<p>2- Motion is relative, the moving observer thinks the stationary observer is moving.</p>
<p>I cannot understand how it is possible then. Because each observer claims the other observer's clock works more slowly which obviously cannot be true. <strong>It is like to have two numbers each smaller than the other, it is not possible</strong>.</p>
<p>Suppose observers A and B are at rest in the origin and simultaneously move in two opposite direction with the same acceleration until they reach the velocity 0.9c and then stop accelerating.
Then after some time one observer in the origin send two flashes of light in opposite direction toward A and B. When they receive the light, A and B record what their clock shows and send the result to the origin. </p>
<p>Because the situation is symmetric their clock must show the same number, but because each of them sees the other as moving one each thinks the other clock ticks more slowly. </p>
<p>What is wrong in this paradox?</p>
<p><a href="https://i.sstatic.net/7XcPL.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/7XcPL.jpg" alt="enter image description here"></a></p>
|
<blockquote>
<p>It is like to have two numbers each smaller than the other, it is not
possible.</p>
</blockquote>
<p>It's not like that at all. Consider the following true statement:</p>
<ul>
<li>B observes A's clock to run slowly while A observes B's clock to
run slowly.</li>
</ul>
<p>This is not a contradiction due to the crucial word <em>observes</em>. Consider an additional true statement:</p>
<ul>
<li>The observer at the origin observes A's clock and B's clock to run at
the same rate as each other but slower than his clock while both A &
B observe the origin's clock to run slower.</li>
</ul>
<p>To see that this is not a contradiction requires clearly thinking about what it means to <em>observe</em> (which doesn't mean <em>see</em>) - how <em>does</em> one determine the rate of a <em>moving</em> clock?</p>
<p>Here's one way:</p>
<ul>
<li>Position two clocks at rest in your lab and spatially separated along
the line of motion of the moving clock.</li>
<li><a href="https://en.wikipedia.org/wiki/Einstein_synchronisation" rel="nofollow noreferrer"><em>Synchronize</em> the two clocks</a></li>
<li>As the moving clock passes the first clock, record both the time on
the moving clock and the first clock</li>
<li>As the moving clock passes the second clock, record both the time on
the moving clock and the second clock</li>
</ul>
<p>Carefully note that it requires two, <em>synchronized</em> clocks in the lab to <em>observe</em> the elapsed time on the moving clock.</p>
<p>According to the lab technician, the elapsed time in the lab is $\Delta t = (t_2 - t_1)$ (since the two lab clocks are synchronized) which is <em>greater</em> than the elapsed time of the moving clock. Thus, the lab technician can validly claim that the moving clock is running slowly compared to the lab clocks.</p>
<p>However, while the two lab clocks are synchronized in the lab frame, <em>the two lab clocks are not synchronized in the moving clock's rest frame</em> (<a href="https://en.wikipedia.org/wiki/Relativity_of_simultaneity" rel="nofollow noreferrer">relativity of simultaneity</a>).</p>
<p>This is the key to the 'paradox'; according to the moving clock, the lab clocks are not synchronized and thus, $(t_2 - t_1)$ is <em>not</em> a valid elapsed time.</p>
<p>To better understand how this works 'both ways', carefully consider the cute spacetime diagram from <a href="https://physics.stackexchange.com/a/111089/9887">this answer</a>:</p>
<p><a href="https://i.sstatic.net/2RVcd.gif" rel="nofollow noreferrer"><img src="https://i.sstatic.net/2RVcd.gif" alt="enter image description here"></a></p>
<p><a href="http://www.phys.vt.edu/~takeuchi/relativity/notes/section12.html" rel="nofollow noreferrer">Image credit</a></p>
| 467
|
special relativity
|
Regarding synchronization of clocks in special relativity
|
https://physics.stackexchange.com/questions/203920/regarding-synchronization-of-clocks-in-special-relativity
|
<p>I am trying to read of synchronization of two clocks in same inertial frame in special relativity. Suppose we have two synchronized clocks in an inertial frame placed at positions $x_1$ and $x_2$ in that frame. Suppose two observers at $x_1$ and and $x_2$ try to measure speed of some object moving in between $x_1$ and $x_2$(not at the midpoint or any close) with a constant speed. Now both observers, watch the object travel the distance $ \Delta x$. Let the object be closer to $x_1$. The observer at $x_1$ records times at which the object enters the $\Delta x$ region and another when it leaves the region. Lets call them $t_1$ and $t'_1$ and similarly the observer at $x_2$ records $t_2$ and $t'_2$.
<a href="https://i.sstatic.net/EOgG8.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/EOgG8.jpg" alt="enter image description here"></a></p>
<p>Now the time difference between $t_1$ and $t'_1$ is should not be same as the difference between $t2$ and $t'2$ since the light travels at a finite speed. So the velocities as measure by $x_1$ and $x_2$ observers should be different because the time intervals measured by them are different. This would imply both the observers measure different speeds. But I am wondering how is this possible? How can an object have two different speeds in same inertial frames? Am I missing something?</p>
<p>(PS: sorry for bad figure)</p>
|
<p>There is a fundamental flaw in your setup. While the actual times the entrance signals may reach clocks 1 and 2 could be different, $t_1'-t_1$ will be the same as $t_2'-t_2$, because the clocks are at rest with respect to each other.</p>
<p>In fact, consider there are two events: entrance and exit. SR says that $$c^2\Delta t^2 - \Delta x ^2 = c^2\Delta t'^2-\Delta x'^2$$ for all inertial reference frames observing the two events. You agree that your observers in the same reference frame as each other measure the same $\Delta x$ as each other. Then they <em>must</em> measure the same $\Delta t$ or the equality wouldn't hold.</p>
| 468
|
special relativity
|
Why do objects traveling faster appear shortened rather than elongated?
|
https://physics.stackexchange.com/questions/220921/why-do-objects-traveling-faster-appear-shortened-rather-than-elongated
|
<p>I'm sure this seems like a stupid question but I hope someone can try to explain it as if to a child...</p>
<p>This relates to Lorentz contraction. Why do objects in motion relative to the observer not get <em>longer</em>?</p>
<p>In the train example. A stationary observer is at the middle of a train car which is traveling from the observer's left to his right. A flashbulb goes off in the middle of the train car as it passes the observer.</p>
<p>Now, I understand the speed of light is constant from the observer's POV regardless of the speed of the source. So the observer sees the light hit the left side of the car before it hits the right. </p>
<p>If the car is going 0.99x the speed of light, the observer at rest sees the light reach the left side of the car in (let's say?) 1/100th the time it would take if the car was at rest relative to the observer -- so since we're measuring the distance by the constant speed of light, the left end is shorter. That is essentially what the Lorentz contraction is, isn't it? So my question is: If that's true, then why wouldn't the observer also see it take 99x longer for the light to catch up with the right end of the car -- making the car appear exponentially longer until, if it were going at the speed of light, it would appear infinitely long to the right of the observer and infinitely short to his left?</p>
<p>In trying to understand this, I found these two diagrams on Wikipedia, and what I noticed was that the bottoms of the vertical sticks aren't in the same place. And if they were, then more speed would make the left side shorter approaching zero, and the right side longer approaching infinity.</p>
<p><a href="https://i.sstatic.net/yi1EQ.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/yi1EQ.png" alt="enter image description here"></a>
<a href="https://i.sstatic.net/EWDGZ.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/EWDGZ.png" alt="enter image description here"></a></p>
|
<p>Actually your perception is correct in regards to the events you consider, but the conclusion does not contradict the concept of length contraction. </p>
<blockquote>
<p>In short, what you are looking at are <em>simultaneous events in the train's rest frame</em>. These do measure the length of the train in the rest frame, but <strong>not</strong> its contracted length as seen by the 'stationary' observer. The reason is that with respect to the stationary observer these events are <strong>no longer simultaneous</strong> due to relativity of simultaneity. In fact, as you correctly observe, to the stationary observer they appear separated by an interval of time that goes to infinity as the speed of the train approaches that of light. Correspondingly, the distance between them as seen by the stationary observer also goes to infinity, and there is nothing wrong with that! </p>
<p>So where does length contraction come from? Actually it comes from the fact that the stationary observer measures the train length, as he observes it, <em>by using events that are simultaneous in his stationary frame</em>, but not simultaneous in the train's frame. </p>
</blockquote>
<p>At this point I prefer to back up my statements with a bit of math to avoid misunderstandings, since there seem to be plenty when it comes to this topic. </p>
<p>Let O be the stationary observer and O' be the train's observer (at rest in the train's frame) in relative motion at velocity $v = \beta c$ ($\beta = \frac{v}{c}$, $\gamma = \frac{1}{\sqrt{1-\beta^2}}$). Take the origin of O' in the center of the train as usual and let A be the left (rear) end of the train, and B the right (forward) end. Let the length of the train be $2L$, such that the positions of A and B wrt O' are $x_A' = -L$ and respectively $x_B'=L$. </p>
<ul>
<li><p>In his frame, O' measures the train's length by using events that are simultaneous at A and B. We could argue that this is not necessary since A and B are at rest wrt O', but the procedure must be the same regardless of whether it is applied to objects at rest or in motion. Since there is no better way to properly define it for objects in motion, we also extend the procedure as such for objects at rest. So lets say that O' measures the distance between A and B using events $\mathcal{A} = (-L, ct'_0)$ and $\mathcal{B} = (L, ct'_0)$ at his time $ct'_0$. What does the stationary observer O see instead? To him event $\mathcal{A}$ has coordinates
$$
x_A = \gamma(x_A' + \beta ct'_0) = \gamma(-L + \beta ct'_0)\\
ct_A = \gamma(ct'_0 +\beta x_A') = \gamma(ct'_0 -\beta L)
$$
while event $\mathcal{B}$ occurs at
$$
x_B = \gamma(L + \beta ct'_0) = x_A + 2 \gamma L\\
ct_B = \gamma(ct'_0 + \beta L) = ct_A + 2 \beta \gamma L
$$
Obviously he observes $\mathcal{A}$ and $\mathcal{B}$ occurring at a distance
$$
x_B - x_A = 2 \gamma L \equiv \gamma( x_B' - x_A') > ( x_B' - x_A') = 2L
$$
and <em>separated in time</em> by
$$
c\Delta t = ct_B - ct_A = 2 \beta \gamma L \equiv \beta (x_B - x_A) > 0
$$
These are the two events you were considering and, again obviously, they don't show a length contraction, but a length dilation! Yep, but this length dilation occurs between non-simultaneous events, so it cannot measure the train length as it is observed by O.</p></li>
<li><p>Let's see how O <em>really</em> measures the train's length. For this purpose O has to consider events or observations $\bar{\mathcal{A}} = (\bar{x}_A, ct_0)$ and $\bar{\mathcal{B}} = (\bar{x}_B, ct_0)$ that are <em>simultaneous with respect to him</em> at some time $ct_0$, although in O' they still occur at the same locations $x_A' = -L$ and $x_B' = L$, at the train's ends. This means that the relation between the coordinates of $\bar{\mathcal{A}}$ and $\bar{\mathcal{B}}$ in the two frames must be
$$
\bar{x}_A = \gamma ( -L + \beta c\bar{t}_A')\\
ct_0 = \gamma (c\bar{t}_A' -\beta L)
$$
and
$$
\bar{x}_B = \gamma ( L + \beta c\bar{t}_B')\\
ct_0 = \gamma (c\bar{t}_B' + \beta L)
$$
If we eliminate the O' times $c\bar{t}_A'$ and $c\bar{t}_B'$, and then look at the distance between $\bar{\mathcal{A}}$ and $\bar{\mathcal{B}}$ as observed in O we obtain
$$
\bar{x}_B - \bar{x}_A = \gamma(L + \frac{\beta}{\gamma}ct_0 - \beta^2 L - (-L) - \frac{\beta}{\gamma}ct_0 - \beta^2 L) = \gamma (1- \beta^2) 2L
$$
or
$$
\bar{x}_B - \bar{x}_A = \frac{2L}{\gamma}
$$</p></li>
</ul>
<blockquote>
<p>In other words, O does see the train length contracted by a factor of $\frac{1}{\gamma}$, although he sees distances between events that are simultaneous in O' as <em>dilated</em> by a factor of $\gamma$.</p>
</blockquote>
| 469
|
special relativity
|
Lorentz contraction of object in circular motion
|
https://physics.stackexchange.com/questions/231614/lorentz-contraction-of-object-in-circular-motion
|
<p>Apparently a bunch of people totally misunderstood my <a href="https://physics.stackexchange.com/questions/231505/what-happens-to-wheels-of-a-car-moving-near-speed-of-light">previous question</a> and choose to ignore the clarifying comments. Let me change the conditions to remove all the confusion.</p>
<p><a href="https://i.sstatic.net/WZIuJ.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/WZIuJ.png" alt="enter image description here"></a></p>
<p>A group of particles forming a circle of uniform density is moving in a circular path (following that circle shape), at a moderate speed $v_1$ (as observed from a frame of reference bound to the center of the circular path, O1). Let's say this is a cloud of ions filling a cyclotron that had accelerated them to 0.2c and then just maintains their circular trajectory with magnetic field without accelerating them any further. (depicted is momentary speed vector of one of these particles)n accelerator. </p>
<p>An observer (O2) moving at $v_2$ near speed of light (say, 0.9c relative to the previous observer (O1) / the cyclotron) observes the positions and movement of these particles. How will they appear from that observer's point of view - what is their apparent their trajectory and how does their density vary from that observer's point of view? </p>
|
<p>I tried to put it into a figure the results from reasoning in the framework of Special Theory of Relativity, this is what the observer would see:</p>
<p><a href="https://i.sstatic.net/asKtf.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/asKtf.png" alt="enter image description here"></a></p>
<p>Where you should notice several effects:</p>
<ul>
<li><strong>length contraction</strong> along observer's motion causes the ring to look like an elipse</li>
<li><strong>equal velocities on equator</strong> where particles are crossing from top to bottom or vice-versa their velocities are perpendicular to observer's hence not affected (0.2c on either side)</li>
<li><strong>fast transit through "bottlenecks"</strong> the apparent concentration of particles on top and bottom, which I mean by bottlenecks, are just an effect of length contraction, but particles are not really stuck there. Actually they are faster than in the rest of the ring (see velocity colorbar). Also the peak velocity on top is higher (.93c) than on bottom (.85c) because on the former they travel towards observer, while on the latter they travel away from him. But the observer sees all of them travelling towards him in both cases with the mentioned peak velocities.</li>
<li><strong>spacing asymmetry top vs bottom</strong> the different speeds top vs bottom will affect the spacing between particles, the top looking more crowded than the bottom. However the time contraction will counter this effect and and at all times there should be the same amount of them on either side.</li>
<li><strong>slanting</strong> also due to speed differences top vs bottom there will be a kind of slanting/forward-fall effect on the shape of the circle.</li>
</ul>
<hr>
<p><strong>Note</strong>
The figure is a representation of the effects and might exaggerate some of them, since is not made by calculating the actual values. The speeds-colorbar only shows the value of horizontal component and is also exaggerated, since minimum values should be .9c instead of 0.</p>
| 470
|
special relativity
|
Feynman Lectures on Physics, Michelson Morley question about angle of light
|
https://physics.stackexchange.com/questions/238535/feynman-lectures-on-physics-michelson-morley-question-about-angle-of-light
|
<p>My question is similar, if not identical, to <a href="https://physics.stackexchange.com/questions/214247/feynman-lectures-on-physics-the-michelson-morley-experiment">this one</a>, but I don't find the answer satisfying, given the context of experiment.</p>
<p>First, here is an outline my understanding of the motivation behind the experiment:</p>
<p><a href="https://en.wikipedia.org/wiki/Galileo%27s_ship#The_proposal" rel="nofollow noreferrer">Galilean invariance</a> states that the laws of motion are identical across inertial frames of reference.</p>
<p>However, it was also known (or suspected), that the speed of light is independent of the speed of its source (as is the case in sound waves). </p>
<p>This presented the apparent opportunity to violate Galilean invariance, and the Michelson Morley experiment is a famous example of an attempt to do just this.</p>
<p>Based on the assumptions and working knowledge of the scientists of the time, if the interferometer was traveling through the ether, then the interference pattern should differ when the contraption was rotated, and this would confirm that the contraption was moving with respect to the ether (i.e. with respect to absolute space), and allow a calculation of the velocity of the Earth (assuming the contraption and the Earth were in the same inertial frame).</p>
<p>The key motif of the Lorentz transformation can be derived from this analysis, which is based upon the Pythagorean theorem:</p>
<p>$1/\sqrt{1-v^2/c^2}$. </p>
<p>Here is a schematic of the Michelson-Morley apparatus, borrowed from <a href="http://www.feynmanlectures.caltech.edu/I_15.html#Ch15-S3" rel="nofollow noreferrer">here</a></p>
<p><a href="https://i.sstatic.net/S4tzR.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/S4tzR.png" alt="Interferometer"></a></p>
<p>The idea here is that, since we had not yet discovered that the speed of light is invariant across all observers within and across all inertial frames, we would <em>expect</em> two particular round trip times for both beams of light, and these times would depend upon which direction the contraption was moving, and at what speed.</p>
<p>I can fully follow and understand the calculation for the <strong>B</strong> - <strong>E'</strong> - <strong>B'</strong> roundtrip. </p>
<p>I can also fully follow the calculation for the <strong>B</strong> - <strong>C'</strong> - <strong>B'</strong> roundtrip, but I can't <em>understand</em> it. In particular, I don't understand why the first half of this round trip strikes the center of the mirror <strong>C'</strong>. Remember, those scientists already understood that the speed of light is independent of the speed of its source, so it wouldn't be like bouncing a ball against the roof of a moving vehicle, where the ball inherits the forward speed of the vehicle. Rather, the beam should strike a point behind the center of <strong>C'</strong> (i.e. to its left). In other words, to an outside observer, the beam would be expected to go straight up and down.</p>
<p>The only way I can reconcile this is that Feynman here is not talking about a laser beam, but rather a point source where the light spreads out in all directions. And in that case, the analysis is done for the particular photon whose angled path through space was such that it struck the center of <strong>C'</strong>. However, I don't think this is the correct answer to my question, as the same issue crops up later in the discussion of a moving light clock that involves a single photon.</p>
<p>Another way of asking this question is as follows:</p>
<p><em>According to the way of thinking at the time of Michelson-Morley</em></p>
<p>If I were to aim a laser beam at a very distant target, and both myself and the target were moving rapidly in a direction perpendicular to the line between myself and the target, then if the distance between myself and the target was great enough, then, by the time the laser beam reaches the target, it will have dodged the laser beam.</p>
<p>Yet, according to my reading of Feynman, their expected calculations imply that the light is carried along by the velocity of the source, which contradicts what they apparently already knew at the time.</p>
<p>What am I missing here?</p>
|
<p>I may be wrong, but I think the effect of the ether would just be to alter the alignment of the mirrors slightly. It is kind of analogous to hitting a golf ball or kicking a football in the wind. If the wind is blowing one way, you have to aim a little the other way to correct for the wind.</p>
<p>For example, in the figure you give, the apparatus is moving to the right through the ether, or, in the apparatus's frame, the ether is moving to the left. When you are aligning the optics, you will try to get the spot to land on the mirror $C$. If the ether is still, you can just have the mirror at a 45 degree angle, but when the either is moving to the left, you will have to tilt the mirror so as to aim the beam a little more to the right. Similarly, for the return trip you would have to alter the alignment of mirror $C$ to correct for the either. </p>
<p>However, since you align the mirrors by looking at where the spot from the light is instead of trying to precisely set the angle of the mirror to 45 or 90 degrees, you wouldn't really be able to tell during the alignment whether there is an ether. The ether doesn't really affect your alignment procedure.</p>
<p>Hopefully this answers your question.</p>
| 471
|
special relativity
|
Does light still slow down in the direction of motion?
|
https://physics.stackexchange.com/questions/240143/does-light-still-slow-down-in-the-direction-of-motion
|
<p>This is quite a naive question however I hope to learn from this - I had always learnt a light clock in a space ship is placed like so :</p>
<p><a href="https://i.sstatic.net/YK2wc.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/YK2wc.png" alt="enter image description here"></a></p>
<p>That said as the light moves the light will seem to slow down due to the velocity of light compensating for the horizontal movement of the space ship. </p>
<p>Now if I were to keep the light like so would it still slow down:</p>
<p><a href="https://i.sstatic.net/MRQIM.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/MRQIM.png" alt="enter image description here"></a></p>
<p>I feel that the light would not slow down because of the fact that light is simply bouncing about and the ship also contracts in the direction leading to the interval of bounce not changing, but somehow I feel I am incorrect so could someone explain why even a horizontal light clocks light slows down. </p>
<p>Thanks</p>
|
<p>Let's calculate it then to gain some intuition. For a stationary horizontal light clock, the time for one tick will be given by $$t=\dfrac{2l}{c}$$ where $l$ is the distance between its two ends.</p>
<p>What about the moving one? First, as you pointed out $l$ will be length contracted by a factor of $\gamma=\dfrac{1}{\sqrt{1-v^2/c^2}}$</p>
<p>So the time taken for one tick(as measured in the frame at rest) will be given by the sum of two times $t_1$ and $t_2$, $t_1$ being the time the light took from the start to the end, $t_2$ is from the end to the start </p>
<p>$$t'=t_1 +t_2 =\dfrac{l/ \gamma}{c-v} + \dfrac{l/ \gamma}{c+v}=\dfrac{2lc}{\gamma (c^2-v^2)}$$</p>
<p>Now $$\dfrac{t}{t'}=\dfrac{\gamma (c^2-v^2)}{c^2}=\dfrac{1-v^2/c^2}{\sqrt{1-v^2/c^2}}=\sqrt{1-v^2/c^2}$$ so $\dfrac{t}{\sqrt{1-v^2/c^2}}= \gamma t =t'$ and since $\gamma$ is always greater than one,then the one tick for $t'$ will always be greater than that of $t$.</p>
<p>To give you a feeling for this, if $\gamma=2$ and the clock at rest reads $t=10$ secs, if we were to look at the moving clock, its reading would be $t'=5$ secs, since its 'tick' is half as slow as $t$, that is $t'$ always needs double the time $t$ needs to register one tick(or 1 sec), so $t'$ will always lag behind $t$ by a factor of 2. </p>
| 472
|
special relativity
|
Direction of the lightbeam in SRT thought experiments
|
https://physics.stackexchange.com/questions/244701/direction-of-the-lightbeam-in-srt-thought-experiments
|
<p>Extending on the <a href="https://physics.stackexchange.com/questions/241792/time-dilation-diagram-on-wikipedia">question about the time dilation on Wikipedia</a>, namely this diagram in the <a href="https://physics.stackexchange.com/a/241807/56376">accepted answer</a>:</p>
<p><a href="https://i.sstatic.net/Ue5Xi.gif" rel="nofollow noreferrer"><img src="https://i.sstatic.net/Ue5Xi.gif" alt="Stationary vs moving frame of reference" /></a>
<a href="http://www.pitt.edu/%7Ejdnorton/teaching/HPS_0410/chapters/Special_relativity_clocks_rods/index.html" rel="nofollow noreferrer"><sub>Image credit: John D. Norton</sub></a></p>
<p>The thing that I don't understand is, <strong>why</strong> does the light appear to be travelling at an angle for the observer outside the moving frame (the one containg mirrors)?</p>
<p>Physically speaking, what forces the light beam emitted by a moving source not to go simply <strong>straight up vertically</strong>, missing the upper mirror completely?</p>
<p>I am fully aware why this works in Minkowski spacetime; all observers must see the spacetime event of the beam hitting the upper mirror, and I understand relativistic aspects like relativity of simultaneity, time dilation and comparing discrete spacetime events in Minkowski spacetime diagrams.</p>
<p>But in this case I am confused:</p>
<ol>
<li>presuming there is vacuum between the plates, what properties of the "empty space" between these mirrors cause it to be slanted, once it leaves the emitter?</li>
<li>is the light supposed to be treated like a particle, and it's actually affected by the train speed?</li>
<li>is the light supposed to be treated like a wave radiating from the emitter, and the point where it hits the mirror is actually not the top of this sphere wave?</li>
</ol>
<p><strong>(Update)</strong></p>
<p>My confusion is deepened by the following statement by @JohnDuffield in <a href="https://physics.stackexchange.com/a/228843/56376">this answer</a>:</p>
<blockquote>
<p>(OP's question:) The reason a wave such as sound would have the trajectory shown in this example is that the medium inside the rocket, air, is moving at the speed of the rocket and the sound wave would take on that velocity as it left it's source. Light does not use a medium to move.</p>
<p>Answer: <strong>It does.</strong> Have a look at Nobel Laureate Robert B Laughlin <a href="https://en.wikipedia.org/wiki/Aether_theories#Quantum_vacuum" rel="nofollow noreferrer">here</a>: "It is ironic that Einstein's most creative work, the general theory of relativity, should boil down to conceptualizing space as a medium when his original premise [in special relativity] was that no such medium existed".</p>
</blockquote>
<p>So, given the link to the <a href="https://en.wikipedia.org/wiki/Aether_theories#Quantum_vacuum" rel="nofollow noreferrer">wikipedia article on aether</a>, this answer seems to imply that aether in fact not only exists, but is also being <em>dragged</em> by the train?</p>
|
<p>Forthcoming answers will, I'm sure, invoke equivalency and other Relativity theory to explain the observed: here I intend to begin with observables and derive the theory.</p>
<p>A photon emitted from a particle is emitted in no preferred direction: it has an equal chance of going any direction in space. In order to generate a directed beam such as a laser you have to have some way to block or direct the emitted photons such that the ones that make it out are going the same way. A laser in particular does this with a tube that has an aperture at one end, so that the photons going one direction get out and the photons going any other direction stay in.</p>
<p>Now, if you have a laser pointed straight up but moving to the right, the photons emitted inside the laser that are going straight up <em>do not</em> escape the aperture: they are intercepted by the side of the tube. However, some of the photons going up <em>and</em> to the right do get out. They were emitted toward the side of the tube, but then the side kept moving until, by the time the photon gets to the boundary of the tube, the aperture is there and the photon escapes. And since it was moving somewhat to the right inside the tube, it continues moving to the side as it goes up, which to the stationary observer looks like a diagonal instead of "straight up."</p>
| 473
|
special relativity
|
Relativity paradox with mirrors and light pulses
|
https://physics.stackexchange.com/questions/259199/relativity-paradox-with-mirrors-and-light-pulses
|
<p><a href="https://i.sstatic.net/1tQFG.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/1tQFG.png" alt="Experiment seen from two different inertial frames."></a></p>
<p>Consider two very short light pulses emitted from the centre (C) of two mirrors A and B (as shown in the diagram).
From the point of view of the lab frame, the apparatus is all moving to the left at velocity v.
Imagine there is also an electron near the centre of the apparatus, which is stationary in the apparatus frame and therefore also moving with velocity v to the left according to the lab frame.</p>
<p>The short light pulses (much shorter than the apparatus length) bounce off mirrors A and B and return and strike the electron.
This situation has similarities with the Michelson-Morley experiment.</p>
<p>According to the frame moving with the apparatus, the pulses take an equal time to bounce off the mirrors and arrive back at C. Therefore the EM waves cancel and there is no net radiation pressure exerted on the electron.</p>
<p>According to the lab frame, the light pulse emitted to the left has less distance to travel overall and so arrives at C before the pulse that was emitted to the right. Therefore the first pulse accelerates the electron by exerting a radiation pressure on it.</p>
<p>Does the electron accelerate or not? :)</p>
<p>(I'm looking for derivations/proofs showing both frames' interpretations)</p>
|
<p>If $t_{CA}$ refers to the time it takes in the lab frame for the light to reach C from A, and the same with $t_{AC}$, $t_{CB}$ and $t_{BC}$ then we have:</p>
<p>$t_{CA}=\frac{L/2+v t_{CA}}{c}$</p>
<p>$t_{AC}=\frac{L/2-v t_{AC}}{c}$</p>
<p>$t_{CB}=\frac{L/2-v t_{CB}}{c}$</p>
<p>$t_{BC}=\frac{L/2+v t_{BC}}{c}$</p>
<p>Thus $t_{CA}=t_{BC}$ and $t_{AC}=t_{CB}$</p>
<p>Finally: $t_{CA}+t_{AC}=t_{CB}+t_{BC}$ </p>
| 474
|
special relativity
|
Special relativity: is this a known paradox, or one at all?
|
https://physics.stackexchange.com/questions/262076/special-relativity-is-this-a-known-paradox-or-one-at-all
|
<p>Two ships of the same proper length $L$ move towards each other, as in the diagram below (which shows it in the reference frame where the ship at the left is at rest). The fronts (noses) are pointing to each other.</p>
<p><a href="https://i.sstatic.net/Dw0QB.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/Dw0QB.jpg" alt="enter image description here"></a></p>
<p>Now, when both noses pass each other, they synchronize their clocks to zero.
Then, when the ships are at the end of passing each other their backs meet, and they stop the watches. </p>
<p>From the point of view of the ship at rest, its clock shows </p>
<p>$\Delta t=(L+L')/v$</p>
<p>From the point of view of the moving ship, its clock shows the same:</p>
<p>$\Delta t'=(L'+L)/v$</p>
<p>This is expected because of the symmetry of the problem. </p>
<p>On the other hand, because of time dilation, we expect that each ship sees the other clock running slower that its own, so each expects that the clock on the other ship will stop showing a smaller time interval than its own. Which I believe it does not happen. None of the ships accelerates or decelerates during the process, thus none experience a change in inertial frames. </p>
<p>Question: which of the two arguments is wrong (the "symmetry" or the "time dilation") and why? </p>
|
<p>I said this in comments already, but I suppose it should be an answer.</p>
<p>Everything depends on where the clocks are. If they're at the front of the ships, then they get synchronized when the fronts pass, and both pilots agree that it's noon. An hour later (according to each of them), the backs of the ships pass. Each pilot says "We passed at 1PM. At that moment, my clock turned off. Five minutes later, at 1:05, the other guy's clock turned off --- but at that moment his clock said 1:00 because it always did run slow."</p>
<p>If the clocks are somewhere else (say the backs of the ships), the story will be somewhat different but will have the same flavor. Likewise, if you want to change the assumption about exactly what happens at 1:00 according to Pilot A (I assumed that he says his own clock turned off; you could instead assume that he says Pilot B's clock turned off), you'll get a somewhat different story with the same flavor.</p>
<p>You seem (in the comments) to suggest that each ship can have two clocks, one at the front and one at the back, that are synchronized with each other. But of course if one pilot says they're synchronized, the other must disagree --- so you have to be very careful about who says which pair is synchronized. Once you specify that carefully, everything comes clear.</p>
<p>And as always --- if you stop to draw a spacetime diagram, you'll avoid getting confused in the first place.</p>
| 475
|
special relativity
|
Special Relativity with no short cuts
|
https://physics.stackexchange.com/questions/277520/special-relativity-with-no-short-cuts
|
<p>I am currently a 3rd year undergraduate electronic engineering student. I have completed a course in dynamics, calculus I, calculus II and calculus III. I have decided to self study a basic introduction to special relativity as it was not part of my physics syllabus and is not included in any of my other modules. The source I am using is the textbook "Fundamentals Of Physics(9th Edition)" by Halliday and Resnick.</p>
<p>I understand the concept of proper time and length, but I am interested in solving problems using the Lorentz transformations rather than using the length contraction and time dilation "short cuts."</p>
<p>For example, the following textbook sample problem:</p>
<p><a href="https://i.sstatic.net/tD3G5.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/tD3G5.png" alt="Sample problem 3(a)"></a></p>
<p><a href="https://i.sstatic.net/HEFqx.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/HEFqx.png" alt="Sample problem 3(b)"></a></p>
<p>I understand how the the answer was obtained, however I am wondering if it is possible to tackle the problem as follows; using just the Lorentz transformation equations and the equation of motion; x = vt and defining 2 events, i.e. [event 1] = point B passes Sally & [event 2] = point A passes Sally. I have tried this but I keep getting the wrong answer unless I use the length contraction formula. Is there a way to work out this problem in this manner based on the given information?</p>
<p>All answers on this will be greatly appreciated.</p>
|
<p>First set up space and time origins for each frame. Say the spatial origin of Sally's frame is right in point A, and her time origin is right when Sam's point B passes by her. For Sam, say his space origin is in point B and his time origin is also when his point B passes Sally at her point A.</p>
<p>Now write down event coordinates for both frames, keeping just the relevant axis (x) along the direction of motion. </p>
<p>In Sally's frame, if she observes Sam's ship to have length L and pass by at velocity $v >0$, then</p>
<p>Event 1 = $(x_1 = 0, \;t_1 = 0)$</p>
<p>Event 2 = $(x_2 = 0, \;t_2 = t_1 + \Delta t = \frac{L}{v})$ </p>
<p>In Sam's frame, his ship is known to have rest length $L_0$ and he observes Sally pass by at velocity $-v < 0$, so the same events must read</p>
<p>Event 1 = $(x'_1 = 0, t'_1 = 0)$</p>
<p>Event 2 = $(x'_2 = - L_0, t'_2 = t'_1 + L_0/v)$ </p>
<p>The Lorentz transformation from Sally's frame to Sam's frame must take $(x_1, t_1)$ into $(x'_1, t'_1)$ and $(x_2, t_2)$ into $(x'_2, t'_2)$. The way we chose coordinate origins renders the first set trivial, $(0,0) \rightarrow (0,0)$, but the 2nd set yields $L$ in terms of $L_0$:
$$
x'_2 \equiv - L_0 = \gamma(x_2 - v t_2) = \gamma \left(0 - v\frac{L}{v}\right) = -\gamma L\\
t'_2 \equiv \frac{L_0}{v} = \gamma\left(t_2 - \frac{v}{c^2}x_2\right) = \gamma\left(\frac{L}{v} - \frac{v}{c^2}0\right) = \gamma \frac{L}{v}
$$
which give of course the length contracted length seen by Sally as
$$
L = \frac{L_0}{\gamma}
$$
Then the ship speed observed by Sally follows from $L \equiv L_0/\gamma = v\Delta t$ etc.</p>
| 476
|
special relativity
|
Relativity of Simultaneity - Issue with the train car example
|
https://physics.stackexchange.com/questions/285760/relativity-of-simultaneity-issue-with-the-train-car-example
|
<p>One of the classic examples to describe that simultaneity is relative is the following:</p>
<p>"Imagine a freight car, traveling at a constant speed along a smooth, straight track. In the very center of the car there hangs a light bulb. When someone switches it on, the light spreads out in all directions at speed c. Because the lamp is equidistant from the two ends, and observer on the train will find that the light reaches the front end at the same instant it reaches the back end... However, to an observer on the ground these same two events are not simultaneous. For as the light travels out from the bulb( going at speed c in both directions) the train itself moves forward, so the beam going to the back has a shorter distance to travel than the one going forward. According to this observer, therefore, [the light hits the back] <em>before</em> [the light hits the front]." - Griffiths, Introduction to Electrodynamics</p>
<p>Here is a picture of the thought experiment:
<a href="https://i.sstatic.net/lAnHM.gif" rel="nofollow noreferrer"><img src="https://i.sstatic.net/lAnHM.gif" alt="enter image description here"></a></p>
<p>My lack of understanding stems from the fact that I am thinking about how fast the distance between the wall and the light beam decreases to 0. Take for example the back wall:</p>
<p>In the reference frame of someone in the train, it makes sense that the distance between the back wall and the light beam decreases at the rate of c, the speed of light, because in this frame the back wall is not moving and the light is moving at the speed of light.</p>
<p>However in the reference of the outside observer looking into the moving train, he/she will see the back wall moving to the right (as in the picture above) and the light beam moving to the left, but the rate that the distance between the back wall and the light beam decreases can't be greater than c, because nothing can move faster than the speed of light c. (If two observers are moving towards each other at the speed of light, they don't see each other moving at twice the speed of light, rather they still observe the other moving at the speed of light). So if the light bulb and the back wall were observed to start at the same distance away in both observers' frames than they both have to see the events happening at the same time.</p>
<p>Where am I tripping up/ where is my logic inconsistent?</p>
|
<blockquote>
<p>but the rate that the distance between the back wall and the light beam decreases can't be greater than c, because nothing can move faster than the speed of light</p>
</blockquote>
<p>You are mistaken. Yes, nothing can move faster than light in a sense that physical motion is fundamentally bounded from above by $c$. But an abstract mathematical quantity, which is the distance between the light beam and the back wall of the wagon, can decrease with a slope higher than $c$. There is nothing in Special Relativity that forbids it from happening.</p>
<p>On the contrary, we know that the light beam is traveling to the left exactly with the speed of light $c$, because this is physical motion we are talking about. And since the wall is moving to the right, the distance decreases faster than $c$. Again, this is completely normal.</p>
<p>Consider another thought experiment: two particles flying in the opposite directions with almost the speed of light each. The distance between them obviously increases with speed $2c$. There is nothing wrong with this! However, in the rest frame of one of the particles, the other one travels not with $2c$, but approximately with $c$ because it is the way the relativistic velocity addition functions.</p>
| 477
|
special relativity
|
A question about the relativity of simultaneity
|
https://physics.stackexchange.com/questions/286536/a-question-about-the-relativity-of-simultaneity
|
<p><a href="https://i.sstatic.net/GULKM.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/GULKM.png" alt="enter image description here"></a></p>
<p>In the above figure Observer A is detecting the events P and Q by sending and receiving lights to the continuous paths. Observer B is also trying to locate the events by sending and receiving lights RQU and SPV respectively. My question is observer B would illuminate the events via other light rays; why s/he is taking the same light rays of observer A?</p>
|
<p>The figure is a <a href="https://en.wikipedia.org/wiki/Minkowski_diagram" rel="nofollow">Minkowski diagram</a>, where light rays always are lines with a 45 degrees slope. The diagram also shows only one space dimension. Wordlines of observers with different speeds are represented by differently inclined lines all passing through $O$.</p>
<p>Because of the above, the light reaching $P$ or $Q$ has to follow the 45 degrees line coming from downward, whatever event (observer) is its source; similarly, the light emitted from $P$ and $Q$, whatever observer it reaches, also must follow the single upward 45 degrees line.</p>
<p>Note that the relativity of simulaneity is illustrated by the upper part of the diagram, by considering at what times the light emitted from $P$ and $Q$ reaches both observers: for $A$, both light rays arrive at the same time. Since the spatial distances $OP$ and $OQ$ are the same, this means $P$ and $Q$ are simultaneous for $A$. On the other hand, for $B$ the light from $Q$ arrives first (at $U$), followed by the light from $P$ at $V$. Since again the spatial distances $OP$ and $OQ$ are the same for $B$* then it means that for $B$, $Q$ happens before $P$.</p>
<p>*It is not obvious from the diagram than spatial $OQ$ is the same than spatial $OP$ for $B$ (although that distance is not the same as for $A$!). To see this you would need to draw the tilted spatial $t=0$ for $B$, which is not horizontal but rises from left to right (see the diagram for a Lorentz boost in the Wikipedia page linked above).</p>
| 478
|
special relativity
|
Bell's spaceship paradox - Special relativity
|
https://physics.stackexchange.com/questions/287428/bells-spaceship-paradox-special-relativity
|
<p>I will refer to this wikipedia page: <a href="https://en.wikipedia.org/wiki/Bell%27s_spaceship_paradox" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Bell%27s_spaceship_paradox</a>
and especially this diagram:</p>
<p><a href="https://i.sstatic.net/8KZdj.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/8KZdj.png" alt="enter image description here"></a>
It appears clearly in frame S' that the length of the rope is longer than at rest in that referential, therefore it breaks.
Nevertheless, I can't see why in the S frame the rope should break as for the S frame point of view, the length of the rope stays the same (L).</p>
<p>Can someone explain how in this diagram the S frame should conclude as well that rope breaks ?</p>
<p>The argument given in the text is that the rope moves so it experiences Lorentz contraction from a frame S point of view. But this does not seem to appear on the diagram, or does it ?</p>
<p>Thanks for your help.</p>
|
<p>From a frame S point of view, one has to compare to what the accelerating rope profile would look like in that diagram and would realize that it would look smaller in frame S. Therefore frame S would conclude that rope should snap.
That information is not in the diagram as it is.</p>
<p>If we were used to look at relativistic phenomenons we would always have seen ropes reducing in size when they accelerate. And if forcing it to maintain the same length when accelerating, we would therefore naturally conclude that it should snap.</p>
| 479
|
special relativity
|
Is a photon going through a center of mass affected by time dilation more then another going around?
|
https://physics.stackexchange.com/questions/289234/is-a-photon-going-through-a-center-of-mass-affected-by-time-dilation-more-then-a
|
<p>Please don't mark as duplicate. My specific question was not answered in other posts. And this question here <a href="https://physics.stackexchange.com/questions/289025/which-photon-would-win-the-race">Which Photon would win the race?</a></p>
<p>is also about neutrinos, electrons. </p>
<p>My question now is only about photons, and the shapiro delay.</p>
<p>then if you try to make analogy to the shapiro delay, then comes my question:</p>
<ol>
<li>according to the shapiro delay, a photon going around the mass arrives later then it should.</li>
<li>there is no experiment where they would somehow shoot a photon also through the center of mass (maybe between two close stars or an artificial mass with a whole through it, or something else)</li>
<li>so shapiro delay only talks about the photon going around, and that photon needed more time to arrive from point A to B, but it's speed had to stay c, so the distance it traveled had to be longer (which it really is in 3D)</li>
<li>so a photon going through a tunnel through the center of mass (or between the two SUNs here) has to travel at speed c too, but its distance is shorter in 3D so it's time need to be shorter too. So that photon has to arrive from A to B faster.</li>
<li>Imagine this set:</li>
</ol>
<p><a href="https://i.sstatic.net/YPIQR.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/YPIQR.jpg" alt="enter image description here"></a></p>
<ol start="6">
<li>We shoot two photons, one around one of the stars, and one inbetween two stars.</li>
<li>We shoot the one going around in an angle that it will cross point B too.</li>
</ol>
<p>Question:</p>
<ol>
<li>so was there any experiment like this?</li>
<li>Am I right that time dilation would affect the photon going between two stars more and it would arrive first?</li>
</ol>
|
<p>You would think that a particle moving in the centre of mass experiences the same time as moving through space wthout any mass in it. In both cases no gravity is felt, so you would think time proceeds at it's fastest rate. The pace of time is nevertheless dependent on the gravitational potential, which is different in both cases, so time is moving at different paces for the two photons (coordinate time, because for the photon itself the pace of time is zero). So if you compensate for the differences, the two photons don't arrive at the same time.</p>
| 480
|
special relativity
|
Analysis of simultaneity in Special Theory of Relativity
|
https://physics.stackexchange.com/questions/299459/analysis-of-simultaneity-in-special-theory-of-relativity
|
<p>Set up: A moving train with two flash lights on the wall, a person in the middle of the train. A mid point on the tracks where second person/observer stands. Flash lights flash a light when mid-point on the tracks and of the train have the same coordinate in both frames of reference. Suppose 1D problem plus time of course :-)
When midpoints of both frames coincide flash lights flash. Observer on the train has to see this as a simultaneous events because obviously it has to be that way. Here we see same thing as in non relativistic case where we fire a bullets from a gun. But what does the person on the tracks see? First of all, he can see light signals meet at the train midpoint where the moving observer is placed. Is it possible that he sees light signals meet at his midpoint? Well that would be imposible because it would suggest that event happend in both places and that is just crazy right? Signals should meet at the trtains mid point. Just like with the bullets. But for bullets we can explain this by simple velocity adition (two bullets do not have the same speed in the track frame of reference when fired). But light has to have the same speed for both observers! So when the light is emited it can either act like it is glued to a train frame or to the track frame. Nature of this light propagation is such that in order to make everything have sense light has to be emitied in different moments for the rtrack observer, it is the only way. My question is this: if you emit a light signal like this here described, how does the light know that this signal has originated from the flashlights that are stationary on the train? What mechanism takes care of this? </p>
<p>I am adding this piture as o ilustrate my troubles and my wondering..are we forced to say that somehow we have two versions of one reality because of constancy of speed of light?
<a href="https://i.sstatic.net/0wdRW.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/0wdRW.png" alt="enter image description here"></a></p>
|
<p>It should be clarified, what means “an observer sees”. To make proper conclusion, at what time certain event took place in certain coordinate of reference frame, you have to have clock in that place. Procedure for every observer looks like this thought experiment. An observer in a train and observer on the platform allocate (place) clocks along platform and along the train. He can put clock every centimeter. Then each observer synchronizes all clocks in his own reference system by means of Einstein technique, admitting that speed of light in each direction for every inertial frame is the same. Then all clocks in his reference system show the same time, a clock on observer’s wrist and a clock a million kilometers away from the observer. If event happens in that place one million kilometers away, now he knows time of event in his frame. Since frames are in relative motion, that leads to so-called relativity of simultaneity. Simultaneous evens in reference frame K will be not simultaneous in reference frame K’. Adjacent clocks of different frames will show different time. This way of thinking puts thoughts in order.</p>
| 481
|
special relativity
|
With quad spaceship setup, will RF time periods between all ships remain the same?
|
https://physics.stackexchange.com/questions/310673/with-quad-spaceship-setup-will-rf-time-periods-between-all-ships-remain-the-sam
|
<p>The red dots shown below, are RF transmitters.<br>
Clocks are located nearby each of these RF transmitters.<br>
All spaceships are identical and thus have identical rocket engines.</p>
<p><a href="https://i.sstatic.net/nb4Sn.gif" rel="nofollow noreferrer"><img src="https://i.sstatic.net/nb4Sn.gif" alt="http://www.outersecrets.com/real/image/4sp.gif"></a></p>
<p>Assume that all 4 spaceships were originally close together and during this time they synchronized their clocks. Next, they moved the spaceships apart such that each spaceships RF transmitter, the red dot, is positioned 300,000 km horizontally and vertically from two other spaceships. Crisscross distances between A and D, and, B and C, therefore become 424,264 km each.</p>
<p>Once in these positions, they send RF pulses to each other at preplanned times. As expected, the spaceships positioned 300,000 km apart receive the RF pulse 1 second after the preplanned pulse release time. From A to D, and B to C, they measure 1.4142 seconds pass to receive the pulse, just as expected. So all the clocks seem to be synchronized, and the spatial distances seem to be verified. </p>
<p>Next, at another preplanned time, all spaceships fire their identical rocket engines and maintain a preplanned acceleration G-force until reaching 260,000 km/s. At this speed, the engines are shut down and thus the velocity of 260,000 km/s is maintained.</p>
<p><strong>Question:</strong> Will everything seem the same, meaning from their point of view, will pulses sent from ship to ship give the same results as before, meaning the 1 second and 1.4142 second results ?</p>
|
<p>After the engines are shut off, they will still be at rest relative to each other, so it will look exactly the same as before. Even before they fired their engines, they were traveling at great speed relative to other galaxies and this is no different. Stuff that is at rest relative to each other and not too distant from each other will always look Newtonian. (Too distant is hand wavy, but one light second should fine.)</p>
| 482
|
special relativity
|
Relativity and a Rotating Rod
|
https://physics.stackexchange.com/questions/318013/relativity-and-a-rotating-rod
|
<p>I was doing a problem on relativity and I was thinking about the problem in terms of the postulates of Special Relativity. Special Relativity only considers motion in inertial reference frames and I was doing this problem over here; </p>
<p>Essentially, if I have a rod which is tilted at an angle (theta)0, with respect to the x-axis, and moving along the x-axis at speed v, and if that rod has a length L0 observed from its frame of reference, what is the length of the rod as observed by an observer in a stationary frame with respect to the frame of the rod. Also, what is the angle, as observed by that stationary observer, of the rod from the x-axis?</p>
<p>I got the answer correct and it's presented below.
n<a href="https://i.sstatic.net/M0OJM.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/M0OJM.jpg" alt="enter image description here"></a></p>
<p>The problem that I with this is that the angle itself changes, which makes me think that rotation has occurred while observing the rod from one frame to the next. And if the rod itself is the second reference frame, in that case, would it not be a rotating reference frame and therefore, not within the realm of special relativity? I mean, if it is a rotating reference frame as observed from other reference frames, then it's going to have an acceleration no? </p>
<p>Sorry if this is a stupid question, I'm just really confused and, yet, excited about this cos it's so interesting.</p>
|
<p><strong>interpretation of your result</strong></p>
<p>You should not interpret this is a rotation of the angle. It is merely a result of there only being length contraction along the motion of the observer.</p>
<p>The y length of the rod changes but it's x length will change due to length contraction (as you correctly derived.) which changes the angle.</p>
<p><strong>extra "riddle"</strong></p>
<p>You mentioned that you are really interested in this effect so let me give you a related problem that leads directly to general relativity:</p>
<p>Imagine a really fast merry go round with some observer riding the attraction.</p>
<ul>
<li>The length along his motion (circumference of the attraction) will get contracted</li>
<li>The length perpendicular to his motion (radius of the attraction) will remain unchanged</li>
</ul>
<p>The result is that this observer will measure a circle with $Circumference \neq 2\pi radius$!</p>
<p>This is not possible in flat spacetime such that the observer must be observing curved spacetime! </p>
| 483
|
special relativity
|
Time lines of observers meeting each other - doubts about their graphical representation
|
https://physics.stackexchange.com/questions/320581/time-lines-of-observers-meeting-each-other-doubts-about-their-graphical-repres
|
<p>I am trying to make sense of Fig. 2.12, page 23 of Introducing Einstein's Relativity (D'Inverno, Oxford University Press). There it goes:</p>
<p><a href="https://i.sstatic.net/wiLWO.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/wiLWO.jpg" alt="fIG.2.12, PAGE 23"></a></p>
<p>The book picture is in black. Scales are such that light rays are inclined by 45deg.</p>
<p>Observer A sees events P and Q happen simultaneously at equal opposite distance. According to the book, observer B (riding his own BLACK time line) meets A at the same moment as events P and Q happen <strong>according to A</strong>.</p>
<p>This does not convince me. In my view an observer that meets A the very moment A observes P and Q must be travelling on the RED line.</p>
<p>The book says that A sees P, Q and O happen at the same time. I'd say that A sees P, Q and O' happen at the same time.</p>
<p>Which is the correct time line for B? The black or the red one?</p>
|
<p>The complete diagram from d'Inverno (p. 23) is shown below.
It appears that observer-A has performed radar experiments on events P and Q. Observer-A assigns time-coordinates to events as the halfway time between emission and reception of the radar signal. So, Observer-A assigns P and Q the same time coordinate---to Observer-A, P and Q are simultaneous. Further, it appears that event O is the midpoint-event between emission and reception, and thus O is simultaneous with P and Q.</p>
<p>Granted, at the meeting event O, observer-A doesn't yet have the information needed to assign those time-coordinates to P and Q yet. </p>
<p>As others have pointed out, "sees" (or observes) is an imperfect term since one might not distinguish (1)
a spacelike-relation of simultaneity with [i.e., assigning the same t-coordinates to] a distant event
from (2)
a past-lightlike-relation with [i.e., visually seeing] a distant event.</p>
<p>Referring to your statements:</p>
<ul>
<li><p>"A sees P, Q and O happen at the same time." really means that A assigns the same time-coordinate to P, Q, and O.</p></li>
<li><p>"A sees P, Q and O' happen at the same time." would be correct if it means that light-signals from P, Q, and O' reach A at the same time... according to A. If you said, with the meaning just given, "A sees P, Q and O' at the same event", then everyone would agree to that.</p></li>
</ul>
<p>B's worldine passes through event O, as the book has drawn.</p>
<p>The purpose of the diagram is that Observer-B will assign distinct time-coordinates to events P, O, and Q.</p>
<p><a href="https://i.sstatic.net/O2YMn.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/O2YMn.png" alt="dInverno diagram"></a></p>
| 484
|
special relativity
|
Rotation matrix and referencials in relativity
|
https://physics.stackexchange.com/questions/594933/rotation-matrix-and-referencials-in-relativity
|
<p>I am reading the book "Gravity in a nutshell" in which the author talks about rotation in a plane and different frames, one rotated with respect to the other. The rotated frame is denoted with a prime. We are trying to find the matrix relating the differences of coordinates in both frame. It is strange, he finds <span class="math-container">$$\begin{bmatrix}
\Delta x'\\ \Delta y'
\end{bmatrix}
=
\begin{bmatrix}
\cos\theta & -\sin\theta \\
\sin\theta & \cos\theta
\end{bmatrix}
\begin{bmatrix}
\Delta x\\ \Delta y
\end{bmatrix}
$$</span></p>
<p>while I found the following matrix.</p>
<p><span class="math-container">$$\begin{bmatrix}
\Delta x'\\ \Delta y'
\end{bmatrix}
=
\begin{bmatrix}
\cos\theta & \sin\theta \\
-\sin\theta & \cos\theta
\end{bmatrix}
\begin{bmatrix}
\Delta x\\ \Delta y
\end{bmatrix}
$$</span></p>
<p>I am not sure what is happening (I don't think it is a matter of convention, he was using the same matrix I found until one line before the solution he found, as follows)</p>
<blockquote>
<p>Let Ms. Unprime assign the coordinates <span class="math-container">$r_P = (x, y)$</span> and <span class="math-container">$r_Q = (X, Y)$</span> to <span class="math-container">$P$</span> and <span class="math-container">$Q$</span>, respectively.
Then Mr. Prime’s coordinates <span class="math-container">$r_{P'} = (x', y')$</span> for <span class="math-container">$P$</span> and <span class="math-container">$r_{Q'} = (X', Y')$</span> for <span class="math-container">$Q$</span> are then given by <span class="math-container">$r_{P'} = \mathbf{R}(θ)r_P$</span> and <span class="math-container">$r_{Q'} = \mathbf{R}(θ)r_Q$</span>. Subtracting the first equation from the second and defining <span class="math-container">$\Delta x = X − x, \Delta y = Y − y$</span>, and the corresponding primed quantities, we obtain...</p>
</blockquote>
<p><span class="math-container">$R(θ)$</span> is the matrix I am using... So, what is the problem?</p>
| 485
|
|
special relativity
|
In relativity, if units of length contracts and time dilates then does unit of velocity or speed also change?
|
https://physics.stackexchange.com/questions/574856/in-relativity-if-units-of-length-contracts-and-time-dilates-then-does-unit-of-v
|
<p>I'm just starting to learn special relativity, and I'm having trouble with the following concept:</p>
<p>In relativity, units of length and time of moving frame are related to that of stationary one through <span class="math-container">$$x’=\frac{x}{\gamma}\quad \quad \text{ and }\quad \quad t’=t\times \gamma$$</span> respectively, where <span class="math-container">$\gamma$</span> is Lorentz Factor.</p>
<p>Does this also mean that units of velocity or speed, i.e. length/time are related as <span class="math-container">$$v’=\frac{x’}{t’}=\frac{x}{t}\times\frac{1}{\gamma^2}=\frac{v}{\gamma^2}?$$</span></p>
<p>Note: By unit, I mean scale of axes in a respective coordinate system and I am not asking about addition or subtraction of velocities, I am enquiring about mutual “scale” difference between the quantity called velocity as measured in two different frames in uniform relative motion to each other.</p>
<p>Why scale of length contracts and not expands while that of time dilates, i.e. expands when the two are symmetrical for Lorentz transformations! Only if length expands with dilation of time can the “scale” of velocity or speed in general and speed of light in particular can remain truly invariant, I guess.</p>
|
<p>No. Velocities do transform in a non-intuitive way in special relativity, but not in the way you're describing. This is sadly a very common misunderstanding due to the fact that Relativity is usually first introduced using time-dilation and length contraction, without actually explaining under which conditions they are applicable. The best way to begin understanding the subject (and to also avoid all these "paradoxes") is to work with the Lorentz Transformations. In one dimension, if the frame <span class="math-container">$S'$</span> is moving with respect to <span class="math-container">$S$</span> with a velocity <span class="math-container">$v$</span>, then</p>
<p><span class="math-container">\begin{aligned}
x' &= \gamma\left( x - vt \right)\\
t' &= \gamma \left( t - \frac{v}{c^2}x\right)
\end{aligned}</span></p>
<p>provided that <span class="math-container">$x=x'=0$</span> when <span class="math-container">$t=t'=0$</span>. Remember, though, that these are <em>coordinates</em>, not intervals. To find how intervals of length and intervals of time are related, we need to take <em>differences</em> of the coordinates, and since <span class="math-container">$v$</span> and <span class="math-container">$\gamma$</span> are constants, it's easy to show that the intervals satisfy similar equations:</p>
<p><span class="math-container">\begin{aligned}
\Delta x' &= \gamma\left( \Delta x - v \Delta t \right)\\
\Delta t' &= \gamma\left( \Delta t - \frac{v}{c^2}\Delta x\right)
\end{aligned}</span></p>
<p>We can use the above equations to easily calculate how the velocities transform between the frames <span class="math-container">$S$</span> and <span class="math-container">$S'$</span>. Remember that an observer in <span class="math-container">$S$</span> will calculate the velocity of an object to be <span class="math-container">$$u = \frac{\Delta x}{\Delta t},$$</span> and one in <span class="math-container">$S'$</span> will calculate it to be <span class="math-container">$$u' = \frac{\Delta x'}{\Delta t'}.$$</span></p>
<p>We can now divide <span class="math-container">$\Delta x'$</span> by <span class="math-container">$\Delta t'$</span> to show that:</p>
<p><span class="math-container">$$u' = \frac{\Delta x'}{\Delta t'} = \frac{u - v}{1 - \frac{uv}{c^2}}.$$</span></p>
<h3>Why doesn't your argument work?</h3>
<p>Length contraction and time dilation are special cases of the general formulae that I have given above. They hold when certain conditions are satisfied, and these conditions are more certainly not satisfied <em>simultaneously</em>. Which means dividing the equations is not going to give you anything sensible. In special relativity, it is best to think in terms of "events" which occur at spacetime points <span class="math-container">$(t, x)$</span> to avoid such false "paradoxes". My answer <a href="https://physics.stackexchange.com/a/563609/157014">here</a>, and the links at the end, should explain it in more detail.</p>
| 486
|
special relativity
|
How can I resolve this version of the twin paradox in special relativity?
|
https://physics.stackexchange.com/questions/603880/how-can-i-resolve-this-version-of-the-twin-paradox-in-special-relativity
|
<p>I have a version of the twin paradox which I am completely stumped by. There is a similar question on the forum but this particular version is unanswered. I really hope someone a lot better at physics than me is able to solve it!</p>
<p>Imagine two twins (Max and Tony) which are accelerated at birth in two opposite directions equally (at exactly the same rate of acceleration) for a long time, and then decelerated (at the exact same rate) until they are at rest with respect to each other far, far apart. Since they both accelerated the same amount, they are now the same age (say 20) in the same inertial reference frame.</p>
<p>Then, they accelerate towards each other until their relative speed is .99c, at which point they stop accelerating and are both travelling at constant velocity in an inertial (non-accelerating) reference frame (special relativity now applies).</p>
<p>At .99c, let's say they pass each other after say 30 years (or we can just stipulate a number of years x). The time dilation factor for .99c is about 1/7, so, when they pass, Max will see Tony at age 20+(30/7) = 24.3 years, and Max will see himself (Max) at age 20+30 = 50 years.</p>
<p>However, by the symmetry of special relativity the same goes for Tony: when they pass, each twin will see the other as 50 - 24.3 or about 26 years younger (or just (6/7x) years younger where x is the number of years travelled at .99c).</p>
<p>Finally, suppose they both accelerate (by decelerating, accelerating and decelerating equally and oppositely) into the same reference frame and come to rest together. Since they accelerate equally, they should age equally (say by y years) to each other. All other acceleration throughout their lives has been equal so no aging differences should have occurred at any point other than when they are travelling at constant speed of 0.99c (when special relativity applies). However, this means that both twins are now both 24.3+y and 50+y years at the same time by the symmetry of special relativity. How can this possibly be?</p>
<p>I’m sure there is a very simple explanation for this but I have no idea how to resolve it. I would expect that both twins are exactly the same age but how do you get round the fact that under special relativity each twin can absolutely legitimately claim the other is younger?</p>
|
<blockquote>
<p>they stop accelerating and are both travelling at constant velocity in an inertial (non-accelerating) reference frame (special relativity now applies).</p>
</blockquote>
<p>Special relativity applies to the whole problem, including the accelerating parts, since there's no gravity.</p>
<blockquote>
<p>Max will see Tony at age 20+(30/7) = 24.3 years, and Max will see himself (Max) at age 20+30 = 50 years.</p>
</blockquote>
<p>That's incorrect. By symmetry they are the same age when they meet, and they'll see each other as being the same age.</p>
<p>Note that when Max is at his farthest from Earth at age 20, he doesn't see Tony at age 20. He sees an infant Tony not long after Tony leaves Earth, because light from most of Tony's trip hasn't had time to reach him yet. While Max travels back toward Earth, he sees Tony age to 20, accelerate toward him, and then <em>very quickly</em> age to 50. The blueshift factor for this last part is <span class="math-container">$(1+.99)/(1-.99) = 199$</span>, so Tony appears to age 30 years in about 2 months. That means he appears to age from infancy to age 20 in about 29 years and 10 months.</p>
<p>When Max is at his farthest from Earth at age 20, if you draw a spacetime plane perpendicular to his worldline and take the intersection of that plane with Tony's worldline, the intersection will be at the symmetric point where Tony is 20. While people are very fond of drawing these so-called "planes of simultaneity", it's important to understand that they have no physical significance whatsoever, and won't help you understand the nature of special relativity. Except in simple cases, they aren't even helpful in doing calculations. Nevertheless, you can analyze this problem using Max's planes of simultaneity if you really want to. In this case, the resolution of the paradox is that when Max accelerates back toward Earth (an acceleration is a rotation in spacetime), the plane of simultaneity also rotates, sweeping over a large part of Tony's worldline (roughly 26 years of Tony's proper time) "during" the acceleration. Once Max is traveling at his constant speed of <span class="math-container">$.99c$</span>, the plane moves along Tony's worldline <span class="math-container">$1/\sqrt{1-.99^2}\approx 7$</span> times slower than Max's proper time, so Tony ages roughly another <span class="math-container">$30/7$</span> years "during" Max's trip back in this unphysical sense of "during". In total Tony ages exactly 30 years "during" Max's acceleration and trip back.</p>
<p>Everything in the last two paragraphs is also true if you swap the names Max and Tony.</p>
| 487
|
special relativity
|
If an object is orbiting another body, don't both the orbiting object and an observer "know" that the orbiting object is moving?
|
https://physics.stackexchange.com/questions/611376/if-an-object-is-orbiting-another-body-dont-both-the-orbiting-object-and-an-obs
|
<p>In many relativity illustrations, it is mentioned that someone in a spaceship believes he is standing still and the observers are moving and vice versa. The example of a light-beam clock slowing down for an observer but not for the passenger of the ship is often how relativity is explained.</p>
<p>But if the ship is in orbit about another body, the I think the passengers can measure this speed. Does the mirror clock argument still work?</p>
|
<p>Well Yes, but actually no. You know that Earth is orbiting the Sun, because you can observe relative motion of Sun to the other (fixed) stars on the sky. But let Say now, that we don't have any other Stars or reference objects. And let's introduce 2D polar coordinates. Let say, that we are on the Sun, and we believe We are stationary. We can then expres position of Earth as:</p>
<p><span class="math-container">$$
\textbf{r}=r_0(\cos \phi, \sin \phi)=r_0(\cos \omega t, \sin \omega t).
$$</span></p>
<p>But we can se, that if we set the coordinate system to the system of Earth we se <span class="math-container">$\textbf{r}_{Sun-reference}=-\textbf{r}_{Earth-reference}$</span>, ergo:
<span class="math-container">$$
\textbf{r}=r_0(\cos \phi, -\sin \phi)=r_0(\cos \omega t, -\sin \omega t)
$$</span>.</p>
<p>We can observe, that we don't have any special effect in the motion, which will show us, which object is actually orbiting around other. But it can be seen, that the rotation changes sign (clockwise to counterclockwise and vice versa.</p>
<p>Bear in mind, that this is only explanation, why you can't know if you are moving or not, if you observe orbiting. I would be happy to see any comments if this is really valid explanation if this case, or should I've been using Lorentz transformations.</p>
| 488
|
special relativity
|
Why do we take measurments of the end points of a rod at the same time in an inertial frame moving with uniform velocity?
|
https://physics.stackexchange.com/questions/612498/why-do-we-take-measurments-of-the-end-points-of-a-rod-at-the-same-time-in-an-ine
|
<p>I have a question regarding the concept of length contraction.
Consider a rod of rest length <span class="math-container">$L_o=x_2-x_1$</span> in frame <span class="math-container">$S$</span>. Now if we want to measure the length of the rod in <span class="math-container">$S^{'}$</span> i.e <span class="math-container">$L^{'}_o=x^{'}_2-x^{'}_1$</span> in this case we use the inverse Lorentz transformation.</p>
<p>My question is why must we take measurements <span class="math-container">$x^{'}_2$</span> and <span class="math-container">$x^{'}_1$</span> at the same time <span class="math-container">$t^{'}$</span> in the <span class="math-container">$S^{'}$</span> frame which is in uniform motion w.r.t the <span class="math-container">$S$</span> frame? Why can't we take the measurements of the end points <span class="math-container">$x^{'}_2$</span> and <span class="math-container">$x^{'}_1$</span> at different times in the <span class="math-container">$S^{'}$</span> frame?</p>
|
<p>Consider a rod of length 1 metre, aligned with and on the X axis, moving at 10 m/s in the +X direction. Clearly, this speed is non-relativistic, so the length contraction is negligible.</p>
<p>At time <span class="math-container">$t=0 s$</span>, we measure that the back of the rod is at <span class="math-container">$x_1=0 m$</span>. At <span class="math-container">$t=10 s$</span>, we measure that the front of the rod is at <span class="math-container">$x_2=101 m$</span>. But the length of the rod is 1 metre, not <span class="math-container">$x_2-x_1=101$</span> metres. So to get the actual length of the rod we must determine the positions of the front and back of the rod at the same instant of time.</p>
<p>If we <em>can't</em> perform those position measurements simultaneously, then we need to take the rod's velocity into account.</p>
<p>If the rod has a high velocity relative to our frame, so that relativistic effects aren't negligible, we need to be very careful that our position sensors record the exact time of their measurements, using clocks that are synchronised in our frame. And if those sensors measure the rod's position from a distance, they also need to compensate for the time it takes light to travel (from the part of the rod that they're measuring) to the sensor.</p>
| 489
|
special relativity
|
Don't twin paradox explanations imply universal velocity/time?
|
https://physics.stackexchange.com/questions/573254/dont-twin-paradox-explanations-imply-universal-velocity-time
|
<p>On the Wikipedia page for the Twin Paradox, the example lays out the perspective of each twin in turn. Both twins are portrayed as understanding the ship's velocity as v, and the travelling twin's sense of time is then explained by saying that the earth-distant star system, being in effect one giant object, undergoes length contraction ie the twin travels less distance, ie less time went by for the twin. But if the entire rest of the twin's system, ie. the twin's whole sense of distance, has shrunk, shouldn't the twin's sense of velocity shrink too? How can they agree on velocity but not distance, when v = d/t ? The v is relative to what?</p>
<p>Or are we saying that our usual sense of reality is wrong, when we think of an object's location at a given time (absolute or otherwise) as an intrinsic truth upon which we define velocity as Δlocation/unit time, when in fact it's the other way around and an object's velocity is intrinsic. Are we saying velocity actually IS absolute even though time and distance aren't?</p>
|
<p>Suppose the stationary twin occupies the origin of our coordinates, <span class="math-container">$x=0$</span>, and observes that the velocity of the traveling twin is <span class="math-container">$v$</span>, in their frame. In other words, suppose that the traveling twin covers a certain distance <span class="math-container">$d$</span> in a certain time <span class="math-container">$t$</span>, such that <span class="math-container">$v=d/t$</span>. Let's find out what the travelling twin measures the <em>stationary</em> twin's velocity to be, by directly applying a Lorentz transformation, boosting to a frame with velocity <span class="math-container">$v$</span>:</p>
<p><span class="math-container">$$x'=\gamma(x-vt)$$</span></p>
<p><span class="math-container">$$t'=\gamma(t-vx/c^2)$$</span></p>
<p>The stationary twin is at the origin of our original coordinates, so plugging in <span class="math-container">$x=0$</span>, we get the coordinates of the stationary twin in the traveling twin's frame:</p>
<p><span class="math-container">$$x'=-\gamma vt$$</span></p>
<p><span class="math-container">$$t'=\gamma t$$</span></p>
<p>Now let's calculate the velocity of the stationary twin in this frame:</p>
<p><span class="math-container">$$v'=\frac{x'}{t'}=\frac{-\gamma v t}{\gamma t}=-v$$</span></p>
<p>In other words, the velocities observed by each twin for the other are equal and opposite. The effects of time dilation and length contraction exactly cancel out.</p>
| 490
|
special relativity
|
Lorentz transformation of a trajectory
|
https://physics.stackexchange.com/questions/578953/lorentz-transformation-of-a-trajectory
|
<p>I am having trouble understanding how to apply the Lorentz transformation to a particle trajectory.</p>
<p>Suppose we have a body moving in one dimension in one frame where the position is given by <span class="math-container">$x(t) = f(t)$</span>. Then if we apply a Lorentz transformation with velocity <span class="math-container">$v$</span> in the <span class="math-container">$\hat x$</span> direction, the trajectory in the boosted frame is <span class="math-container">$x'(t) = \gamma(v)[ x(t) - vt ] = \gamma(v) [ f(t) - vt ]$</span>.</p>
<p>What I'm having trouble understanding is how to properly parametrize the time coordinate in the boosted frame. According to the Lorentz transformation, we have <span class="math-container">$ t' = \gamma(v) [ t - v f(t) / c^2 ]$</span>. This implicitly defines <span class="math-container">$ t = g(t')$</span>. Is the correct trajectory in the boosted frame then given by <span class="math-container">$x'(t') = \gamma(v)[ f(g(t')) - vg(t') ]$</span>? Is this the correct procedure to generally Lorentz transform a given trajectory?</p>
|
<p>Yes, your explanation is right. Moreover, the function you call <span class="math-container">$g$</span> always exists because of the implicit function theorem. You can easily see that <span class="math-container">$t'=F(t)$</span> satisfies <span class="math-container">$F'(t)\neq 0$</span>. Here <span class="math-container">$F(t)=\gamma(v)[t-\frac{vf(t)}{c^{2}}]$</span>, so <span class="math-container">$F'(t)=\gamma(v)[1-\frac{v}{c}\frac{f'(t)}{c}]$</span>, but remembering that <span class="math-container">$f'(t)$</span> is the velocity of the body in the first sistem so must be less than <span class="math-container">$c$</span> an consequently the fractions are less than <span class="math-container">$1$</span>.</p>
<p>Explicitly find <span class="math-container">$g$</span> depends on the trayectory, but may be in simple examples you can do it.</p>
| 491
|
special relativity
|
What happens with the number of protons inside a wire in movement?
|
https://physics.stackexchange.com/questions/581380/what-happens-with-the-number-of-protons-inside-a-wire-in-movement
|
<p>In Special Relativity we solve the problem of the moving wire with an electrical current saying the protons inside the wire have a higher density: hence the wire is not neutral anymore (and we exchange a force caused by a B field with a force caused by an E field):</p>
<p><a href="https://phys.libretexts.org/Bookshelves/University_Physics/Book%3A_Introductory_Physics_-_Building_Models_to_Describe_Our_World_(Martin_Neary_Rinaldo_and_Woodman)/24%3A_The_Theory_of_Special_Relativity/24.05%3A_Electric_and_magnetic_fields_and_Special_Relativity" rel="nofollow noreferrer">https://phys.libretexts.org/Bookshelves/University_Physics/Book%3A_Introductory_Physics_-<em>Building_Models_to_Describe_Our_World</em>(Martin_Neary_Rinaldo_and_Woodman)/24%3A_The_Theory_of_Special_Relativity/24.05%3A_Electric_and_magnetic_fields_and_Special_Relativity</a></p>
<p><a href="https://phys.libretexts.org/@api/deki/files/16021/clipboard_e35d001b9d784182accd394cf30f1dbf8.png?revision=1" rel="nofollow noreferrer"><img src="https://phys.libretexts.org/@api/deki/files/16021/clipboard_e35d001b9d784182accd394cf30f1dbf8.png?revision=1" alt="enter image description here" /></a></p>
<p>But if the density of protons is higher, then the number of protons must be higher too, so changing the frame of reference is also creating new protons?</p>
<p>How can we explain the higher density without adding extra protons?</p>
|
<p>Normally this idea is presented in the case of a wire that is infinitely long. That gives the problem a translational symmetry and makes everything simpler. For an infinite wire, the number of protons is infinite. If you change to a different frame, the number is still infinite, but with a different density. This is similar to Hilbert's hotel: <a href="https://en.wikipedia.org/wiki/Hilbert%27s_paradox_of_the_Grand_Hotel" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Hilbert%27s_paradox_of_the_Grand_Hotel</a></p>
| 492
|
special relativity
|
Compute the vertical motion of the Rindler's length contraction paradox
|
https://physics.stackexchange.com/questions/584984/compute-the-vertical-motion-of-the-rindlers-length-contraction-paradox
|
<p>Suppose a rod with length <span class="math-container">$L$</span> moving toward a hole whose radius is <span class="math-container">$L$</span>. The paradox is, from the ground frame, the length of the rod is contracted hence rod will fall into the hole. However, from the rod's frame, it is the length of the hole contracted, so the rod will not fall. Now, ask to find the vertical motion of the rod within these two perspectives.</p>
<p>From the gound frame:
<span class="math-container">$$
y(t)=\frac{1}{2}gt^2
$$</span></p>
<p>From the rod's frame, due to the Lorentz transformation, the vertical motion of the front of the rod is
<span class="math-container">$$
y(t)=y'_{front}(t')=\frac{1}{2}g\gamma^2(t'+\frac{vx'}{c^2})^2
$$</span>
However, it is where I begin to confuse. For example, how to compute the <span class="math-container">$y'_{back}(t')$</span>? Obviously <span class="math-container">$\Delta t'\neq0$</span>, and it is equaled to
<span class="math-container">$$
\Delta t'=-\gamma\frac{v\Delta x'}{c^2}=-\gamma\frac{vL}{c^2}
$$</span>
Can anyone tell me if I am correct so far and if so how to compute <span class="math-container">$y'_{back}(t')$</span> properly?</p>
|
<p>I am afraid that your second equation belongs not only to the front end of the rod but also to any point on the rod at a distance <span class="math-container">$x'$</span> from the observer located at the hind end of the rod. For the front end, you had better substitute <span class="math-container">$x'=L'$</span>, and for the hind end where the observer is located use <span class="math-container">$x'=0$</span>. Therefore, we have:</p>
<p><span class="math-container">$$y'_{front}(t')=\frac{1}{2}g\gamma^2(t'+\frac{vL'}{c^2})^2\space,$$</span></p>
<p><span class="math-container">$$y'_{hind}(t')=\frac{1}{2}g\gamma^2t'^2\space.$$</span></p>
| 493
|
special relativity
|
Confusion about Lorentz Coordinate Transformation
|
https://physics.stackexchange.com/questions/584040/confusion-about-lorentz-coordinate-transformation
|
<p>A normal Lorentz coordinate problem might say:</p>
<p>At <span class="math-container">$t=t'=0$</span>, two coordinate systems <span class="math-container">$S$</span> and <span class="math-container">$S'$</span> have their origins coincide with the <span class="math-container">$S'$</span> system moving with speed <span class="math-container">$v$</span> in the <span class="math-container">$+x$</span> direction relative to <span class="math-container">$S$</span>. If event 1 happens at <span class="math-container">$x=a$</span>, <span class="math-container">$t=0$</span> in the <span class="math-container">$S$</span> system then when/where does this event happen in the <span class="math-container">$S'$</span> system. I know that <span class="math-container">$x'=\gamma(x-vt)$</span> and I have no confusion with that equation, but in problems like these, I am confused specifically about the time. Since <span class="math-container">$t=t'=0$</span> and <span class="math-container">$t=0$</span>, I would imagine that the time in the prime system would also be 0. But when one uses the Lorentz transformation <span class="math-container">$t'=\gamma(t-vx/c^2)$</span>, <span class="math-container">$t'$</span> comes out as a non-zero negative number. How do we reconcile this?</p>
|
<p>Consider Einstein's train thought experiment that demonstrates non-simultaneity. Let's say the train moves left to right. The lightning strikes at the front and back of the train occur simultaneously for the platform observer (S frame) but for the train observer (S' frame) the strike at the front of the train occurred first. If S and S' are set up so that the two frames coincide with t = t' = 0 at the back of the train when the strike there occurs, we have t = 0 in the S frame for both strikes but in the S' frame the strike at the front would have occurred at t' < 0. In your example, the two two events in the S frame are (x,t) = (0,0) and (a,0). These are simultaneous in the S frame so we know they are not simultaneous in the S' frame and the Lorentz transformation give us the t' values in S'. The (0,0) event also occurs at (0,0) in S', but the (a,0) event does not occur at t' = 0.</p>
| 494
|
special relativity
|
About relativity and train
|
https://physics.stackexchange.com/questions/590415/about-relativity-and-train
|
<p>Imagine a train moving with velocity v wrt a reference frame S, there is a clock in the rear and in the front of the train, and them are synchronized in the train reference, let's call this reference frame S'.
If you, the reference frame S, take a look simultaneously wrt your reference on the clocks, how will the time be relationed?
OBS: This is not a question, it is a doubt about the theory.
This is a well know fact in special relativity, the rear will be beyond the front by <span class="math-container">$Lv/c^{2}$</span>, but i am not sure why!!
See:
To both clocks being synchronized in the train reference, i will imagine a clock on the middle of the train of length proper L. When the light get in each side, the clock of its side starts to "click".</p>
<p>But, in the ground frame, the photon will need <span class="math-container">$$L/2(c-v)$$</span> to get in the rear, and <span class="math-container">$$L/2(c+v)$$</span> to came in the front. That is <span class="math-container">$$\delta t = Lv/(c^2-v^2)$$</span> and not Lv/c²!</p>
<p>Where is the error in this assumption?</p>
|
<p>You are comparing things in different coordinates, which always causes problems in relativity, and you do it twice:</p>
<p>(1) The train is not <span class="math-container">$L$</span> in <span class="math-container">$S$</span>, it is <span class="math-container">$L/\gamma$</span>, which brings your formula to:</p>
<p><span class="math-container">$$ \delta t = \frac L {\gamma}\frac v {c^2-v^2} $$</span></p>
<p>Note that:</p>
<p><span class="math-container">$$ \frac 1 {c^2-v^2} = \frac 1 {c^2} \gamma^2 $$</span></p>
<p>so that :</p>
<p><span class="math-container">$$ \delta t = \frac L {\gamma}\frac v {c^2}\gamma^2 = \gamma\delta t' $$</span></p>
<p>(2)The offset between the front and back clocks, <span class="math-container">$\delta t'$</span>, observed in <span class="math-container">$S$</span> is not a time measured in <span class="math-container">$S$</span>, it is a coordinate difference observed in <span class="math-container">$S'$</span>. You can related it to a time in <span class="math-container">$S$</span> per the prescription in the OP:</p>
<p><span class="math-container">$$ \delta t = \gamma \delta t'$$</span>.</p>
| 495
|
special relativity
|
Lorentz transformation and rotation matrix
|
https://physics.stackexchange.com/questions/590829/lorentz-transformation-and-rotation-matrix
|
<p><span class="math-container">$$\begin{pmatrix}
1& 0 & 0 & 0\\
0& \cos\theta & \sin\theta &0 \\
0& -\sin\theta & \cos\theta &0 \\
0 & 0 & 0 &1
\end{pmatrix}$$</span></p>
<p>See this matrix, it represents a rotation in the xy plane for the Lorentz transformation.</p>
<p>I am a little confused, is impression mine or it is a rotation measure as clockwise positive? Why this convention? It is always counterclock positive.</p>
|
<p>The rotation in the opposite direction (with <span class="math-container">$-\theta$</span>) is also a Lorentz transformation. Conventions about which one to use as an example have no physical significance.</p>
<p>This is similar to asking why we use a right-hand rule rather than a left-hand rule to define a cross product of two vectors. The answer has nothing to do with physical laws.</p>
| 496
|
special relativity
|
Light perpendicular to spaceship constant relativistic speed - different points where it hits the wall as seen inside and from outside?
|
https://physics.stackexchange.com/questions/591364/light-perpendicular-to-spaceship-constant-relativistic-speed-different-points
|
<p>I googled that question and found this answer:</p>
<p><a href="https://www.quora.com/If-I-had-a-laserpoint-inside-a-very-fast-moving-spaceship-and-I-point-at-the-wall-in-front-of-me-normal-to-the-direction-of-motion-will-the-light-hit-a-bit-lower-due-to-the-speed" rel="nofollow noreferrer">https://www.quora.com/If-I-had-a-laserpoint-inside-a-very-fast-moving-spaceship-and-I-point-at-the-wall-in-front-of-me-normal-to-the-direction-of-motion-will-the-light-hit-a-bit-lower-due-to-the-speed</a></p>
<p>Also sometime ago I watched the video on SR paradoxes explanations, where train fits to tunnel and paradox is explained by train not being rigid body:</p>
<p><a href="https://www.youtube.com/watch?v=Xrqj88zQZJg" rel="nofollow noreferrer">https://www.youtube.com/watch?v=Xrqj88zQZJg</a></p>
<p>In the quora answer the picture shows that light beam will go diagonally as seen from outside, however person answering claims for insider will see light hit the wall exactly in opposite point. The video above claims observers should agree on event (e.g. result of train hold in the tunnel). How then observers can agree what point the laser light will point to in the ship?</p>
<p>The ship length will diminish for outside observer, but opposite point on the wall should remain the same as I see it. That laser beam may trigger some other events like Schrodinger cat.</p>
<p>I found only questions regarding simultaneity or exceeding speed of light, however my question is different.</p>
|
<p>For the laser being inside the spaceship, the answer to the question , <em>"will the light hit a bit lower due to the speed"</em>, the correct answer is, "What speed?". The spaceship isn't moving, so a horizontal laser moves across the capsule at a constant height above the floor.</p>
<p>If the laser is 3' above the floor, then so is the spot on the wall (hull?). Of course an external observer for which the s/c is moving upwards at relativistic speed, the height isn't 3', but 3'<span class="math-container">$/\gamma$</span>, but the source and spot height must be the same distance from the floor.</p>
<p>All that means is that laser isn't aimed horizontally in the external frame. You can consider it a Doppler shift, or a type of stellar aberration, or, if you're a laser expert, you will realize which is a constant phase across the opening of the laser is the s/c frame is a phase ramp to a moving observer, and phase ramps steer coherent beams off the normal of the emitter.</p>
| 497
|
special relativity
|
Two light sources are a distance D apart. Show that in an inertial frame O' the photons are separated by a constant distance
|
https://physics.stackexchange.com/questions/590998/two-light-sources-are-a-distance-d-apart-show-that-in-an-inertial-frame-o-the
|
<p>Two light sources are at rest and at a distance D apart on the x-axis of some inertial frame, O. They emit photons simultaneously in that frame in the positive x-direction. Show that in an inertial frame, O', in which the sources have a velocity v along the x-axis, the photons are separated by a constant distance <span class="math-container">$$D\sqrt{\frac{c-u}{c+u}}$$</span></p>
<p>I let E1 be the event where source 1 emits the photon and E2 for the second source with the respective coordinates in O as <span class="math-container">$(x_1, t_1$</span>) and <span class="math-container">$(x_2,t_2)$</span> such that <span class="math-container">$t_2=t_1 \because$</span> simultaneous and <span class="math-container">$x_2-x_1 =D$</span>.</p>
<p>Using Lorentz transformation I obtained that in O', <span class="math-container">$$x'_2-x'_1 = \gamma (D-v(t_2-t_1))=\gamma D \ \because t_2 = t_1$$</span> In the solutions <span class="math-container">$t_2-t_1 = \frac{D}{c} $</span> however I cannot see where this came from.</p>
|
<p>Note that in <span class="math-container">$O$</span>, if the photons have wavelength <span class="math-container">$\lambda$</span>, then the number of cycles (of phase) between the two photons is:</p>
<p><span class="math-container">$$ N = D/\lambda $$</span></p>
<p>Since <span class="math-container">$N$</span> is a Lorentz invariant and the Doppler shift says:</p>
<p><span class="math-container">$$ \lambda' = \sqrt{\frac{c-v}{c+v}} $$</span></p>
<p>then:</p>
<p><span class="math-container">$$ D' = D\sqrt{\frac{c-v}{c+v}} $$</span></p>
<p>So that <em>is</em> the correct answer.</p>
<p>If you set the <span class="math-container">$O$</span> origin so that</p>
<p><span class="math-container">$$x_1 = 0$$</span>
<span class="math-container">$$x_2 = D$$</span>
<span class="math-container">$$t_1 = t_2 = 0$$</span></p>
<p>and make the <span class="math-container">$O'$</span> origin the same event as the <span class="math-container">$O$</span> origin:</p>
<p><span class="math-container">$$ x_1' = 0$$</span>
<span class="math-container">$$ x_2' = \gamma D$$</span>
<span class="math-container">$$ t_1' = 0$$</span>
<span class="math-container">$$ t_2' = \gamma(0-\frac{vD}{c^2}) = -\gamma \frac{vD}{c^2} $$</span></p>
<p>so that <span class="math-container">$E_2$</span> occurs before <span class="math-container">$E_1$</span> in <span class="math-container">$O'$</span>.</p>
<p>If you add that "head-start" onto <span class="math-container">$x_2'-x_1'$</span>:</p>
<p><span class="math-container">$$ D' = (x_2'-x_1') + c(t_2' - t_1') $$</span>
<span class="math-container">$$ D' = (\gamma D) + c(-\gamma \frac{vD}{c^2})$$</span>
<span class="math-container">$$ D' = D(\gamma -\gamma\frac{v}{c}) = \gamma D(1-\frac v c)$$</span>
<span class="math-container">$$ D' = D\frac{1-v/c}{\sqrt{(1-v/c)(1+v/c)}} = D\sqrt{\frac{c-v}{c+v}} $$</span></p>
<p>So that is the position of the "2" photon when the "1" photon is emitted in <span class="math-container">$O'$</span>:</p>
<p><span class="math-container">$$ (t', x')_{O'} = (0, D')_{O'} $$</span></p>
<p>It is <em>not</em> the same as <span class="math-container">$t'_2$</span>.</p>
| 498
|
special relativity
|
Does clock really run slow inside a high velocity rocket ship?
|
https://physics.stackexchange.com/questions/592924/does-clock-really-run-slow-inside-a-high-velocity-rocket-ship
|
<p>In time dilation, does clock really run slow when it’s in high speed? Is that means the mechanical parts of the clock run slow?
Or , It’s just the other inertial observer who thinks that the other one’s clock is running slow.(because the successive light signal is taking more time to reach the destination)</p>
<p>And, If the biological/atomic process is not running slow for the traveling twin in twin paradox, isn’t it impossible that he will be younger when they meet?</p>
|
<p>At high speeds, time really does slow down as you get close to the speed of light. All processes (not just mechanical clocks) slow down, including things like radioactive decay as well as everything biological going on inside your body. This means you yourself do not notice your clocks running slow inside the spaceship.</p>
<p>The twin paradox is a more complicated thing to wrap your head around and I recommend reading the reference provided by John Rennie.</p>
| 499
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.