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= Sνµρ , (1.64) we say that Sµνρ is symmetric in its ﬁrst two indices, while if Sµνρ = Sµρν = Sρµν = Sνµρ = Sνρµ = Sρνµ , (1.65) we say that Sµνρ is symmetric in all three of its indices. Similarly, a tensor is antisymmetric (or “skew-symmetric”) in any of its indices if it changes sign when those indices are exchange...
erior derivative of.) It is striking that information about the topology can be extracted in this way, which essentially involves the solutions to diﬀerential equations. The dimension bp of the space H p(M) is − 1 SPECIAL RELATIVITY AND FLAT SPACETIME 23 called the pth Betti number of M, and the Euler characteristic is...
ity is automatically normalized: − ηµνU µU ν = 1 . − 27 (1.98) (1.99) (It will always be negative, since we are only deﬁning it for timelike trajectories. You could deﬁne an analogous vector for spacelike paths as well; null paths give some extra problems since the norm is zero.) In the rest frame of a particle, its fo...
etric is the same, but only basic notions of analysis like open sets, functions, and coordinates.) The entire manifold is constructed by smoothly sewing together these local regions. Examples of manifolds include: • • • Rn itself, including the line (R), the plane (R2), and so on. This should be obvious, since Rn looks...
d has an individual existence independent of any embedding. We have no reason to believe, for example, that four-dimensional spacetime is stuck in some larger space. (Actually a number of people, string theorists and so forth, believe that our four-dimensional world is part of a ten- or eleven-dimensional spacetime, bu...
bitrary manifold, and we have speciﬁed our basis vectors to be ˆe(µ) = ∂µ. This particular basis (ˆe(µ) = ∂µ) is known as a coordinate basis for Tp; it is the formalization of the notion of setting up the basis vectors to point along the coordinate axes. There is no reason why we are limited to coordinate bases when we...
have x = r sin θ cos φ y = r sin θ sin φ z = r cos θ , which leads directly to ds2 = dr2 + r2 dθ2 + r2 sin2 θ dφ2 . (2.31) (2.32) Obviously the components of the metric look diﬀerent than those in Cartesian coordinates, but all of the properties of the space remain unaltered. Perhaps this is a good time to note that mo...
e weight of the density a density into an honest tensor — multiply by (the absolute value signs are there because g < 0 for Lorentz metrics). The result will transform according to the tensor transformation law. Therefore, for example, we can deﬁne the Levi-Civita tensor as g \| \| − ǫµ1µ2 µn = g ˜ǫµ1µ2 µn . (2.42) ··· \|...
covariant derivative of a one-form can also be expressed as a partial derivative plus some linear transformation. But there is no reason as yet that the matrices representing this transformation should be related to the coeﬃcients Γν In general we could write something like µλ. ∇µων = ∂µων + Γλ µνωλ , (3.7) Γλ e µν is...
hat the Christoﬀel connection satisﬁes µλV λ . (3.26) and we therefore obtain Γµ µλ = 1 g \| \| q ∂λ , g \| \| q ∇µV µ = 1 g \| \| q ∂µ( \| q g V µ) . \| (3.27) (3.28) 3 CURVATURE 62 There are also formulas for the divergences of higher-rank tensors, but they are generally not such a great simpliﬁcation. As the last factoid we...
right triangle, or n-simplex. 3 CURVATURE 67 λ λ0 Z A(η1) dη1 λ η2 λ0 Z λ0 Z A(η2)A(η1) dη1dη2 λ η3 η2 λ0 Z Z λ0 Z λ0 A(η3)A(η2)A(η1) d3η It would simplify things if we could consider such an integral to be over an n-cube instead of an n-simplex; is there some way to do this? There are n! such simplices in each cube, s...
paths we would have derived the same equation, since the only diﬀerence is an overall minus sign in the ﬁnal answer. There are also null geodesics, which satisfy the same equation, except that the proper time cannot be used as a parameter (some set of allowed parameters will exist, related to each other by linear tran...
ed from non-tensorial elements; you can check that the transformation laws all work out to make this particular combination a legitimate tensor. The antisymmetry of Rρ its derivation. σµν in its last two indices is immediate from this formula and We constructed the curvature tensor completely from the connection (no me...
hich we could not set to zero by a clever choice of coordinates. This should reinforce your conﬁdence that the Riemann tensor is an appropriate measure of curvature. In addition to the algebraic symmetries of the Riemann tensor (which constrain the number of independent components at any point), there is a diﬀerential ...
t cross. The entire surface is the set of points xµ(s, t) M. We have two natural vector ﬁelds: the tangent vectors to the geodesics, ∈ ∈ and the “deviation vectors” Sµ = ∂xµ ∂s T µ = ∂xµ ∂t , . (3.107) (3.108) This name derives from the informal notion that Sµ points from one geodesic towards the neighboring ones. The ...
3.128) (The name “spin connection” comes from the fact that this can be used to take covariant derivatives of spinors, which is actually impossible using the conventional connection coeﬃcients.) In the presence of mixed Latin and Greek indices we get terms of both kinds. The usual demand that a tensor be independent of...
therefore deﬁne a connection on the ﬁber bundle to be an object Aµ B, with two “group indices” and one spacetime index. Under GCT’s it transforms as a one-form, while under gauge transformations it transforms as A A′ B′ = OA′ Aµ AOB′ BAµ A OB′ C∂µOA′ C . B − (3.146) (Beware: our conventions are so drastically diﬀerent...
ysics. It is the EEP which implies (or at least suggests) that we should attribute the action of gravity to the curvature of spacetime. Remember that in special relativity a prominent role is played by inertial frames — while it was not possible to single out some frame of reference as uniquely “at rest”, it was possib...
(4.11) (4.12) Finally, the weakness of the gravitational ﬁeld allows us to decompose the metric into the Minkowski form plus a small perturbation: gµν = ηµν + hµν , hµν\| \| << 1 . (4.13) (We are working in Cartesian coordinates, so ηµν is the canonical form of the metric. The “smallness condition” on the metric perturb...
tion of the mass density is; it’s the energy-momentum tensor Tµν. The gravitational potential, meanwhile, should get replaced by the metric tensor. We might therefore guess that our new equation will have Tµν set proportional to some tensor which is second-order in derivatives of the metric. In fact, using (4.21) for t...
ade from second derivatives of the metric, and we argued earlier that the only independent scalar we could construct from the Riemann tensor was the Ricci scalar R. What we did not show, but is nevertheless true, is that any nontrivial tensor made from the metric and its ﬁrst and second derivatives can be expressed in ...
however, the cosmological constant has proven diﬃcult to kill oﬀ. If we like we can move the additional term in (4.74) to the right hand side, and think of it as a kind of energy-momentum tensor, with Tµν = Λgµν (it is automatically conserved by metric compatibility). Then Λ can be interpreted as the “energy density of...
= 8πGTµν do involve second derivatives of the metric with respect to time (since the connection involves ﬁrst derivatives of the metric and the Einstein tensor involves ﬁrst derivatives of the connection), ∇µGµν = 0. so we seem to be on the right track. However, the Bianchi identity tells us that We can rewrite this e...
point, then none of the points in the region containing closed timelike curves are in the domain of dependence of Σ, since the closed timelike curves themselves do not intersect Σ. This is obviously a worse problem than the previous one, since a well-deﬁned initial value problem does not seem to × 4 GRAVITATION 127 id...
ν = = ∂yα ∂yβ ∂xν gαβ ∂xµ 1 0 0 sin2 θ , (5.12) (5.13) as you can easily check. Once again, the answer is the same as you would get by naive substitution, but now we know why. We have been careful to emphasize that a map φ : M N can be used to push certain things forward and pull other things back. The reason why it ge...
vative derived from gµν. ∗ ∗ gµν, φ Let’s put some of these ideas into the context of general relativity. You will often hear it proclaimed that GR is a “diﬀeomorphism invariant” theory. What this means is that, if the universe is represented by a manifold M with metric gµν and matter ﬁelds ψ, and φ : M → M is a diﬀeom...
d as our ability to decompose the metric into the ﬂat Minkowski metric plus a small perturbation, gµν = ηµν + hµν , hµν\| \| << 1 . (6.1) We will restrict ourselves to coordinates in which ηµν takes its canonical form, ηµν = 1, +1, +1, +1). The assumption that hµν is small allows us to ignore anything that is diag( highe...
e gauge freedom remaining, since we can change our coordinates by (inﬁnitesimal) harmonic functions. ∂λh = 0 . (6.17) ∂µhµ λ − 6 WEAK FIELDS AND GRAVITATIONAL RADIATION 147 In this gauge, the linearized Einstein equations Gµν = 8πGTµν simplify somewhat, to 2hµν − 1 2 ηµν2h = 16πGTµν , − while the vacuum equations Rµν =...
get a feeling for the physical eﬀects due to gravitational waves, it is useful to consider the motion of test particles in the presence of a wave. It is certainly insuﬃcient to solve for the trajectory of a single particle, since that would only tell us about the values of the coordinates along the world line. (In fact...
ymmetric in i and j, while the third and fourth lines are simply repetitions of reverse integration by parts and conservation of T µν. It is conventional to deﬁne the quadrupole moment tensor of the energy density of the source, e qij(t) = 3 yiyjT 00(t, y) d3y , Z (6.83) a constant tensor on each surface of constant ti...
elativity for energy loss through gravitational radiation. Hulse and Taylor were awarded the Nobel Prize in 1993 for their eﬀorts. December 1997 Lecture Notes on General Relativity Sean M. Carroll 7 The Schwarzschild Solution and Black Holes We now move from the domain of the weak-ﬁeld limit to solutions of the full no...
we get the following nonvanishing components of the Riemann tensor: R0 R0 R0 R0 R0 R1 R1 R2 − re− re− re− re− 101 = e2(β 202 = 303 = 212 = 313 = 212 = re− 313 = re− 323 = (1 − − − − − 0β + (∂0β)2 α)[∂2 2β∂1α 2β sin2 θ ∂1α 2α∂0β 2α sin2 θ ∂0β 2β∂1β 2β sin2 θ ∂1β e− 2β) sin2 θ . ∂0α∂0β] + [∂1α∂1β − ∂2 1α − − (∂1α)2] (7....
dλ  dr dλ − sin θ cos θ 2 dφ dλ ! = 0 , and d2φ dλ2 + 2 r dφ dλ dr dλ + 2 cos θ sin θ dθ dλ dφ dλ = 0 . (7.35) (7.36) (7.37) There does not seem to be much hope for simply solving this set of coupled equations by inspection. Fortunately our task is greatly simpliﬁed by the high degree of symmetry of the Schwarzschild...
a stable circular orbit in the Schwarzschild metric. There are also unbound orbits, which come in from inﬁnity and turn around, and bound but noncircular ones, which oscillate around the stable circular radius. Note that such orbits, which would describe exact conic sections in Newtonian gravity, will not do so in GR,...
e let ˜u = t + r∗ 7 THE SCHWARZSCHILD SOLUTION AND BLACK HOLES 184 t r = 2GM r* = - 8 r* ˜v = t r∗ , − (7.68) then infalling radial null geodesics are characterized by ˜u = constant, while the outgoing ones satisfy ˜v = constant. Now consider going back to the original radial coordinate r, but replacing the timelike co...
the past 7 THE SCHWARZSCHILD SOLUTION AND BLACK HOLES 190 event horizon, while the boundary of region II is called the future event horizon. Region IV, meanwhile, cannot be reached from our region I either forward or backward in time (nor can anybody from over there reach us). It is another asymptotically ﬂat region o...
I It is nice to be able to ﬁt all of Minkowski space on a small piece of paper, but we don’t really learn much that we didn’t already know. Penrose diagrams are more useful when we want to represent slightly more interesting spacetimes, such as those for black holes. The original use of Penrose diagrams was to compare ...
child, due to the extra term in the function ∆(r) (which can be thought of as measuring “how much the light cones tip over”). One thing remains the same: at r = 0 there is a true curvature singularity (as could be checked by computing the curvature scalar RµνρσRµνρσ). Meanwhile, the equivalent of r = 2GM will be the ra...
ing tensor. This is any symmetric (0, n) tensor ξµ1 µn which satisﬁes ··· ∇(σξµ1 µn) = 0 . ··· (7.119) Simple examples of Killing tensors are the metric itself, and symmetrized tensor products of Killing vectors. Just as a Killing vector implies a constant of geodesic motion, if there exists a Killing tensor then along...
n into the external universe. Since your energy is conserved along the way, at the end we will have E(1) > E(0) . (7.141) Thus, you have emerged with more energy than you entered with. 7 THE SCHWARZSCHILD SOLUTION AND BLACK HOLES 214 (top view) ergosphere Killing horizon µ (2) p (1) µ p (0) µ p There is no such thing a...
and as a result we see a dipole anisotropy in the cosmic microwave background as a result of the conventional Doppler eﬀect. − Our interest is therefore in maximally symmetric Euclidean three-metrics γij. We know that maximally symmetric metrics obey (3)Rijkl = k(γikγjl − γilγjk) , (8.2) where k is some constant, and ...
Ω is greater than, equal to, or less than one. We have ↔ ↔ ↔ The density parameter, then, tells us which of the three Robertson-Walker geometries describes our universe. Determining it observationally is an area of intense investigation. ↔ ↔ ↔ ρ < ρcrit ↔ ρ = ρcrit ↔ ρ > ρcrit ↔ open ﬂat closed . It is possible to solv...
OLOGY 230 fact that in ﬂat space, for a source at distance d the ﬂux over the luminosity is just one over the area of a sphere centered around the source, F/L = 1/A(d) = 1/4πd2. In an FRW universe, however, the ﬂux will be diluted. Conservation of photons tells us that the total number of photons emitted by the source ...
l to propagate from one to the other. This particular diﬃculty may be resolved by introducing the concept of the electromagnetic ﬁeld. Then we may suppose that one charge does not act directly on the other, but on the ﬁeld in its immediate vicinity; this in turn aﬀects the ﬁeld further out, and so on. By supposing that...
law of universal gravitation, provide the equations from which we can determine the motion of any dynamical system. Problems Note. Here and in later chapters, starred problems are somewhat harder. 1. An object A moving with velocity v collides with a stationary object B. After the collision, A is moving with velocity 1...
ver come to rest: it will continue moving in the same direction for ever, with decreasing speed as it approaches x = 0 and then increasing speed. The force corresponding to the potential energy function (2.10) is, by (2.7), F (x) = kx. − (2.12) Linear Motion 21 V E (a) −a a x Fig. 2.2 V E (1) E (2) (b) x It is an attra...
damped oscillator, as in (2.18). The λ ˙x2. If λ were negative, rate at which work is done by the force the particle would therefore be gaining, rather than losing, energy. So we shall assume that λ is positive. λ ˙x is − − 28 Classical Mechanics Large damping If λ is so large that γ > ω0, then both roots for p are rea...
ar, a simple periodic force may be written as a sum of two terms proportional to eiωt and e− iωt.) Because of the linearity of (2.34), the corresponding solution is x = Areiωrt + transient, (2.44) r where each Ar is related to the corresponding Fr by (2.37). This may easily be veriﬁed by direct substitution in (2.34). ...
te the energy and momentum conservation equations in the forms m1v2 1 − m1v1 − 1 = m2u2 m1u2 2 − m1u1 = m2u2 − m2v2 2, m2v2. We can then divide the ﬁrst equation by the second, to obtain or, equivalently, v1 + u1 = u2 + v2, v2 − v1 = u1 − u2. (2.59) This shows that the relative velocity is just reversed by the collisio...
m of period 2 s, and ﬁnd the angular velocity of the pendulum when it reaches the downward vertical. 19. *The particle of Problem 11 starts from rest at x = a, and is given a small push to start it moving to the right. What is its velocity when it reaches the point x? Given that t = 0 is the instant when it reaches x =...
to the velocity, so that Energy and Angular Momentum 53 the equation of motion becomes m¨r = − λ ˙r + mg. (3.16) The actual dependence of atmospheric drag on velocity is deﬁnitely nonlinear, but this equation nevertheless gives a reasonable qualitative picture of the motion. Deﬁning γ = λ/m, we may write (3.16) as ¨x =...
g them is represented by an equation y = y(x) such that the function satisﬁes the boundary conditions y(x0) = y0, y(x1) = y1. (3.33) Consider two neighbouring points on this curve. The distance dl between them is given by dl = dx2 + dy2 = 1 + y 2 dx, 60 Classical Mechanics where y = dy/dx. Thus the total length of the ...
n of 3.4 that the angular momentum vector J in this case is a constant, and that the motion is conﬁned to a plane. The equations of motion are considerably simpliﬁed by an appropriate choice of the polar axis θ = 0. If it is chosen so that ˙ϕ = 0 initially, then by the third of Eqs. (3.48), ˙ϕ will always be zero. The ...
ϕ(θ). To perform the integration, use the substitution x = cot θ.] 18. *Parabolic co-ordinates (ξ, η) in a plane are deﬁned by ξ = r + x, η = x. Find x and y in terms of ξ and η. Show that the kinetic energy r of a particle of mass m is − T = m 8 (ξ + η) ˙ξ2 ξ + ˙η2 η . Hence ﬁnd the equations of motion. 19. Write dow...
.14). If the particle is initially at a distance r0 from the origin, and moving with velocity v0 in a direction making an angle α with the radius vector (as in Fig. 4.1), then the values of E and J are ≤ ≤ r E = 1 2 mv2 0 + 1 2 kr2 0, J = mr0v0 sin α. Thus the equation whose roots are a and b becomes (on multiplying by...
θ θ0) − − 1] = l, (k > 0) (4.28) and in the attractive case r[e cos(θ − θ0) + 1] = l. (k < 0) (4.29) These are the polar equations of conic sections, referred to a focus as origin (see Appendix B). The constant e, the eccentricity, determines the shape of the orbit; l, called the semi-latus rectum, determines its scale...
useful to note an alternative deﬁnition of the diﬀerential crosssection, which is applicable even if we cannot follow the trajectory of each individual particle, and therefore cannot say just which of the incoming particles are those that emerge in a particular direction. We may deﬁne dσ/dΩ to be the ratio of scattere...
nd 27 days, respectively. The mass of Jupiter is 318 times that of the Earth. The semi-major axis of the Earth’s orbit, or astronomical unit (au) is 1.50 108 km.] 3. The semi-major axis of Jupiter’s orbit is 5.20 au. Find its orbital period in years, and its mean (time-averaged) orbital speed. (Mean orbital speed of Ea...
for particles of mass m and initial veπ/n (see locity v scattered by a repulsive inverse-cube-law force is π Problem 25). Hence ﬁnd the diﬀerential cross-section. − 27. *The potential energy of a particle of mass m is V (r) = k/r + c/3r3, where k < 0 and c is a small constant. (The gravitational potential energy in th...
lar methods can be used, however, even at the highest energies. 110 Classical Mechanics E E ~ Fig. 5.4 If there are both electric and magnetic ﬁelds present, a charged particle B). An interesting case is experiences the Lorentz force, F = q(E + v that of crossed ﬁelds. ∧ Example: Crossed electric and magnetic ﬁelds A p...
elative to the inertial Rotating Frames 117 h x (b) x h (a) Fig. 5.8 observer. As it falls, the angular momentum about the Earth’s axis remains constant, and therefore its angular velocity increases, so that it gets ahead of the ground beneath it. Figs. 5.8(a) and 5.8(b) show the experiment as it appears (but much exag...
angular momentum is ∧ n and b = n a. Then the ∧ J = mr ∧ v = mrvn. (5.28) If we specify the position of the particle in its orbit by the angle ψ between a and r, then in terms of the axes a, b, n, the components of r, v and B Rotating Frames 123 are r = (r cos ψ, r sin ψ, 0), v = ( v sin ψ, v cos ψ, 0), − B = (0, B sin...
ight h above the equator by using an inertial frame, and verify that the answer agrees with that (Hint : Use equations (3.48). Recall found using a rotating frame. Fig. 5.8.) 17. Find the equations of motion for a particle in a frame rotating with variable angular velocity ω, and show that there is another apparent for...
mple, as a point mass in discussing the motion of the planets. Of course, any deviation from spherical symmetry will lead to a modiﬁcation in the inverse square law. We shall discuss the nature of this correction later. Potential Theory 137 6.4 Expansion of Potential at Large Distances It is only for simple cases that ...
quatorial radii. As expected, the value of J2 is somewhat less than 2 5 . The relation between the two is determined by the density distribution within the Earth. (See Problem 15.) One consequence of the fact that the Earth’s surface is approximately an equipotential surface should be noted. Because of the general prop...
y seem rather small, especially when further reduced by the eﬀect just mentioned. However, it is important to remember that they refer to the unrealistic case of an Earth without continents. The observed tidal range in mid-ocean (which can be measured by ranging from satellites) is in fact quite small, normally less th...
s distribution. 9. *Two equal charges q are located at the points ( a, 0, 0), and two a, 0). Find the leading term in the potential at charges large distances, and the corresponding electric ﬁeld. 2 cos2 θ q at (0, − ± ± − 10. Find the gravitational potential at large distances of a thin circular 3. Find also the loop ...
) The mass µ is called the reduced mass, because it is always less than either m1 or m2. These two equations are now completely separate. Equation (7.4) shows that the centre of mass moves with uniform acceleration g. In the case g = 0 it is equivalent to the law of conservation of momentum, M ˙R = m1 ˙r1 + m2 ˙r2 = P ...
e interesting quantity is the fraction of the total kinetic energy which is transferred. This is T2 T = 4m1m2 M 2 sin2 1 2 θ∗. (7.22) The maximum possible kinetic energy transfer occurs for a head-on collision (θ∗ = π), and is T2/T = 4m1m2/(m1 + m2)2. Clearly this can be close to unity only if m1 and m2 are comparable ...
ondition. We shall now calculate the rate at which recoiling target particles enter the detector. In a collision in which the incoming particle has impact parameter b, and is moving in a plane speciﬁed by the angle ϕ, the target particle will emerge in a direction speciﬁed by the polar angles α, ψ, where α is related t...
the sum of their masses. It is also clear that if Newton’s third law applies to each pair of particles from two composite bodies, then it will apply to the bodies as a whole. Thus, with a suitable (and very natural) interpretation of the concepts involved, Newton’s three basic laws may be applied to composite bodies as...
f gradually slowing the Earth’s rotation. We can picture this eﬀect as follows: as the Earth rotates, it tries to carry with it the tidal ‘bulges’, while the Moon’s attraction is pulling them back into line. (See Fig. 8.3.) Thus there is a couple acting to slow the Earth’s § Moon Fig. 8.3 rotation, and a corresponding ...
e external forces do work. We can also ﬁnd the rate of change of energy relative to the centre of mass. From (8.6), we have d dt 2 2 M ˙R ( 1 ) = M ˙R ¨R = ˙R · F i. · i Hence, subtracting from (8.32) and using (8.29) and (8.16), we ﬁnd d dt (T ∗ + Vint) = F i. ˙r∗i · i (8.33) This equation is the analogue of (8.21). N...
sive force of impulse I delivered once every orbit, at perigee. By considering changes in energy and angular momentum, ﬁnd the changes in the parameters a and l. Show that e)2, and hence that the eﬀect is to decrease the period and δl = δa(1 apogee distance, while leaving the perigee distance unaﬀected. (The orbit ther...
dulum of length l = I/M R. (This is the length of the equivalent simple pendulum) Thus the period of small oscillations is 2π I/M gR. As we shall see, l is always larger than the distance R from the pivot to the centre of mass. l/g = 2π The energy conservation equation is T + V = 1 2 I ˙ϕ2 − M gR cos ϕ = E = constant. ...
= I1, the symmetry axis e3 is of course uniquely determined, but the I3 axes e1 and e2 are not. We may choose them to be any pair of orthogonal axes in the plane normal to e3. Indeed, they need not even be ﬁxed in the body, so long as they always remain perpendicular to e3 and to each other. Equation (9.23) will still ...
elocity ω. Thus, to minimize the eﬀect of a given force, we should use a fat, rapidly spinning body. 10.3 and § § The great stability of rapidly rotating bodies is the basis of the gyroscope. Essentially, this consists of a spinning body suspended in such a way that its axis is free to rotate relative to its support. T...
oking for solutions of the form ω1 = a1ept, ω2 = a2ept, where a1, a2 and p are constants. (This is essentially the method we used to treat the the problem of stability of equilibrium in one dimension in 2.2, 2.3. We shall discuss the general problem of stability by a similar §§ method in the next chapter.) Substituting...
d its angular velocity when it reaches the lowest point. 2. An insect of mass 100 mg is resting on the edge of a ﬂat uniform disc of mass 3 g and radius 50 mm, which is rotating at 60 r.p.m. about a smooth pivot. The insect crawls in towards the centre of the disc. Find the angular velocity when it reaches it, and the ...
e represented by algebraic conditions on the co-ordinates (e.g., X = Y = Z = 0), which may be used to eliminate some of the co-ordinates. In this particular case, the three Euler angles alone suﬃce to ﬁx the position of every particle. The second type of constraint is represented by conditions on the velocities rather ...
+ I3( ˙ψ + ˙ϕ cos θ) cos θ] = 0, d dt [I3( ˙ψ + ˙ϕ cos θ)] = 0. (10.13) Lagrangian Mechanics 237 Note that this last equation tells us that the component of angular velocity ω3 about the axis of symmetry is constant ω3 = ˙ψ + ˙ϕ cos θ = constant. (10.14) From the equations (10.13), given that θ remains constant, we lea...
= ∇ Lagrangian function is ∧ A using (A.55).) Thus the L = 1 2 m( ˙ρ2 + ρ2 ˙ϕ2 + ˙z2) + 1 2 qBρ2 ˙ϕ. Lagrange’s equations are therefore m¨ρ = mρ ˙ϕ2 + qBρ ˙ϕ, d dt (mρ2 ˙ϕ + 1 2 qBρ2) = 0, m¨z = 0. (10.30) (10.31) In particular, in the case where ρ is a constant, we learn from the last two equations that ˙ϕ and ˙z are ...
mum angular velocity ω3 for which steady precession at this angle is possible. 9. A simple pendulum of mass m and length l hangs from a trolley of mass M running on smooth horizontal rails. The pendulum swings in a plane parallel to the rails. Using the position x of the trolley and the angle of inclination θ of the pe...
e sake of symmetry we have written k21 = k12 in the second equation. k12q2, k22q2, ¨q1 = ¨q2 = (11.10) − − Small Oscillations and Normal Modes 257 In the general case, V may be taken to be a homogeneous quadratic function of the co-ordinates, which can be written V = n n α=1 β=1 1 2 kαβqαqβ, (11.11) with kβα = kαβ. (No...
. 11.3(a)). Since the spring is neither expanded nor compressed in this motion, it is not surprising that the frequency is just that of the uncoupled pendulums. In the second normal mode, which has a somewhat higher frequency, the pendulums swing in opposite directions, alternately expanding and compressing the spring ...
chapter. According to Small Oscillations and Normal Modes 271 Fig. 11.7 (10.38), f (x) is a periodic function of x, with period 2l. Hence it may be expanded in a Fourier series (see Eq. (2.45)), + ∞ f (x) = fneinπx/l. n= −∞ Thus the solution (10.37) is + ∞ y(x, t) = fn einπ(ct+x)/l einπ(ct − x)/l − n= −∞ + ∞ n= −∞ = 2...
in the form ˙pα = ∂L ∂qα , where the generalized momenta are deﬁned by pα = ∂L ∂ ˙qα . (12.1) (12.2) Here and in the following equations, α runs over 1, 2, . . . , n. The instantaneous position and velocity of every part of our system may be speciﬁed by the values of the 2n variables q and ˙q. However, we 277 278 Clas...
ow that for the symmetric top two of the three Euler angles are ignorable co-ordinates, and ﬁnd the ‘eﬀective potential energy function’ for the remaining co-ordinate. We start from the Lagrangian function (10.11): 2 I3( ˙ψ + ˙ϕ cos θ)2 2 I1 ˙ϕ2 sin2 θ + 1 2 I1 ˙θ2 + 1 L = 1 M gR cos θ. − The corresponding generalized ...
ition between one type and another as the parameters of the system are varied, is now one of the most active ﬁelds of mathematical physics. It has revealed an astonishing range of possibilities. 12.6 Symmetries and Conservation Laws §§ In 12.2 and 12.3, we found some examples of conserved quantities, but so far we have...
ery reasonable. We know from our earlier work that momentum is conserved for an isolated system, but not for a system subjected to external forces which determine a preferred origin (for example, a centre of force). Rotations An inﬁnitesimal rotation through an angle δϕ about the z-axis yields δxi = δpxi = yi δϕ, pyi δ...
ion when ω2 > g/l. 3. A light, inextensible string passes over a small pulley and carries a mass 2m on one end. On the other end is a mass m, and beneath it, supported by a spring with spring constant k, a second mass m. Find the Hamiltonian function, using the distance x of the ﬁrst mass beneath the pulley, and the ex...
ll trajectories in this space for various initial prescribed values of x, together with the velocity directions upon them, constitutes the phase portrait of the system. The evolutions of a continuous dynamical system pictured in this manner are often described as a ﬂow, by analogy with ﬂuid motion (see Fig. 13.1). The ...
stant multiples of eλt. The analysis appears similar to that involved in the calculation of the frequencies of the normal modes in 11.3, but we are dealing with matrices M which are not necessarily symmetric, so that the roots of the quadratic characteristic equation for λ need not be real. Evidently, the signs of the ...
ersed (again, see Problem 7). In human conﬂict there are mathematical theories of war. The simplest such model involves forces x(t), y(t) which are eﬀectively isolated apart from the confrontation between them. Example: Combat model The simple system ˙x = ˙y = ay, bx, − − where a, b are positive constants, models the a...
e precession rate for the Chandler wobble of the Earth is about Ω (i.e. once per day), while the precession rate is about 1 2 Ω for a spun pass with an American football and about 2Ω for a spun coin, discus or dinner plate! 9.9. § The problem of rotational motion, when the external force moment G is not special, can be...
g. 13.24 The parameter λ in the exponent is called a Lyapunov exponent . When a typical small volume of initial states in phase space is considered, it is evident that dynamical evolution will result, in general, in stretching and squeezing of this volume. There will be a number of principal deformations, which is equa...
susceptibles, moved/recovered and a, b are positive constants. infectives, re- (a) Show that the overall population N = S +I +R remains constant, so that we may consider (S, I) in a projected phase plane. Hence show that a trajectory with initial values (S0, I0) has equation I(S) = I0 + S0 − (b) Using the function I(S)...
conditions then lead to tori which are nested within the phase space. 350 Classical Mechanics On a particular torus M the separate vector ﬁelds correspond to n independent types of linking circuit (see Fig. 14.1(a),(b)), and the integrable motion is then exactly equivalent to the combination of n separate corresponding...
we have 2.1 and \\\\ § § H(q, p) = p2 2m + V (q) = E, (14.7) so that p = 2m[H ± I = 1 2π − % V (q)] and p dq = 2√2m 2π q2 q1 H − V (q) dq. (14.8) 356 Classical Mechanics Example: Simple harmonic oscillator Find action/angle variables for the simple harmonic oscillator. Here V (q) = 1 2 kq2, so that I = H/ω0, where ω0 ...
at a point on its axis of symmetry. The problem has three degrees of freedom and there are three independent conserved quantities H, pϕ, pψ, so that the system is integrable and the motions detailed are regular and ordered. More complicated systems abound when the equations of rotational motion of an asymmetric body ar...

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