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advantage that from now on we need not write #hree laws every tỉme we write
Newton”s equations or other laws of physics. We write what looks like one law,
but really, of course, it is the three laws for any particular set of axes, because
any vector equation involves the statement that cach oƒ the components is cqudl.
Fig. 11-7. A curved trajectory.
The fact that the acceleration is the rate of change of the vector velocity
helps us to calculate the acceleration in some rather complicated circumstances.
Suppose, for instance, that a particle is moving on some complicated curve
(Fig. 11-7) and that, at a given instant ứ, it had a certain velocity ơi, but that
when we go to another instant £a a little later, it has a diferent velocity 0a. What
is the acceleration? Answer: Acceleration is the diference in the velocity divided
by the small time interval, so we need the diference of the two velocities. How
do we get the diference of the velocities? '[o subtract two vectors, we put the
vector across the ends of 0a and 0; that is, we draw Ao as the diference of the
two vectors, right? /o/ That only works when the #øÏs of the vectors are in the
same placel It has no meaning if we move the vector somewhere else and then
--- Trang 226 ---
= ~Í\
V2
Fig. 11-8. Diagram for calculating the acceleration.
draw a line across, so watch outl We have to draw a new diagram to subtract
the vectors. In Fig. 11-8, 0 and 0a are both drawn parallel and equal to their
counterparts in Fig. 11-7, and now we can discuss the acceleration. Of course the
acceleration is simply Aø/Af. Tt is interesting to nobe that we can compose the
velocity diference out of two parts; we can think of acceleration as having #uo
componenis, A0||, in the direction tangent to the path and Aø_ at right angles
to the path, as indicated in Eig. 11-8. 'Phe acceleration tangent to the path is, of
course, just the change in the lengfh of the vector, i.e., the change in the speed 0:
địị = du/dt. (11.15)
The other component of acceleration, at ripght angles to the curve, is easy %O
calculate, using Eigs. I1-7 and 11-8. In the short time Af let the change in angle
bebween Øø¡ and 0a; be the small angle A0. If the magnitude of the velocity is
called ø, then of course
AUL =uA0
and the acceleration ø will be
ø¡ = 0(A0/At).
NÑow we need to know A6/A¿, which can be found thìs way: TẾ, at the given
mmoment, the curve is approximated as a circle of a certain radius #, then in a
time A£ the distance s is, of course, 0A, where 0 is the speed.
A0 =(uAt)/R, Or A0/At = u/R.
'Therefore, we find
ai =02/R, (11.16)
as we have seen before.
11-7 Scalar product of vectors
Now let us examine a little further the properties of vectors. Ï% is easy to see
that the lengfh of a step In space would be the same in any coordinate system.
--- Trang 227 ---
'That 1s, if a particular step 7 is represented by z#, , z, In one coordinate system,
and by 4,0,2” in another coordinate system, surely the distance z = |r| would
be the same in both. Ñow
r=VW#2+ 2+ z2
and also
+ = \/„2 +2 -+- z2.
So what we wish to verify is that these two quantities are equal. It is mụch more
convenient not to bother to take the square root, so let us talk about the square
of the distance; that ïs, let us fnd out whether
z2? +?2+z?=z^2+^2+ z2. (11.17)
It had better be—and if we substitute Eq. (11.5) we do indeed ñnd that it is.
So we see that there are other kinds of equations which are true for any ÿWO
coordinate systems.
Something new is involved. We can produce a new quantity, a function of
z, , and z, called a scalar ƒunctlion, a quantity which has no direction but which
1s the same in both systems. Out of a vector we can make a scalar. We have to
ñnd a general rule for that. It is clear what the rule is for the case just considered:
add the squares of the components. Let us now define a new thing, which we
call œ- œ. 'This is not a vector, but a scalar; it is a number that is the same in all
coordinate systems, and it is defned to be the sum of the squares of the three
components of the vector:
qŒ-d = d2 + d2 + đệ. (11.18)
Now you say, “But with what axes?” It does not depend on the axes, the answer is
the same in euer set of axes. So we have a new kznởd of quantity, a new ?nuariant
or scalar produced by one vector “squared.” IÝÍ we now defñne the following quantity
for any two vectors œ and b:
œ-b= q„bÙ„ + aub„ + azÐz, (11.19)
we fñnd that this quantity, calculated in the primed and unprimed systems, also
stays the same. To prove it we note that it is true of ø - ø, b- b, and e- c, where
--- Trang 228 ---
c=øœ+b. Therefore the sum of the squares (a„ + b„)” + (œy + b„)Ÿ + (a; + b;)?
will be invarlant:
(a„ + b„)Ÿ + (ay + bụ)Ÿ + (ay + by)” = (a„ + bại)”
+ (dự; + bự)Ÿ + (az + b„.)Š. (11.20)
Tf both sides of this equation are expanded, there will be cross produects of Jjust the
type appearing in Eq. (11.19), as well as the sums of squares oŸ the components
of Ͽ and b. The invariance of terms of the form of Eq. (11.18) then leaves the
cross product terms (11.19) invariant also.
The quantity œ - b is called the scalar product of two vectors, œ and b, and ït
has many interesting and useful properties. For instance, it is easily proved that
œ-(b+c)=a-b+eœ-c. (11.21)
AIlso, there is a simple geometrical way to calculate ø - b, without having to
calculate the components of œ and b: ø- b is the product of the length of œ and
the length of b times the cosine of the angle between them. Why? Suppose
that we choose a special coordinate system in which the z-axis lies along œ; in
those circumstances, the only component of œ that will be there 1s ø„, which is
of course the whole length of œ. Thus Eq. (11.19) reduces to ø- Ð = a„b„ for this
case, and this is the length of œ times the component of b in the direction of œ,
that is, bcos ổ:
œ-b = abcos 0.
Therefore, in that special coordinate system, we have proved that œ - b ¡is the
length of œ times the length of b times cosØ. But ?ƒ ?# ¡s truc ?ím one coordinate
sustem, tt 1s true ím œÏÏ, because œ - b is independent of the coordinate system;