TEST BANK FOR A Unified Grand Tour of Theoretical Physics 3RD Edition By Lawrie I.

A Unied Grand Tour of
Theoretical Physics
Solution Manual
Ian D Lawrie
September 2012
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Ian D Lawrie 2012
It may be freely shared, but may not be altered or sold.
Chapter 2
Chapter 2
Geometry
Exercise 2.1
Consider a Cartesian coordinate system S and and a second one, S′, which
is obtained by giving S a velocity v, without rotating its axes. Then the
origin of S′ moves with constant velocity v relative to S, and we take the
two origins to coincide at t = t′ = 0. Assume that the relation between the
two sets of coordinates is linear and that space is isotropic. The most general
form of the transformation law can then be written as
x′ = α
[
(1 − λv2)x + (λv · x − βt)v
]
t′ = γ
[
t − (δ/c2)v · x
]
where α, β, γ, δ and λ are functions of v2. For the case that v is in the
positive x direction, write out the transformations for the four coordinates.
Write down the trajectory of the S′ origin as seen in S and that of the S origin
as seen in S′ and show that β = 1 and α = γ.Write down the trajectories seen
in S and S′ of a light ray emitted from the origin at t = t′ = 0 that travels
in the positive x direction, assuming that it is observed to travel with speed
c in each case. Show that δ = 1. The transformation from S′ to S should be
the same as the transformation from S to S′, except for the replacement of
v by −v. Use this to complete the derivation of the Lorentz transformation
[2.2] by finding γ and λ.
Solution
First, a few words about the way this question is set up. Once we have learned
enough about the geometry of Minkowski spacetime, the best way of arriving
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Chapter 2
at the Lorentz transformation is to ask about coordinate transformations that
preserve the metric or, equivalently, the form of the proper time interval [2.6].
Thus, an inertial Cartesian frame of reference is a set of coordinates such that
c2(dτ )2 = c2(dt)2−(dx)2−(dy)2−(dz)2 and, using xμ for (ct, x, y, z) and xμ′
for (ct′, x′, y′, z′), we look for a constant matrix Λμ′
μ such that, if xμ′
= Λμ′
μxμ,
then
c2(dt′)2 − (dx′)2 − (dy′)2 − (dz′)2 = c2(dt)2 − (dx)2 − (dy)2 − (dz)2.
By studying the most general matrix that satisfies this requirement, we find
that the change of coordinates can be interpreted as a combination of a relative
velocity of the origins and a rotation of the spatial axes. (There is some
further discussion in §3.5.) For the purposes of this question, we are taking a
more primitive point of view, by simply trying to find a transformation rule
that works, without any insight into its geometrical meaning. For that reason,
the terminology needs to be considered carefully in the light of the later
theory. In particular, a ‘rotation of spatial axes’ turns out to mean different
things in two frames of reference that are in relative motion.
In setting up the problem, I assumed that space (more accurately, spacetime)
is isotropic. That means that there is no naturally-occurring vector
that distinguishes one direction from any other direction. Consequently, the
new 3-dimensional vector x′ must be constructed from the only vectors we
have to hand, namely x and v. That is, x′ = Ax + Bv. The coefficients A
and B can depend only on scalar quantities that are unchanged by spatial
rotations, namely t and the dot products of vectors, x · x = |x|2, v · v = v2
and v · x. Since we also assume that the transformation is linear in x and
t, we find that A can be a function only of v2, while B can only have the
form B = B1(v2)v · x + B2(v2)t. Similarly, t′, which is a scalar from the 3-
dimensional point of view, can only have the form t′ = C1(v2)t+C2(v2)v ·x.
I traded in the five functions A, B1, B2, C1 and C2 for five other functions α,
β, γ, δ and λ because I happen to know that this will simplify the algebra.
Now for the problem itself. Say that v = (v, 0, 0). Then the Lorentz
transformation given above reads explicitly
x′ = α[(1 − λv2)x + (λvx − βt)v] = α(x − βvt) (2.1)
y′ = α(1 − λv2)y (2.2)
z′ = α(1 − λv2)z (2.3)
t′ = γ(t − δvx/c2). (2.4)
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Chapter 2
The origin of the S′ system is at x′ = y′ = z′ = 0, so its coordinates in S are
at x = βvt and y = z = 0. It is supposed to be moving along the x axis with
speed v, so we must have β = 1 . The origin of S is at x = y = z = 0, so with
β = 1, its coordinates in S′ are y′ = z′ = 0 and x′ = −αvt = −(αv/γ)t′. It
must be moving in the negative x′ direction with speed v, so we find α = γ .
A light ray that sets out from x = 0 at t = 0 finds itself at x = ct at
time t relative to S. Substituting x = ct in (2.1) and (2.4), we find that its
position in S′ when t′ = γ(1−δv/c) is x′ = γ(c−v)t = [(c−v)/(1−δv/c)]t′.
If the light ray also travels with speed c relative to S′, this position must be
x′ = ct′, so we conclude that δ = 1 .
At this point, two of the transformation equations read x′ = γ(x − vt)
and t′ = γ(t − vx/c2), and these equations can be solved to give
x =
x′ + vt′
γ(1 − v2/c2)
and t =
t′ + vx′/c2
γ(1 − v2/c2)
. (2.5)
This transformation from S′ to S should have the same form as the original
transformation from S to S′, if we replace v with −v, so we conclude that
γ = (1 − v2/c2)−1/2 . For the same reason, we conclude from (2.2) and (2.3),
with α = γ, that γ(1 − λv2) = 1, or λ = (γ − 1)/γv2 .
Clearly, these results give the special form of the Lorentz transformation
[2.2], which applies when v is in the x direction. But since the functions
α, β, . . . depend only on the magnitude of v, they remain valid when v is in
any direction, and we get the more general result
x′ = x +
(γ − 1)(v · x)
v2 v − γvt (2.6)
t′ = γ(t − v · x/c2), (2.7)
with γ = (1 − v2/c2)−1/2.
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Chapter 2
Exercise 2.2
Two coordinate frames are related by the Lorentz transformation (2.2). A
particle moving in the x direction passes their common origin at t = t′ = 0
with velocity u and acceleration a as measured in S. Show that its velocity
and acceleration as measured in S′ are
u′ =
u − v
1 − uv/c2 , a′ =
(1 − v2/c2)3/2
(1 − uv/c2)3 a .
Solution
The particle’s trajectory as seen in S is x = ut+ 1
2at2. Substituting this into
the Lorentz transformation equations gives
x′ = γ[(u − v)t + 1
2at2] (2.8)
t′ = γ[(1 − uv/c2)t − (av/2c2)t2]. (2.9)
One way of proceeding would be to solve (2.9) for t as a function of t′ and
substitute this into (2.8) to get the trajectory x′(t′) as seen in S′. We would
then differentiate to find u′ = dx′/dt′ and a′ = d2x′/dt′2. A neater way is to
treat these two equations as a parametric form of the trajectory. Then we
can calculate the velocity and acceleration as
u′ =
dx′
dt′ =
dx′/dt
dt′/dt
(2.10)
a′ =
du′
dt′ =
du′/dt
dt′/dt
=
d2x′/dt2
(dt′/dt)2
− (dx′/dt)(d2t′/dt2)
(dt′/dt)3 . (2.11)
Evaluating all the derivatives at t = 0, we get
dx′
dt
= γ(u − v),
d2x′
dt2 = γa,
dt′
dt
= γ(1 − uv/c2),
d2t′
dt2 = −γav/c2,
(2.12)
and substituting these results into the two previous equations gives the advertised
answers.
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Chapter 2
Exercise 2.3
A rigid rod of length L is at rest in S′, with one end at x′ = 0 and the
other at x′ = L. Find the trajectories of the two ends of the rod as seen
in S and show that the length of the rod as measured in S is L/γ, where
γ = (1 − v2/c2)−1/2. This is the Fitzgerald contraction. If the rod lies along
the y′ axis of S′, what is its apparent length in S? A clock is at rest at the
origin of S′. It ticks at t′ = 0 and again at t′ = τ . Show that the interval
between these ticks as measured in S is γτ . This is time dilation.
Solution
Clearly, the rod points in the x′ direction, and we might as well take it to be
on the x′ axis. Then its two ends are at (x′, y′, z′) = (0, 0, 0) and (x′, y′, z′) =
(L, 0, 0). Substituting these values into the Lorentz transformation equations,
we find that at time t as seen in S, the two ends are at (x, y, z) = (vt, 0, 0)
and (x, y, z) = (L/γ + vt, 0, 0). Thus, as seen in S at time t, the differences
in coordinates of the two ends of the rod are (Δx,Δy,Δz) = (L/γ, 0, 0) and
the length of the rod is L/γ. This is always ≤ L because γ is always ≥ 1.
The transformation equation t′ = γ(t − vx/c2) is irrelevant to the above
calculation, but it tells us something that is worth noting. For example, the
right-hand end of the rod (the one at x′ = L) is seen by an observer in S
to pass the point x = L/γ at t = 0. This is an event that occurs at one
definite point in space and time. For an observer in S′, this event happens
at t′ = −vL/c2.
Now suppose that the two ends of the rod are at (x′, y′, z′) = (0, 0, 0) and
(x′, y′, z′) = (0, L, 0). As seen in S, the corresponding positions are (x, y, z) =
(vt, 0, 0) and (vt, L, 0). We get (Δx,Δy,Δz) = (0, L, 0), so the observer in S
sees a rod of length L.
The first tick of the clock occurs at (x′, y′, z′, t′) = (0, 0, 0, 0) as seen in
S′. Substituting these values into the transformation equations gives four
simultaneous equations to solve for the corresponding coordinates of this
event as seen in S, and the solution is (x, y, z, t) = (0, 0, 0, 0). The second
tick occurs at (x′, y′, z′, t′) = (0, 0, 0, τ ), and this again gives four simultaneous
equations. Two of them are 0 = y and 0 = z, which are quite easy to solve.
The other two are 0 = γ(x−vt) and τ = γ(t−vx/c2). Using the first one to
eliminate x, we get τ = γ(t − v2t/c2) = γ(1 − v2/c2)t = t/γ, so the solution
is t = γτ . Then we also have x = vt = γvτ . Thus, for the observer in S,
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Chapter 2
the time interval between the two ticks is γτ . This is a longer time than τ ,
so it is often said that a moving clock appears to run slow. Obviously, the
observer in S sees the clock tick for the first time when it is at x = 0 and for
the second time when it is at x = γvτ .
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Chapter 2
Exercise 2.4
As seen in S, a signal is emitted from the origin at t = 0, travels along the x
axis with speed u, and is received at time τ at x = uτ. Show that, if u > c2/v
then, as seen in S′, the signal is received before being sent. Show that if such
paradoxes are to be avoided, no signal can travel faster than light.
Solution
The y and z coordinates are irrelevant to this problem, so I will ignore them.
As seen in S, the signal is emitted at (x, t) = (0, 0) and received at (x, t) =
(uτ, τ ). As seen in S′, the Lorentz transformation tells us that it is emitted
at (x′, t′) = (0, 0) and received at x′ = γ(u − v)τ and t′ = γ(1 − uv/c2)τ . So
if uv > c2, an observer in S′ sees the signal being received before it is sent.
Since this seems to defy our usual expectation that a cause should precede
its effect, we suspect that there must be some maximum speed, say umax with
which any signal can travel. Now, the relative speed v of S and S′ cannot
be greater than c, because this would lead to an imaginary value of γ, and
hence imaginary values of x′ and t′. So the maximum value of uv is umaxc,
and since this maximum value is supposed to be c2, we find that umax = c.
It is important to check that this maximum speed applies equally to any
frame of reference, and we can do this by using the result of exercise 2.2.
If the signal has velocity u relative to S, then its velocity relative to S′ is
u′ = (u − v)/(1 − uv/c2). With a short calculation, we can work out that
u′2 − c2 =
(u2 − c2)(1 − v2/c2)
(1 − uv/c2)2 . (2.13)
Clearly, if the signal travels with velocity u = ±c relative to some frame S,
it also travels with velocity u′ = ±c relative to any other frame. This is just
as well, since the constancy of the speed of light was a basic assumption of
the theory. We also see, though, that if |u| < c, then the right-hand side of
(2.13) is negative, and thus |u′| < c. Therefore, if a signal travels with speed
less than c relative to any one inertial frame, it also travels with speed less
than c in any other frame; the statement that a signal cannot travel faster
than c is independent of which frame of reference we use.
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Chapter 2
Exercise 2.5
A wheel has a perfectly rigid circular rim connected by unbreakable joints
to perfectly rigid spokes. When measured at rest, its radius is r and its
circumference is 2πr. When the wheel is set spinning with angular speed
ω, what, according to exercise 2.3, is the apparent circumference of its rim
and the apparent length of its spokes? What is the speed of sound in a solid
material of density ρ whose Young’s modulus is Y ? Is the notion of a perfectly
rigid material consistent with the conclusion of exercise 2.4?
Solution
Let S be an inertial frame relative to which an observer sees the wheel spinning
with angular velocity ω. A small element of the rim travels with speed
v = ωr along its own length, and we can treat it like the rod in exercise 2.3
if, at some instant of time we choose the x axis in S to lie in the direction
in which the element moves, and we take S′ to be the frame in which this
element is instantaneously at rest. The apparent length of this tiny rod as
seen in S is smaller than its rest length by a factor 1/γ = 1/
√
1 − v2/c2.
Consequently, the circumference is seen in S to be 2πr/
√
1 − v2/c2. An element
of a spoke of the wheel travels in a direction perpendicular to its length,
so, according to exercise 2.3, its apparent length in S is the same as its rest
length, namely r. Thus the circumference of the spinning wheel is smaller
than 2π times the length of the spokes, which means that the spokes must
be compressed or bent. They can’t be compressed or bent, however, because
they are perfectly rigid.
We thus find a contradiction, which means that the notion of a perfectly
rigid material is inconsistent with special relativity. A general way of seeing
this is to note that, as shown in any first-year physics text, the speed of
longitudinal sound waves in a solid rod is cs =
√
Y/ρ, where Y is Young’s
modulus and ρ is the density. For a perfectly rigid material, Y , and therefore
cs, is infinite. This is inconsistent with the conclusion of exercise 2.4, that
no signal can travel faster than c. Happily, perfectly rigid materials do not
occur in nature. The highest speed of sound I know of (in beryllium) is about
12, 500ms−1, which is much smaller than c.
9
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Ian D Lawrie 2012
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