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

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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Ian D Lawrie 2012
It may be freely shared, but may not be altered or sold.
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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This document can be downloaded from http://www.unifiedgrandtours.org. c⃝
Ian D Lawrie 2012
It may be freely shared, but may not be altered or sold.
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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This document can be downloaded from http://www.unifiedgrandtours.org. c⃝
Ian D Lawrie 2012
It may be freely shared, but may not be altered or sold.
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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