Ch12 four vector (optional)

Chapter 12
4-vectors
We now come to a most useful concept in relativity, namely that of 4-vectors. Al-
though it is possible to derive everything in special relativity without the use of
4-vectors (which is the route, give or take, that we took in the previous two chap-
ters), they are extremely helpful in making calculations and concepts much simpler
and far more transparent.
I have chosen to postpone their introduction in order to make it clear that
everything in relativity can be derived without them. In encountering relativity for
the first time, it’s nice to know that no “advanced” techniques are required. But
now that you’ve seen everything once, let’s go back and derive some things in an
easier way.
This situation, where 4-vectors are helpful but not necessary, is more pronounced
in general relativity, where the concept of tensors (the generalization of 4-vectors) is,
for all practical purposes, completely necessary for an understanding of the subject.
We won’t have time to go very deeply into GR in Chapter 13, so you’ll have to just
accept this fact. But suffice it to say that an eventual understanding of GR requires
a firm understanding of special-relativity 4-vectors.
12.1 Definition of 4-vectors
Definition 12.1 The 4-tuplet A = (A0, A1, A2, A3) is a “4-vector” if the Ai trans-
form under a Lorentz transformation in the same way that (cdt, dx, dy, dz) do. In
other words, they must transform like (assuming the LT is along the x-direction; see
Fig. 12.1):
x x'
S S'
v
Figure 12.1
A0 = γ(A0 + (v/c)A1),
A1 = γ(A1 + (v/c)A0),
A2 = A2,
A3 = A3. (12.1)
Remarks:
1. Similar equations must hold, of course, for Lorentz transformations in the y- and
z-directions.
XII-1

XII-2 CHAPTER 12. 4-VECTORS
2. Additionally, the last three components have to be a vector in 3-space (that is, they
have to transform like a usual vector under rotations in 3-space).
3. We’ll use a capital roman letter to denote a 4-vector. A bold-face letter will denote,
as usual, a vector in 3-space.
4. The Ai may be functions of the dxi, the xi, their derivatives, and also v and any
invariants (that is, frame-independent quantities) like the mass m.
5. Lest we get tired of writing the c’s over and over, we will henceforth work in units
where c = 1.
6. The first component of a 4-vector is sometimes referred to as the “time component”.
The other three are the “space components”.
7. 4-vectors are the obvious generalization of vectors in regular space. A vector in 3-
dimensions, after all, is something that transforms under a rotation just like (dx, dy, dz)
does. We have simply generalized a 3-D rotation to a 4-D Lorentz transformation. ♣
12.2 Examples
So far, we have only one 4-vector at our disposal, namely (dt, dx, dy, dz). What are
some others? Well, (7dt, 7dx, 7dy, 7dz) certainly works, as does any other constant
multiple of (dt, dx, dy, dz). Indeed, m(dt, dx, dy, dz) is also a 4-vector, since m is an
invariant (that is, independent of frame).
How about A = (dt, 2dx, dy, dz)? No, this isn’t a 4-vector, because on one hand
it has to transform like
dt ≡ A0 = γ(A0 + vA1) ≡ γ(dt + v(2dx )),
2dx ≡ A1 = γ(A1 + vA0) ≡ γ((2dx ) + vdt ),
dy ≡ A2 = A2 ≡ dy ,
dz ≡ A3 = A3 ≡ dz , (12.2)
from the definition of a 4-vector. But on the other hand, it has to transform like
dt = γ(dt + vdx ),
2dx = 2γ(dx + vdt ),
dy = dy ,
dz = dz , (12.3)
because this is how the dxi transform. The two preceding sets of equations are
inconsistent, so A = (dt, 2dx, dy, dz) is not a 4-vector. Note that if we had instead
considered the 4-tuplet A = (dt, dx, 2dy, dz), then the two preceding equations would
have been consistent. But if we had then looked at how A transforms under an LT
in the y-direction, we would have found that it is not a 4-vector.
The moral of this story is that the above definition of a 4-vector is nontrivial
because there are two possible ways a 4-tuplet can transform. We can transform it
according to the 4-vector definition, as in eq. (12.2). Or, we can simply transform
each Ai separately (knowing how the dxi transform), as in eq. (12.3). Only for

12.2. EXAMPLES XII-3
certain special 4-tuplets do these two methods give the same result. By deﬁnition,
we call these 4-vectors.
Let us now construct some less trivial examples of 4-vectors. In constructing
these, we will make abundant use of the fact that the proper-time interval, dτ ≡√
dt2 − dr2, is an invariant.
• Velocity 4-vector: We can divide (dt, dx, dy, dz) by dτ, where dτ is the
proper time between two events (the same two events that yielded the dt,
etc.). The result is indeed a 4-vector, because dτ is independent of the frame
in which it is measured. Using dτ = dt/γ, we see that
V ≡
1
dτ
(dt, dx, dy, dz) = γ 1,
dx
dt
,
dy
dt
,
dz
dt
= (γ, γv) (12.4)
is a 4-vector. This is known as the velocity 4-vector. (With the c’s, we have
V = (γc, γv).) In the rest frame of the object, V reduces to V = (1, 0, 0, 0).
• Energy-momentum 4-vector: If we multiply the velocity 4-vector by the
invariant m, we obtain another 4-vector,
P ≡ mV = (γm, γmv) = (E, p), (12.5)
which is known as the energy-momentum 4-vector (or the 4-momentum for
short), for obvious reasons. (With the c’s, we have P = (γmc, γmv) =
(E/c, p).) In the rest frame of the object, P reduces to P = (m, 0, 0, 0).
• Acceleration 4-vector: We can also take the derivative of the velocity
4-vector with respect to τ. The result is indeed a 4-vector, because taking
the derivative is essentially taking the diﬀerence between two 4-vectors (which
results in a 4-vector because eq. (12.1) is linear), and then dividing by the
invariant dτ (which again results in a 4-vector). We obtain
A ≡
dV
dτ
=
d
dτ
(γ, γv) = γ
dγ
dt
,
d(γv)
dt
. (12.6)
Using dγ/dt = v ˙v/(1 − v2)3/2 = γ3v ˙v, we have
A = (γ4
v ˙v , γ4
v ˙vv + γ2
a), (12.7)
where a ≡ dv/dt. A is known as the acceleration 4-vector. In the rest frame
of the object (or, rather, the instantaneous inertial frame), A reduces to A =
(0, a).
As we always do, we will pick the relative velocity, v, to point in the x-
direction. Hence, v = (vx, 0, 0), v = vx, and ˙v = ˙vx ≡ ax. We then have
A = (γ4
vxax , γ4
v2
xax + γ2
ax , γ2
ay , γ2
az)
= (γ4
vxax , γ4
ax , γ2
ay , γ2
az). (12.8)
We can keep taking derivatives with respect to τ to create other 4-vectors, but
they aren’t very relevant to the real world.

• Force 4-vector: We deﬁne the force 4-vector as
F ≡
dP
dτ
= γ
dE
dt
,
dp
dt
= γ
dE
dt
, f , (12.9)
where f is the usual 3-force. (We’ll use f instead of F in this chapter, to avoid
confusion with the 4-force, F.)
In the case where m is constant,1 F can be written as F = d(mV )/dτ =
m dV/dτ = mA. We therefore still have a nice “F equals mA” law of physics,
but it’s now a 4-vector equation instead of the old 3-vector one. In terms of
the acceleration 4-vector, we may write (if m is constant)
F = mA = (γ4
mv ˙v , γ4
mv ˙vv + γ2
ma). (12.10)
In the rest frame of the object (or, rather, the instantaneous inertial frame), F
reduces to F = (0, f), and mA reduces to mA = (0, ma). So F = mA becomes
f = ma.
12.3 Properties of 4-vectors
The appealing thing about 4-vectors is that they have many useful properties. Let’s
look at these.
• Linear combinations: If A and B are 4-vectors, then C ≡ aA + bB is also a
4-vector (as we noted above when deriving the acceleration 4-vector). This is
true because the transformations in eq. (12.1) are linear; so the transformation
of, say, the time component is
C0 ≡ (aA + bB)0 = aA0 + bB0 = a(A0 + vA1) + b(B0 + vB1)
= (aA0 + bB0) + v(aA1 + bB1)
≡ C0 + vC1, (12.11)
which is the proper transformation for the time component of a 4-vector.
Likewise for the other components. This property holds, of course, just as it
does for linear combinations of vectors in 3-space.
• Inner product invariance: Consider two arbitrary 4-vectors, A and B.
Deﬁne their inner product to be
A · B ≡ A0B0 − A1B1 − A2B2 − A3B3 ≡ A0B0 − A · B. (12.12)
Then A · B is invariant (that is, independent of the frame in which it is calcu-
lated). This is easily shown by direct calculation, using the transformations
1
m would not be constant if the object were being heated, or if extra mass were being added to
it. We won’t concern ourselves with such cases here.

12.3. PROPERTIES OF 4-VECTORS XII-5
in eq. (12.1).
A · B ≡ A0B0 − A1B1 − A2B2 − A3B3
= γ(A0 + vA1) γ(B0 + vB1) − γ(A1 + vA0) γ(B1 + vB0)
−A2B2 − A3B3
= γ2
A0B0 + v(A0B1 + A1B0) + v2
A1B1
−γ2
A1B1 + v(A1B0 + A0B1) + v2
A0B0
−A2B2 − A3B3
= A0B0(γ2
− γ2
v2
) − A1B1(γ2
− γ2
v2
) − A2B2 − A3B3
= A0B0 − A1B1 − A2B2 − A3B3
≡ A · B . (12.13)
The importance of this result cannot be overstated. This invariance is analo-
gous, of course, to the invariance of the inner product A · B for rotations in
3-space. The above inner product is also clearly invariant under rotations in
3-space, since it involves the combination A · B.
The minus signs in the inner product may seem a little strange. But the goal
was to ﬁnd a combination of two arbitrary vectors that is invariant under a
Lorentz transformation (because such combinations are very useful in seeing
what’s going on in a problem). The nature of the LT’s demands that there be
opposite signs in the inner product, so that’s the way it is.
• Norm: As a corollary to the invariance of the inner product, we can look at
the inner product of a 4-vector with itself (which is by deﬁnition the square
of the norm). We see that
|A|2
≡ A · A ≡ A0A0 − A1A1 − A2A2 − A3A3 = A2
0 − |A|2
(12.14)
is invariant. This is analogous, of course, to the invariance of the norm
√
A · A
for rotations in 3-space.
• A theorem: Here’s a nice little theorem:
If a certain one of the components of a 4-vector is 0 in every frame, then all
four components are 0 in every frame.
Proof: If one of the space components (say, A1) is 0 in every frame, then the
other space components must also be 0 in every frame (otherwise a rotation
would make A1 = 0). Also, the time component A0 must be 0 in every frame
(otherwise a Lorentz transformation in the x-direction would make A1 = 0).
If the time component, A0, is 0 in every frame, then the space components
must also be 0 in every frame (otherwise a Lorentz transformation in the
appropriate direction would make A0 = 0).

12.4 Energy, momentum
12.4.1 Norm
Many useful things arise from the simple fact that the P in eq. (12.5) is a 4-vector.
The invariance of the norm implies that P · P = E2 − |p|2 is invariant. If we are
dealing with only one particle, we may ﬁnd the value of P2 by conveniently picking
the rest-frame of the particle (so that v = 0), to obtain
E2
− p2
= m2
, (12.15)
or E2 −p2c2 = m2c4, with the c’s. We already knew this, of course, from just writing
out E2 − p2 = γ2m2 − γ2m2v2 = m2.
For a collection of particles, knowledge of the norm is very useful. If a process
involves many particles, then we can say that for any subset of the particles,
E
2
− p
2
is invariant, (12.16)
because this is simply the norm of the sum of the energy-momentum 4-vectors of
the chosen particles, and the sum is again a 4-vector, due to the linearity of eqs.
(12.1).
What is the value of this invariant? The most concise description (which is
basically a tautology) is that it is the square of the energy in the CM frame (that
is, in the frame where p = 0). For one particle, this reduces to m2.
Note that the sums are taken before squaring in eq. (12.16). Squaring before
adding would simply give the sum of the squares of the masses.
12.4.2 Transformation of E,p
We already know how the energy and momentum transform (see Section 11.2), but
we’ll derive the transformation again in a very quick and easy manner.
We know that (E, px, py, pz) is a 4-vector. So it must transform according to eq.
(12.1). Therefore (for an LT in the x-direction),
E = γ(E + vpx),
px = γ(px + vE ),
py = py,
pz = pz. (12.17)
That’s all there is to it.
Remark: The fact that E and p are part of the same 4-vector provides an easy way to
see that if one of them is conserved, then the other is also. Consider an interaction among a
set of particles. Look at the 4-vector ∆P ≡ Pafter −Pbefore. If E is conserved in every frame,
then the time component of ∆P is 0 in every frame. But then the theorem in the previous
section says that all four components of ∆P are 0 in every frame. Hence, p is conserved.
Likewise for the case where one of the pi is known to be conserved. ♣

12.5. FORCE AND ACCELERATION XII-7
12.5 Force and acceleration
Throughout this section, we will deal with objects with constant mass (which we
will call “particles”). The treatment can be generalized to cases where the mass
changes (for example, the object is being heated, or extra mass is being dumped on
it), but we won’t concern ourselves with these.
12.5.1 Transformation of forces
Let us first look at the force 4-vector in the instantaneous inertial frame of a given
particle (frame S ). Eq. (12.9) gives
F = γ
dE
dt
, f = (0, f ). (12.18)
The first component is zero because dE /dt = d(m/
√
1 − v 2 )/dt carries a factor of
v , which is zero in this frame. (Equivalently, you can just use eq.(12.10), with a
speed of zero.)
We may now write down two expressions for the 4-force, F, in another frame,
S. First, since F is a 4-vector, it transforms according to eq. (12.1). So we have
(using eq. (12.18))
F0 = γ(F0 + vF1) = γvfx,
F1 = γ(F1 + vF0) = γfx,
F2 = F2 = fy,
F3 = F3 = fz. (12.19)
And second, from the definition in eq. (12.9), we have
F0 = γdE/dt,
F1 = γfx,
F2 = γfy,
F3 = γfz. (12.20)
Eqs. (12.19) and (12.20) give
dE/dt = vfx,
fx = fx,
fy = fy/γ,
fz = fz/γ. (12.21)
We therefore recover the results of Section 11.5.3. The longitudinal force is the same
in both frames, but the transverse forces are larger by a factor of γ in the particle’s
frame. Hence, fy/fx decreases by a factor of γ when going from the particle’s frame
to the lab frame (see Fig. 12.2 and Fig. 12.3).
f'
f'
f'
x
y
S
frame S'
Figure 12.2
fx
fy
f
f'
f'
x
y
S
frame S
γ
=
=
__
Figure 12.3

As a bonus, the F0 component tells us (after multiplying through by dt) that
dE = fxdx, which is the work-energy result. (In other words, we just proved again
the result dE/dx = dp/dt from Section 11.5.1.)
As noted in Section 11.5.3, we can’t switch the S and S frames and write
fy = fy/γ. When talking about the forces on a particle, there is indeed one preferred
frame of reference, namely that of the particle. All frames are not equivalent here.
When forming all of our 4-vectors in Section 12.2, we explicitly used the dτ, dt, dx,
etc., from two events, and it was understood that these two events were located at
the particle.
12.5.2 Transformation of accelerations
The procedure here is similar to the above treatment of the force.
Let us first look at the acceleration 4-vector in the instantaneous inertial frame
of a given particle (frame S ). Eq. (12.7) or eq. (12.8) gives
A = (0, a ), (12.22)
since v = 0 in S .
We may now write down two expressions for the 4-acceleration, A, in another
frame, S. First, since A is a 4-vector, it transforms according to eq. (12.1). So we
have (using eq. (12.22))
A0 = γ(A0 + vA1) = γvax,
A1 = γ(A1 + vA0) = γax,
A2 = A2 = ay,
A3 = A3 = az. (12.23)
And second, from the definition in eq. (12.8), we have
A0 = γ4
vax,
A1 = γ4
ax,
A2 = γ2
ay,
A3 = γ2
az. (12.24)
Eqs. (12.23) and (12.24) give
ax = ax/γ3
,
ax = ax/γ3
,
ay = ay/γ2
,
az = az/γ2
. (12.25)
We see that ay/ax increases by a factor of γ when going from the particle’s frame
to the lab frame (see Fig. 12.4 and Fig. 12.5). This is the opposite of the effect
a'
a'
a'
x
y
S
frame S'
Figure 12.4
ax
ay
a
a'
a'
x
y
S
frame S
γ
=
=
__
γ
__
2
3
Figure 12.5

12.6. THE FORM OF PHYSICAL LAWS XII-9
on fy/fx.2 This makes it clear that an f = ma law wouldn’t make any sense. If it’s
true in one frame, it might not be true in another.
Note also that the increase in ay/ax in going to the lab frame is consistent with
length contraction, as the Bead-on-a-rod example in Section 11.5.3 showed.
Example (Acceleration for circular motion): A particle moves with constant
speed v along the circle x2
+ y2
= r2
, z = 0, in the lab frame. At the instant the
particle crosses the negative y-axis (see Fig. 12.6), find the three-acceleration and x
y
v
Figure 12.6
four-acceleration in both the lab frame and the instantaneous rest frame of the particle
(with axes chosen parallel to the lab’s axes).
Solution: Let the lab frame be S, and let the particle’s instantaneous inertial frame
be S when it crosses the negative y-axis. Then S and S are related by a boost in
the x-direction.
The 3-acceleration in S is simply
a = (0, v2
/r, 0). (12.26)
Eq. (12.7) or (12.8) then gives the 4-acceleration in S as
A = (0, 0, γ2
v2
/r, 0). (12.27)
To find the vectors in S , we use the fact that the transformation between S and S
involves a boost in the x-direction. Therefore, the A2 component is unchanged. So
the 4-acceleration in S is the same,
A = A = (0, 0, γ2
v2
/r, 0). (12.28)
In the particle’s frame, a is simply the space part of A (using eq. (12.7) or (12.8),
with v = 0 and γ = 1), so the 3-acceleration in S is
a = (0, γ2
v2
/r, 0). (12.29)
Remark: We can also arrive at the two factors of γ in a by using a simple time-dilation
argument. We have
ay =
d2
y
dτ2
=
d2
y
d(t/γ)2
= γ2 d2
y
dt2
= γ2 v2
r
, (12.30)
where we have used the fact that transverse lengths are the same in the two frames. ♣
12.6 The form of physical laws
One of the postulates of special relativity is that all inertial frames are equivalent.
Therefore, if a physical law holds in one frame, then it must hold in all frames
(otherwise it would be possible to differentiate between frames).
2
In a nutshell, the difference is due to the fact that γ changes with time. When talking about
acceleration, there are γ’s that we have to differentiate. This isn’t the case with forces, because the
γ is absorbed into the definition of p = γmv.

As noted in the previous section, the statement “f = ma” cannot be a physical
law. The two sides transform differently when going from one frame to another, so
the statement cannot be true in all frames.
If a statement has any chance of being true in all frames, it must involve only
4-vectors. Consider a 4-vector equation (say, “A = B”) which is true in frame S.
Then if we apply to this equation a Lorentz transformation from S to another frame
S (call it M), we have
A = B,
=⇒ MA = MB,
=⇒ A = B .
(12.31)
The law is therefore true in frame S , also.
Of course, there are many 4-vector equations that are simply not true (for ex-
ample, F = P). Only a small set of such equations (for example, F = dP/dτ)
correspond to the real world.
Physical laws may also take the form of scalar equations, such as P · P = m2. A
scalar is by definition a quantity that is frame-independent (as we have shown the
inner product to be). So if this statement is true in one frame, then it is true in
any other. (Physical laws may also be higher-rank “tensor” equations, such as arise
in electromagnetism and general relativity. We won’t discuss such things here, but
suffice it to say that tensors may be thought of as things built up from 4-vectors.
Scalars and 4-vectors are special cases of tensors.)
This is exactly analogous, of course, to the situation in 3-D space. In Newtonian
mechanics, f = ma is a possible law, because both sides are 3-vectors. But f =
m(2ax, ay, az) is not a possible law, because the right-hand side is not a 3-vector; it
depends on which axis you label as the x-axis. Another example is the statement
that a given stick has a length of 2 meters. This is fine, but if you say that the stick
has an x-component of 1.7 meters, then this cannot be true in all frames.
God said to his cosmos directors,
“I’ve added some stringent selectors.
One is the clause
That your physical laws
Shall be written in terms of 4-vectors.”

12.7. PROBLEMS XII-11
12.7 Problems
1. Velocity addition
In A’s frame, B moves to the right with speed v, and C moves to the left with
speed u. What is the speed of B with respect to C? (In other words, use
4-vectors to derive the velocity addition formula.)
2. Relative speed *
In the lab frame, two particles move with speed v along the paths shown in
Fig. 12.7. The angle between the trajectories is 2θ. What is the speed of one
θ
θ
v
v
Figure 12.7
particle, as viewed by the other?
3. Another relative speed *
In the lab frame, two particles, A and B, move with speeds u and v along the
paths shown in Fig. 12.8. The angle between the trajectories is θ. What is
θ
u
v
A
B
Figure 12.8
the speed of one particle, as viewed by the other?
4. Acceleration for linear motion *
A spaceship starts at rest with respect to S and accelerates with constant
proper acceleration a. In section 10.8, we showed that the speed of the space-
ship w.r.t. S is given by v(τ) = tanh(aτ), where τ is the spaceship’s proper
time (and c = 1).
Let V be the spaceship’s 4-velocity, and let A be the spaceship’s 4-acceleration.
In terms of the proper time τ (it’s easier to do the problem in terms of τ than
in terms of the t of frame S),
(a) Find V and A in frame S (by explicitly using v(τ) = tanh(aτ)).
(b) Write down V and A in the spaceship’s frame, S .
(c) Verify that V and A transform like 4-vectors between the two frames.

12.8 Solutions
1. Velocity addition
Let the desired speed of B with respect to C be w (see Fig. 12.9).
v uB C
wB C
A's frame
C's frame
Figure 12.9
In A’s frame, the 4-velocity of B is (γv, γvv), and the 4-velocity of C is (γu, −γuu)
(suppressing the y and z components).
In C’s frame the 4-velocity of B is (γw, γww), and the 4-velocity of C is (1, 0).
The invariance of the inner product implies
(γv, γvv) · (γu, −γuu) = (γw, γww) · (1, 0)
=⇒ γuγv(1 + uv) = γw
=⇒
1 + uv
√
1 − u2
√
1 − v2
=
1
√
1 − w2
. (12.32)
Squaring, and solving for w gives
w =
u + v
1 + uv
. (12.33)
2. Relative speed
In the lab frame, the 4-velocities of the particles are (suppressing the z component)
(γv, γvv cos θ, −γvv sin θ) and (γv, γvv cos θ, γvv sin θ). (12.34)
Let w be the desired speed of one particle as viewed by the other. Then in the frame
of one particle, the 4-velocities are (suppressing two spatial components)
(γw, γww) and (1, 0). (12.35)
(We have rotated the axes so that the relative motion is along the x-axis in this
frame.) Since the 4-vector inner product is invariant under Lorentz transformations
and rotations, we have (using cos 2θ = cos2
θ − sin2
θ)
(γv, γvv cos θ, −γvv sin θ) · (γv, γvv cos θ, γvv sin θ) = (γw, γww) · (1, 0)
=⇒ γ2
v(1 − v2
cos 2θ) = γw. (12.36)
Using the deﬁnitions of the γ’s, squaring, and solving for w gives
w =
2v2(1 − cos 2θ) − v4 sin2
2θ
1 − v2 cos 2θ
. (12.37)
If desired, this can be rewritten (using some double-angle formulas) in the form
w =
2v sin θ
√
1 − v2 cos2 θ
1 − v2 cos 2θ
. (12.38)
Remark: If 2θ = 180◦
, then w = 2v/(1 + v2
), as it should. And if θ = 0◦
, then w = 0,
as it should. If θ is very small, then the result reduces to w ≈ 2v sin θ/
√
1 − v2, which is
simply the relative speed in the lab frame, multiplied by the time dilation factor between
the frames. (The particles’ clocks run slow, and transverse distances don’t change, so the
motion is faster in a particle’s frame.) ♣

12.8. SOLUTIONS XII-13
3. Another relative speed
For you to do.
4. Acceleration for linear motion
(a) Using v(τ) = tanh(aτ), we have γ = 1/
√
1 − v2 = cosh(aτ). Therefore (sup-
pressing the two transverse components of V , which are 0),
V = (γ, γv) = (cosh(aτ), sinh(aτ)), (12.39)
and so
A =
dV
dτ
= a(sinh(aτ), cosh(aτ)). (12.40)
(b) The spaceship is at rest in its instantaneous inertial frame, so
V = (1, 0). (12.41)
In the rest frame, we also have
A = (0, a). (12.42)
Equivalently, these are obtained by setting τ = 0 in the results above (because
the spaceship hasn’t started moving at τ = 0, as is always the case in the
instantaneous rest frame).
(c) The Lorentz transformation matrix from S to S is
M =
γ γv
γv γ
=
cosh(aτ) sinh(aτ)
sinh(aτ) cosh(aτ)
. (12.43)
We must check that
V0
V1
= M
V0
V1
and
A0
A1
= M
A0
A1
. (12.44)
These are easily seen to be true.

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