8Special relativity
IA Dynamics and Relativity
8.1 The Lorentz transformation
Consider again inertial frames
S
and
S
0
whose origins coincide at
t
=
t
0
= 0. For
now, neglect the
y
and
z
directions, and consider the relationship between (
x, t
)
and (x
0
, t
0
). The general form is
x
0
= f(x, t), t
0
= g(x, t),
for some functions f and g. This is not very helpful.
In any inertial frame, a free particle moves with constant velocity. So straight
lines in (
x, t
) must map into straight lines in (
x
0
, t
0
). Therefore the relationship
must be linear.
Given that the origins of
S
and
S
0
coincide at
t
=
t
0
= 0, and
S
0
moves with
velocity v relative to S, we know that the line x = vt must map into x
0
= 0.
Combining these two information, the transformation must be of the form
x
0
= γ(x − vt), (1)
for some factor
γ
that may depend on
|v|
(not
v
itself. We can use symmetry
arguments to show that γ should take the same value for velocities v and −v).
Note that Galilean transformation is compatible with this – just take
γ
to
be always 1.
Now reverse the roles of the frames. From the perspective
S
0
,
S
moves with
velocity −v. A similar argument leads to
x = γ(x
0
+ vt
0
), (2)
with the same factor
γ
, since
γ
only depends on
|v|
. Now consider a light ray
(or photon) passing through the origin
x
=
x
0
= 0 at
t
=
t
0
= 0. Its trajectory in
S is
x = ct.
We want a γ such that the trajectory in S
0
is
x
0
= ct
0
as well, so that the speed of light is the same in each frame. Substitute these
into (1) and (2)
ct
0
= γ(c − v)t
ct = γ(c + v)t
0
Multiply the two equations together and divide by tt
0
to obtain
c
2
= γ
2
(c
2
− v
2
).
So
γ =
r
c
2
c
2
− v
2
=
1
p
1 −(v/c)
2
.
Definition (Lorentz factor). The Lorentz factor is
γ =
1
p
1 − (v/c)
2
.
Note that
– γ ≥ 1 and is an increasing function of |v|.
– When v c, then γ ≈ 1, and we recover the Galilean transformation.
– When |v| → c, then γ → ∞.
–
If
|v| ≥ c
, then
γ
is imaginary, which is physically impossible (or at least
weird).
–
If we take
c → ∞
, then
γ
= 1. So Galilean transformation is the transfor-
mation we will have if light is infinitely fast. Alternatively, in the world of
Special Relativity, the speed of light is “infinitely fast”.
v
c
γ
1
For the sense of scale, we have the following values of γ at different speeds:
– γ = 2 when v = 0.866c.
– γ = 10 when v = 0.9949.
– γ = 20 when v = 0.999c.
We still have to solve for the relation between
t
and
t
0
. Eliminate
x
between
(1) and (2) to obtain
x = γ(γ(x − vt) + vt
0
).
So
t
0
= γt − (1 − γ
−2
)
γx
v
= γ
t −
v
c
2
x
.
So we have
Law
(Principle of Special Relativity)
.
Let
S
and
S
0
be inertial frames, moving
at the relative velocity of v. Then
x
0
= γ(x − vt)
t
0
= γ
t −
v
c
2
x
,
where
γ =
1
p
1 − (v/c)
2
.
This is the Lorentz transformations in the standard configuration (in one spatial
dimension).
The above is the form the Lorentz transformation is usually written, and is
convenient for actual calculations. However, this lacks symmetry between space
and time. To display the symmetry, one approach is to use units such that
c
= 1.
Then we have
x
0
= γ(x − vt),
t
0
= γ(t − vx).
Alternatively, if we want to keep our
c
’s, instead of comparing
x
and
t
, which
have different units, we can compare x and ct. Then we have
x
0
= γ
x −
v
c
(ct)
,
ct
0
= γ
ct −
v
c
x
.
Symmetries aside, to express
x, t
in terms of
x
0
, t
0
, we can invert this linear
mapping to find (after some algebra)
x = γ(x
0
+ vt
0
)
t = γ
t
0
+
v
c
2
x
0
Directions perpendicular to the relative motion of the frames are unaffected:
y
0
= y
z
0
= z
Now we check that the speed of light is really invariant:
For a light ray travelling in the x direction in S:
x = ct, y = 0, z = 0.
In S
0
, we have
x
0
t
0
=
γ(x − vt)
γ(t − vx/c
2
)
=
(c − v)t
(1 − v/c)t
= c,
as required.
For a light ray travelling in the Y direction in S,
x = 0, y = ct, z = 0.
In S
0
,
x
0
t
0
=
γ(x − vt)
γ(t − vx/c
2
)
= −v,
and
y
0
t
0
=
y
γ(t − vx/c
2
=
c
γ
,
and
z
0
= 0.
So the speed of light is
p
x
02
+ y
02
t
0
=
p
v
2
+ γ
−2
c
2
= c,
as required.
More generally, the Lorentz transformation implies
c
2
t
02
− r
02
= c
2
t
02
− x
02
− y
02
− z
02
= c
2
γ
2
t −
v
c
2
x
2
− γ
2
(x − vt)
2
− y
2
− z
2
= γ
2
1 −
v
2
c
2
(c
2
t
2
− x
2
) − y
2
− z
2
= c
2
t
2
− x
2
− y
2
− z
2
= c
2
t
2
− r
2
.
We say that the quantity c
2
t
2
− x
2
− y
2
− z
2
is Lorentz-invariant.
So if
r
t
= c, then
r
0
t
0
= c also.