8Special relativity
IA Dynamics and Relativity
8.3 Relativistic physics
Now we can look at all sorts of relativistic weirdness!
Simultaneity
The first relativistic weirdness is that different frames disagree on whether two
evens are simultaneous
Definition
(Simultaneous events)
.
We say two events
P
1
and
P
2
are simultane-
ous in the frame S if t
1
= t
2
.
They are represented in the following spacetime diagram by horizontal dashed
lines.
However, events that are simultaneous in
S
0
have equal values of
t
0
, and so
lie on lines
ct −
v
c
x = constant.
x
ct
P
1
P
2
x
0
ct
0
The lines of simultaneity of
S
0
and those of
S
are different, and events simulta-
neous in
S
need not be simultaneous in
S
0
. So simultaneity is relative.
S
thinks
P
1
and P
2
happened at the same time, while S
0
thinks P
2
happens first.
Note that this is genuine disagreement. It is not due to effects like, it takes
time for the light conveying the information to different observers. Our account
above already takes that into account (since the whole discussion does not involve
specific observers).
Causality
Although different people may disagree on the temporal order of events, the
consistent ordering of cause and effect can be ensured.
Since things can only travel at at most the speed of light,
P
cannot affect
R
if
R
happens a millisecond after
P
but is at millions of galaxies away. We can
draw a light cone that denotes the regions in which things can be influenced
by
P
. These are the regions of space-time light (or any other particle) can
possibly travel to.
P
can only influence events within its future light cone, and
be influenced by events within its past light cone.
x
ct
P
Q
R
All observers agree that
Q
occurs after
P
. Different observers may disagree on
the temporal ordering of
P
and
R
. However, since nothing can travel faster
than light,
P
and
R
cannot influence each other. Since everyone agrees on how
fast light travels, they also agree on the light cones, and hence causality. So
philosophers are happy.
Time dilation
Suppose we have a clock that is stationary in
S
0
(which travels at constant
velocity
v
with respect to inertial frame
S
) ticks at constant intervals ∆
t
0
. What
is the interval between ticks in S?
Lorentz transformation gives
t = γ
t
0
+
v
c
2
x
0
.
Since x
0
= constant for the clock, we have
∆t = γ∆t
0
> ∆t
0
.
So the interval measured in S is greater! So moving clocks run slowly.
A non-mathematical explanation comes from Feynman (not lectured): Sup-
pose we have a very simple clock: We send a light beam towards a mirror, and
wait for it to reflect back. When the clock detects the reflected light, it ticks,
and then sends the next light beam.
Then the interval between two ticks is the distance 2
d
divided by the speed
of light.
d
From the point of view of an observer moving downwards, by the time light
reaches the right mirror, it would have moved down a bit. So S sees
d
a
However, the distance travelled by the light beam is now
p
(2d)
2
+ a
2
>
2
d
.
Since they agree on the speed of light, it must have taken longer for the clock to
receive the reflected light in S. So the interval between ticks are longer.
By the principle of relativity, all clocks must measure the same time dilation,
or else we can compare the two clocks and know if we are “moving”.
This is famously evidenced by muons. Their half-life is around 2 microseconds
(i.e. on average they decay to something else after around 2 microseconds). They
are created when cosmic rays bombard the atmosphere. However, even if they
travel at the speed of light, 2 microseconds only allows it to travel
600 m
, certainly
not sufficient to reach the surface of Earth. However, we observe lots of muons
on Earth. This is because muons are travelling so fast that their clocks run
really slowly.
The twin paradox
Consider two twins: Luke and Leia. Luke stays at home. Leia travels at a
constant speed
v
to a distant planet
P
, turns around, and returns at the same
speed.
In Luke’s frame of reference,
x
ct
Luke
cT
2cT
Leia: x = vt
A (Leia’s arrival)
R
Leia’s arrival (A) at P has coordinates
(ct, x) = (cT, vT ).
The time experienced by Leia on her outward journey is
T
0
= γ
T −
v
c
2
T
=
T
γ
.
By Leia’s return
R
, Luke has aged by 2
T
, but Leia has aged by
2T
γ
<
2
T
. So
she is younger than Luke, because of time dilation.
The paradox is: From Leia’s perspective, Luke travelled away from her at
speed and the returned, so he should be younger than her!
Why is the problem not symmetric?
We can draw Leia’s initial frame of reference in dashed lines:
x
ct
Leia
A
Han
R
ct
0
x
0
X
Z
In Leia’s frame, by the time she arrives at
A
, she has experienced a time
T
0
=
T
γ
as shown above. This event is simultaneous with event
X
in Leia’s frame. Then
in Luke’s frame, the coordinates of X are
(ct, x) =
cT
0
γ
, 0
=
cT
γ
2
, 0
,
obtained through calculations similar to that above. So Leia thinks Luke has
aged less by a factor of 1
/γ
2
. At this stage, the problem is symmetric, and Luke
also thinks Leia has aged less by a factor of 1/γ
2
.
Things change when Leia turns around and changes frame of reference. To
understand this better, suppose Leia meets a friend, Han, who is just leaving
P
at speed
v
. On his journey back, Han also thinks Luke ages
T/γ
2
. But in his
frame of reference, his departure is simultaneous with Luke’s event
Z
, not
X
,
since he has different lines of simultaneity.
So the asymmetry between Luke and Leia occurs when Leia turns around.
At this point, she sees Luke age rapidly from X to Z.
Length contraction
A rod of length L
0
is stationary in S
0
. What is its length in S?
In
S
0
, then length of the rod is the distance between the two ends at the
same time. So we have
x
0
ct
0
L
0
L
0
In S, we have
x
ct
L
x
0
L
0
The lines
x
0
= 0 and
x
0
=
L
0
map into
x
=
vt
and
x
=
vt
+
L
0
/γ
. So the length
in
S
is
L
=
L
0
/γ < L
0
. Therefore moving objects are contracted in the direction
of motion.
Definition
(Proper length)
.
The proper length is the length measured in an
object’s rest frame.
This is analogous to the fact that if you view a bar from an angle, it looks
shorter than if you view it from the front. In relativity, what causes the
contraction is not a spatial rotation, but a spacetime hyperbolic rotation.
Question: does a train of length 2
L
fit alongside a platform of length
L
if it
travels through the station at a speed v such that γ = 2?
For the system of observers on the platform, the train contracts to a length
2L/γ = L. So it fits.
But for the system of observers on the train, the platform contracts to length
L/γ = L/2, which is much too short!
This can be explained by the difference of lines of simultaneity, since length
is the distance between front and back at the same time.
x
ctback of train front of train
back of
platform
front of
platform
L
fits in S
doesn’t fit in S
0
Composition of velocities
A particle moves with constant velocity
u
0
in frame
S
0
, which moves with velocity
v relative to S. What is its velocity u in S?
The world line of the particle in S
0
is
x
0
= u
0
t
0
.
In S, using the inverse Lorentz transformation,
u =
x
t
=
γ(x
0
+ vt
0
)
γ(t
0
+ (v/c
2
)x
0
)
=
u
0
t
0
+ vt
0
t
0
+ (v/c
2
)u
0
t
0
=
u
0
+ v
1 + u
0
v/c
2
.
This is the formula for the relativistic composition of velocities.
The inverse transformation is found by swapping
u
and
u
0
, and swapping the
sign of v, i.e.
u
0
=
u − v
1 − uv/c
2
.
Note the following:
–
if
u
0
v c
2
, then the transformation reduces to the standard Galilean
addition of velocities u ≈ u
0
+ v.
– u
is a monotonically increasing function of
u
for any constant
v
(with
|v| < c).
–
When
u
0
=
±c
,
u
=
u
0
for any
v
, i.e. the speed of light is constant in all
frames of reference.
–
Hence
|u
0
| < c
iff
|u| < c
. This means that we cannot reach the speed of
light by composition of velocities.