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.