8Special relativity

IA Dynamics and Relativity

8.5 Relativistic kinematics

In Newtonian mechanics, we describe a particle by its position

x

(

t

), with its

velocity being u(t) =

dx

dt

.

In relativity, this is unsatisfactory. In special relativity, space and time can

be mixed together by Lorentz boosts, and we prefer not to single out time from

space. For example, when we write the 4-vector

X

, we put in both the time and

space components, and Lorentz transformations are 4

×

4 matrices that act on

X.

In the definition of velocity, however, we are differentiating space with respect

to time, which is rather weird. First of all, we need something to replace time.

Recall that we defined “proper length” as the length in the item in its rest frame.

Similarly, we can define the proper time.

Definition (Proper time). The proper time τ is defined such that

∆τ =

∆s

c

τ

is the time experienced by the particle, i.e. the time in the particles rest frame.

The world line of a particle can be parametrized using the proper time by

t(τ) and x(τ).

x

ct

τ

1

τ

2

Infinitesimal changes are related by

dτ =

ds

c

=

1

c

p

c

2

dt

2

− |dx|

2

=

r

1 −

|u|

2

c

2

dt.

Thus

dt

dτ

= γ

u

with

γ

u

=

1

q

1 −

|u|

2

c

2

.

The total time experienced by the particle along a segment of its world line is

T =

Z

dτ =

Z

1

γ

u

dt.

We can then define the position 4-vector and 4-velocity.

Definition (Position 4-vector and 4-velocty). The position 4-vector is

X(τ) =

ct(τ)

x(τ)

.

Its 4-velocity is defined as

U =

dX

dτ

=

c

dt

dτ

dx

dτ

=

dt

dτ

c

u

= γ

u

c

u

,

where u =

dx

dt

.

Another common notation is

X = (ct, x), U = γ

u

(c, u).

If frames

S

and

S

0

are related by

X

0

= Λ

X

, then the 4-velocity also transforms

as U

0

= ΛU.

Definition

(4-vector)

.

A 4-vector is a 4-component vectors that transforms in

this way under a Lorentz transformation, i.e. X

0

= ΛX.

When using suffix notation, the indices are written above (superscript) instead

of below (subscript). The indices are written with Greek letters which range

from 0 to 3. So we have

X

µ

instead of

X

i

, for

µ

= 0

,

1

,

2

,

3. If we write

X

µ

instead, it means a different thing. This will be explained more in-depth in the

electromagnetism course (and you’ll get more confused!).

U

is a 4-vector because

X

is a 4-vector and

τ

is a Lorentz invariant. Note

that dX/dt is not a 4-vector.

Note that this definition of 4-vector is analogous to that of a tensor — things

that transform nicely according to our rules. Then

τ

would be a scalar, i.e.

rank-0 tensor, while t is just a number, not a scalar.

For any 4-vector

U

, the inner product

U · U

=

U

0

· U

0

is Lorentz invariant,

i.e. the same in all inertial frames. In the rest frame of the particle,

U

= (

c,

0).

So U · U = c

2

.

In any other frame, Y = γ

u

(c, u). So

Y · Y = γ

2

u

(c

2

− |u|

2

) = c

2

as expected.

Transformation of velocities revisited

We have seen that velocities cannot be simply added in relativity. However, the

4-velocity does transform linearly, according to the Lorentz transform:

U

0

= ΛU.

In frame

S

, consider a particle moving with speed

u

at an angle

θ

to the

x

axis

in the

xy

plane. This is the most general case for motion not parallel to the

Lorentz boost.

Its 4-velocity is

U =

γ

u

c

γ

u

u cos θ

γ

u

u sin θ

0

, γ

u

=

1

p

1 − u

2

/c

2

.

With frames

S

and

S

0

in standard configuration (i.e. origin coincide at

t

= 0,

S

0

moving in x direction with velocity v relative to S),

U

0

=

γ

u

0

c

γ

u

0

u

0

cos θ

0

γ

u

0

u

0

sin θ

0

0

=

γ

v

−γ

v

v/c 0 0

−γ

v

v/c γ

v

0 0

0 0 1 0

0 0 0 1

γ

u

c

γ

u

u cos θ

γ

u

u sin θ

0

Instead of evaluating the whole matrix, we can divide different rows to get useful

results.

The ratio of the first two lines gives

u

0

cos θ

0

=

u cos θ − v

1 −

uv

c

2

cos θ

,

just like the composition of parallel velocities.

The ratio of the third to second line gives

tan θ

0

=

u sin θ

γ

v

(u cos θ − v)

,

which describes aberration, a change in the direction of motion of a particle due

to the motion of the observer. Note that this isn’t just a relativistic effect! If

you walk in the rain, you have to hold your umbrella obliquely since the rain

seems to you that they are coming from an angle. The relativistic part is the

γ

v

factor in the denominator.

This is also seen in the aberration of starlight (

u

=

c

) due to the Earth’s

orbital motion. This causes small annual changes in the apparent positions of

stars.

4-momentum

Definition (4-momentum). The 4-momentum of a particle of mass m is

P = mU = mγ

u

c

u

The 4-momentum of a system of particles is the sum of the 4-momentum of the

particles, and is conserved in the absence of external forces.

The spatial components of P are the relativistic 3-momentum,

p = mγ

u

u,

which differs from the Newtonian expression by a factor of

γ

u

. Note that

|p| → ∞

as |u| → c.

What is the interpretation of the time component

P

0

(i.e. the first time

component of the P vector)? We expand for |u| c:

P

0

= mγc =

mc

p

1 − |u|

2

/c

2

=

1

c

mc

2

+

1

2

m|u|

2

+ ···

.

We have a constant term

mc

2

plus a kinetic energy term

1

2

m|u|

2

, plus more

tiny terms, all divided by

c

. So this suggests that

P

0

is indeed the energy for

a particle, and the remaining

···

terms are relativistic corrections for our old

formula

1

2

m|u|

2

(the mc

2

term will be explained later). So we interpret P as

P =

E/c

p

Definition

(Relativistic energy)

.

The relativistic energy of a particle is

E

=

P

0

c

.

So

E = mγc

2

= mc

2

+

1

2

m|u|

2

+ ···

Note that E → ∞ as |u| → c.

For a stationary particle, we obtain

E = mc

2

.

This implies that mass is a form of energy.

m

is sometimes called the rest mass.

The energy of a moving particle,

mγ

u

c

2

, is the sum of the rest energy

mc

2

and kinetic energy m(γ

u

− 1)c

2

.

Since

P ·P

=

E

2

c

2

−|p|

2

is a Lorentz invariant (lengths of 4-vectors are always

Lorentz invariant) and equals

m

2

c

2

in the particle’s rest frame, we have the

general relation between energy and momentum

E

2

= |p|

2

c

2

+ m

2

c

4

In Newtonian physics, mass and energy are separately conserved. In relativity,

mass is not conserved. Instead, it is just another form of energy, and the total

energy, including mass energy, is conserved.

Mass can be converged into kinetic energy and vice versa (e.g. atomic bombs!)

Massless particles

Particles with zero mass (

m

= 0), e.g. photons, can have non-zero momentum

and energy because they travel at the speed of light (γ = ∞).

In this case,

P · P

= 0. So massless particles have light-like (or null)

trajectories, and no proper time can be defined for such particles.

Other massless particles in the Standard Model of particle physics include

the gluon.

For these particles, energy and momentum are related by

E

2

= |p|

2

c

2

.

So

E = |p|c.

Thus

P =

E

c

1

n

,

where n is a unit (3-)vector in the direction of propagation.

According to quantum mechanics, fundamental “particles” aren’t really

particles but have both particle-like and wave-like properties (if that sounds

confusing, yes it is!). Hence we can assign it a de Broglie wavelength, according

to the de Broglie relation:

|p| =

h

λ

where h ≈ 6.63 × 10

−34

m

2

kg s

−1

is Planck’s constant.

For massless particles, this is consistent with Planck’s relation:

E =

hc

λ

= hν,

where ν =

c

λ

is the wave frequency.

Newton’s second law in special relativity

Definition (4-force). The 4-force is

F =

dP

dτ

This equation is the relativistic counterpart to Newton’s second law.

It is related to the 3-force F by

F = γ

u

F · u/c

F

Expanding the definition of the 4-force componentwise, we obtain

dE

dτ

= γ

u

F · u ⇒

dE

dt

= F · u

and

dp

dτ

= γ

u

F ⇒

dp

dt

= F

Equivalently, for a particle of mass m,

F = mA,

where

A =

dU

dτ

is the 4-acceleration.

We have

U = γ

u

c

u

So

A = γ

u

dU

dt

= γ

u

˙γ

u

c

γ

u

a + ˙γ

u

u.

where a =

du

dt

and ˙γ

u

= γ

3

u

a·u

c

2

.

In the instantaneous rest frame of a particle, u = 0 and γ

u

= 1. So

U =

c

0

, A =

0

a

Then

U · A

= 0. Since this is a Lorentz invariant, we have

U · A

= 0 in all

frames.