5Electromagnetism and relativity

IB Electromagnetism

5.5 The Lorentz force law

The final aspect of electromagnetism is the Lorentz force law for a particle with

charge q moving with velocity u:

dp

dt

= q(E + v × B).

To write this in relativistic form, we use the proper time

τ

(time experienced by

particle), which obeys

dt

dτ

= γ(u) =

1

p

1 −u

2

/c

2

.

We define the 4-velocity

U

=

dX

dτ

=

γ

c

u

, and 4-momentum

P

=

E/c

p

,

where E is the energy. Note that E is the energy while E is the electric field.

The Lorentz force law can be written as

dP

µ

dτ

= qF

µν

U

ν

.

We show that this does give our original Lorentz force law:

When µ = 1, 2, 3, we obtain

dp

dτ

= qγ(E + v × B).

By the chain rule, since

dt

dτ

= γ, we have

dp

dt

= q(E + v × B).

So the good, old Lorentz force law returns. Note that here

p

=

mγv

, the

relativistic momentum, not the ordinary momentum.

But how about the µ = 0 component? We get

dP

0

dτ

=

1

c

dE

dτ

=

q

c

γE ·v.

This says that

dE

dt

= qE · v,

which is our good old formula for the work done by an electric field.

Example

(Motion in a constant field)

.

Suppose that

E

= (

E,

0

,

0) and

u

=

(v, 0, 0). Then

m

d(γu)

dt

= qE.

So

mγu = qEt.

So

u =

dx

dt

=

qEt

p

m

2

+ q

2

E

2

t

2

/c

2

.

Note that u → c as t → ∞. Then we can solve to find

x =

mc

2

q

r

1 +

q

2

E

2

t

2

mc

2

− 1

!

For small t, x ≈

1

2

qEt

2

, as expected.

Example

(Motion in constant magnetic field)

.

Suppose

B

= (0

,

0

, B

). Then

we start with

dP

0

dτ

= 0 ⇒ E = mγc

2

= constant.

So |u| is constant. Then

m

∂(γu)

∂t

= qu × B.

So

mγ

du

dt

= qu × B

since

|u|

, and hence

γ

, is constant. This is the same equation we saw in Dynamics

and Relativity for a particle in a magnetic field, except for the extra

γ

term.

The particle goes in circles with frequency

ω =

qB

mγ

.