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
=
v
, 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
u = 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
dP
0
dτ
= 0 E = c
2
= constant.
So |u| is constant. Then
m
(γu)
t
= qu × B.
So
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
.