5Electromagnetism and relativity

IB Electromagnetism

5.3 Gauge potentials and electromagnetic fields
Recall that when solving electrostatic and magentostatic problems, we introduced
the potentials ϕ and A to help solve two of Maxwell’s equations:
×E = 0 E = −∇φ
· B = 0 B = × A.
However, in general, we do not have × E = 0. Instead, the equations are
× E +
B
t
= 0, · B = 0.
So we cannot use
E
=
−∇φ
directly if
B
E
and
B
in terms of φ and A as follows:
E = −∇φ
A
t
B = × A.
It is important to note that φ and A are not unique. We can shift them by
A 7→ A + χ
φ 7→ φ
χ
t
for any function
χ
(
x, t
). These are known as gauge transformations. Plug them
into the equations and you will see that you get the same E and B.
Now we have ended up with four objects, 1 from
φ
and 3 from
A
, so we can
put them into a 4-vector gauge potential:
A
µ
=
φ/c
A
We will assume that this makes sense, i.e. is a genuine 4-vector.
Now gauge transformations take a really nice form:
A
µ
7→ A
µ
µ
χ.
Finally, we define the anti-symmetric electromagnetic tensor
F
µν
=
µ
A
ν
ν
A
µ
.
Since this is antisymmetric, the diagonals are all 0, and
A
µν
=
A
νµ
. So this
thing has (4
×
4
4)
/
2 = 6 independent components. So it could encapsulate
information about the electric and magnetic fields (and nothing else).
We note that F is invariant under gauge transformations:
F
µν
7→ F
µν
µ
ν
χ +
ν
µ
χ = F
µν
.
We can compute components of F
µν
.
F
01
=
1
c
t
(A
x
)
φ/c
x
=
E
x
c
.
Note that we have
A
x
A
x
since
F
µν
was defined in terms of
A
µ
with
indices down. We can similarly obtain F
02
= E
y
/c and F
03
= E
z
/c.
We also have
F
12
=
x
(Ay)
y
(A
x
) = B
z
.
So
F
µν
=
0 E
x
/c E
y
/c E
z
/c
E
x
/c 0 B
z
B
y
E
y
/c B
z
0 B
x
E
z
/c B
y
B
x
0
Raising both indices, we have
F
µν
= η
µρ
η
νσ
F
ρσ
=
0 E
x
/c E
y
/c E
z
/c
E
x
/c 0 B
z
B
y
E
y
/c B
z
0 B
x
E
z
/c B
y
B
x
0
So the electric fields are inverted and the magnetic field is intact (which makes
sense, since moving indices up and down will flip the sign of the spacial com-
ponents. The electric field has one, while the magnetic field has two. So the
magnetic field is swapped twice and is intact).
Both F
µν
and F
µν
are tensors. Under a Lorentz transformation, we have
F
0µν
= Λ
µ
ρ
Λ
ν
σ
F
ρσ
.
For example under rotation
Λ =
1 0 0 0
0
0 R
0
,
we find that
E
0
= RE
B
0
= RB.
Under a boost by v in the x-direction , we have
E
0
x
= E
x
E
0
y
= γ(E
y
vB
z
)
E
0
z
= γ(E
z
+ vB
y
)
B
0
x
= B
x
B
0
y
= γ
B
y
+
v
c
2
E
z
B
0
z
= γ
B
z
v
c
2
E
y
Example
(Boosted line charge)
.
An infinite line along the
x
direction with
uniform charge per unit length, η, has electric field
E =
η
2πε
0
(y
2
+ z
2
)
0
y
z
.
The magnetic field is
B
= 0. Plugging this into the equation above, an observer
in frame S
0
boosted with v = (v, 0, 0), i.e. parallel to the wire, sees
E =
ηγ
2πε
0
(y
2
+ z
2
)
0
y
z
=
ηγ
2πε
0
(y
02
+ z
02
)
0
y
0
z
0
.
Also,
B =
ηγv
2πε
0
σ
2
(y
2
+ z
2
)
0
z
y
=
ηγv
2πε
0
σ
2
(y
02
+ z
02
)
0
z
0
y
0
.
In frame
S
0
, the charge density is Lorentz contracted to
γη
. The magnetic field
can be written as
B =
µ
0
I
0
2π
p
y
02
+ z
02
ˆϕ
0
,
where
ˆϕ
0
=
1
p
y
02
+ z
02
0
z
0
y
0
is the basis vector of cylindrical coordinates, and
I
0
= γηv is the current.
This is what we calculated from Ampere’s law previously. But we didn’t use
Ampere’s law here. We used Gauss’ law, and then applied a Lorentz boost.
We see that magnetic fields are relativistic effects of electric fields. They are
what we get when we apply a Lorentz boost to an electric field. So relativity is
not only about very fast objects
TM
. It is there when you stick a magnet onto
Example
(Boosted point charge)
.
A boosted point charge generates a current,
but is not the steady current we studied in magnetostatics. As the point charge
moves, the current density moves.
A point charge Q, at rest in frame S has
E =
Q
4πε
0
r
2
ˆ
r =
Q
4πε
2
0
(x
2
+ y
2
+ z
2
)
3/2
x
y
z
,
and B = 0.
In frame S
0
, boosted with v = (v, 0, 0), we have
E
0
=
Q
4πε
0
(x
2
+ y
2
+ z
2
)
3/2
x
γy
γz.
.
We need to express this in terms of x
0
, y
0
, z
0
, instead of x, y, z. Then we get
E
0
=
4πε
0
(γ
2
(x
0
+ vt
0
)
2
+ y
02
+ z
02
)
3/2
x
0
+ vt
0
y
0
z
0
.
Suppose that the particle sits at (
vt
0
,
0
,
0) in
S
0
. Let’s look at the electric at
t
0
= 0. Then the radial vector is r
0
=
x
0
y
0
z
0
. In the denominator, we write
γ
2
x
02
+ y
02
+ z
02
= (γ
2
1)x
02
+ r
02
=
v
2
γ
2
c
2
x
02
+ r
02
=
v
2
γ
2
c
2
cos
2
θ + 1
!
r
02
= γ
2
1
v
2
c
2
sin
2
θ
r
02
where θ is the angle between the x
0
axis and r
0
.
So
E
0
=
1
γ
2
1
v
2
c
2
sin
2
θ
3/2
Q
4πε
0
r
2
ˆ
r
0
.
The factor 1
2
1
v
2
c
2
sin
2
θ
3/2
squashes the electric field in the direction of
motion.
This result was first discovered by Lorentz from solving Maxwell’s equations
directly, which lead to him discovering the Lorentz transformations.
There is also a magnetic field
B =
µ
0
v
4π(γ
2
(x
0
+ vt
0
)
2
+ y
02
+ z
02
)
3/2
0
z
0
y
0
.
Lorentz invariants
We can ask the question “are there any combinations of
E
and
B
that all
observers agree on?” With index notation, all we have to do is to contract all
indices such that there are no dangling indices.
It turns out that there are two such possible combinations.
The first thing we might try would be
1
2
F
µν
F
µν
=
E
2
c
2
+ B
2
,
which works great.
To describe the second invariant, we need to introduce a new object in
Minkowski space. This is the fully anti-symmetric tensor
ε
µνρσ
=
+1 µνρσ is even permutation of 0123
1 µνρσ is odd permutation of 0123
0 otherwise
This is analogous to the
ε
ijk
in
R
3
, just that this time we are in 4-dimensional
spacetime. Under a Lorentz transformation,
ε
0µνρσ
= Λ
µ
κ
Λ
ν
λ
Λ
ρ
α
Λ
σ
β
ε
κλαβ
.
Since
ε
µνρσ
is fully anti-symmetric, so is
ε
0µνρσ
. Similar to what we did in
R
3
,
we can show that the only fully anti-symmetric tensors in Minkowski space are
multiples of ε
µνρσ
. So
ε
0µνρσ
=
µνρσ
for some constant a. To figure out what a is, test a single component:
ε
00123
= Λ
0
κ
Λ
1
λ
Λ
2
β
Λ
3
β
ε
κλαβ
= det Λ.
Lorentz transformations have
det
Λ = +1 (rotations, boosts), or
det
Λ =
1
(reflections, time reversal).
We will restrict to Λ such that
det
Λ = +1. Then
a
= 1, and
ε
µνρσ
is
invariant. In particular, it is a tensor.
We can finally apply this to electromagnetic fields. The dual electromagnetic
tensor is defined to be
˜
F
µν
=
1
2
ε
µνρσ
F
ρσ
. =
0 B
x
B
y
B
z
B
x
0 E
z
/c E
y
/c
B
y
E
z
/c 0 E
x
/c
B
z
E
y
/c E
x
/c 0
.
Why do we have the factor of a half? Consider a single component, say
˜
F
12
. It
gets contributions from both
F
03
and
F
30
, so we need to average the sum to
avoid double-counting.
˜
F
µν
is yet another antisymmetric matrix.
This is obtained from
F
µν
through the substitution
E 7→ cB
and
B 7→ E/c
.
Note the minus sign!
Then
˜
F
µν
is a tensor. We can construct the other Lorentz invariant using
˜
F
µν
. We don’t get anything new if we contract this with itself, since
˜
F
µν
˜
F
µν
=
F
µν
F
µν
. Instead, the second Lorentz invariant is
1
4
˜
F
µν
F
µν
= E · B/c.