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

changes. Instead, we define

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

instead of

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

your fridge!

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

=

Qγ

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

/γ

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

Qγv

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

= aε

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.