32.3 The Lagrangian for Electromagnetic Fields
There are not many ways to make ascalar Lagrangian from the field tensor. We already know
that
∂Fμν
∂xν
=
jμ
cand we need to make our Lagrangian out of the fields, not just the current. Again,xμcannot
appear explicitlybecause that violates symmetries of nature. Also we want a linear equation and
so higher powers of the field should not occur. A term of the formmAμAμis a mass term and would
cause fields to fall off faster than^1 r. So, the only reasonable choice is
FμνFμν= 2(B^2 −E^2 ).One might consider
eμνλσFμνFλσ=B~·E~
but that is a pseudo-scalar, not a scalar. That is, it changes sign under a parity transformation.
The EM interaction is known to conserve parity so this is not a real option. As with the scalar field,
we need toadd an interaction with a source term. Of course, we know electromagnetism well,
so finding the right Lagrangian is not really guess work. The source of the field is the vectorjμ, so
the simple scalar we can write isjμAμ.
TheLagrangian for Classical Electricity and Magnetismwe will try is.
LEM=−
1
4
FμνFμν+1
cjμAμIn working with this Lagrangian, we willtreat each component ofAas an independent field.
The next step is to check what the Euler-Lagrange equation gives us.
∂
∂xν(
∂L
∂(∂Aμ/∂xν))
−
∂L
∂Aμ= 0
L = −
1
4
FμνFμν+1
cjμAμ=−1
4
(
∂Aν
∂xμ−
∂Aμ
∂xν)(
∂Aν
∂xμ−
∂Aμ
∂xν)
+
1
cjμAμ∂L
∂(∂Aμ/∂xν)= −
1
4
∂
∂(∂Aμ/∂xν)(
∂Aσ
∂xλ−
∂Aλ
∂xσ)(
∂Aσ
∂xλ−
∂Aλ
∂xσ)
= −
1
4
∂
∂(∂Aμ/∂xν)(
2
∂Aσ
∂xλ∂Aσ
∂xλ− 2
∂Aσ
∂xλ∂Aλ
∂xσ)
= −
1
4
4
(
∂Aμ
∂xν−
∂Aν
∂xμ)
= −Fνμ=Fμν
∂
∂xν
Fμν−∂L
∂Aμ= 0
∂
∂xνFμν−jμ
c= 0
∂
∂xνFμν =jμ
c