Topics in Classical Electrodynamics

where we have added a new term S field , which represents the interaction
between the particles and the field that they have produced themselves. We
will write is as
∫
∫
S field ∼ F µν F µν d 4 x = F µν F µν cdt d 3 x .
Then adding the proportionality constants the total action is written as
∫
S = −mc ds − e ∫
A µ dx µ − 1 ∫
F µν F µν cdt d 3 x ,
c
16πc
where we have adopted the Gauss system of units, i.e. µ 0 = 4π and ε 0 = 1 . 4π
Note that we can rewrite the second term as
e
c
∫
∫
A µ dx µ = 1 c
= 1 ∫
c
∫
ρA µ dx µ dV = 1 c
j µ A µ dV dt = 1 ∫
c 2
dx µ
ρA µ dV dt
dt
j µ A µ d 4 x , (48)
where in the second line we have introduced, the current j i = ρ dxi = (ρc, ρ⃗v).
dt
Including this, we can now write the action of the moving test charge as
∫
S = −mc ds − 1 ∫
j ρ A
c 2 ρ d 4 x − 1 ∫
F µν F µν cdtd 3 x .
16πc
Keeping sources constant and the path unchanged (i.e. δj µ = 0 and δs = 0),
we can write the deviation from the action as follows
δS = − 1 ∫
j ρ δA
c 2 ρ d 4 x − 1 ∫
F µν δF µν cdtd 3 x
8πc
= − 1 [ ∫ 1
j ρ δA ρ d 4 x + 1 ∫ ]
∂F
µν
c c
4π ∂x δA µcdtd 3 x , (49)
ν
where in the last term in the first line, we have used that
δF µν = ∂ µ δA ν − ∂ ν δA µ .
To find the extremum, we need to satisfy δS = 0, which due to eq.(49),
is equivalent
− 1 c 2 jµ − 1
4πc ∂µ F µν = 0 .
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