2.5. PRINCIPLE OF LAGRANGIAN DYNAMICS (PLD) 83
It is clear that
d
dλ
∣
∣
∣
λ= 0
L 2 (Aμ+λA ̃μ) =
1
c
∫
QT
d
dλ
(Aμ+λA ̃μ)Jμdxdt=
1
c
∫
QT
JμA ̃μdxdt.
We infer from (2.5.16) that
(2.5.17) δLμ 2 =
1
c
Jμ.
Noting thatgμ ν=gν μ, we have
d
dλ
∣
∣
∣
λ= 0
L 1 (Aμ+λA ̃μ) =
1
8 π
∫
QT
gμ αgν βFα β
d
dλ
∣
∣
∣
λ= 0
(Fμ ν+λF ̃μ ν)dxdt
=
1
8 π
∫
QT
gμ αgν βFα β
(
∂A ̃μ
∂xν
−
∂A ̃ν
∂xμ
)
dxdt
By the Gauss formula,
∫
QT
gμ αgν βFα β
∂A ̃μ
∂xν
dxdt=−
∫
QT
gμ αgν β
∂Fα β
∂xν
A ̃μdxdt,
∫
QT
gμ αgν βFα β
∂A ̃ν
∂xμ
dxdt=−
∫
QT
gμ αgν β
∂Fα β
∂xμ
A ̃νdxdt
=(by the permutation ofμandν)
=−
∫
QT
gν αgμ β
∂Fα β
∂xν
A ̃μdxdt
=
∫
QT
gμ αgν β
∂Fα β
∂xν
A ̃μdxdt,
where a permutation onαandβis performed, andFβ α=−Fα β.
Thus, we obtain that
d
dλ
∣
∣
∣
λ= 0
L 1 (Aμ+λA ̃μ) =−
1
4 π
∫
QT
gμ αgν β
∂Fα β
∂xν
A ̃μdxdt.
We infer then from (2.5.16) that
(2.5.18) δL 1 =−
1
4 π
gμ αgν β
∂Fα β
∂xν
=−
1
4 π
∂Fμ ν
∂xν
.
Hence it follows from (2.5.17) and (2.5.18) that
δL=−
1
4 π
∂Fμ ν
∂xν
+
1
c
Jμ,
which implies that the equationδL=0 takes the form:
(2.5.19)
∂Fμ ν
∂xν
=
4 π
c
Jμ.