the equation of continuity in tensor form

December 07, 2014

ρ = Σ a e a δ (r - r a)

$\rho = \Sigma_a e_a \delta(r-r_a)$ or

d e = ρ d V

$de = \rho dV$

d x i d e = ρ d x i d V

$dx^i de = \rho dx^idV$

d x i d e = ρ d x i d t d V d t

$dx^i de = \rho \frac{dx^i}{dt}dV dt$

d x i d e = ρ v i d Ω c

$dx^i de = \rho v^i\frac{d\Omega}{c}$

d x i d e = j i d Ω c

$dx^i de = j^i \frac{d\Omega}{c}$
where

ji=(ρc,ρvi) $j^i=(\rho c, \rho v^i)$
now, consider the term

∫ecAkdxk $\int \frac{e}{c}A_kdx^k$ from the action term

\int e c A k d x k = \int 1 c ρ d V A k d x k

$\int \frac{e}{c}A_kdx^k = \int \frac{1}{c}\rho dV A_k dx^k$

= \int 1 c ρ A k d x k d t d V d t

$= \int \frac{1}{c}\rho A_k \frac{dx^k}{dt} dV dt$

= \int 1 c 2 j k A k d Ω

$= \int \frac{1}{c^2}j^k A_k d\Omega$
substituting this in the action equation

S = - \int b a m c d s - \int b a e c A μ d x μ - 1 16 π c \int b a d Ω F μ ν F μ ν

$S = -\int_a^b mcds - \int_a^b\frac{e}{c}A_{\mu} dx^{\mu} - \frac{1}{16\pi c}\int_a^b d\Omega F_{\mu\nu}F^{\mu\nu}$

S = - \int m c d s - \int 1 c 2 j k A k d Ω - 1 16 π c \int d Ω F μ ν F μ ν

$S = -\int mcds - \int \frac{1}{c^2}j^k A_k d\Omega - \frac{1}{16\pi c}\int d\Omega F_{\mu\nu}F^{\mu\nu}$

δ S = - \int m c d v μ δ x μ - \int 1 c 2 δ j k A k d Ω - \int 1 c 2 j k δ A k d Ω - 1 16 π c \int d Ω δ (F μ ν F μ ν)

$\delta S = -\int mc dv^{\mu}\delta x_{\mu} - \int \frac{1}{c^2}\delta j^k A_k d\Omega - \int \frac{1}{c^2}j^k \delta A_k d\Omega -\frac{1}{16\pi c}\int d\Omega \delta (F_{\mu\nu}F^{\mu\nu})$

δ (F μ ν F μ ν) = 2 F μ ν δ F μ ν

$\delta (F_{\mu\nu}F^{\mu\nu}) = 2 F_{\mu\nu}\delta F^{\mu\nu}$

= 2 F μ ν δ (d μ A ν - d ν A μ)

$=2 F_{\mu\nu}\delta (d^{\mu}A^{\nu}-d^{\nu}A^{\mu})$

= 2 F μ ν (2 δ d μ A ν)

$=2 F_{\mu\nu} (2 \delta d^{\mu}A^{\nu})$

= 4 F μ ν d μ δ A ν

$=4 F_{\mu\nu} d^{\mu}\delta A^{\nu}$

= 4 d μ (F μ ν δ A ν) - 4 d μ F μ ν δ A ν

$=4 d^{\mu}(F_{\mu\nu} \delta A^{\nu}) -4 d^{\mu}F_{\mu\nu} \delta A^{\nu}$

δ S = - \int m c d v μ δ x μ - \int 1 c 2 δ j k A k d Ω - \int 1 c 2 j k δ A k d Ω - 1 16 π c \int 4 (d μ (F μ ν δ A ν) - d μ F μ ν δ A ν) d Ω

$\delta S = -\int mc dv^{\mu}\delta x_{\mu} - \int \frac{1}{c^2}\delta j^k A_k d\Omega - \int \frac{1}{c^2}j^k \delta A_k d\Omega -\frac{1}{16\pi c}\int 4( d^{\mu}(F_{\mu\nu} \delta A^{\nu}) -d^{\mu}F_{\mu\nu} \delta A^{\nu})d\Omega$
considering the terms with a variation in

Ak $A_k$

δ S = - \int 1 c 2 j k δ A k d Ω + 1 4 π c \int d μ F μ ν δ A ν d Ω

$\delta S = - \int \frac{1}{c^2}j^k \delta A_k d\Omega +\frac{1}{4\pi c}\int d^{\mu}F_{\mu\nu} \delta A^{\nu}d\Omega$
since

δS=0 $\delta S =0$

- \int 1 c 2 j k δ A k d Ω + 1 4 π c \int d μ F μ ν δ A ν d Ω = 0

$- \int \frac{1}{c^2}j^k \delta A_k d\Omega +\frac{1}{4\pi c}\int d^{\mu}F_{\mu\nu} \delta A^{\nu}d\Omega = 0$

- \int 4 π c j k δ A k d Ω + \int d μ F μ ν δ A ν d Ω = 0

$- \int \frac{4\pi}{c}j^k \delta A_k d\Omega +\int d^{\mu}F_{\mu\nu} \delta A^{\nu}d\Omega = 0$

\int (- 4 π c j ν δ A ν d Ω + d μ F μ ν δ A ν d Ω) = 0

$\int (-\frac{4\pi}{c}j^{\nu} \delta A_{\nu} d\Omega + d^{\mu}F_{\mu\nu} \delta A^{\nu}d\Omega) = 0$

\int δ A ν d Ω (- 4 π c j ν + d μ F μ ν) = 0

$\int \delta A^{\nu}d\Omega (-\frac{4\pi}{c}j^{\nu} + d^{\mu}F_{\mu\nu}) = 0$

- 4 π c j ν + d ν F ν μ = 0

$-\frac{4\pi}{c}j^{\nu} + d^{\nu}F_{\nu\mu} = 0$

d ν F ν μ = 4 π c j ν

$d^{\nu}F_{\nu\mu} = \frac{4\pi}{c}j^{\nu}$

d ν F μ ν = - 4 π c j ν

$d_{\nu}F_{\mu\nu} = -\frac{4\pi}{c}j_{\nu}$

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