the lienard-wiechert potentials

A μ (x) = 4 π c \int d 4 x' G (x', t', x, t) j μ (x')

$A^{\mu}(x) = \frac{4\pi}{c}\int d^4 x'G(x',t',x,t)j^{\mu}(x')$ where

j μ (x') = c e \int d τ v μ (τ) δ (4) (x' - r μ (τ))

$j^{\mu}(x') = ce \int d\tau v^{\mu}(\tau)\delta^{(4)}(x'-r^{\mu}(\tau))$ and

G (x', t', x, t) = 1 2 π θ (c t - c t') δ (4) [(x - x') 2]

$G(x',t',x,t) = \frac{1}{2\pi}\theta(ct-ct')\delta^{(4)}[(x-x')^2]$

A μ (x) = 4 π c \int d 4 x' 1 2 π θ (c t - c t') δ (4) [(x - x') 2] c e \int d τ v μ (τ) δ (4) (x' - r μ (τ))

$A^{\mu}(x) = \frac{4\pi}{c} \int d^4 x' \frac{1}{2\pi}\theta(ct-ct')\delta^{(4)}[(x-x')^2] ce \int d\tau v^{\mu}(\tau)\delta^{(4)}(x'-r^{\mu}(\tau))$

A μ (x) = 2 e \int d 4 x' θ (c t - c t') δ (4) [(x - x') 2] \int d τ v μ (τ) δ (4) (x' - r μ (τ))

$A^{\mu}(x) = 2e \int d^4 x' \theta(ct-ct')\delta^{(4)}[(x-x')^2] \int d\tau v^{\mu}(\tau)\delta^{(4)}(x'-r^{\mu}(\tau))$

A μ (x) = 2 e \int d τ v μ (τ) \int d 4 x' θ (c t - c t') δ (4) [(x - x') 2] δ (4) (x' - r μ (τ))

$A^{\mu}(x) = 2e \int d\tau v^{\mu}(\tau) \int d^4 x' \theta(ct-ct')\delta^{(4)}[(x-x')^2] \delta^{(4)}(x'-r^{\mu}(\tau))$

A μ (x) = 2 e \int d τ v μ (τ) θ (c t - r 0 (τ)) δ (4) [(x - r μ (τ)) 2]

$A^{\mu}(x) = 2e \int d\tau v^{\mu}(\tau) \theta(ct-r_0(\tau))\delta^{(4)}[(x-r^{\mu}(\tau))^2]$
where

ct′=r0(τ) $ct' = r_0(\tau)$
using the identity

δ (n) (f (x)) = δ ( n ) ( x ) d f d x

$\delta^{(n)}(f(x))=\frac{\delta^{(n)}(x)}{\frac{df}{dx}}$
in our case,

x=τ $x=\tau$ and

f(τ)=(x−rμ(τ))2 $f(\tau) = (x-r^{\mu}(\tau))^2$

d f d τ = 2 (x - r μ (τ)) (- v μ (τ))

$\frac{df}{d\tau} = 2(x-r^{\mu}(\tau))(-v^{\mu}(\tau))$

= - 2 (x - r (τ)) μ v μ (τ)

$= -2(x-r(\tau))_{\mu}v^{\mu}(\tau)$
substituting this in the above equation gives

A μ (x) = 2 e \int d τ v μ (τ) θ (c t - r 0 (τ)) δ ( 4 ) ( τ - τ 0 ) - 2 ( x - r ( τ ) ) μ v μ ( τ )

$A^{\mu}(x) = 2e \int d\tau v^{\mu}(\tau) \theta(ct-r_0(\tau)) \frac{\delta^{(4)}(\tau-\tau_0)}{-2(x-r(\tau))_{\mu}v^{\mu}(\tau)}$

A μ (x) = - e \int d τ v μ (τ) θ (c t - r 0 (τ)) δ ( 4 ) ( τ - τ 0 ) [ ( x - r ( τ ) ) \cdot v ( τ ) ]

$A^{\mu}(x) = -e \int d\tau v^{\mu}(\tau) \theta(ct-r_0(\tau)) \frac{\delta^{(4)}(\tau-\tau_0)}{[(x-r(\tau))\cdot v(\tau)]}$

A μ (x) = - e [v μ ( τ ) [ ( x - r ( τ ) ) \cdot v ( τ ) ]] τ 0

$A^{\mu}(x) = -e [\frac{v^{\mu}(\tau)}{[(x-r(\tau))\cdot v(\tau)]}]_{\tau_0}$

A μ (x) = 2 e \int d τ v μ (τ) θ (c t - r 0 (τ)) δ (4) [(x - r μ (τ)) 2]

$A^{\mu}(x) = 2e \int d\tau v^{\mu}(\tau) \theta(ct-r_0(\tau))\delta^{(4)}[(x-r^{\mu}(\tau))^2]$

d α A β (x) = 2 e \int d τ d α [v β (τ) θ (c t - r 0 (τ)) δ (4) [(x - r μ (τ)) 2]]

$d^{\alpha}A^{\beta}(x) = 2e \int d\tau d^{\alpha}[v^{\beta}(\tau) \theta(ct-r_0(\tau))\delta^{(4)}[(x-r^{\mu}(\tau))^2]]$

= 2 e \int d τ v β (τ) θ (c t - r 0 (τ)) d α δ (4) [(x - r μ (τ)) 2]]

$= 2e \int d\tau v^{\beta}(\tau) \theta(ct-r_0(\tau))d^{\alpha}\delta^{(4)}[(x-r^{\mu}(\tau))^2]]$

= 2 e \int d τ v β (τ) θ (c t - r 0 (τ)) d d x α δ (4) [(x - r μ (τ)) 2]]

$= 2e \int d\tau v^{\beta}(\tau) \theta(ct-r_0(\tau))\frac{d}{dx^{\alpha}} \delta^{(4)}[(x-r^{\mu}(\tau))^2]]$

= 2 e \int d τ v β (τ) θ (c t - r 0 (τ)) d d x α d f d f δ (4) [(x - r μ (τ)) 2]]

$= 2e \int d\tau v^{\beta}(\tau) \theta(ct-r_0(\tau))\frac{d}{dx^{\alpha}} \frac{df}{df} \delta^{(4)}[(x-r^{\mu}(\tau))^2]]$

= 2 e \int d τ v β (τ) θ (c t - r 0 (τ)) d f d x α d d f δ (4) [(x - r μ (τ)) 2]]

$= 2e \int d\tau v^{\beta}(\tau) \theta(ct-r_0(\tau))\frac{df}{dx^{\alpha}} \frac{d}{df} \delta^{(4)}[(x-r^{\mu}(\tau))^2]]$

= 2 e \int d τ v β (τ) θ (c t - r 0 (τ)) d f d x α d d f d τ d τ δ (4) [(x - r μ (τ)) 2]]

$= 2e \int d\tau v^{\beta}(\tau) \theta(ct-r_0(\tau))\frac{df}{dx^{\alpha}} \frac{d}{df} \frac{d\tau}{d\tau} \delta^{(4)}[(x-r^{\mu}(\tau))^2]]$

= 2 e \int d τ v β (τ) θ (c t - r 0 (τ)) d f d x α d τ d f d d τ δ (4) [(x - r μ (τ)) 2]]

$= 2e \int d\tau v^{\beta}(\tau) \theta(ct-r_0(\tau))\frac{df}{dx^{\alpha}} \frac{d\tau}{df} \frac{d}{d\tau} \delta^{(4)}[(x-r^{\mu}(\tau))^2]]$
where

dfdτ=−2(x−r(τ))μvμ(τ)=−2(x−r(τ))⋅v(τ) $\frac{df}{d\tau} = -2(x-r(\tau))_{\mu}v^{\mu}(\tau) = -2(x-r(\tau))\cdot v(\tau)$ and

dfdxα=2(x−r(τ))α $\frac{df}{dx^{\alpha}} = 2(x-r(\tau))^{\alpha}$

d α A β (x) = 2 e \int d τ [2 ( x - r ( τ ) ) α v β ( τ ) - 2 ( x - r ( τ ) ) \cdot v ( τ )] θ (c t - r 0 (τ)) d d τ δ (4) [(x - r μ (τ)) 2]]

$d^{\alpha}A^{\beta}(x)= 2e \int d\tau [\frac{2(x-r(\tau))^{\alpha}v^{\beta}(\tau)}{-2(x-r(\tau))\cdot v(\tau)}] \theta(ct-r_0(\tau)) \frac{d}{d\tau} \delta^{(4)}[(x-r^{\mu}(\tau))^2]]$

= - 2 e \int d τ d d τ [( x - r ( τ ) ) α v β ( τ ) ( x - r ( τ ) ) \cdot v ( τ )] θ (c t - r 0 (τ)) δ (4) [(x - r μ (τ)) 2]]

$= -2e \int d\tau \frac{d}{d\tau} [\frac{(x-r(\tau))^{\alpha}v^{\beta}(\tau)}{(x-r(\tau))\cdot v(\tau)}] \theta(ct-r_0(\tau)) \delta^{(4)}[(x-r^{\mu}(\tau))^2]]$

F α β = - 2 e \int d τ d d τ [( x - r ( τ ) ) α v β ( τ ) - ( x - r ( τ ) ) β v α ( τ ) ( x - r ( τ ) ) \cdot v ( τ )] θ (c t - r 0 (τ)) δ (4) [(x - r μ (τ)) 2]]

$F^{\alpha\beta}= -2e \int d\tau \frac{d}{d\tau} [\frac{(x-r(\tau))^{\alpha}v^{\beta}(\tau)-(x-r(\tau))^{\beta}v^{\alpha}(\tau)}{(x-r(\tau))\cdot v(\tau)}] \theta(ct-r_0(\tau)) \delta^{(4)}[(x-r^{\mu}(\tau))^2]]$
is similar to

A μ (x) = 2 e \int d τ v μ (τ) θ (c t - r 0 (τ)) δ (4) [(x - r μ (τ)) 2]

$A^{\mu}(x) = 2e \int d\tau v^{\mu}(\tau) \theta(ct-r_0(\tau))\delta^{(4)}[(x-r^{\mu}(\tau))^2]$
therefore, we can say that

F α β = - 2 e [ ( x - r ) \cdot v ] d d τ [( x - r ) α v β - ( x - r ) β v α ( x - r ) \cdot v]

$F^{\alpha\beta}= \frac{-2e}{[(x-r)\cdot v]}\frac{d}{d\tau}[\frac{(x-r)^{\alpha}v^{\beta}-(x-r)^{\beta}v^{\alpha}}{(x-r)\cdot v}]$

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