### the lienard-wiechert potentials

Aμ(x)=4πcd4xG(x,t,x,t)jμ(x)
where
jμ(x)=cedτvμ(τ)δ(4)(xrμ(τ))
and
G(x,t,x,t)=12πθ(ctct)δ(4)[(xx)2]

Aμ(x)=4πcd4x12πθ(ctct)δ(4)[(xx)2]cedτvμ(τ)δ(4)(xrμ(τ))

Aμ(x)=2ed4xθ(ctct)δ(4)[(xx)2]dτvμ(τ)δ(4)(xrμ(τ))

Aμ(x)=2edτvμ(τ)d4xθ(ctct)δ(4)[(xx)2]δ(4)(xrμ(τ))

Aμ(x)=2edτvμ(τ)θ(ctr0(τ))δ(4)[(xrμ(τ))2]

where ct=r0(τ)$ct' = r_0(\tau)$
using the identity
δ(n)(f(x))=δ(n)(x)dfdx

in our case, x=τ$x=\tau$ and f(τ)=(xrμ(τ))2$f(\tau) = (x-r^{\mu}(\tau))^2$
dfdτ=2(xrμ(τ))(vμ(τ))

=2(xr(τ))μvμ(τ)

substituting this in the above equation gives
Aμ(x)=2edτvμ(τ)θ(ctr0(τ))δ(4)(ττ0)2(xr(τ))μvμ(τ)

Aμ(x)=edτvμ(τ)θ(ctr0(τ))δ(4)(ττ0)[(xr(τ))v(τ)]

Aμ(x)=e[vμ(τ)[(xr(τ))v(τ)]]τ0

Aμ(x)=2edτvμ(τ)θ(ctr0(τ))δ(4)[(xrμ(τ))2]

dαAβ(x)=2edτdα[vβ(τ)θ(ctr0(τ))δ(4)[(xrμ(τ))2]]

=2edτvβ(τ)θ(ctr0(τ))dαδ(4)[(xrμ(τ))2]]

=2edτvβ(τ)θ(ctr0(τ))ddxαδ(4)[(xrμ(τ))2]]

=2edτvβ(τ)θ(ctr0(τ))ddxαdfdfδ(4)[(xrμ(τ))2]]

=2edτvβ(τ)θ(ctr0(τ))dfdxαddfδ(4)[(xrμ(τ))2]]

=2edτvβ(τ)θ(ctr0(τ))dfdxαddfdτdτδ(4)[(xrμ(τ))2]]

=2edτvβ(τ)θ(ctr0(τ))dfdxαdτdfddτδ(4)[(xrμ(τ))2]]

where dfdτ=2(xr(τ))μvμ(τ)=2(xr(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(xr(τ))α$\frac{df}{dx^{\alpha}} = 2(x-r(\tau))^{\alpha}$
dαAβ(x)=2edτ[2(xr(τ))αvβ(τ)2(xr(τ))v(τ)]θ(ctr0(τ))ddτδ(4)[(xrμ(τ))2]]

=2edτddτ[(xr(τ))αvβ(τ)(xr(τ))v(τ)]θ(ctr0(τ))δ(4)[(xrμ(τ))2]]

Fαβ=2edτddτ[(xr(τ))αvβ(τ)(xr(τ))βvα(τ)(xr(τ))v(τ)]θ(ctr0(τ))δ(4)[(xrμ(τ))2]]

is similar to
Aμ(x)=2edτvμ(τ)θ(ctr0(τ))δ(4)[(xrμ(τ))2]

therefore, we can say that
Fαβ=2e[(xr)v]ddτ[(xr)αvβ(xr)βvα(xr)v]