### Maxwell's equations in tensor form

we know that the action
S=ba(mcdsecAμdxμ)

and since δS=0$\delta S = 0$, we can rewrite the above equation as
δS=ba(mcδdsecδ(Aμdxμ))

δS=ba(mcdδsec(dxμδAμ+Aμδ(dxμ))

δS=ba(mcdδsec(dxμδAμ+Aμdδxμ)

using ds=(dxμdxμ)12$ds = (dx_{\mu}dx^{\mu})^{\frac{1}{2}}$, we can calculate that δds=vμδ(dxμ)$\delta ds = v_{\mu}\delta(dx_{\mu})$ i.e
\ dδs=vμdδxμ$d \delta s = v_{\mu}d \delta x_{\mu}$. We can substitute this term in the previous equation which therefore becomes
δS=ba(mcvμdδxμec(dxμδAμ+Aμdδxμ)

We can rewrite vμdδxμ=d(vμδxμ)δxμdvμ$v_{\mu}d \delta x_{\mu} = d (v_{\mu} \delta x_{\mu}) - \delta x_{\mu}dv_{\mu}$
\ similarly Aμdδxμ=d(Aμδxμ)δxμdAμ$A_{\mu} d \delta x^{\mu} = d(A_{\mu}\delta x^{\mu}) - \delta x^{\mu}dA_{\mu}$
substituting the above results in the previous equation leads to
δS=ba(mc(d(vμδxμ)δxμdvμ)ec(dxμδAμ+d(Aμδxμ)δxμdAμ)

δS=ba(mcd(vμδxμ)ecd(Aμδxμ))+ba(mcδxμdvμ)ec(dxμδAμδxμdAμ)

let us rewrite δAμ$\delta A_{\mu}$ as dAμdxνδxν$\frac{dA_{\mu}}{dx^{\nu}}\delta x^{\nu}$
and similarly dAμ$dA_{\mu}$ as dAμdxνdxν$\frac{dA_{\mu}}{dx^{\nu}}dx^{\nu}$
correcting for the above changes in the equation above gives
δS=ba(mcδxμdvμ)ec(dxμdAμdxνδxνδxμdAμdxνdxν)

δS=ba(mcδxμdvμ)ec(dxνdAνdxμδxμδxμdAμdxνdxν)

δS=baδxμ((mcdvμ)ec(dxνdAνdxμdAμdxνdxν))

and since δS=0$\delta S = 0$
baδxμ((mcdvμ)ec(dxνdAνdxμdAμdxνdxν))=0

mcdvμec(dxνdAνdxμdAμdxνdxν)=0

mcdvμ=ec(dxνdAνdxμdAμdxνdxν)

mcdvμ=ec(dAνdxμdAμdxν)dxν

dividing the whole equation by ds$ds$ and remembering that dxμds=vμ$\frac{dx_{\mu}}{ds} = v_{\mu}$, we can rewrite the above equation as
mcdvμds=ec(dAνdxμdAμdxν)dxνds

mcdvμds=ec(dAνdxμdAμdxν)vν

we now define
Fμν=dAνdxμdAμdxν

mcdvμds=Fμνvν

We can understand from the construction of Fμν$F_{\mu \nu}$ that it is an antisymmetric matrix i.e Fνμ=Fμν$F_{\nu \mu} = -F_{\mu \nu}$

S=ba(mcdsecAμdxμ)

specifically, let’s look at Aμdxμ$A_{\mu}dx^{\mu}$. we know that Aμ=(ϕ/c,A⃗ )$A_{\mu} = (\phi /c,\vec{A})$ and that dxμ=(cdt,dr⃗ )$dx_{\mu} = (cdt,-d\vec{r})$
therefore Aμdxμ=ϕdtA⃗ dr⃗ $A_{\mu}dx^{\mu} = \phi dt - \vec{A}\cdot d\vec{r}$
S=ba(mcdsec(ϕdtA⃗ dr⃗ ))

S=ba(mcdsecϕdt+ecA⃗ dr⃗ ))

we know that ds=cdτ$ds = cd\tau$ and that dτ=dtγ$d\tau = \frac{dt}{\gamma}$ substituting these changes changes the above equation to
S=ba(mc2dtγecϕdt+ecA⃗ dr⃗ ))

S=ba(mc2dtγecϕdt+ecA⃗ dr⃗ dtdt))

where
L=mc21γecϕ+ecA⃗ dr⃗ dt

L=mc21γecϕ+ecA⃗ V⃗

we can get the equations of motion from
LrddtLr˙=0

Lr=ecϕ+ec⃗ (A⃗ V⃗ )

⃗ (A⃗ V⃗ )=V⃗ ×(⃗ ×A⃗ )A⃗ (⃗ V⃗ )

since A⃗ (⃗ V⃗ )=0$\vec{A}(\vec{\nabla}\cdot\vec{V}) = 0$, we can rewrite the earlier equation as
Lr=ecϕ+ecV⃗ ×(⃗ ×A⃗ )

similarly
Lr˙=r˙(mc21γecA⃗ V⃗ )

Lr˙=r˙(mc21v2c2ecA⃗ V⃗ )

Lr˙=γmV⃗ +ecA⃗

going back to
LrddtLr˙=0

ecϕ+ecV⃗ ×(⃗ ×A⃗ )ddt(γmV⃗ +ecA⃗ )=0

ecϕ+ecV⃗ ×(⃗ ×A⃗ )dp⃗ dt+ecdA⃗ dt=0

where p⃗ =γmV⃗ $\vec{p}=\gamma m\vec{V}$
dp⃗ dt=ecϕ+ecdA⃗ dt+ecV⃗ ×(⃗ ×A⃗ )

dp⃗ dt=e(1cϕ+1cdA⃗ dt)+ecV⃗ ×(⃗ ×A⃗ )

we’re defining
E⃗ =1cϕ1cdA⃗ dt
and
B⃗ =⃗ ×A⃗

dp⃗ dt=eE⃗ +ecV⃗ ×B⃗

if we assume Aμ=(ϕ,A⃗ )$A_{\mu} = (\phi,\vec{A})$,
E⃗ =ϕ1cdA⃗ dt