Maxwell's equations in tensor form

we know that the action
S=ba(mcdsecAμdxμ)

and since δS=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, we can calculate that δds=vμδ(dxμ) i.e
\ dδs=vμdδxμ. We can substitute this term in the previous equation which therefore becomes
δS=ba(mcvμdδxμec(dxμδAμ+Aμdδxμ)

We can rewrite vμdδxμ=d(vμδxμ)δxμdvμ
\ similarly Aμdδxμ=d(Aμδxμ)δxμdAμ
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 as dAμdxνδxν
and similarly dAμ as dAμdxνdxν
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
baδxμ((mcdvμ)ec(dxνdAνdxμdAμdxνdxν))=0

mcdvμec(dxνdAνdxμdAμdxνdxν)=0

mcdvμ=ec(dxνdAνdxμdAμdxνdxν)

mcdvμ=ec(dAνdxμdAμdxν)dxν

dividing the whole equation by ds and remembering that dxμds=vμ, 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 that it is an antisymmetric matrix i.e Fνμ=Fμν

S=ba(mcdsecAμdxμ)

specifically, let’s look at Aμdxμ. we know that Aμ=(ϕ/c,A⃗ ) and that dxμ=(cdt,dr⃗ )
therefore Aμdxμ=ϕdtA⃗ dr⃗ 
S=ba(mcdsec(ϕdtA⃗ dr⃗ ))

S=ba(mcdsecϕdt+ecA⃗ dr⃗ ))

we know that ds=cdτ and that dτ=dtγ substituting these changes changes the above equation to
S=ba(mc2dtγecϕdt+ecA⃗ dr⃗ ))

S=ba(mc2dtγecϕdt+ecA⃗ dr⃗ dtdt))

S=badt(mc21γecϕ+ecA⃗ dr⃗ dt))

S=badtL

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, 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 
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 ),
E⃗ =ϕ1cdA⃗ dt

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