### Maxwell's equations in tensor form

we know that the action
S=ba(mcdsecAÎ¼dxÎ¼)

and since Î´S=0$\delta S = 0$, we can rewrite the above equation as
Î´S=ba(mcÎ´dsecÎ´(AÎ¼dxÎ¼))

Î´S=ba(mcdÎ´sec(dxÎ¼Î´AÎ¼+AÎ¼Î´(dxÎ¼))

Î´S=ba(mcdÎ´sec(dxÎ¼Î´AÎ¼+AÎ¼dÎ´xÎ¼)

using ds=(dxÎ¼dxÎ¼)12$ds = (dx_{\mu}dx^{\mu})^{\frac{1}{2}}$, we can calculate that Î´ds=vÎ¼Î´(dxÎ¼)$\delta ds = v_{\mu}\delta(dx_{\mu})$ i.e
\ dÎ´s=vÎ¼dÎ´xÎ¼$d \delta s = v_{\mu}d \delta x_{\mu}$. We can substitute this term in the previous equation which therefore becomes
Î´S=ba(mcvÎ¼dÎ´xÎ¼ec(dxÎ¼Î´AÎ¼+AÎ¼dÎ´xÎ¼)

We can rewrite vÎ¼dÎ´xÎ¼=d(vÎ¼Î´xÎ¼)Î´xÎ¼dvÎ¼$v_{\mu}d \delta x_{\mu} = d (v_{\mu} \delta x_{\mu}) - \delta x_{\mu}dv_{\mu}$
\ similarly AÎ¼dÎ´xÎ¼=d(AÎ¼Î´xÎ¼)Î´xÎ¼dAÎ¼$A_{\mu} d \delta x^{\mu} = d(A_{\mu}\delta x^{\mu}) - \delta x^{\mu}dA_{\mu}$
substituting the above results in the previous equation leads to
Î´S=ba(mc(d(vÎ¼Î´xÎ¼)Î´xÎ¼dvÎ¼)ec(dxÎ¼Î´AÎ¼+d(AÎ¼Î´xÎ¼)Î´xÎ¼dAÎ¼)

Î´S=ba(mcd(vÎ¼Î´xÎ¼)ecd(AÎ¼Î´xÎ¼))+ba(mcÎ´xÎ¼dvÎ¼)ec(dxÎ¼Î´AÎ¼Î´xÎ¼dAÎ¼)

let us rewrite Î´AÎ¼$\delta A_{\mu}$ as dAÎ¼dxÎ½Î´xÎ½$\frac{dA_{\mu}}{dx^{\nu}}\delta x^{\nu}$
and similarly dAÎ¼$dA_{\mu}$ as dAÎ¼dxÎ½dxÎ½$\frac{dA_{\mu}}{dx^{\nu}}dx^{\nu}$
correcting for the above changes in the equation above gives
Î´S=ba(mcÎ´xÎ¼dvÎ¼)ec(dxÎ¼dAÎ¼dxÎ½Î´xÎ½Î´xÎ¼dAÎ¼dxÎ½dxÎ½)

Î´S=ba(mcÎ´xÎ¼dvÎ¼)ec(dxÎ½dAÎ½dxÎ¼Î´xÎ¼Î´xÎ¼dAÎ¼dxÎ½dxÎ½)

Î´S=baÎ´xÎ¼((mcdvÎ¼)ec(dxÎ½dAÎ½dxÎ¼dAÎ¼dxÎ½dxÎ½))

and since Î´S=0$\delta S = 0$
baÎ´xÎ¼((mcdvÎ¼)ec(dxÎ½dAÎ½dxÎ¼dAÎ¼dxÎ½dxÎ½))=0

mcdvÎ¼ec(dxÎ½dAÎ½dxÎ¼dAÎ¼dxÎ½dxÎ½)=0

mcdvÎ¼=ec(dxÎ½dAÎ½dxÎ¼dAÎ¼dxÎ½dxÎ½)

mcdvÎ¼=ec(dAÎ½dxÎ¼dAÎ¼dxÎ½)dxÎ½

dividing the whole equation by ds$ds$ and remembering that dxÎ¼ds=vÎ¼$\frac{dx_{\mu}}{ds} = v_{\mu}$, we can rewrite the above equation as
mcdvÎ¼ds=ec(dAÎ½dxÎ¼dAÎ¼dxÎ½)dxÎ½ds

mcdvÎ¼ds=ec(dAÎ½dxÎ¼dAÎ¼dxÎ½)vÎ½

we now define
FÎ¼Î½=dAÎ½dxÎ¼dAÎ¼dxÎ½

mcdvÎ¼ds=FÎ¼Î½vÎ½

We can understand from the construction of FÎ¼Î½$F_{\mu \nu}$ that it is an antisymmetric matrix i.e FÎ½Î¼=FÎ¼Î½$F_{\nu \mu} = -F_{\mu \nu}$

S=ba(mcdsecAÎ¼dxÎ¼)

specifically, let’s look at AÎ¼dxÎ¼$A_{\mu}dx^{\mu}$. we know that AÎ¼=(Ï•/c,A⃗ )$A_{\mu} = (\phi /c,\vec{A})$ and that dxÎ¼=(cdt,dr⃗ )$dx_{\mu} = (cdt,-d\vec{r})$
therefore AÎ¼dxÎ¼=Ï•dtA⃗ dr⃗ $A_{\mu}dx^{\mu} = \phi dt - \vec{A}\cdot d\vec{r}$
S=ba(mcdsec(Ï•dtA⃗ dr⃗ ))

S=ba(mcdsecÏ•dt+ecA⃗ dr⃗ ))

we know that ds=cdÏ„$ds = cd\tau$ and that dÏ„=dtÎ³$d\tau = \frac{dt}{\gamma}$ substituting these changes changes the above equation to
S=ba(mc2dtÎ³ecÏ•dt+ecA⃗ dr⃗ ))

S=ba(mc2dtÎ³ecÏ•dt+ecA⃗ dr⃗ dtdt))

where
L=mc21Î³ecÏ•+ecA⃗ dr⃗ dt

L=mc21Î³ecÏ•+ecA⃗ V⃗

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

Lr=ecÏ•+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=ecÏ•+ecV⃗ ×(⃗ ×A⃗ )

similarly
Lr˙=r˙(mc21Î³ecA⃗ V⃗ )

Lr˙=r˙(mc21v2c2ecA⃗ V⃗ )

Lr˙=Î³mV⃗ +ecA⃗

going back to
LrddtLr˙=0

ecÏ•+ecV⃗ ×(⃗ ×A⃗ )ddt(Î³mV⃗ +ecA⃗ )=0

ecÏ•+ecV⃗ ×(⃗ ×A⃗ )dp⃗ dt+ecdA⃗ dt=0

where p⃗ =Î³mV⃗ $\vec{p}=\gamma m\vec{V}$
dp⃗ dt=ecÏ•+ecdA⃗ dt+ecV⃗ ×(⃗ ×A⃗ )

dp⃗ dt=e(1cÏ•+1cdA⃗ dt)+ecV⃗ ×(⃗ ×A⃗ )

we’re defining
E⃗ =1cÏ•1cdA⃗ dt
and
B⃗ =⃗ ×A⃗

dp⃗ dt=eE⃗ +ecV⃗ ×B⃗

if we assume AÎ¼=(Ï•,A⃗ )$A_{\mu} = (\phi,\vec{A})$,
E⃗ =Ï•1cdA⃗ dt

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