Maxwell's equations in tensor form

we know that the action
$$S = \int_a^b(-mcds - \frac{e}{c}A_{\mu} dx^{\mu})$$
and since $\delta S = 0$, we can rewrite the above equation as
$$\delta S = \int_a^b(-mc\delta ds - \frac{e}{c}\delta(A_{\mu} dx^{\mu}))$$
$$\delta S = \int_a^b(-mcd\delta s - \frac{e}{c}(dx^{\mu} \delta A_{\mu} + A_{\mu} \delta (dx^{\mu}))$$
$$\delta S = \int_a^b(-mcd\delta s - \frac{e}{c}(dx^{\mu} \delta A_{\mu} + A_{\mu} d \delta x^{\mu})$$
using $ds = (dx_{\mu}dx^{\mu})^{\frac{1}{2}}$, we can calculate that $\delta ds = v_{\mu}\delta(dx_{\mu})$ i.e
\ $d \delta s = v_{\mu}d \delta x_{\mu}$. We can substitute this term in the previous equation which therefore becomes
$$\delta S = \int_a^b(-mcv_{\mu}d \delta x_{\mu} - \frac{e}{c}(dx^{\mu} \delta A_{\mu} + A_{\mu} d \delta x^{\mu})$$
We can rewrite $v_{\mu}d \delta x_{\mu} = d (v_{\mu} \delta x_{\mu}) - \delta x_{\mu}dv_{\mu}$
\ similarly $A_{\mu} d \delta x^{\mu} = d(A_{\mu}\delta x^{\mu}) - \delta x^{\mu}dA_{\mu}$
substituting the above results in the previous equation leads to
$$\delta S = \int_a^b(-mc(d (v_{\mu} \delta x_{\mu}) - \delta x_{\mu}dv_{\mu}) - \frac{e}{c}(dx^{\mu} \delta A_{\mu} + d(A_{\mu}\delta x^{\mu}) - \delta x^{\mu}dA_{\mu})$$
$$\delta S = \int_a^b(-mc d (v_{\mu} \delta x_{\mu}) - \frac{e}{c}d(A_{\mu}\delta x^{\mu})) + \int_a^b (mc\delta x_{\mu}dv_{\mu}) - \frac{e}{c}(dx^{\mu} \delta A_{\mu} - \delta x^{\mu}dA_{\mu})$$
let us rewrite $\delta A_{\mu}$ as $\frac{dA_{\mu}}{dx^{\nu}}\delta x^{\nu}$
and similarly $dA_{\mu}$ as $\frac{dA_{\mu}}{dx^{\nu}}dx^{\nu}$
correcting for the above changes in the equation above gives
$$\delta S = \int_a^b (mc\delta x_{\mu}dv_{\mu}) - \frac{e}{c}(dx^{\mu}\frac{dA_{\mu}}{dx^{\nu}}\delta x^{\nu} - \delta x^{\mu}\frac{dA_{\mu}}{dx^{\nu}}dx^{\nu})$$
$$\delta S = \int_a^b (mc\delta x_{\mu}dv_{\mu}) - \frac{e}{c}(dx^{\nu}\frac{dA_{\nu}}{dx^{\mu}}\delta x^{\mu} - \delta x^{\mu}\frac{dA_{\mu}}{dx^{\nu}}dx^{\nu})$$
$$\delta S = \int_a^b \delta x^{\mu}((mcdv_{\mu}) - \frac{e}{c}(dx^{\nu}\frac{dA_{\nu}}{dx^{\mu}} - \frac{dA_{\mu}}{dx^{\nu}}dx^{\nu}))$$
and since $\delta S = 0$
$$\int_a^b \delta x^{\mu}((mcdv_{\mu}) - \frac{e}{c}(dx^{\nu}\frac{dA_{\nu}}{dx^{\mu}} - \frac{dA_{\mu}}{dx^{\nu}}dx^{\nu})) = 0$$
$$mcdv_{\mu} - \frac{e}{c}(dx^{\nu}\frac{dA_{\nu}}{dx^{\mu}} - \frac{dA_{\mu}}{dx^{\nu}}dx^{\nu})= 0$$
$$mcdv_{\mu} = \frac{e}{c}(dx^{\nu}\frac{dA_{\nu}}{dx^{\mu}} - \frac{dA_{\mu}}{dx^{\nu}}dx^{\nu})$$
$$mcdv_{\mu} = \frac{e}{c}(\frac{dA_{\nu}}{dx^{\mu}} - \frac{dA_{\mu}}{dx^{\nu}})dx^{\nu}$$
dividing the whole equation by $ds$ and remembering that $\frac{dx_{\mu}}{ds} = v_{\mu}$, we can rewrite the above equation as
$$mc\frac{dv_{\mu}}{ds} = \frac{e}{c}(\frac{dA_{\nu}}{dx^{\mu}} - \frac{dA_{\mu}}{dx^{\nu}})\frac{dx^{\nu}}{ds}$$
$$mc\frac{dv_{\mu}}{ds} = \frac{e}{c}(\frac{dA_{\nu}}{dx^{\mu}} - \frac{dA_{\mu}}{dx^{\nu}})v^{\nu}$$
we now define $$F_{\mu \nu} = \frac{dA_{\nu}}{dx^{\mu}} - \frac{dA_{\mu}}{dx^{\nu}}$$
$$mc\frac{dv_{\mu}}{ds} = F_{\mu \nu}v^{\nu}$$
We can understand from the construction of $F_{\mu \nu}$ that it is an antisymmetric matrix i.e $F_{\nu \mu} = -F_{\mu \nu}$

$$S = \int_a^b(-mcds - \frac{e}{c}A_{\mu} dx^{\mu})$$
specifically, let’s look at $A_{\mu}dx^{\mu}$. we know that $A_{\mu} = (\phi /c,\vec{A})$ and that $dx_{\mu} = (cdt,-d\vec{r})$
therefore $A_{\mu}dx^{\mu} = \phi dt - \vec{A}\cdot d\vec{r}$
$$S = \int_a^b(-mcds - \frac{e}{c}(\phi dt - \vec{A}\cdot d\vec{r}))$$
$$S = \int_a^b(-mcds - \frac{e}{c}\phi dt + \frac{e}{c}\vec{A}\cdot d\vec{r}))$$
we know that $ds = cd\tau$ and that $d\tau = \frac{dt}{\gamma}$ substituting these changes changes the above equation to
$$S = \int_a^b(-mc^2\frac{dt}{\gamma} - \frac{e}{c}\phi dt + \frac{e}{c}\vec{A}\cdot d\vec{r}))$$
$$S = \int_a^b(-mc^2\frac{dt}{\gamma} - \frac{e}{c}\phi dt + \frac{e}{c}\vec{A}\cdot \frac{d\vec{r}}{dt}dt))$$
$$S = \int_a^b dt(-mc^2\frac{1}{\gamma} - \frac{e}{c}\phi + \frac{e}{c}\vec{A}\cdot \frac{d\vec{r}}{dt}))$$
$$S = \int_a^b dt\mathcal{L}$$
where $$\mathcal{L} = -mc^2\frac{1}{\gamma} - \frac{e}{c}\phi + \frac{e}{c}\vec{A} \cdot \frac{d\vec{r}}{dt}$$
$$\mathcal{L} = -mc^2\frac{1}{\gamma} - \frac{e}{c}\phi + \frac{e}{c}\vec{A}\cdot\vec{V}$$
we can get the equations of motion from $$\frac{\partial\mathcal{L}}{\partial r} -\frac{d}{dt} \frac{\partial \mathcal{L}}{\partial \dot{r}} = 0$$
$$\frac{\partial\mathcal{L}}{\partial r} = -\frac{e}{c}\nabla\phi + \frac{e}{c}\vec{\nabla}(\vec{A}\cdot\vec{V})$$
$$\vec{\nabla}(\vec{A}\cdot\vec{V}) = \vec{V}\times(\vec{\nabla}\times\vec{A}) - \vec{A}(\vec{\nabla}\cdot\vec{V})$$
since $\vec{A}(\vec{\nabla}\cdot\vec{V}) = 0$, we can rewrite the earlier equation as
$$\frac{\partial\mathcal{L}}{\partial r} = -\frac{e}{c}\nabla\phi + \frac{e}{c} \vec{V}\times(\vec{\nabla}\times\vec{A})$$
similarly $$\frac{\partial\mathcal{L}}{\partial\dot{r}} = \frac{\partial}{\partial\dot{r}}(-mc^2\frac{1}{\gamma} - \frac{e}{c}\vec{A}\cdot\vec{V})$$
$$\frac{\partial\mathcal{L}}{\partial\dot{r}} = \frac{\partial}{\partial\dot{r}}(-mc^2\sqrt{1-\frac{v^2}{c^2}} - \frac{e}{c}\vec{A}\cdot\vec{V})$$
$$\frac{\partial\mathcal{L}}{\partial\dot{r}} = \gamma m\vec{V} + \frac{e}{c}\vec{A}$$
going back to $$\frac{\partial\mathcal{L}}{\partial r} -\frac{d}{dt} \frac{\partial \mathcal{L}}{\partial \dot{r}} = 0$$
$$-\frac{e}{c}\nabla\phi + \frac{e}{c} \vec{V}\times(\vec{\nabla}\times\vec{A}) - \frac{d}{dt}(\gamma m\vec{V} + \frac{e}{c}\vec{A}) = 0$$
$$-\frac{e}{c}\nabla\phi + \frac{e}{c} \vec{V}\times(\vec{\nabla}\times\vec{A}) - \frac{d\vec{p}}{dt} + \frac{e}{c}\frac{d\vec{A}}{dt} = 0$$
where $\vec{p}=\gamma m\vec{V}$
$$\frac{d\vec{p}}{dt} = -\frac{e}{c}\nabla\phi + \frac{e}{c}\frac{d\vec{A}}{dt}+ \frac{e}{c} \vec{V}\times(\vec{\nabla}\times\vec{A})$$
$$\frac{d\vec{p}}{dt} = -e(\frac{1}{c}\nabla\phi + \frac{1}{c}\frac{d\vec{A}}{dt})+ \frac{e}{c} \vec{V}\times(\vec{\nabla}\times\vec{A})$$
we’re defining $$\vec{E} = -\frac{1}{c}\nabla\phi - \frac{1}{c}\frac{d\vec{A}}{dt}$$ and $$\vec{B} = \vec{\nabla}\times\vec{A}$$
$$\frac{d\vec{p}}{dt} = e\vec{E}+ \frac{e}{c} \vec{V}\times\vec{B}$$
if we assume $A_{\mu} = (\phi,\vec{A})$, $$\vec{E} = -\nabla\phi - \frac{1}{c}\frac{d\vec{A}}{dt}$$