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Given

$ F_{\mu\nu} = \partial_{\mu}A_{\nu} - \partial_{\nu}A_{\mu} $

It is obvious that the diagonals are zero, as

$ F_{ii} =\partial_{i}A_{i} - \partial_{i}A_{i} = 0 $

And, setting $0$ as time and $1,2,3$ as $x,y,z$ respectively, then

$F_{01} = F_{0x} =\frac{\partial}{\partial x^{0}}A_{1}-\frac{\partial}{\partial x^{1}}A_{0} = E_{x} $

continuing with $F_{03}$ we obtain $E_{x} ...E_{z}$ in the first row of $F$

$F_{12} = F_{xy} = \partial_{x}A_{y}-\partial_{y}A_{x} = B_{z}$

also continuing with $F_{23} = F_{yz} $ and $F_{31} = F_{zx}$ we obtain $B_{x}$ and $ >B_{y}$

Thus we get $\triangledown \times \vec{A} = \vec{B}$

So getting the signs straight, we finally have this.

$F_{\mu\nu} = \begin{pmatrix} 0 & E_{x}&E_{y} &E_{z} \\ -E_{x} & 0 & B_{z} &-B_{y} \\ -E_{y}& -B_{z} & 0 &B_{x} \\ -E_{z} & B_{y} & -B_{x} & 0 \end{pmatrix}$

I understand there should be "over c" for $E$ components.

Two Questions:

  1. Why does

$F_{01} = F_{0x} =\frac{\partial}{\partial x^{0}}A_{1}-\frac{\partial}{\partial x^{1}}A_{0}$ result in $ E_{x}$?

$A$ is a vector potential, and I've learned that $\vec{E}$ can be represented by a $-\triangledown \phi$ where $\phi$ is a scalar potential.

  1. I don't understand what I am supposed to do to with this matrix to get the two Maxwell's equations below.

$\triangledown \cdot \vec{E} = \rho$

and

$\triangledown \times \vec{B} = \vec{J} + \frac{\partial\vec{E}}{\partial t}$

Apparently, this can be solved by

$\sum_{\mu}^{3}\partial_{\mu}F^{\mu\nu} = J^{\nu}$

where,

$\nu = 0, \triangledown \cdot \vec{E} = \rho$

and

$\nu = i, \triangledown \times \vec{B} = \vec{J} + \frac{\partial\vec{E}}{\partial t}$

But where did $\sum_{\mu}^{3}\partial_{\mu}F^{\mu\nu} = J^{\nu}$ come from?

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  • $\begingroup$ Should probably note I've just made $4\pi$ and all those $c$ constants equal to 1 just for the simplicity's sake. $\endgroup$ – VladeKR Dec 6 '14 at 20:57
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The most general form of Maxwell's equations are (setting $\mu_0 = \varepsilon_0 = 1$) \begin{align} \vec{\nabla} \cdot \vec{B} &= 0 \\ \vec{\nabla} \times \vec{E} &= - \frac{ \partial \vec{B} }{ \partial t} \\ \vec{\nabla} \cdot \vec{E} &= \rho \\ \vec{\nabla} \times \vec{B} &= \vec{J} + \frac{ \partial \vec{E} }{ \partial t} \end{align} The first equation implies $$ \boxed{ \vec{B} = \vec{\nabla} \times \vec{A} } $$ Plugging this into the second equation, we find $$ \vec{\nabla} \times \left( \vec{E} + \frac{ \partial \vec{A} }{ \partial t} \right) = 0 $$ This equation then solves to $$ \boxed{ \vec{E} = - \vec{\nabla} \phi - \frac{ \partial \vec{A} }{ \partial t} } $$ Plugging the boxed equations into the last two Maxwell's equations, we get $$ \nabla^2 \phi + \frac{ \partial }{ \partial t} (\vec{\nabla} \cdot \vec{A} ) = - \rho ~~~~~~~~ ...... (1) $$ and $$ \frac{ \partial^2 \vec{A} }{ \partial t^2} + \vec{\nabla} \times ( \vec{\nabla} \times \vec{A} ) + \frac{\partial }{\partial t} (\vec{\nabla}\phi) = \vec{J} ~~~~~~~~ ...... (2) $$ Note that we have a total of 4 equations. In the covariant formalism, we have the define the 4-vectors $$ A^\mu : = ( \phi , \vec{A}),~~~ J^\mu : = (\rho, \vec{J}) $$ All you have to do is show that the equation $$ \partial_\mu F^{\mu\nu} = J^\nu $$ are in fact identical to (1) and (2). [The Minkowski sign convention is here assumed to be $(+,-,-,-)$.] For instance, if I choose $\nu = 0$ in the equation above, I find $$ J^0 = \partial_\mu F^{\mu0} = \partial_0 F^{00} + \partial_i F^{i0} = \partial_i ( \partial^i A^0 - \partial^0 A^i ) $$ Using the definitions above, we find $$ \rho = -\nabla^2 \phi - \frac{ \partial }{ \partial t} (\vec{\nabla} \cdot \vec{A} ) $$ which is precisely (1).

I will leave it to you to show that if I choose $\nu = i = 1,2,3$, then I reproduce the 3 equations (2).

Thus, the equation $\partial_\mu F^{\mu\nu} = J^\nu$ "comes from" the Maxwell equations themselves. They are simply a convenient rewriting of the Maxwell equations.

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  • $\begingroup$ I see. The title should've been the other way around. $\endgroup$ – VladeKR Dec 6 '14 at 22:15
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Maxwell equations are not "derived" from this tensor.

1) This is a definition of $E_x$ via $\partial A_x/\partial t -(\nabla\phi)_x$.

The other equations are derived from the corresponding action or postulated, if you like. In the former case you can use the equations for $A_{\mu}$ if you have them from the action.

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  • $\begingroup$ so E is not only the $\triangledown \phi$ but also the time-derivative of vector potential A? $\endgroup$ – VladeKR Dec 6 '14 at 21:53
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    $\begingroup$ Exactly. For a static case it is reduced to the gradient. $\endgroup$ – Vladimir Kalitvianski Dec 6 '14 at 21:54

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