Deriving Maxwell's Equations from Electromagnetic Tensor

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?

• Should probably note I've just made $4\pi$ and all those $c$ constants equal to 1 just for the simplicity's sake. – VladeKR Dec 6 '14 at 20:57

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.

• I see. The title should've been the other way around. – VladeKR Dec 6 '14 at 22:15

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.

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