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Lets start by considering the electromagnetic tensor $F^{\mu \nu}$: $$F^{\mu \nu}=\begin{bmatrix}0 & -E_x/c & -E_y/c & -E_z/c \\ E_x/c & 0 & -B_z & B_y \\ E_y/c & B_z & 0 & -B_x \\ E_z/c & -B_y & B_x & 0\end{bmatrix}$$ And now consider Maxwell's equation: $$\nabla \cdot \vec{E}=\frac{\rho}{\varepsilon _0}$$ $$\nabla \cdot \vec{B}=0$$ $$\nabla \times \vec{E}=-\frac{\partial \vec{B}}{\partial t}$$ $$\nabla \times \vec{B}=\mu _0 \vec{j}+\mu _0 \varepsilon _0 \frac{\partial \vec{E}}{\partial t}$$ The claim is that the first and the fourth equations are equivalent to the following tensor equation: $$\partial _{\mu}F^{\mu \nu}=\mu _0 j^{\nu}$$ (where: $j^{\nu}=(c\rho , \vec{j})$) and that the second and the third equations are also equivalent to: $$dF=0$$ where the $dF$ is simply a shortcut to write: $$\partial _{\lambda}F_{\mu \nu}+\partial _{\nu}F_{\lambda \nu}+\partial _\mu F_{\nu \lambda}$$ My objective is to prove, using tensor algebra, that this statement is indeed correct: Lets begin, the first part of the statement is easy; if we think about the first term: $$\partial _{\mu}F^{1}=\mu _0 j^{1}$$ we get: $$\frac{1}{c}\left(\frac{\partial E_x}{\partial x}+\frac{\partial E_y}{\partial y}+\frac{\partial E_z}{\partial z}\right)=\mu_0 c \rho \ \Rightarrow \ \nabla \cdot \vec{E}=\mu _0 c^2 \rho \ \Rightarrow \ \nabla \cdot \vec{E}=\frac{\rho}{\varepsilon _0}$$ Wonderful! By applying the same process to the other terms we can see that this tensor equation is also equal to the fourth Maxwell's equation.
Lets now think about the second part of the statement, the one about $dF$; this time we can see that the left hand side of the tensor equation is a tensor of rank three, we can think about it as a 3D matrix. Now: all the terms of the matrix, according to the equation, are equal to zero, so we get $4^3=64$ scalar equation that together should be equivalent to the remaining two Maxwell's equations. However this seems to me like a gargantuan amount of algebra.

Question: Is there a better, or faster, way to prove the correctness of the statement in question?

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  • $\begingroup$ Related : Derivation of Maxwell's equations from field tensor lagrangian. $\endgroup$ – Frobenius Jun 26 at 16:47
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    $\begingroup$ we get 4×3=12 scalar equation I don’t follow your logic. You probably meant $4^3=64$ equations. But $dF=0$ is actually only four equations. To get something nontrivial, you have to choose $\lambda$, $\mu$, and $\nu$ to be different indices, and there are only four ways to do this. Just write out these four equations and see that they are the other Maxwell equations. Convince yourself that if any if the indices are the same, $dF=0$ reduces to $0=0$. $\endgroup$ – G. Smith Jun 26 at 16:54
  • $\begingroup$ You are absolutely right, I have already edited my question. $\endgroup$ – Noumeno Jun 26 at 16:56
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Basically you are right: there is a lot of redundancy in the expression and what happens is that it leads to many copies of the homogeneous Maxwell equations. But the index notation can help you. If you keep the indices general as much as possible, then you don't need to obtain all the repeated versions one by one. You obtain all of them all at once.

I would recommend proceeding by first picking a specific value for two of the indices, say pick $\lambda = 0$, and $\mu = 1$, and see what you get. After that don't just move to another pair of values. Rather, sit and think. Argue that the cyclic symmetry among $\lambda, \mu, \nu$ immediately generates some further results without the need for calculation. Also, when you pick a spatial value for an index, say $\mu = 1$, then the fact that you are dealing with a tensor expression guarantees that the results for $2$ and $3$ will have an outcome such that the vector character of the fields is maintained.

I think you will learn more by this approach than by invoking fancy mathematical concepts which you have not learned yet.

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The fastest way is perhaps by using exterior algebra: start by writing Faraday's tensor as

$$F = E_x\ \text dt\wedge\text dx + E_y\ \text dt\wedge\text dy + E_z\ \text dt\wedge\text dz + B_x\ \text dy\wedge\text dz + B_y\ \text dz\wedge\text dx + B_z\ \text dx\wedge\text dy$$

and then take the exterior derivative $\text dF$ to obtain

$$\text dF = \frac{\partial E_x}{\partial y}\ \text dt\wedge\text dx\wedge dy + \cdots + \frac{\partial B_x}{\partial t}\ \text dt\wedge\text dy\wedge\text dz + \frac{\partial B_x}{\partial x}\ \text dx\wedge\text dy\wedge\text dz + \cdots$$

and by equating $\text dF$ to 0, you see that, term by term, that gives you

$$\nabla\times\mathbf E + \frac{\partial\mathbf B}{\partial t} = 0\qquad\wedge\qquad \nabla\cdot\mathbf B=0,$$

To get the other two equations, do the same with $\text d\star F+J=0$, where $\star$ denotes the Hodge-dual (that is, if you set $G=\star F$, then $G$ has the components of $\mathbf B$ where those of $\mathbf E$ were and the components of $-\mathbf E$ where those of $\mathbf B$ were, or opposite signs, can't remember).

You can abstract from this and represent a 2-form in terms of its polar and axial parts, say $F=(\mathbf E,\mathbf B)$. Then the exterior derivative gives you a density 3-form which is dual of the 1-vector $(\nabla\cdot\mathbf B,\nabla\times\mathbf E + \frac{\partial \mathbf B}{\partial t})$. With the swap $(\mathbf E,\mathbf B)\mapsto(\mathbf B,-\mathbf E)$ you then get the 3-form $\text dG$ with dual 1-vector $(-\nabla\cdot\mathbf E, \nabla\times\mathbf B - \frac{\partial\mathbf E}{\partial t})$.

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The question becomes more linear if we consider the genesis of the electromagnetic tensor. Homogeneous Maxwell's equations (written here in the Gauss system)

$$ \nabla\cdot\boldsymbol{B}=0 \qquad \qquad \nabla\times\boldsymbol{E} + \frac{1}{c}\frac{\partial\boldsymbol{B}}{\partial t}=0 $$

allow the definition of electromagnetic potentials (unless a gauge transformation)

$$ \boldsymbol{B} = \nabla \times \boldsymbol{A} \qquad \qquad \boldsymbol{E} = - \nabla \Phi - \frac{1}{c}\,\frac{\partial \boldsymbol{A}}{\partial t} $$

With these quantities the electromagnetic potential quadrivector can be formed

$$ A_\mu = (\Phi,\boldsymbol{-A}) $$

By definition the electromagnetic tensor is the curl of the electromagnetic potential $A_\mu$

$$ F_{\alpha\beta} = \partial_\alpha A_\beta - \partial_\beta A_\alpha $$

In terms of components it takes the well known matrix form

$$ F_{\alpha\beta} = \begin{pmatrix} 0 & +E_x & +E_y & +E_z \\ -E_x & 0 & -B_z & +B_y \\ -E_y & +B_z & 0 & -B_x \\ -E_z & -B_y & +B_x & 0 \end{pmatrix} $$

Therefore you can write that

$ \partial_\gamma F_{\alpha\beta} = \partial_\gamma\partial_\alpha A_\beta - \partial_\gamma\partial_\beta A_\alpha \\ \partial_\beta F_{\gamma\alpha} = \partial_\beta\partial_\gamma A_\alpha - \partial_\beta\partial_\alpha A_\gamma \\ \partial_\alpha F_{\beta\gamma} = \partial_\alpha\partial_\beta A_\gamma - \partial_\alpha\partial_\gamma A_\beta $

By adding the three relations and taking into account the invertibility of the order of the derivations, you obtain the required tensor relationship

$$ \partial_\alpha F_{\beta\gamma} + \partial_\beta F_{\gamma\alpha} + \partial_\gamma F_{\alpha\beta} = 0 $$

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  • $\begingroup$ Suppose we have A, and we can prove that A implies B. That is good to know. But the question here is asking, is the relationship one to one? Is B a sufficient as well as necessary condition? In other words, does B imply A? In order to answer such a question it is not sufficient merely to show that A implies B. $\endgroup$ – Andrew Steane Jul 2 at 13:32
  • $\begingroup$ You are right, but my aim was only to answer the question at the bottom "Is there a better, or faster, way to prove the correctness of the statement in question?" $\endgroup$ – Pangloss Jul 2 at 14:42

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