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Since $\mathbf{E}\cdot\mathbf{B}$ is a Lorentz invariant of the electromagnetic fields it seems like an interesting thing to plug into a Lagrangian to see what happens. However, this ends up disappearing, and I'm told this should be obvious because it is a total derivative.

This however is not obvious to me. Is there an easy way to see that

$$ \frac{1}{2}\epsilon_{\alpha\beta\gamma\delta}F^{\alpha\beta} F^{\gamma\delta} = - \frac{4}{c}\left( \mathbf{B} \cdot \mathbf{E} \right) $$

is actually a total derivative?

I'd also appreciate if someone can show what it is the derivative of, so that I can work out the derivative to help it sink in.

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    $\begingroup$ If you know the differential forms, $\mathbf{E}\cdot\mathbf{B}$ can be written as $F\wedge F$. Then you can easily find that it can be written as $d(A\wedge F)$ where $F=dA$. This is exactly same explanation with Javier's one. $\endgroup$ – Frame Oct 3 '17 at 2:41
  • $\begingroup$ @Minkyu I'm not conversant with differential forms, so I did not see that as $F \wedge F$, but with $F=dA$ and the identity $dF = ddA = 0$, the result is pretty straight forward. I can see now why it would be obvious to those that are fluent in differential forms. Thanks! $\endgroup$ – PPenguin Oct 3 '17 at 3:02
  • $\begingroup$ I have to say that what I wrote above is not quite same as the Javier's answer after I realize the problem mentioned by Sean E. Lake. But my comment is still valid after fixing gauge mentioned by Sean E. Lake. $\endgroup$ – Frame Oct 3 '17 at 3:41
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    $\begingroup$ @Minkyu the best reason for not to put this term into a Lagrangian is not because it is a total derivative, but because it is a psuedoscalar, so it will end up that your action 1) if contains only this term is not invariant under symmetry transformation ($x\rightarrow -x$), but change only its sign 2) if contains other terms that are real scalar it has not clear transformation properties under symmetry an it can changes a lot $\endgroup$ – Annibale Oct 3 '17 at 6:19
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    $\begingroup$ @Minkyu Can you post your comment as an answer, so I can accept it? It really feels like the best answer. It is so short and sweet it really captures why this would be "obvious" to some people, and explains the "magic" behind what is going on in the manipulations of the other answers. $\endgroup$ – PPenguin Oct 3 '17 at 7:16
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First note that you can rewrite $\mathbf{E}\cdot\mathbf{B}$ as $$ \mathbf{E}\cdot\mathbf{B}\propto F\wedge F $$ using a field strength $2$-form $F$ where $\mathbf{E}$ and $\mathbf{B}$ are defined as \begin{align} F_{0i}&=E_i ,\\ F_{ij}&=\epsilon_{ijk}B_k. \end{align} More specifically, \begin{align} F\wedge F&=\frac{1}{4}F_{\mu\nu}F_{\rho\sigma}\, dx^\mu\wedge dx^\nu\wedge dx^\rho\wedge dx^\sigma \\ &=-\frac{1}{4}\epsilon^{\mu\nu\rho\sigma}F_{\mu\nu}F_{\rho\sigma}\,\text{vol.}\\ &\propto \epsilon^{ijk}F_{0i}F_{jk}=\mathbf{E}\cdot\mathbf{B}. \end{align} Then it is easy to show that $\mathbf{E}\cdot\mathbf{B}$ is a total derivative using $F=dA$, i.e., $$ \mathbf{E}\cdot\mathbf{B}\propto F\wedge F=d(A\wedge F). $$


As a side comment, $F\wedge F$ contains volume form but it is absent in $\mathbf{E}\cdot\mathbf{B}$. So the correct way to write is $$ \int d^4x \,\mathbf{E}\cdot\mathbf{B}\propto\int F\wedge F. $$

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For completeness, here's the tensor notation version.

First rewriting:

$$ (\mathbf{E}\cdot\mathbf{B})\ \propto \ \epsilon_{\alpha\beta\gamma\delta}F^{\alpha\beta} F^{\gamma\delta} = \epsilon_{\alpha\beta\gamma\delta} \left( \partial^\alpha A^\beta - \partial^\beta A^\alpha \right)F^{\gamma\delta} = 2 \ \epsilon_{\alpha\beta\gamma\delta} \left(\partial^\alpha A^\beta \right) F^{\gamma\delta}$$

where the last step uses relabeling and the anti-symmetry of $\epsilon_{\alpha\beta\gamma\delta}$.

Similarly

$$\epsilon_{\alpha\beta\gamma\delta} \left(\partial^\alpha A^\beta \right) F^{\gamma\delta} = 2 \ \epsilon_{\alpha\beta\gamma\delta} \left(\partial^\alpha A^\beta \right) \left(\partial^\gamma A^\delta \right).$$

Now moving one of the derivatives to the front

$$\epsilon_{\alpha\beta\gamma\delta} \left(\partial^\alpha A^\beta \right) \left(\partial^\gamma A^\delta \right) = \partial^\alpha \left( \epsilon_{\alpha\beta\gamma\delta} A^\beta \left(\partial^\gamma A^\delta \right)\right) - \epsilon_{\alpha\beta\gamma\delta} A^\beta \left(\partial^\alpha \partial^\gamma A^\delta \right)$$

and note that the last term is zero because the derivatives commute and so are symmetric in the those labels, while $\epsilon_{\alpha\beta\gamma\delta}$ is anti-symmetric.

All together this gives: $$ (\mathbf{E}\cdot\mathbf{B})\ \propto \ \partial^\alpha \left( \epsilon_{\alpha\beta\gamma\delta} A^\beta \left(\partial^\gamma A^\delta \right)\right) $$

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The easiest way to examine $\mathbf{E}\cdot\mathbf{B}$ is to look at the fields in terms of the vector potential in the Weyl gauge, where $A^0=0$. In that gauge, you get: \begin{align} \mathbf{E} &= - \frac{\partial \mathbf{A}}{\partial t},\ \mathrm{and} \\ \mathbf{B} &= \nabla \times \mathbf{A}. \end{align}

Thus you get: \begin{align} \mathbf{E}\cdot\mathbf{B} &= -\epsilon_{ijk}\frac{\partial A_i}{\partial t} \left(\frac{\partial A_k}{\partial x_j}\right) \end{align} Now let's examine the space-time integral of this quantity: \begin{align} \int \mathbf{E}\cdot\mathbf{B} \operatorname{d}x^4 & = -\int \epsilon_{ijk}\frac{\partial A_i}{\partial t} \left(\frac{\partial A_k}{\partial x_j}\right) \operatorname{d}x^4 \\ & = \int\left[- \frac{\partial }{\partial t} \left(\epsilon_{ijk} A_i \frac{\partial A_k}{\partial x_j}\right) + \epsilon_{ijk} A_i \frac{\partial^2 A_k}{\partial x_j \partial t} \right] \operatorname{d}x^4 && \mathrm{integrate\ by\ parts\ in\ }t \\ & = \int\left[- \frac{\partial }{\partial t} \left(\epsilon_{ijk} A_i \frac{\partial A_k}{\partial x_j}\right) + \frac{\partial }{\partial x_j} \left(\epsilon_{ijk} A_i \frac{\partial A_k}{\partial t}\right) - \epsilon_{ijk}\frac{\partial A_k}{\partial t} \left(\frac{\partial A_i}{\partial x_j}\right) \right] \operatorname{d}x^4 && \mathrm{integrate\ by\ parts\ in\ space} \\ & = -\int \epsilon_{kji}\frac{\partial A_i}{\partial t} \left(\frac{\partial A_k}{\partial x_j}\right) \operatorname{d}x^4 && \mathrm{drop\ surface\ terms} \\ &&& \mathrm{and\ relabel\ dummy\ variables} \\ & = \int \epsilon_{ijk}\frac{\partial A_i}{\partial t} \left(\frac{\partial A_k}{\partial x_j}\right) \operatorname{d}x^4 && \mathrm{exchange\ indices\ of\ }\epsilon \end{align} Notice that we just proved that the integral equals minus itself, and therefore must vanish if the surface terms vanish.

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  • $\begingroup$ +1 I'd never heard of the Weyl gauge before; now I know! $\endgroup$ – WetSavannaAnimal Oct 3 '17 at 4:02

protected by Qmechanic Oct 3 '17 at 5:43

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