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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.

Qmechanic
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PPenguin
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3 Answers3

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

PPenguin
  • 1,299
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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.

Sean E. Lake
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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 $$ \int d^4x \,\mathbf{E}\cdot\mathbf{B}\propto\int F\wedge F. $$

Frame
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