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My homework assignment is

Prove that $$\partial_\beta F_{\gamma \delta} + \partial_\gamma F_{\delta \beta} + \partial_\delta F_{\beta \gamma} = 0$$ with the electromagnetic tensor

$$F_{\alpha \beta} = \begin{pmatrix} 0 & -E_1 & - E_2 & -E_3 \\ E_1 & 0 & B_3 & -B_2 \\ E_2 & -B_3 & 0 & B_1 \\ E_3 & B_2 & -B_1 & 0 \end{pmatrix}$$

leads to the following Maxwell's equations $$\operatorname{div} \mathbf{B} = 0$$ $$\operatorname{rot} \mathbf{E} = - \frac{1}{c} \frac{\partial \mathbf{B}}{\partial t}$$

I know that the first equation is equivalent to $\partial_\beta \tilde{F}^{\beta \alpha} = 0$ with $\tilde{F}^{\mu\nu} := \frac{1}{2}\, \varepsilon^{\mu\nu\alpha\beta}\,F_{\alpha\beta}$ and from there it's easy to prove. But in the form $ \partial_\beta F_{\gamma \delta} + \partial_\gamma F_{\delta \beta} + \partial_\delta F_{\beta \gamma} = 0$ I cannot think of something else than to calculate all 64 equations, which would be rather lengthy.

Any advice? Is there an easy proof for that?

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closed as off-topic by ACuriousMind, Neuneck, Jim, Kyle Kanos, Brandon Enright Nov 10 '14 at 16:11

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    $\begingroup$ Set $(\beta,\gamma,\delta)=(0,i,j)$ and $(\beta,\gamma,\delta)=(i,j,k)$ where $i,j,k=1,2,3$. Further use the identifications, $F_{0i}=-E_i$ and $B_i = \tfrac12 \epsilon_{ijk} F_{jk}$. $\endgroup$ – suresh Nov 10 '14 at 10:54
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    $\begingroup$ a teacher we had used to say "this is just a brutal substitution into the formula" $\endgroup$ – Nikos M. Nov 10 '14 at 10:59
  • $\begingroup$ Why not use symmetry? The fact that you have rotated indices, just solve $\partial_\beta F_{\gamma\delta}$ once for all indices and then swap out terms as needed. $\endgroup$ – Kyle Kanos Nov 10 '14 at 14:35
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There are several way to prove this formula.

1)Set $\beta=0$. In this case it has following form: $$\partial_0 F_{\gamma \delta} + \partial_\gamma F_{\delta 0} + \partial_\delta F_{0 \gamma} = 0$$ And use that $F_{0 i}=-E_i$ we and set$\gamma=i$, $\delta=j$ (in another case it is trivial identity). After that times this equation on $\varepsilon_{ijk}$ you will get $$\operatorname{rot} \mathbf{E} = - \frac{1}{c} \frac{\partial \mathbf{B}}{\partial t}$$

If we set ($\beta,\gamma,\delta)=(i,j,k)$ and times on $\varepsilon_{ijk}$ we will get $$\operatorname{div} \mathbf{B} = 0$$ The identity with another index is trivial.But it is check not prove.

2)But I prefer another way. Let's start with equation $\tilde{F}^{\mu\nu} := \frac{1}{2}\, \varepsilon^{\mu\nu\alpha\beta}\,F_{\alpha\beta}$ We will use identity $$\varepsilon^{\mu\nu\alpha\beta}\varepsilon_{\mu\gamma\delta\sigma}=C(\delta^\nu_\gamma\delta^\alpha_\delta\delta^\beta_\sigma-\delta^\nu_\gamma\delta^\alpha_\sigma\delta^\beta_\delta-\delta^\nu_\delta\delta^\alpha_\gamma\delta^\beta_\sigma-\delta^\nu_\sigma\delta^\alpha_\delta\delta^\beta_\gamma+\delta^\nu_\sigma\delta^\alpha_\gamma\delta^\beta_\delta+\delta^\nu_\delta\delta^\alpha_\sigma\delta^\beta_\gamma)$$ The constant $C$ can be found using identity $\varepsilon^{\mu\nu\alpha\beta}\varepsilon_{\mu\nu\alpha\beta}=4!$. But it does not matter for this problem. If we times our identity $\partial_\beta \tilde{F}^{\beta \alpha} = 0$ on $\varepsilon^{\mu\nu\alpha\gamma} $ and using this identity and that $F_{\mu\nu}=-F_{\nu\mu}$ it is easy to obtain our identity.

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