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It is know that

$$\epsilon{^\mu} {^\nu} {^\rho} {^\sigma} \partial_{\nu} F_{\rho} {_\sigma} = 0$$

How can one deduce from this equation that

$$ \partial_{\mu}F_{\nu} {_\lambda} + \partial_{\lambda}F_{\mu} {_\nu} + \partial_{\nu}F_{\lambda} {_\mu} = 0$$

How many terms does this equation have? My professor told me that we choose only permutations: $(2, 3, 4)$, $(3, 4, 2)$, $(4, 2, 3)$. Why only these?

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  • $\begingroup$ Do you understand which components of the Levi-Civita tensor $\epsilon$ are nonzero? $\endgroup$ – G. Smith Jun 21 at 4:18
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    $\begingroup$ Your professor was probably giving just one example. (If he wasn’t, he told you something that’s not correct.) $\endgroup$ – G. Smith Jun 21 at 4:23
  • $\begingroup$ This is known as Bianchi identity and holds for all permutations. I am not sure what your professor means by choosing a particular set of permutations. $\endgroup$ – Abhay Hegde Jun 21 at 4:28
  • $\begingroup$ Related: 296164 $\endgroup$ – Abhay Hegde Jun 21 at 4:29
  • $\begingroup$ You should look at what happens if you don’t take $\mu$, $\nu$, and $\lambda$ to be all different in the second equation. $\endgroup$ – G. Smith Jun 21 at 4:30
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The antisymmetric tensor $\epsilon^{\mu\nu\rho\sigma}$ by definition only has a nonzero value when the indices are a permutation of 1,2,3,4. Only those terms will survive when you do the sum, all the others drop out. Even permutations will get a plus sign, odd ones will get a minus sign.

Note that in the equation $\epsilon^{\mu\nu\rho\sigma} \partial_\nu F_{\rho\sigma} = 0$, $\mu$ is a free index (not summed over), so this is actually four equations, one for each value of $\mu$. If you take the $\mu = 1$ part for instance, then the remaining indices can only vary over permutations of 2,3,4.

Now recall the field tensor $F_{\rho\sigma}$ is also antisymmetric. So, for any term in the sum, there is another one with the last two indices swapped—for example, $\partial_2 F_{34}$ and $\partial_2 F_{43}$. These terms will get opposite signs from the $\epsilon^{\mu\nu\rho\sigma}$, since the two permutations have opposite parity. But then they will get opposite signs again from the antisymmetry of $F_{\rho\sigma}$, so they will contribute the same: $\partial_2 F_{34} - \partial_2 F_{43} = 2\partial_2 F_{34}$. An overall factor of 2 doesn't matter here (it can be divided away), so the end result is we only need to consider even permutations in this equation; the odd permutations just give another copy of the same terms. That's probably what your professor was referring to with only choosing permutations 2,3,4, 3,4,2, and 4,2,3.

That was only one of the four equations; the other three work the same way and give you the other even permutations of three distinct indices.

OK, now let's look at to $\partial_{\mu}F_{\nu\lambda} + \partial_{\lambda}F_{\mu\nu} + \partial_{\nu}F_{\lambda\mu} = 0$. This looks like it has three free indices and thus would represent $4^3 = 64$ different equations! But look what happens if any two of the indices are equal. Let's take $\mu = \nu$ for example: $\partial_{\mu}F_{\mu\lambda} + \partial_{\lambda}F_{\mu\mu} + \partial_{\mu}F_{\lambda\mu} = 0$. Then $F_{\mu\mu} = 0$ , and $F_{\mu\lambda} = -F_{\lambda\mu}$, from the antisymmetry of $F$. So the equations with two (or more) equal indices reduce to just $0 = 0$. We're left with the ones whose indices are all distinct. Now notice that if you apply an even permutation to the indices, you get the same equation back again, as it generates the same three terms (in a different order). If you apply an odd permutation to the indices, you get the same equation but negative, again due to the antisymmetry of $F$. After all that, there are only 4 of the 64 equations left. They're the same 4 that are generated by $\epsilon^{\mu\nu\rho\sigma} \partial_\nu F_{\rho\sigma} = 0$.

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