# Extracting Maxwell's equations

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## Homework Statement

Show one can obtain Maxwell's equations from $$\partial^{\mu} F_{\mu \nu} = 0\,\,\,; \,\, \partial_{\mu}F_{\nu \rho} + \partial_{\nu} F_{\rho \mu} + \partial_{\rho}F_{\mu \nu} = 0,$$ where $F_{ij} = \epsilon_{ijk}B_k$ and $F_{i0} = E_i$ with $F_{\mu \nu} = - F_{\nu \mu}$.
From these two equations, show that one also has $\partial^{\mu}T_{\mu \nu} = 0$, where $T_{\mu \nu} = F_{\mu \rho}F_{\nu \sigma} \eta^{\rho \sigma} - \frac{1}{4} \eta_{\mu \nu} F_{\alpha \beta}F^{\alpha \beta}$

## Homework Equations

All given in problem statement

## The Attempt at a Solution

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I've done the first part of the question.

My working for the second part is shown below. Insert $T_{\mu \nu}$ and contract with $\partial^{\mu}$. This gives

$$(\partial^{\mu}F_{\mu \rho}) F_{\nu \sigma} \eta^{\rho \sigma} + F_{\mu \rho}(\partial^{\mu} F_{\nu \sigma}) \eta^{\rho \sigma} - \frac{1}{4}\eta_{\mu \nu}(\partial^{\mu}F_{\alpha \beta})F^{\alpha \beta} - \frac{1}{4} \eta_{\mu \nu} F_{\alpha \beta}(\partial^{\mu}F^{\alpha \beta})$$
The first term is zero by Maxwell eqn. In the last term, lower the index on the $\partial^{\mu}$ term and then use the fact that $\eta_{\mu \nu} \eta^{\mu \gamma} = \delta^{\gamma}_{\nu}$ to reduce the last term to $\frac{1}{4} F_{\alpha \beta} \partial_{\nu} F^{\alpha \beta}$. The same analysis can be applied on third term. Then use the fact that $F^{\alpha \beta} = F_{\rho \sigma} \eta^{\alpha \rho} \eta^{\beta \sigma}$. The third and fourth terms are then identical and the whole expression reduces to $$F_{\mu \rho} \eta^{\mu \alpha} \partial_{\alpha} F_{\nu \sigma} \eta^{\rho \sigma} - \frac{1}{2} (\partial_{\nu}F_{\alpha \beta})F^{\alpha \beta} = F^{\alpha \sigma} \partial_{\alpha} F_{\nu \sigma} - \frac{1}{2} (\partial_{\nu}F_{\alpha \beta})F^{\alpha \beta}$$. I'm unsure of a next step. Possibly rearrange $\partial_{\alpha}F_{\nu \sigma}$ using the second Maxwell eqn above but not sure if this would help.

Many thanks.

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TSny
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$$F^{\alpha \sigma} \partial_{\alpha} F_{\nu \sigma} - \frac{1}{2} (\partial_{\nu}F_{\alpha \beta})F^{\alpha \beta}$$. I'm unsure of a next step. Possibly rearrange $\partial_{\alpha}F_{\nu \sigma}$ using the second Maxwell eqn above but not sure if this would help.
Yes, that would be a good idea. Also, note that in the first term, the index $\sigma$ is a dummy index. Think about renaming this index so that the term looks more similar to the last term.

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Yes, that would be a good idea. Also, note that in the first term, the index $\sigma$ is a dummy index. Think about renaming this index so that the term looks more similar to the last term.
Hi TSny, yes that would give $$\frac{1}{2 } ( \partial_{\nu} F_{\alpha \beta })F^{\alpha \beta} - (\partial_{\beta}F_{\alpha \nu})F^{\alpha \beta}$$ Still a bit unclear as to how to progress, thanks.

TSny
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Hi TSny, yes that would give $$\frac{1}{2 } ( \partial_{\nu} F_{\alpha \beta })F^{\alpha \beta} - (\partial_{\beta}F_{\alpha \nu})F^{\alpha \beta}$$ Still a bit unclear as to how to progress, thanks.
Not sure of all the things you did here. Especially the sign changes. Let's see. You had $$F^{\alpha \sigma} \partial_{\alpha} F_{\nu \sigma} - \frac{1}{2} (\partial_{\nu}F_{\alpha \beta})F^{\alpha \beta}$$
And, for now, you just want to rename the dummy index $\sigma$ in the first term to something more appropriate. If that's all you do, then what do you get?

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Not sure of all the things you did here.
I used the second Maxwell equation to rearrange the $\partial_{\alpha}F_{\nu \sigma}$ term and collected terms by setting $\sigma \rightarrow \beta$.
And, for now, you just want to rename the dummy index $\sigma$ in the first term to something more appropriate. If that's all you do, then what do you get?
This would give $$F^{\alpha \beta}\partial_{\alpha} F_{\nu \beta} - \frac{1}{2}(\partial_{\nu} F_{\alpha \beta})F^{\alpha \beta}$$ where I relabelled $\sigma$ to $\beta$.

TSny
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I used the second Maxwell equation to rearrange the $\partial_{\alpha}F_{\nu \sigma}$ term and collected terms by setting $\sigma \rightarrow \beta$.
OK. For me, it seems to work out quicker if I use the 2nd Maxwell equation on the term with the 1/2 factor.

This would give $$F^{\alpha \beta}\partial_{\alpha} F_{\nu \beta} - \frac{1}{2}(\partial_{\nu} F_{\alpha \beta})F^{\alpha \beta}$$ where I relabelled $\sigma$ to $\beta$.
Now try the 2nd ME on the term with the 1/2.

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Now try the 2nd ME on the term with the 1/2.
Ok, so reexpress the second term as $-\frac{1}{2} (-\partial_{\alpha}F_{\beta \nu} - \partial_{\beta} F_{\nu \alpha})$ and altogether this gives $$F^{\alpha \beta} \left\{ \frac{1}{2}\partial_{\alpha} F_{\nu \beta} + \frac{1}{2}\partial_{\beta}F_{\nu \alpha} \right\}.$$ Then interchange $\alpha$ and $\beta$ to give $$\frac{1}{2}F^{\beta \alpha} ( \partial_{\beta} F_{\nu \alpha} + \partial_{\alpha}F_{\nu \beta}) = - \frac{1}{2}F^{\alpha \beta} (\partial_{\beta}F_{\nu \alpha} + \partial_{\alpha}F_{\nu \beta})$$ So we have something like $X=-X$ and this conclude $X=0$ as required. Right?

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TSny
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Ok, so reexpress the second term as $-\frac{1}{2} (-\partial_{\alpha}F_{\beta \nu} - \partial_{\beta} F_{\nu \alpha})$ and altogether this gives $$F^{\alpha \beta} \left\{ \frac{1}{2}\partial_{\alpha} F_{\nu \beta} + \frac{1}{2}\partial_{\beta}F_{\nu \alpha} \right\}.$$ Then interchange $\alpha$ and $\beta$ to give $$\frac{1}{2}F^{\beta \alpha} ( \partial_{\beta} F_{\nu \alpha} + \partial_{\alpha}F_{\nu \beta}) = - \frac{1}{2}F^{\alpha \beta} (\partial_{\beta}F_{\nu \alpha} + \partial_{\alpha}F_{\nu \beta})$$ So we have something like $X=-X$ and this conclude $X=0$ as required. Right?
Yes, that will get it. Good.

Or, note that if you contract $F^{\alpha \beta}$ with any tensor $B_{\mu \nu \sigma}$, say, to get the expression $F^{\alpha \beta}B_{\alpha \beta \nu}$ , then due to the antisymmetry of $F^{\alpha \beta}$, you get the identity $$F^{\alpha \beta}B_{\alpha \beta \nu} = -F^{\beta \alpha}B_{\alpha \beta \nu} = -F^{\alpha \beta}B_{\beta \alpha \nu}$$ where the last equality is just renaming dummy indices.

Thus interchanging the contracted indices on B in the expression $F^{\alpha \beta}B_{\alpha \beta \nu}$ just changes the overall sign of the expression.

You have $$F^{\alpha \beta} \left\{ \frac{1}{2}\partial_{\alpha} F_{\nu \beta} + \frac{1}{2}\partial_{\beta}F_{\nu \alpha} \right\} = \frac{1}{2}F^{\alpha \beta}\partial_{\alpha} F_{\nu \beta} + \frac{1}{2}F^{\alpha \beta}\partial_{\beta}F_{\nu \alpha}$$ How does the second term compare to the first term?

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You have $$F^{\alpha \beta} \left\{ \frac{1}{2}\partial_{\alpha} F_{\nu \beta} + \frac{1}{2}\partial_{\beta}F_{\nu \alpha} \right\} = \frac{1}{2}F^{\alpha \beta}\partial_{\alpha} F_{\nu \beta} + \frac{1}{2}F^{\alpha \beta}\partial_{\beta}F_{\nu \alpha}$$ How does the second term compare to the first term?
The second term is minus the first term. $\frac{1}{2}F^{\alpha \beta}\partial_{\beta}F_{\nu \alpha} = \frac{1}{2}F^{\beta \alpha}\partial_{\alpha}F_{\nu \beta} = -\frac{1}{2}F^{\alpha \beta}\partial_{\alpha}F_{\nu \beta}$ and so the whole expression reduces to 0.

TSny
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Yes, that's right.