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Extracting Maxwell's equations

  1. Jan 17, 2015 #1

    CAF123

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    1. The problem statement, all variables and given/known data
    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}##

    2. Relevant equations

    All given in problem statement


    3. The attempt at a solution

    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.
     
    Last edited: Jan 17, 2015
  2. jcsd
  3. Jan 17, 2015 #2

    TSny

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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.
     
  4. Jan 17, 2015 #3

    CAF123

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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.
     
  5. Jan 17, 2015 #4

    TSny

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    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?
     
  6. Jan 18, 2015 #5

    CAF123

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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##.
    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##.
     
  7. Jan 18, 2015 #6

    TSny

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    OK. For me, it seems to work out quicker if I use the 2nd Maxwell equation on the term with the 1/2 factor.

    Now try the 2nd ME on the term with the 1/2.
     
  8. Jan 18, 2015 #7

    CAF123

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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?
     
    Last edited: Jan 18, 2015
  9. Jan 18, 2015 #8

    TSny

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    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?
     
  10. Jan 18, 2015 #9

    CAF123

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    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.
     
  11. Jan 18, 2015 #10

    TSny

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    Yes, that's right.
     
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