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Four Tensor Derivatives -- EM Field Lagrangian Density

  1. Mar 24, 2015 #1
    1. The problem statement, all variables and given/known data
    Given the Lagrangian density

    [tex] \Lambda = -\frac{1}{c}j^lA_l - \frac{1}{16 \pi} F^{lm}F_{lm} [/tex]

    and the Euler-Lagrange equation for it

    [tex] \frac{\partial }{\partial x^k}\left ( \frac{\partial \Lambda}{\partial A_{i,k}} \right )- \frac{\partial \Lambda}{\partial A_{i}} =0 [/tex]

    derive the inhomogeneous, manifestly covariant, field equations.
    2. Relevant equations
    In class we were told

    [tex] \frac{\partial \Lambda}{\partial A_{i,k}} = -\frac{1}{4 \pi} F^{kl} [/tex]

    as a starting point, and I am trying to show this.

    3. The attempt at a solution
    I have that

    [tex] F_{lm}=\frac{\partial A_m}{\partial x^l}-\frac{\partial A_l}{\partial x^m} [/tex]

    and

    [tex] F^{lm}=\frac{\partial A^m}{\partial x_l}-\frac{\partial A^l}{\partial x_m} [/tex]

    I think I can see how inseting these into the above, possibly taking one of the field tensors to be constant, and taking derivatives (I think the first term does not depend on [itex] A_{i,k} [/itex]) leads to the correct form or we were shown in class. I'm not great at four-tensor manipulation or keeping my indices straight, however.

    I am trying to understand how to correctly take these derivatives.

    [tex] \frac{\partial }{\partial A_{i,k}} \left ( F^{lm}F_{lm} \right )=0 [/tex]
     
  2. jcsd
  3. Mar 25, 2015 #2

    Orodruin

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    Well, to start your equation does not make sense, you need to have the same indices on both sides of the equation.

    Did you try simply applying the product rule?
     
  4. Mar 25, 2015 #3
    I copied the equation incorrectly, it should be:
    [tex]
    \frac{\partial \Lambda}{\partial A_{l,m}} = -\frac{1}{4 \pi} F^{ml}
    [/tex]

    Using the product rule, I get:
    [tex] \left (\frac{\partial A^m}{\partial x_l}-\frac{\partial A^l}{\partial x_m}\right ) +
    \left( \frac{\partial A_m}{\partial x^l}-\frac{\partial A_l}{\partial x^m} \right)\frac{\partial }{\partial A_{m,l}}
    \left(\frac{\partial A^m}{\partial x_l}-\frac{\partial A^l}{\partial x_m} \right )
    [/tex]

    But am not sure what to do with the derivatives in the second term.
     
  5. Mar 25, 2015 #4

    Orodruin

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    I suggest taking this one step at a time. Without rewriting F in terms of A, what does the product rule give you? (Note that your expression contains too many m and l to make sense. Each should appear twice or not at all in your result. Your free indices are i and k.)
     
  6. Mar 25, 2015 #5
    So the first step is:
    [tex] F^{lm} \frac{\partial F_{lm}}{\partial A_{i,k}} + F_{lm}\frac{\partial F^{lm}}{\partial A_{i,k}} [/tex]
     
  7. Mar 25, 2015 #6

    Orodruin

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    Correct. Now, both terms are equal, so that gives a factor of two (just raise/lower the indices using the metric). This leaves you with the task of computing ##\partial F_{lm}/\partial A_{i,k}##. How do you express ##F_{lm}## in terms of ##A##? (Note that ##A_{i,k} \equiv \partial_k A_i##)
     
  8. Mar 25, 2015 #7
    Like this:

    [tex] F_{lm}=\frac{\partial A_m}{\partial x^l}-\frac{\partial A_l}{\partial x^m}=A_{m,l}-A_{l,m} [/tex]
     
  9. Mar 25, 2015 #8

    Orodruin

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    Yes, so what is the derivative of this with respect to ##A_{i,k}##?
     
  10. Mar 25, 2015 #9
    Is [itex] A_{l,m} [/itex] antisymmetric? Then I could switch the indices on the second term, add them for the extra factor of 2, and use

    [tex] \frac{\partial A_{m,l}}{\partial A_{i,k}}=\delta^{i}_{m}\delta^{l}_{k}=1 [/tex]
     
  11. Mar 25, 2015 #10

    Orodruin

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    No, it is not antisymmetric, but what makes you think you cannot use this relation separately on each term?
     
  12. Mar 25, 2015 #11
    I can use it on each term. I can only see that leaving me with zero instead of adding for the necessary factor of two.
     
  13. Mar 25, 2015 #12

    Orodruin

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    How would it leave you with zero? The terms are also not equal. What do you get if you do the computation?
     
  14. Mar 26, 2015 #13
    It must be simple, but I'm not making the connection. I think the derivative reduces to:

    [tex]
    \delta^{i}_{m}\delta^{l}_{k}-\delta^{m}_{l}\delta^{i}_{k} [/tex]

    Which can only be 0 or 1.
     
  15. Mar 26, 2015 #14

    Orodruin

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    The second term is incorrect, check the indices. The same indices must be up/down in both terms!

    What happens when you contract this expression with the F that was outside?
     
  16. Mar 26, 2015 #15
    I see. The resulting difference ends up as [itex] -4F^{ik} [/itex] after swapping the indices on the (antisym.) field tensor.
     
  17. Mar 26, 2015 #16

    Orodruin

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    Yes, this is correct. So what are the resulting equations of motion?
     
  18. Mar 26, 2015 #17
    They would then be [tex]\partial_kF^{kl} = -\frac{4 \pi}{c}j^l[/tex]
     
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