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Partial derivative with respect to metric tensor

  1. Oct 26, 2014 #1
    [itex]\mathcal{L}_M(g_{kn}) = -\frac{1}{4\mu{0}}g_{kj} g_{nl} F^{kn} F^{jl} \\

    \frac{\partial{\mathcal{L}_M}}{\partial{g_{kn}}}=-\frac{1}{4\mu_0}F^{pq}F^{jl} \frac{\partial}{\partial{g_{kn}}}(g_{pj}g_{ql})=+\frac{1}{4\mu_0} F^{pq} F^{lj} 2 \delta^k_p \delta^n_j g_{ql}[/itex]

    I need to know how [itex]\frac{\partial}{\partial{g_{kn}}}(g_{pj}g_{ql}) =
    2\delta^k_p \delta^n_j g_{ql}
    [/itex]. Can you explain how the final result on the right side was obtained?
     
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  3. Oct 27, 2014 #2

    Orodruin

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    I would suggest that you use the product rule of differentiation. You should be getting two terms and they can be written the same way by clever use of renaming summation indices and the symmetries of the problem.
     
  4. Oct 27, 2014 #3
    I already tried. Here is what I got:

    [itex]
    \frac{\partial}{\partial g_{kn}} (g_{pj}g_{ql}) = \frac{\partial g_{pj}}{ \partial g_{kn}} g_{ql} + g_{pj} \frac{\partial g_{ql}}{\partial g_{kn}}
    [/itex]
    Because the metric tensor is symmetric:

    [itex]
    g_{pj} = \frac{1}{2} (a_{pj} + a_{jp})
    [/itex]

    [itex]
    \frac{\partial}{\partial g_{kn}} (g_{pj}g_{ql}) = \frac{1}{2} (\delta_{p}^{k} \delta_{j}^{n} + \delta_{j}^{k} \delta_{p}^{n}) g_{ql} + \frac{1}{2} (\delta_{q}^{k} \delta_{l}^{n} + \delta_{l}^{k} \delta_{q}^{n}) g_{pj}
    [/itex]

    Now trying to contract the above expression with [itex]
    F^{pq} F^{jl}
    [/itex]

    [itex]
    \begin{align} -\frac{1}{4\mu_0}F^{pq} F^{jl} \frac{\partial}{\partial{g_{kn}}}(g_{pj} g_{ql}) & = -\frac{1}{4\mu_0} \frac{1}{2} \left( F^{pq} F^{jl} \delta^k_p \delta^n_jg_{ql} + F^{pq} F^{jl} \delta^k_j \delta^n_p g_{ql} + F^{pq} F^{jl}\delta^k_q \delta^n_l g_{pj} + F^{pq} F^{jl} \delta^k_l\delta^n_q g_{pj} \right) \\ & = -\frac{1}{4\mu_0} \frac{1}{2} \left( F^{kq} F^{nl}g_{ql} + F^{nq} F^{kl}g_{ql} + F^{pk} F^{jn}g_{pj} + F^{pn} F^{jk}g_{pj} \right) \\ \end{align}
    [/itex]

    Interchanging the dummy indexes in the 3rd and 4th terms
    [itex]p\Longleftrightarrow q[/itex] and [itex]j\Longleftrightarrow l[/itex]

    [itex]
    = -\frac{1}{4\mu_0} \frac{1}{2} \left( F^{kq} F^{nl}g_{ql} + F^{nq} F^{kl}g_{ql} + F^{qk} F^{ln}g_{ql} + F^{qn} F^{lk}g_{ql} \right)
    [/itex]

    and because [itex]F^{ab}[/itex] is an anti-symmetric tensor (i.e. [itex]F^{ab} = -F^{ba}[/itex]):

    [itex]
    \begin{align} & = -\frac{1}{4\mu_0} \frac{1}{2} \left( 2 F^{kq} F^{nl}g_{ql} + 2 F^{nq} F^{kl}g_{ql} \right) \\ & = -\frac{1}{4\mu_0} \left( F^{kq} F^{nl}g_{ql} + F^{nq} F^{kl}g_{ql} \right) \\ & = -\frac{1}{4\mu_0} \left( F^{kq} F^{n}_q + F^{nq} F^{k}_q \right) \\ \end{align}
    [/itex]

    I don't see how this matches the origianl expression in my post.
     
  5. Oct 27, 2014 #4

    Orodruin

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    Well, to start with, both of your terms are the same (use ##A_\mu B^\mu = A^\mu B_\mu## ... and keep track of which order your indices of F are in, it is anti-symmetric so it makes a lot of difference). Then use the ##\delta##s and the metric from your original expression to rewrite it.
     
  6. Oct 27, 2014 #5
    Which terms are the same?
    Can you explain more?
     
  7. Oct 27, 2014 #6

    Orodruin

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    ##F^{kq} F^{n}{}_q + F^{nq} F^{k}{}_q = 2 F^{kq} F^{n}{}_q ##

    This relation should be fairly straight-forward. Then use your ##\delta##s in the first expression.
     
  8. Oct 27, 2014 #7
    I see. Is the following correct?

    [itex]
    F^{kq} F^{n}_q + F^{nq} F^{k}_q = F^{kq} F^{n}_q + \delta^k_n \delta^n_k F^{nq} F^{k}_q \\
    = F^{kq} F^{n}_q + F^{nq} F^{n}_q
    [/itex]
     
    Last edited: Oct 27, 2014
  9. Oct 27, 2014 #8

    Orodruin

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    No, you have three of the same index in the second term in the second step. I suggest that you instead start from your second to last expression, perform the sum over ##\ell## in one of the terms and over ##q## in the other, then you rename the summation index in one of the terms to the same name as the summation index in the other. And again: Be careful with the positions of the indices in ##F## when they are up and down.
     
  10. Oct 27, 2014 #9
    I understand. How about the following?
    [itex]
    \begin{align} & -\frac{1}{4\mu_0} \left( F^{kq} F^{nl}g_{ql} + F^{nq} F^{kl}g_{ql} \right) \\ & = -\frac{1}{4\mu_0} \left( F^{kq} F^{ns}g_{qs} + F^{ns} F^{kl}g_{sl} \right) \\ & = -\frac{1}{4\mu_0} \left( F^{kl} F^{ns}g_{ls} + F^{ns} F^{kl}g_{sl} \right) \\ \end{align}
    [/itex]
     
  11. Oct 27, 2014 #10

    Orodruin

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    That looks fine. You can now use the symmetry of the metric to write both as one term.
     
  12. Oct 27, 2014 #11
    So it was just a matter of exchanging the indices q and l in either one of terms, right?
     
  13. Oct 28, 2014 #12

    Orodruin

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    Basically, or if you will, just changing which of the qs were up and down in the expression where you had used the metric to change indices.
     
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