Orodruin said:
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.
I already tried. Here is what I got:
<br />
\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}}<br />
Because the metric tensor is symmetric:
<br />
g_{pj} = \frac{1}{2} (a_{pj} + a_{jp})<br />
<br />
\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}<br />
Now trying to contract the above expression with <br />
F^{pq} F^{jl}<br />
<br />
\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}<br />
Interchanging the dummy indexes in the 3rd and 4th terms
p\Longleftrightarrow q and j\Longleftrightarrow l
<br />
= -\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)<br />
and because F^{ab} is an anti-symmetric tensor (i.e. F^{ab} = -F^{ba}):
<br />
\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}<br />
I don't see how this matches the origianl expression in my post.
