Partial derivative with respect to metric tensor $\frac{\partial}{\partial{g_{kn}}}(g_{pj}g_{ql})$

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This is a consequence of the symmetry of the metric tensor (which is what I'm assuming $g_{pq}$ is).

Clearly, $$ \frac{\partial}{\partial g_{kn}} (g_{pj}g_{ql}) = \frac{\partial g_{pj}}{ \partial g_{kn}} g_{ql} + g_{pj} \frac{\partial g_{ql}}{\partial g_{kn}} $$

The partial derivative in this expression must give something symmetric. For a non-symmetric tensor $a$, all 9 components are independent and $\dfrac{\partial a_{pq}}{\partial a_{kn}}$ is non-zero only when $p=k$ and $q=n$, i.e. you have

$$ \dfrac{\partial a_{pq}}{\partial a_{kn}} = \delta_{p}^{k} \delta_{q}^{n} $$

For a symmetric tensor, $g_{ij}=g_{ji}$ so you can't do that. Instead, you must symmetrize the argument by rewriting it as

$$g_{pj} = \frac{1}{2} (a_{pj} + a_{jp}) $$

Now take partial derivatives using the formula derived above for the non-symmetric tensor. You'll get

$$\frac{\partial g_{pj}}{\partial g_{kn}} = \frac{1}{2} (\delta_{p}^{k} \delta_{j}^{n} + \delta_{j}^{k} \delta_{p}^{n}) \\ $$

You can proceed similarly for the other term $\dfrac{\partial g_{ql}}{\partial g_{kn}}$. Finally you have $$ \frac{\partial}{\partial g_{kn}} (g_{pj}g_{ql}) = \frac{1}{2} (\delta_{p}^{k} \delta_{j}^{n} + \delta_{j}^{k} \delta_{p}^{n}) g_{ql} + \frac{1}{2} (\delta_{q}^{k} \delta_{l}^{n} + \delta_{l}^{k} \delta_{q}^{n}) g_{pj} $$

Now consider the contraction of this expression with $F^{pq} F^{jl}$ and proceed.

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Updated on October 24, 2020

Comments

  • Nazaf
    Nazaf about 3 years

    $$-\frac{1}{4\mu_0}F^{pq}F^{jl} \frac{\partial}{\partial{g_{kn}}}(g_{pj}g_{ql})=+\frac{1}{4\mu_0} F^{pq} F^{lj} 2 \delta^k_p \delta^n_j g_{ql}$$

    I need to know how to derive $\frac{\partial}{\partial{g_{kn}}}(g_{pj}g_{ql})$. Can you explain how the final result on the right side was obtained?

    • Admin
      Admin about 9 years
      Are you sure about the sign?
    • Robert Lewis
      Robert Lewis about 9 years
      @John: I think the sign is OK since as I recall $F^{lj}$ is skew-symmetric.
    • Nazaf
      Nazaf about 9 years
      Yes, I think $F^{jl}$ is an anti-symmetric tensor, i.e. $F^{jl}=-F^{lj}$
  • Nazaf
    Nazaf about 9 years
    Thanks but I am unable to get to $2\delta^k_p \delta^n_j g_{ql}$ from $$ \frac{\partial}{\partial g_{kn}} (g_{pj}g_{ql}) = \frac{1}{2} (\delta_{p}^{k} \delta_{j}^{n} + \delta_{j}^{k} \delta_{p}^{n}) g_{ql} + \frac{1}{2} (\delta_{q}^{k} \delta_{l}^{n} + \delta_{l}^{k} \delta_{q}^{n}) g_{pj} $$
  • Nazaf
    Nazaf about 9 years
    This is the farthest I could get after the contraction: I had to post multiple comments since it's too long. $$ \begin{align} -\frac{1}{4\mu_0}F^{pq} F^{jl} \frac{\partial}{\partial{g_{kn}}}(g_{pj} g_{ql}) & = -\frac{1}{4\mu_0} \frac{1}{2} \left( F^{pq} F^{jl} \delta^k_p \delta^n_jg_{ql} + F^{pq} F^{jl} \delta^k_j \delta^n_p g_{ql} + F^{pq} F^{jl}\delta^k_q \delta^n_l g_{pj} + F^{pq} F^{jl} \delta^k_l\delta^n_q g_{pj} \right) \\ & = -\frac{1}{4\mu_0} \frac{1}{2} \left( F^{kq} F^{nl}g_{ql} + F^{nq} F^{kl}g_{ql} + F^{pk} F^{jn}g_{pj} + F^{pn} F^{jk}g_{pj} \right) \\ \end{align} $$
  • Nazaf
    Nazaf about 9 years
    Interchanging the dummy indexes in the 3rd and 4th terms $p\Longleftrightarrow q$ and $j\Longleftrightarrow l$: $$ = -\frac{1}{4\mu_0} \frac{1}{2} \left( F^{kq} F^{nl}g_{ql} + F^{nq} F^{kl}g_{ql} + F^{qk} F^{ln}g_{ql} + F^{qn} F^{lk}g_{ql} \right) $$ and because $F^{ab}$ is an anti-symmetric tensor (i.e. $F^{ab} = -F^{ba}$):
  • Nazaf
    Nazaf about 9 years
    $$\begin{align} & = -\frac{1}{4\mu_0} \frac{1}{2} \left( 2 F^{kq} F^{nl}g_{ql} + 2 F^{nq} F^{kl}g_{ql} \right) \\ & = -\frac{1}{4\mu_0} \left( F^{kq} F^{nl}g_{ql} + F^{nq} F^{kl}g_{ql} \right) \\ & = -\frac{1}{4\mu_0} \left( F^{kq} F^{n}_q + F^{nq} F^{k}_q \right) \\ \end{align} $$