1
$\begingroup$

In the book "Tensors, Relativity and Cosmology" the author derived Maxwell's Equation in covariant form using the EM field tensor Lagrangian $L=-\frac{1}{4}F^{jl}F_{jl}$ (source=0). One of the steps was $$ \frac{\partial F^{jl} F_{jl}}{\partial F_{kn}} = F^{jl}(\delta^k_j\delta^n_l-\delta^k_l\delta^n_j) $$ where $F^{jl}=-F^{lj}$ is the EM field tensor. I tried to derive the expression above by converting $F_{ij}$ into $ \partial_i A_j-\partial_j A_i $ but I was not able to yield the RHS?

$\endgroup$
1
$\begingroup$

There is no need to write the field strength tensor $F_{ab}$ in terms of the vector field $A_{i}$. For a tensor without any symmetry constraints

$$ \frac{\partial F^{mn}}{\partial F^{ab}} = \delta^m_a\delta^n_b $$

and one can easily show that

$$ \frac{\partial (F^{mn}F_{mn})}{\partial F_{ab}} = \frac{\partial (F_{mn}F_{rs}g^{mr}g^{ns})}{F_{ab}} = 2F^{ab}$$

Using the antisymmetry of $F_{ab}$ one can see that is the same as what you have written:

$$\frac{\partial F^{jl} F_{jl}}{\partial F_{kn}} = F^{jl}(\delta^k_j\delta^n_l-\delta^k_l\delta^n_j) = 2F^{kn}$$.

However, if we enforce symmetry constraints on $F_{ab}$ from the beginning, then we cannot vary $F_{ab}$ and $F_{ba} = - F_{ab}$ independently. Therefore we must have

$$ \frac{\partial F^{mn}}{\partial F^{ab}} = \frac{1}{2}(\delta^m_a\delta^n_b - \delta^m_b\delta^n_a)$$

$\endgroup$
  • $\begingroup$ The first identity you wrote does not take antisymmetry into account: the rhs already incudes a further addend $-\delta^m_b \delta^n_a$ if you interpret $F$ as the strength field. $\endgroup$ – Valter Moretti May 25 '19 at 13:41
  • $\begingroup$ Thanks, I'll edit my answer to reflect this. $\endgroup$ – ultracoldgrl May 25 '19 at 14:06
  • $\begingroup$ Thanks, that was really helpful. Is there a book that explains all this? (e.g.Differentiating a tensor with respect to a tensor, which was never explained in all the tensor calculus books that I’ve read) I feel like I might have missed some important concepts ... $\endgroup$ – Chern-Simons May 26 '19 at 22:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.