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?


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

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

  • $\begingroup$ 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

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