# Tensor Derivatives (Derivation of the electromagnetic energy stress tensor)

I have the task to derive the electromagnetic energy stress tensor. I'm pretty new to the tensor index notation and I have a problem with an occuring derivative.

I have to verify that $\frac{\partial \mathcal{L}}{\partial (\partial_{\mu} A_{\lambda})}=-F^{\mu \lambda}$ with $\mathcal{L}=-\frac{1}{2}(\partial^{\mu}A^{\nu}\partial_{\mu}A_{\nu}-\partial^{\mu}A^{\nu}\partial_{\nu}A_{\mu})$

Can someone please show me step by step how to do this derivative? I understand the tensor notation itself but I have problems to understand such a derivative, especially when I have a term like $\frac{\partial(\partial^{\mu}A^{\nu})}{\partial(\partial_{\sigma} A_{\lambda})}$.

## 2 Answers

The thing is that when you are talking about the Lagrangian $\mathcal{L}$ it is to be considered as a function of the fields. So it is $\mathcal{L}[A_\mu,\partial_\nu A_\mu]$. One considers all components of $A$, that is $A_\mu$ for $\mu=0,1,2,3$ and all derivatives of all components of $A$, that is $\partial_\nu A_\mu$ for $\mu,\nu=0,1,2,3$ to be independent variables.

Imagine that you are dealing with a function of several variables where the variables are the components of $A$ and the derivatives.

So the derivative

$$\dfrac{\partial}{\partial(\partial_\nu A_\mu)}$$

Is a derivative with respect to the particular coordinate $\partial_\nu A_\mu$ for this $\mu,\nu$.

Remember that for independent coordinates $x^\mu$ the equation

$$\dfrac{\partial x^\mu}{\partial x^\nu}=\delta^\mu_\nu$$

holds. The same happens here. Since $A_\mu$ and $\partial_\alpha A_\beta$ are independent coordinates we have

$$\dfrac{\partial A_\mu}{\partial(\partial_\alpha A_\beta)}=\dfrac{\partial (\partial_\alpha A_\beta)}{\partial A_\mu}=0,\quad \dfrac{\partial A_\mu}{\partial A_\nu}=\delta_{\mu\nu},\quad \dfrac{\partial (\partial_\alpha A_\beta)}{\partial (\partial _\mu A_\nu)}=\delta_{\alpha\mu}\delta_{\beta\nu}.$$

Convince yourself of that, considering the equation above well known in $\mathbb{R}^n$. It is the same equation with just different names for the coordinates!

To deal with the ones with raised index just explicitly write the metric. So we know that $\partial^\mu A^\nu = g^{\mu\lambda}g^{\nu\sigma}\partial_\lambda A_\sigma$.

In particular we have

$$\dfrac{\partial( \partial^\mu A^\nu)}{\partial(\partial_\alpha A_\beta)}=\dfrac{\partial( g^{\mu\sigma}g^{\nu\rho} \partial_\sigma A_\rho)}{\partial(\partial_\alpha A_\beta)}=g^{\mu\sigma}g^{\nu\rho}\dfrac{\partial( \partial_\sigma A_\rho)}{\partial(\partial_\alpha A_\beta)}=g^{\mu\sigma}g^{\nu\rho} \delta_{\sigma\alpha}\delta_{\rho\beta}=g^{\mu\alpha}g^{\nu\beta}.$$

Then just differentiate $\mathcal{L}$ normaly as function of these coordinates. We have

$$\mathcal{L}=-\dfrac{1}{2}(\partial^\mu A^\nu\partial_\mu A_\nu-\partial^\mu A^\nu \partial_\nu A_\mu)$$

To not mix the indices from the contraction with the free index from the derivative you want to perform, use a different one. We will differentiate with respect to $\partial_\alpha A_\beta$

$$\dfrac{\partial \mathcal{L}}{\partial(\partial_\alpha A_\beta)}=-\dfrac{1}{2}\left(\dfrac{\partial (\partial^\mu A^\nu)}{\partial(\partial_\alpha A_\beta)}\partial_\mu A_\nu+\partial^\mu A^\nu \dfrac{\partial(\partial_\mu A_\nu)}{\partial(\partial_\alpha A_\beta)}-\dfrac{\partial(\partial^\mu A^\nu)}{\partial(\partial_\alpha A_\beta)}\partial_\nu A_\mu - \partial^\mu A^\nu \dfrac{\partial(\partial_\nu A_\mu)}{\partial(\partial_\alpha A_\beta)}\right)$$

Now using what is above the result follows.

• Thanks for the detailed answer. If I do the derivative with respect to $\alpha$ and $\beta$ do I just switch them back to $\mu$ and $\lambda$ in the end? – S.P. Nov 9 '17 at 17:08

Once you see how to differentiate one term, you should be set. If you have something like

$$\frac{\partial}{\partial(\partial_a A_b)} \partial^c A^d$$ then you should first recognize that $$\partial^c A^d = g^{ce}g^{df} \partial_e A_f$$ From there, note that we're treating each derivative of each component of $A$ as independent. The derivative of $\partial_e A_f$ with respect to $\partial_1 A_3$ is equal to $1$ if $e=1$ and $f=3$, and zero otherwise - in other words,

$$\frac{\partial}{\partial (\partial_a A_b)} \partial_e A_f = \delta^a_e \delta^b_f$$

And so putting it all together, $$\frac{\partial}{\partial(\partial_a A_b)} \partial^c A^d = g^{ca}g^{db}$$

• Thank you very much that's very helpful. I wrote it down and got the right answer. Needs some time to get a feeling for those notations I think. – S.P. Nov 9 '17 at 14:51