Calculating the energy-momentum tensor for Maxwell Lagrangian

[teddd]

Hi guys, can you help me with this?

I'm supposed to calculate the energy momentum for the classic Maxwell Lagrangian, $\mathcal{L}=-\frac{1}{4}F^{\mu\nu}F_{\mu\nu}$ , where $F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$
with the well known formula:
$$T^{\sigma\rho}=\frac{\delta\mathcal{L}}{\delta \partial_{\sigma} A_\gamma}\partial^\rho A_\gamma-\mathcal{L}g^{\sigma\rho}$$

The point is that I'm not sure on how should I calculate the $$\frac{\delta\mathcal{L}}{\delta\partial_\sigma A_\gamma}\partial^\rho A_\gamma=-\frac{1}{4}\frac{\delta\left[(\partial^\mu A^\nu-\partial^\nu A^\mu)(\partial_\mu A_\nu -\partial_\nu A_\mu)\right]}{\delta\partial_\sigma A_\gamma}\partial^\rho A_\gamma$$ term; i cannot figure out on which component should i derive.Can you help me?

 

[Fredrik]

I don't think that should be a functional derivative, but to be honest, I have never bothered to study functional derivatives, so maybe I'm wrong. Anyway, if you can interpret it as a partial derivative, then it's easy. You just have to understand what function you're talking partial derivatives of. It's just a polynomial in several variables, and you know how to take partial derivatives of polynomial. The expression $$\frac{\partial\mathcal L}{\partial(\partial_\sigma A_\gamma)}$$ looks scary, but it's just the nth partial derivative (for some integer n) of the polynomial $\mathcal L$.

Consider this simpler example first:

For example, in the case of a single particle in 1 dimension, the Lagrangian $L:\mathbb R^3\rightarrow\mathbb R$ is defined by $$L(a,b,c)=\frac{1}{2}mb^2-V(a)$$ for all $a,b,c\in\mathbb R$. $L(q(t),\dot q(t),t)$ is just a number in the range of that function.

The denominator of $$\frac{\partial L}{\partial\dot q}$$ just tells you which partial derivative you're dealing with.
$$\frac{\partial L}{\partial\dot q}=\frac{\partial}{\partial b}\bigg|_{a=q(t),\ b=\dot q(t),\ c=t}L(a,b,c)=(mb)_{a=q(t),\ b=\dot q(t),\ c=t}=m\dot q(t)$$
But maybe you knew all this already. I just explained the part that confuses everyone at first. I realize now that this is a bit tricky even if you know this.

 

[dextercioby]

It is a partial derivative there, don't use the delta, that symbol should stand for the functional (Ga^teaux derivative). So the partial derivative obeys the product law. Start differentiating from the left.

 

[teddd]

Well, that is a functional derivative, being $\partial_\mu \phi$ a function!

 

[samalkhaiat]

Well, that is a functional derivative, being $\partial_\mu \phi$ a function!

Lagrangians are local functions not functionals. So, you are dealing with partial derivatives. Cast the em Lagrangian in the form
$$\mathcal{L}= - \frac{1}{2} \eta^{\rho \sigma} \eta^{\mu \nu}( \partial_{\rho}A_{\mu} \partial_{\nu}A_{\sigma} - \partial_{\rho}A_{\mu}\partial_{\sigma}A_{\nu}).$$
Now use
$$\frac{\partial}{\partial ( \partial_{\lambda}A_{\tau})} \left( \partial_{\alpha}A_{\beta}\right) = \delta^{\lambda}_{\alpha}\delta^{\tau}_{\beta}.$$

Sam

 

[Fredrik]

Well, that is a functional derivative, being $\partial_\mu \phi$ a function!

You need to look at what I said about how Lagrangians are polynomials. In the simple example I posted, $\dot q$ is a function but $\partial/\partial \dot q$ is just the partial derivative operator with respect to the second variable, and the function L that it's supposed to be applied to, is just a polynomial in 3 variables. You might as well write it as $D_2$, or $D_2|_{(q(t),\ \dot q(t),\ t)}$ to indicate at what point the partial derivative of the polynomial is to be evaluated.