Trace of the electromagnetic stress-energy tensor I was looking in wikipedia about the Electromagnetic stress–energy tensor and it's properties, and it is traceless. I was looking at the proof of it, but I do not understand some steps, mainly because I lack the knowledge about the rules in tensor algebra. If we have:
$$\Theta^{\mu \nu}= \frac 1 {\mu_0}(F^{\mu}_{\ \ \lambda}F^{\lambda \nu} +\frac 1 4 g^{\mu \nu}F_{\lambda \tau} F^{\lambda \tau})$$
Then:
$$\Theta^{\mu}_{\ \ \mu}=g_{\mu \nu}\Theta^{\mu \nu}= \frac 1 {\mu_0} (F^{\mu \lambda}F_{\lambda \mu}+\frac 14 g^{\mu}_{\ \ \mu} F_{\lambda \tau} F^{\lambda \tau})$$
So, how do we raise and lower the indicies in the first term at the right side?
I understand that $g_{\mu \nu} g^{\mu \nu}=g^{\mu}_{\ \ \mu}$
But I don't understand how $g_{\mu \nu}F^{\mu}_{\ \ \lambda}F^{\lambda \nu}=F^{\mu \lambda}F_{\lambda \mu}$
I also would like to know about the rules used for this final trasformation.
 A: So you have that the electromagnetic stress-energy tensor is given by$$\Theta^{\mu\nu}=\frac{1}{\mu_0}\left(F^\mu_\lambda F^{\lambda\nu}+\frac{1}{4}g^{\mu\nu}F_{\lambda\tau}F^{\lambda\tau}\right).$$To show that $\Theta^\mu_\mu=0$, I will do the computation step by step, explaining each part of the calculation and why/how it is done. So consider $\Theta^\mu_\mu$, which is given by$$\Theta^\mu_\mu=g_{\mu\nu}\Theta^{\mu\nu}=\frac{1}{\mu_0}\left(g_{\mu\nu}F^\mu_\lambda F^{\lambda\nu}+\frac{1}{4}g_{\mu\nu}g^{\mu\nu}F_{\lambda\tau}F^{\lambda\tau}\right).$$So firstly I applied the definition of $\Theta^\mu_\mu$ and then passed the metric into the brackets and distributed it over each term inside the brackets. Now I can make use of the fact that $g_{\mu\nu}F^\mu_\lambda=g_{\nu\mu}F^\mu_{\lambda}=F_{\nu\lambda}$ (also making use of the symmetric property of $g_{\mu\nu}$, i.e., $g_{\mu\nu}=g_{\nu\mu}$) and $g_{\mu\nu}g^{\mu\nu}=4$ to write$$\Theta^{\mu}_\mu=\frac{1}{\mu_0}\left(F_{\nu\lambda}F^{\lambda\nu}+F_{\lambda\tau}F^{\lambda\tau}\right).$$Now since all of my indices are contracted, I am free to re-label. So in the second term, I will make a relabelling of indices $(\tau\leftrightarrow\nu)$ to get$$\Theta^\mu_\mu=\frac{1}{\mu_0}\left(F_{\nu\lambda}F^{\lambda\nu}+F_{\lambda\nu}F^{\lambda\nu}\right).$$Finally, we use the anti-symmetric property of $F_{\mu\nu}$. This is explicitly given by$$F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu=-\left(\partial_\nu A_\mu-\partial_\mu A_\nu\right)=-F_{\nu\mu},$$with $A_\mu$ being the electromagnetic four potential. So we can write$$\Theta^\mu_\mu=\frac{1}{\mu_0}\left(-F_{\lambda\nu}F^{\lambda\nu}+F_{\lambda\nu}F^{\lambda\nu}\right)=0$$as required.
A: Firstly, the EM stress-energy is written with a minus sign between the two terms$^{1}$ as,
$$\Theta^{\mu \nu}= \frac 1 {\mu_0}(F^{\lambda \mu}  F^{\nu}{}_{\lambda}-\frac 1 4 g^{\mu \nu}F_{\lambda \tau} F^{\lambda \tau}) \ ,$$
which makes the cancelling of terms a little more self-evident.
Next, as you noted, $g^{\mu \nu} g_{\mu \nu} = \delta^{\mu}_{\mu} = n$ where here $n=4$. The main confusion probably stems around tensor algebra itself, which you can find much more information about on other questions here, or Wikipedea, or online notes/textbooks, but for this question what you need to know is that the metric tensor raises and lowers indices on tensors: e.g $g_{\mu \nu} A^{\mu} = A_{\nu}$. (Repeated indices are summer over, so $g_{\mu \nu} A^{\mu}$ and $g_{\lambda \nu} A^{\lambda}$ represent exactly the same thing).
So we have $g_{\mu \nu} F^{\mu}{}_{\lambda} = F_{\nu \lambda}$, we've lowered the first index. Using that and the information above, you can quickly deduce that $\Theta^{\mu}{}_{\mu} = 0$, because both the first and last term will be $F^{\mu \nu}F_{\mu \nu}$ (these are summed indices).
$^1$Using the antisymmetry of $F_{\mu \nu} = - F_{\nu \mu}$ should match any definition of $\Theta^{\mu \nu}$. I've just used the definition given on the Wikipedia, but a different signature gives the same results.
