Derivation of the Electromagnetic Stress-Energy Tensor in Flat Space-time

I am working on deriving the electromagnetic stress energy tensor using the electromagnetic tensor in the $$(-, +, +, +)$$ sign convention. However, I have hit a snag and cannot figure out where I have gone wrong.

$$F^{\mu \alpha}= \begin{bmatrix} 0 & \frac{E_{x}}{c} & \frac{E_{y}}{c} & \frac{E_{z}}{c} \\ -\frac{E_{x}}{c} & 0 & B_{z} & -B_{y} \\ -\frac{E_{y}}{c} & -B_{z} & 0 & B_{x} \\ -\frac{E_{z}}{c} & B_{y} & -B_{x} & 0 \\ \end{bmatrix}$$

$$F^{\mu}_{\alpha} = \begin{bmatrix} 0 & \frac{E_{x}}{c} & \frac{E_{y}}{c} & \frac{E_{z}}{c} \\ \frac{E_{x}}{c} & 0 & B_{z} & -B_{y} \\ \frac{E_{y}}{c} & -B_{z} & 0 & B_{x} \\ \frac{E_{z}}{c} & B_{y} & -B_{x} & 0 \\ \end{bmatrix}$$

$$T^{\mu\nu} = \frac{1}{\mu_0}(F^{\mu \alpha}F^{v}_{\alpha} - \frac{1}{4}\eta^{\mu\nu}F_{\alpha\beta}F^{\alpha \beta})$$

Doing matrix multiplication of the matrices $$F^{\mu \alpha}$$ and $$F^{\nu}_{\alpha}$$ from above gives

$$F^{\mu \alpha}F^{\nu}_{\alpha} = \begin{bmatrix} (\frac{E}{c})^{2} & -B_{z}\frac{E_{y}}{c} + B_{y}\frac{E_{z}}{c} & \frac{E_{x}}{c}B_{z} - \frac{E_{z}}{c}B_{x} & -\frac{E_{x}}{c}B_{y} + \frac{E_{y}}{c}B_{x} \\ B_{z}\frac{E_{y}}{c} - B_{y}\frac{E_{z}}{c} & -B_{z}^{2} - B_{y}^{2} - (\frac{E_{x}}{c})^{2} & -\frac{E_{x}}{c}\frac{E_{y}}{c} + B_{y}B_{x} & \frac{E_{x}}{c}\frac{E_{z}}{c} + B_{z}B_{x} \\ -B_{z}\frac{E_{x}}{c} + B_{x}\frac{E_{z}}{c} & -\frac{E_{y}}{c}\frac{E_{x}}{c} + B_{x}B_{y} & -(\frac{E_{y}}{c})^{2}-B_{z}^{2}-B_{x}^{2} & -\frac{E_{y}}{c}\frac{E_{z}}{c} + B_{z}B_{y} \\ B_{y}\frac{E_{x}}{c} - B_{x}\frac{E_{y}}{c} & -\frac{E_{z}}{c}\frac{E_{x}}{c} + B_{x}B_{z} & -\frac{E_{z}}{c}\frac{E_{y}}{c} + B_{y}B_{z} & -(\frac{E_{z}}{c})^{2}-B_{y}^{2}-B_{x}^{2} \\ \end{bmatrix}$$

Subtracting the $$\frac{1}{4}\eta^{\mu\nu}F_{\alpha\beta}F^{\alpha\beta}= \frac{1}{4}\eta^{\mu\nu}[2(B^{2} - (\frac{E}{c})^{2})]$$ and multiplying by $$\frac{1}{\mu_{0}}$$ gives

$$T^{\mu\nu}=\frac{1}{\mu_{0}} \begin{bmatrix} (\frac{E}{c})^{2} + \frac{1}{2}(B^{2} - (\frac{E}{c})^{2}) & -B_{z}\frac{E_{y}}{c} + B_{y}\frac{E_{z}}{c} & \frac{E_{x}}{c}B_{z} - \frac{E_{z}}{c}B_{x} & -\frac{E_{x}}{c}B_{y} + \frac{E_{y}}{c}B_{x} \\ B_{z}\frac{E_{y}}{c} - B_{y}\frac{E_{z}}{c} & -B_{z}^{2} - B_{y}^{2} - (\frac{E_{x}}{c})^{2} - \frac{1}{2}(B^{2} - (\frac{E}{c})^{2}) & -\frac{E_{x}}{c}\frac{E_{y}}{c} + B_{y}B_{x} & \frac{E_{x}}{c}\frac{E_{z}}{c} + B_{z}B_{x} \\ -B_{z}\frac{E_{x}}{c} + B_{x}\frac{E_{z}}{c} & -\frac{E_{y}}{c}\frac{E_{x}}{c} + B_{x}B_{y} & -(\frac{E_{y}}{c})^{2}-B_{z}^{2}-B_{x}^{2} - \frac{1}{2}(B^{2} - (\frac{E}{c})^{2}) & -\frac{E_{y}}{c}\frac{E_{z}}{c} + B_{z}B_{y} \\ B_{y}\frac{E_{x}}{c} - B_{x}\frac{E_{y}}{c} & -\frac{E_{z}}{c}\frac{E_{x}}{c} + B_{x}B_{z} & -\frac{E_{z}}{c}\frac{E_{y}}{c} + B_{y}B_{z} & -(\frac{E_{z}}{c})^{2}-B_{y}^{2}-B_{x}^{2} - \frac{1}{2}(B^{2} - (\frac{E}{c})^{2}) \\ \end{bmatrix}$$

However, the textbook definition of the electromagnetic stress energy tensor is: $$T^{\mu\nu} = \begin{bmatrix} \frac{1}{2}(\epsilon_{0} |E|^{2} + \frac{1}{\mu_{0}}|B|^{2}) & \frac{S_{x}}{c} & \frac{S_{y}}{c} & \frac{S_{z}}{c} \\ \frac{S_{x}}{c} & -\sigma_{xx} & -\sigma_{xy} & -\sigma_{xz} \\ \frac{S_{y}}{c} & -\sigma_{yx} & -\sigma_{yy} & -\sigma_{yz} \\ \frac{S_{z}}{c} & -\sigma_{zx} & -\sigma_{zy} & -\sigma_{zz} \\ \end{bmatrix}$$ with $$\vec{S} = \frac{1}{\mu_{0}}(\vec{E} \times \vec{B})$$ and $$\sigma_{ij} = \epsilon_{0} E_{i}E_{j} + \frac{1}{\mu_{0}}B_{i}B_{j} - \frac{1}{2}(\epsilon_{0} E^{2} + \frac{1}{\mu_{0}}B^{2})\delta_{ij}$$

So, I know my $$T^{01} = T^{10}$$, $$T^{02} = T^{20}$$, and $$T^{03} = T^{30}$$ but they do not. They are of opposite signs. What did I do incorrectly?

You could do yourself a big favor by working in units with $$c=1$$. It's really cumbersome writing all the factors of $$c$$. You can always reinsert them at the end if you want a result in SI. Or if you want to compare with Wikipedia's equation that's expressed in SI, just drop all the $$c$$'s from WP's version.

You detected the problem because the final result lacked the proper symmetry. So look at your calculation for the first place where that symmetry is lost. This happens at the very first step, where you calculate $$F^{\mu\alpha}F^\mu{}_\alpha$$. This should be the same as $$g_{\beta\alpha}F^{\mu\alpha}F^{\nu\beta}$$, which is manifestly symmetric in $$\mu$$ and $$\nu$$. For example, its 01 component is $$F^{02}F^{12}g_{22}+F^{03}F^{13}g_{33}=E_yB_z-E_zB_y$$, which is the same as its 10 component. I think the problem is that you say you found this result by matrix multiplication. The ordinary rules of matrix multiplication assume that both matrices are written in mixed upper-lower index form.

Also note that in general there is a distinction between $$T^\mu{}_\nu$$ and $$T_\nu{}^\mu$$, so you can't just write $$T^\mu_\nu$$ without ambiguity. If this was just an issue with mathjax, the syntax that works is this: T^\mu{}_\nu.

• How did you get $F^{02}F^{12}g_{22}+F^{03}F^{13}g_{33}$ for the 01 component from $g_{\beta\alpha}F^{\mu\alpha}F^{\nu\beta}$? I found a paper for an introduction to Einstein-Maxwell equations and it seemed to me he just did straight matrix multiplication for that step. So, I am unsure as how you arrived at the formula for that sum. – Jay May 11 at 20:10
• Could I use plain old matrix multiplication with the matrices of $g_{\beta\alpha}F^{\mu\alpha}F^{\nu\beta}$, since $g_{\beta\alpha}$, $F^{\mu\alpha}$, and $F^{\nu\beta}$ are either upper or lower form and not mixed? – Jay May 11 at 21:03
• How did you get F02F12g22+F03F13g33 for the 01 component from gβαFμαFνβ? By applying the Einstein summation convention and writing down the two nonvanishing terms. I found a paper for an introduction to Einstein-Maxwell equations and it seemed to me he just did straight matrix multiplication for that step. Matrix multiplication $AB=C$ is represented in index notation as $A^\mu{}_\nu B^\nu{}_\lambda=C^\mu{}_\lambda$. That is, all three matrices, A, B, and C, have to be in mixed upper-lower form. You can't apply the rules of matrix multiplication to tensors that aren't expressed this way. – Ben Crowell May 12 at 0:56
• Could I use plain old matrix multiplication with the matrices of gβαFμαFνβ, since gβα, Fμα, and Fνβ are either upper or lower form and not mixed? No, that doesn't work. – Ben Crowell May 12 at 0:57
• So, the Einstein summing convention like I did in my other posted answer? Sorry, I had to post as an answer not a comment, as it was too long. – Jay May 12 at 1:01

Ok...I have done a little more work and I think I have it. If $$F^{\mu \alpha}= \begin{bmatrix} 0 & \frac{E_{x}}{c} & \frac{E_{y}}{c} & \frac{E_{z}}{c} \\ -\frac{E_{x}}{c} & 0 & B_{z} & -B_{y} \\ -\frac{E_{y}}{c} & -B_{z} & 0 & B_{x} \\ -\frac{E_{z}}{c} & B_{y} & -B_{x} & 0 \\ \end{bmatrix}$$ and $$T^{mv} = \frac{1}{\mu_0}(F^{\mu \alpha}F^{\nu}{}_{\alpha} - \frac{1}{4}\eta^{\mu\nu}F_{\alpha\beta}F^{\alpha \beta})$$ then letting c = 1 $$F^{\mu \alpha}F^{v}{}_{\alpha} = \eta_{\beta\alpha}F^{\mu\alpha}F^{\nu\beta}$$ If we let $$\mu = 0,1,2,3$$, $$\nu = 0, 1, 2, 3$$, and $$\alpha=\beta$$ summing over repeated indexes gives

for $$\mu = 0$$ and $$\nu = 0$$

$$\eta_{\beta \alpha}F^{0\alpha}F^{0\beta} = \eta_{00}F^{00}F^{00}+\eta_{11}F^{01}F^{01}+ \eta_{22}F^{02}F^{02} + \eta_{33}F^{03}F^{03} = E_{x}^{2} + E_{y}^{2} + E_{z}^{2}$$

for $$\mu = 1$$ and $$\nu = 0$$

$$\eta_{\beta \alpha}F^{1\alpha}F^{0\beta} = \eta_{00}F^{10}F^{00}+\eta_{11}F^{11}F^{01}+ \eta_{22}F^{12}F^{02} + \eta_{33}F^{13}F^{03} = B_{z}E_{y} - B_{y}E_{z}$$

and so on...

Is this correct? When looking at the indexes for the metric will $$\alpha=\beta$$ always be true when working in general relativity?