Show that the energy-momentum tensor has the following matrix structure

[Lambda96]

Homework Statement

Show that the energy-momentum tensor has the following matrix structure (see post)

Relevant Equations

none

Hi,

the task is as follows

I had no problems deriving the expressions $\omega$, $\frac{\textbf{S}}{c}$ and $\frac{\textbf{S}^T}{c}$, but now I have problems showing -{$\sigma_{ij}$}. I assumed the following for the calculation:

$$F^{\mu \sigma} F_{\ \sigma}^{\! \nu}=\sum\limits_{\sigma=0}^{3}F^{\mu \sigma} F_{\ \sigma}^{\! \nu}$$

$$F^{\sigma \rho}F_{\sigma \rho}=\sum\limits_{\sigma=0}^{3}\sum\limits_{\rho=0}^{3} F^{\sigma \rho}F_{\sigma \rho}$$

But if I now calculate $T^{11}$, I get $\frac{1}{4 \pi}(E^2_x+B^2_z+B^2_y-\frac{1}{4} (\textbf{E}^2+\textbf{B}^2))$ according to the definition of the task sheet for -{$\sigma_{ij}$}, i should get the following result $\frac{1}{4 \pi}(E^2_x+B^2_x-\frac{1}{4} (\textbf{E}^2+\textbf{B}^2))$. Is the definition wrong or have I done something wrong?

 

[Orodruin]

You have done something wrong. Without actually being provided with your derivation it is impossible to determine what.

 

[Lambda96]

Sorry, here is my calculation

$$\begin{align*}
T^{11}&=\frac{1}{4 \pi}(F^{10}F_{\ 0}^{\! 1}+F^{11}F_{\ 1}^{\! 1}+F^{12}F_{\ 2}^{\! 1}+F^{13}F_{\ 3}^{\! 1}+\frac{1}{4}\eta^{11}2(B^2-E^2))\\
T^{11}&=\frac{1}{4 \pi}(E_x E_x+B_z B_z+B_yB_y-\frac{1}{4} (\textbf{E}^2+\textbf{B}^2))\\
T^{11}&=\frac{1}{4 \pi}(E^2_x+B^2_z+B^2_y-\frac{1}{4} (\textbf{E}^2+\textbf{B}^2))
\end{align*}$$

With $F^{\mu \nu}=\left( \begin{array}{rrr} 0 & -E_x & -E_y & -E_z\\\ E_x & 0 & -B_z & B_y\\ E_y & B_z & 0 & -B_x\\ E_z & -B_y & B_x & 0\\\ \end{array}\right)$ and $\eta^{\mu \nu}=\left( \begin{array}{rrr} 1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1\\ \end{array}\right)$

 

[Orodruin]

It is unclear what happens with the invariant $B^2 - E^2$ in the second step.

 

[Lambda96]

I don't know why I wrote it this way, but it should actually look like this

$$\begin{align*}
T^{11}&=\frac{1}{4 \pi}(F^{10}F_{\ 0}^{\! 1}+F^{11}F_{\ 1}^{\! 1}+F^{12}F_{\ 2}^{\! 1}+F^{13}F_{\ 3}^{\! 1}+\frac{1}{4}\eta^{11}2(\textbf{B}^2-\textbf{E}^2))\\
T^{11}&=\frac{1}{4 \pi}(E_x E_x+B_z B_z+B_yB_y-\frac{1}{4} (\textbf{B}^2-\textbf{E}^2))\\
T^{11}&=\frac{1}{4 \pi}(E^2_x+B^2_z+B^2_y-\frac{1}{4} (\textbf{B}^2-\textbf{E}^2))
\end{align*}$$

 

[vela]

You dropped the factor of 2, and you seem to have made a sign error calculating $F^{1\mu}F^1{}_\mu$.

 

[Orodruin]

There are quite some sign issues, be very careful of those. Once you get the signs correctly (and the factor of two), you should note that you can rewrite $B_y^2 + B_z^2 = {\bf B}^2 - B_x^2$ and things will fall into place.

 

[Lambda96]

Thank you vela and Orodruin for your help , I was now able to derive the form as shown on the task sheet