It has been a while since I touched tensor so some assistance would certainly be appreciated.
The maxwell stress tensor is given as: $T_{ij}=\epsilon _{0}\left ( E_{i}E_{j}\frac{1}{2}\delta _{ij}E^{2} \right )+\frac{1}{\mu_{0}}\left ( B_{i}B_{j}-\frac{1}{2}\delta _{ij}B^{2} \right ) $
For i=j=x,
$T_{xx}=\frac{1}{2}\epsilon _{0}\left ( E_{x}^{2}+E_{y}^{2}+E_{z}^{2}+ \right )+\frac{1}{2\mu_{0}}\left ( B_{x}^{2}+B_{y}^{2}+B_{z}^{2} \right )$
How did this came to be? When I did direct 'substitution' of i=j=x, I was not able to arrive at the final expression.
Any help is appreciated.
Thanks in advance.
Recall that the Kronecker Delta $\delta_{ij}$ is given by
$$\delta_{ij}=\begin{cases}1&,i=j\\\\0&,i\ne j\end{cases}$$
Therefore, we have
$$\begin{align} \sum_{i=1}^3T_{ii}&=\sum_{i=1}^3\left(\epsilon _{0}\left ( E_{i}E_{i}-\frac{1}{2}\delta _{ii}E^{2} \right )+\frac{1}{\mu_{0}}\left ( B_{i}B_{i}-\frac{1}{2}\delta _{ii}B^{2} \right )\right)\\\\ &=\frac{\epsilon_0}{2}(E_1^2+E_2^2+E_3^2)+\frac{1}{2\mu_0}\left(B_1^2+B_2^2+B_3^2\right) \end{align}$$