Is the Maxwell Stress Tensor Coordinate Dependent?

Question

I am wondering if the Maxwell stress tensor, defined as $$T_{ij} = \epsilon_0 (E_iE_j-\frac{1}{2}\delta_{ij}E^2) + \frac{1}{\mu_0}(B_iB_j-\frac{1}{2}\delta_{ij}B^2) $$ is coordinate dependent. I would imagine that it works well for Cartesian coordinates $x,y,z$, but I am not as sure that I could use spherical coordinates $r,\theta,\phi$ without having to do some type of transformation. If one cannot use this formula for spherical coordinates, how would one go about transforming it to its spherical form? I am new to the field of tensors, so I apologize if the answer is obvious.

Answer 1

The tensor itself is coordinate independent, however its components with respect to a basis in the tensor product space are not. You can switch back and forth between tensor components of the same type (such as 2 times covariant $T_{\mu\nu}$) using the general transformation law for tensor components that you can find in any introductory differential geometry or general relativity text.

Answer 2

This form is also correct if you use spherical coordinates. If you have a vector $\vec{x}\in K^3$, where $K$ is a field, you can adress coordinates to it only after choosing a basis. This can be the typical $\vec{v}_1=\hat{e}_x,\vec{v}_2=\hat{e}_y,\vec{v}_3=\hat{e}_z$ but it can also be for spherical coordinates specifically $$\vec{v}_1=\hat{e}_r=\sin{\theta}\cos{\phi}\hat{e}_x+\sin{\theta}\sin{\phi}\hat{e}_y+\cos{\theta}\hat{e}_z,\\ \vec{v}_2=\hat{e}_{\theta}=\cos{\theta}\cos{\phi}\hat{e}_x+\cos{\theta}\sin{\phi}\hat{e}_y-\sin{\theta}\hat{e}_z,\\ \vec{v}_3=\hat{e}_{\phi}=-\sin{\phi}\hat{e}_x+\cos{\phi}\hat{e}_y$$ After choosing a basis $\vec{v}_1,\vec{v}_2,\vec{v}_3$, you then can adress coordinates $x_1,x_2,x_3$ and write $$\vec{x}=\left(\begin{array}{c} x_1 \\ x_2 \\ x_3 \end{array}\right)=x_1\cdot\vec{v}_1+x_2\cdot\vec{v}_2+x_3\cdot\vec{v}_3$$ It is extremely important to note that these coordinates doesn't have to be cartesian.

For some formulas it is good to be rather general. This is what we can do, when deriving this formula. We choose a "random" basis $\vec{v}_1,\vec{v}_2,\vec{v}_3$ and adress coordinates respectively to these vectors. If you now want to calculate these components from a given electromagnetic field, you have to make sure not to mess up with your basis, i.e given $\vec{E}(\vec{x})$ in another basis than $\vec{B}(\vec{x})$ or $\vec{x}$ and forgetting to transform. After that plug it in and get it into an expression as easy as possible.

If you are a little bit more interested in this tensor I would recommend taking a look into relativistic electrodynamics. With the electromagnetic field strength tensor (in Gaussian units) $$F=\left(\begin{matrix} 0 & E^1 & E^2 & E^3 \\ -E^1 & 0 & -B^3 & B^2 \\ -E^2 & B^3 & 0 & -B^1 \\ -E^3 & -B^2 & B^1 & 0 \end{matrix}\right) $$ you can write $T_{\mu\nu}=\frac{1}{4\pi}(F_{\mu\lambda}F^{\lambda}_{\nu}+\frac{1}{4}F_{\alpha\beta}F^{\alpha\beta}\eta_{\mu\nu})$ for the electromagnetic part of the energy-momentum tensor, where $\mu$ and $\nu$ go from $0$ to $3$. The components $T_{ij}$ with $i,j\in\{1,2,3\}$ then have the expression you wrote in your question. Interestingly, contracting $T$ with an $(1,0)$-tensorfield $Y$ with $Y_\nu=a_\nu+\omega_{\nu\kappa}x^\kappa$ with $\omega_{\mu\nu}=-\omega_{\nu\mu}$ to $T^{\mu\nu}Y_\nu$ and integrating leads to the conservation of the angular momentum $\vec{L}=\frac{1}{4\pi c}\int\mathrm{d^3x} [\vec{x}\times[\vec{E}(\vec{x},t)\times\vec{B}(\vec{x},t)]]$ of the electromagnetic field (in the case where you don't have a mass in your system. If you have, it's the sum of the angular momentum of the electromagnetic field and the "normal" relativistic angular momentum of the masses.)