I am trying to calculate the stress–energy tensor of an electromagnetic wave in curved spacetime, characterized by the diagonal metric $$ g_{\mu\nu} = \begin{pmatrix} -g_{zz} & 0 & 0 & 0 \\ 0 & g_{xx} & 0 & 0 \\ 0 & 0 & g_{yy} & 0 \\ 0 & 0 & 0 & g_{zz} \\ \end{pmatrix} $$ with signature (−, +, +, +), where $g_{xx},g_{yy},g_{zz}$ are arbitrary functions of the coordinates $t,x,y,z$ and $g_{tt}=-g_{zz}$. Assuming that the electric field $\vec{E}$ is oriented along the $x$-axis and the magnetic field $\vec{B}$ along the $y$-axis, the electromagnetic tensor becomes: $$ F_{\mu\nu} = \big|\vec{B}\big| \begin{pmatrix} 0 & -1 & 0 & 0 \\ 1 & 0 & 0 & -1 \\ 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ \end{pmatrix} $$ According to Wikipedia and this question, the stress–energy tensor is then given by: $$ T_{\mu\nu} = -\frac{1}{\mu_0} \left( F_{\mu\alpha} g^{\alpha\beta} F_{\beta\nu} - \frac{1}{4} g_{\mu\nu} F_{\sigma\alpha} g^{\alpha\beta} F_{\beta\rho} g^{\rho\sigma} \right) $$ Calculating the first term yields: $$ F_{\mu\alpha} g^{\alpha\beta} F_{\beta\nu} = \big|\vec{B}\big|^2 \begin{pmatrix} 0 & -1 & 0 & 0 \\ 1 & 0 & 0 & -1 \\ 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ \end{pmatrix} \begin{pmatrix} -g^{zz} & 0 & 0 & 0 \\ 0 & g^{xx} & 0 & 0 \\ 0 & 0 & g^{yy} & 0 \\ 0 & 0 & 0 & g^{zz} \\ \end{pmatrix} \begin{pmatrix} 0 & -1 & 0 & 0 \\ 1 & 0 & 0 & -1 \\ 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ \end{pmatrix} $$ $$ F_{\mu\alpha} g^{\alpha\beta} F_{\beta\nu} = \frac{\big|\vec{B}\big|^2}{g_{xx}} \begin{pmatrix} -1 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & -1 \\ \end{pmatrix} $$ The term $F_{\sigma\alpha} g^{\alpha\beta} F_{\beta\rho} g^{\rho\sigma}$ is calculated by: $$ \big|\vec{B}\big|^2\text{tr}\Bigg[\begin{pmatrix} 0 & -1 & 0 & 0 \\ 1 & 0 & 0 & -1 \\ 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ \end{pmatrix} \begin{pmatrix} -g^{zz} & 0 & 0 & 0 \\ 0 & g^{xx} & 0 & 0 \\ 0 & 0 & g^{yy} & 0 \\ 0 & 0 & 0 & g^{zz} \\ \end{pmatrix} \begin{pmatrix} 0 & -1 & 0 & 0 \\ 1 & 0 & 0 & -1 \\ 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ \end{pmatrix} \begin{pmatrix} -g^{zz} & 0 & 0 & 0 \\ 0 & g^{xx} & 0 & 0 \\ 0 & 0 & g^{yy} & 0 \\ 0 & 0 & 0 & g^{zz} \\ \end{pmatrix}\Bigg] $$ $$ F_{\sigma\alpha} g^{\alpha\beta} F_{\beta\rho} g^{\rho\sigma} = \frac{\big|\vec{B}\big|^2}{g_{zz}g_{xx}}\text{tr} \begin{pmatrix} 1 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ -1 & 0 & 0 & -1 \\ \end{pmatrix} = 0 $$ Therefore the entire second term vanishes and the stress-energy tensor is only given by: $$ T_{\mu\nu} = -\frac{\big|\vec{B}\big|^2}{\mu_0g_{xx}}\begin{pmatrix} -1 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & -1 \\ \end{pmatrix} $$ However, this seems wrong to me, since the stress-energy tensor is only dependent on the $g_{xx}$ component of the metric and not at all on the $g_{yy}$ component. Does that mean that the metric only has an effect on the electric field (in $x$-direction) and not on the magnetic field (in $y$-direction)? Or are there errors in my calculations? Perhaps the electromagnetic tensor $F_{\mu\nu}$ takes a different form in curved spacetime?
1 Answer
When you assumed that $$ F_{\mu\nu} = \big|\vec{B}\big| \begin{pmatrix} 0 & -1 & 0 & 0 \\ 1 & 0 & 0 & -1 \\ 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ \end{pmatrix} $$ you implicitly assumed that $g_{xx} = g_{yy} = 1$. This is because the norm of a spacelike vector (such as $\vec{E}$ or $\vec{B}$), assuming a diagonal metric, is $$ |\vec{B}|^2 = (B^x)^2g_{xx} + (B^y)^2g_{yy} + (B^z)^2g_{zz} = (B_x)^2g^{xx} + (B_y)^2g^{yy} + (B_z)^2g^{zz} \\ \neq (B^x)^2 + (B^y)^2 + (B^z)^2. $$ The coordinate components of the tensor $F_{\mu \nu}$ are expressed in terms of the coordinate components $B_y$ and $E_x$; so if you want to express $F_{\mu \nu}$ in terms of the (spatially) invariant magnitude $|\vec{B}|$, you will have to include factors of the metric components as well.
