# How electromagnetic energy-momentum looks like for arbitrary 4-velocity vector?

I need to expresse the electromagnetic energy-momentum tensor in a vacuum $$T^\nu_{\ \ \ \mu} = F_{\mu\alpha}F^{\nu\alpha} - \frac14 F_{\alpha\beta}F^{\alpha\beta}\delta^\nu_{\ \ \mu}$$ in terms of the electric $$E_\mu$$ and magnetic $$B^\mu$$ 4-vector fields $$E_\mu:=F_{\mu\nu}u^\nu, \quad \text{and} \quad B^\mu:=F^{*\mu\nu}u_\nu,$$ where $$F^{*\mu\nu}=\frac12\varepsilon^{\mu\nu\sigma\rho}F_{\sigma\rho}$$, and $$u^\mu$$ is an arbitrary time-like 4-velocity.

The best I can get so far is $$\begin{multline} T^\nu_{\ \ \ \mu} = (E_\alpha E^\alpha +B_\alpha B^\alpha)\left(h^\nu_{\ \ \mu}-\frac12\delta^\nu_{\ \ \mu}\right)-E_\mu E^\nu - B_\mu B^\nu \\ -\varepsilon^{\nu\alpha\beta\gamma}u_\mu E_\alpha u_\beta B_\gamma -\varepsilon_{\mu\alpha\beta\gamma}u^\nu E^\alpha u^\beta B^\gamma ,\tag{1}\label{tag1} \end{multline}$$ where $$h^\nu_{\ \ \mu} = \delta^\nu_{\ \ \mu} + u^\nu u_\mu$$ is the projector.

I am not sure about this result even though it gives the conventional expression for $$T^\nu_{\ \ \ \mu}$$ if $$u^\mu=(1,0,0,0)$$. The derivation is rather lengthy. Usually, \eqref{tag1} is written for the velocity of a comoving observer but I need for a general 4-velocity. Have you ever tried to derive the energy-momentum for arbitrary 4-velocity or maybe know a suitable reference which may help to verify \eqref{tag1}? Please let me know.

To obtain \eqref{tag1}, I used the identities $$F_{\mu\nu} = u_\mu E_\nu - u_\nu E_\mu - \varepsilon_{\mu\nu\sigma\rho}u^\sigma B^\rho, \quad F^{*\mu\nu} = u^\mu B^\nu - u^\nu B^\mu + \varepsilon^{\mu\nu\sigma\rho}u_\sigma E_\rho,$$ which are valid for arbitrary time-like vector $$u^\mu$$ and skew-symmetric tensor $$F_{\mu\nu}$$.

• Why do you want to use those (noninvariant) four-vectors? They don't look very useful.
– Buzz
Commented Jun 21, 2019 at 16:29
• @Buzz I need them for the numerical simulation where sometimes it is convenient to use the velocity of a non-comoving observer. Commented Jun 21, 2019 at 20:36
• @Buzz those 4-vectors represent the electric and magnetic fields measured by an observer moving with 4-velocity $u^\mu$. They are defined covariantly. Commented Jun 22, 2019 at 1:16

Assuming (1) is correct (I have not checked yet), then you have solved your problem! You have expressed $$T^{\mu\nu}$$ in terms of $$E^\mu$$ and $$B^\mu$$ and $$u^\mu$$. This last dependence is inevitable, since $$E^\mu$$ and $$B^\mu$$ depend on it too.