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 Jun 21 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. – peshenator Jun 21 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. – magma Jun 22 at 1:16

1 Answer

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.

These 4-vectors represent the electric and magnetic fields measured by an observer moving with 4-velocity u and they are standard objects in GR. They are spacelike.

If you already checked (1) in the comoving frame, then the equation is correct since it is covariant. Alternatively you can use a CAS