Relativistic hydrodynamics Consider the hydrodynamic energy-momentum tensor in 3 + 1 dimensions in
at space with
the Minkowski metric $\eta_{\mu\nu}$. The coordinates are chosen as $u^\mu = (t; x; y; z)$. To first order in
derivatives, the energy-momentum tensor can be written as
$T^{\mu\nu}=(\epsilon+P)u^\mu u^\nu+P\eta^{\mu\nu}-\eta\Delta^{\mu\alpha} \Delta^{\nu\beta}(\partial_\alpha u_\beta + \partial_\beta u_\alpha -\frac{2}{3}\eta_{\alpha\beta}\partial_\lambda u^\lambda)$
The
velocity vector field $u^\mu$ is normalised to $u^\mu u_\mu=-1$ and pressure is a function of energy
density, $P(\epsilon)$ (this function is the equation of state). The coeffcient $\eta(\epsilon)$ (also a function of
energy density) is known as shear viscosity. The projector $\Delta^{\mu\nu}$ is defined as $\Delta^{\mu\nu}=u^\mu u^\nu + \eta^{\mu\nu}$.
(a.) Compute the trace $ T\equiv \eta_{\mu\nu}T^{\mu\nu}$
(b.) What would be the trace of the bulk viscosity term, which we set to zero from the start:
$T^{\mu\nu}=\dots \zeta\Delta^{\mu\nu}\partial_\lambda u^\lambda.$
Could anyone lend me some advice?
 A: First, the trace $\eta_{\mu\nu}T^{\mu\nu} = T^{\mu}_{\mu}$. Just swap the $\nu$ index in the Energy-Momentum Tensor to the bottom and set $\nu = \mu$.
With this we have
$$T^{\mu}_{\mu} = (\epsilon + P)u^{\mu}u_{\mu} + P\eta^{\mu}_{\mu} - \eta\Delta^{\mu\alpha}\Delta^{\beta}_{\mu}(\partial_{\alpha}u_{\beta} + \partial_{\beta}u_{\alpha} - \frac{2}{3}\eta_{\alpha\beta}\partial_{\lambda}u^{\lambda})$$
As said, the contraction of the four-velocity $u^{\mu}u_{\mu} = -1$. Similarly we have $\eta^{\mu}_{\mu} = 4$ (just compute the sum manually)
Let's simplify the non-viscuous terms of the tensor first:
$$T^{\mu}_{\mu} = 3P-\epsilon - \eta\Delta^{\mu\alpha}\Delta^{\beta}_{\mu}(\partial_{\alpha}u_{\beta} + \partial_{\beta}u_{\alpha} - \frac{2}{3}\eta_{\alpha\beta}\partial_{\lambda}u^{\lambda})$$
Now, let's compute the giant $\Delta$ term, since it's a bit long:
$$\Delta^{\mu\alpha}\Delta^{\beta}_{\mu} = (u^{\alpha}u^{\mu}+\eta^{\mu\alpha})(u^{\beta}u_{\mu}+\eta^{\beta}_{\mu})$$
$$ = u^{\alpha}u^{\beta}u^{\mu}u_{\mu} + \eta^{\mu\alpha}u^{\beta}u_{\mu} + \eta^{\mu\alpha}\eta^{\beta}_{\mu} + u^{\alpha}u^{\mu}\eta_{\mu}^{\beta}$$
Using the raising and lowering property of the metric,
$$ = -u^{\alpha}u^{\beta} + u^{\alpha}u^{\beta} + \eta^{\alpha\beta} + u^{\alpha}u^{\beta}$$
So
$$\Delta^{\mu\alpha}\Delta^{\beta}_{\mu} = u^{\alpha}u^{\beta}+\eta^{\alpha\beta}$$
Thus
$$T^{\mu}_{\mu} = 3P-\epsilon - \eta(u^{\alpha}u^{\beta}+\eta^{\alpha\beta})(\partial_{\alpha}u_{\beta} + \partial_{\beta}u_{\alpha} - \frac{2}{3}\eta_{\alpha\beta}\partial_{\lambda}u^{\lambda})$$
Here we have the the viscous term expanding:
$$(u^{\alpha}u^{\beta}+\eta^{\alpha\beta})(\partial_{\alpha}u_{\beta} + \partial_{\beta}u_{\alpha} - \frac{2}{3}\eta_{\alpha\beta}\partial_{\lambda}u^{\lambda})$$
$$ = u^{\alpha}u^{\beta}\partial_{\alpha}u_{\beta} +  u^{\alpha}u^{\beta}\partial_{\beta}u_{\alpha} -\frac{2}{3} u^{\alpha}u^{\beta}\eta_{\alpha\beta}\partial_{\lambda}u^{\lambda} + \eta^{\alpha\beta}\partial_{\alpha}u_{\beta}+  \eta^{\alpha\beta}\partial_{\beta}u_{\alpha} -\frac{2}{3}\eta^{\alpha\beta}\eta_{\alpha\beta}\partial_{\lambda}u^{\lambda})$$
$$ = u^{\alpha}u^{\beta}\partial_{\alpha}u_{\beta} +  u^{\alpha}u^{\beta}\partial_{\beta}u_{\alpha} + \frac{2}{3}\partial_{\lambda}u^{\lambda} + \partial_{\alpha}u^{\alpha} + \partial_{\beta}u^{\beta} - \frac{8}{3}\partial_{\lambda}u^{\lambda}$$
From here, the fact that $u^{\mu}u_{\mu} = -1 \implies \partial_{\nu}(u^{\mu}u_{\mu}) = 0 \implies u_{\mu}\partial_{\nu}u^{\mu} = 0$, which kills the first two terms
$$= \frac{2}{3}\partial_{\lambda}u^{\lambda} + \partial_{\alpha}u^{\alpha} + \partial_{\beta}u^{\beta} - \frac{8}{3}\partial_{\lambda}u^{\lambda}$$
From here, since everything's a dummy index, we can all relabel them to all be the same variable ($\mu$) and sum them all up:
$$ = (\frac{2}{3} + 1 + 1 -\frac{8}{3})\partial_{\lambda}u^{\lambda} = 0$$
So, the trace of the viscosity term vanishes, and we are left with
$$T^{\mu}_{\mu} = 3P-\epsilon$$
Now, for computing the extra term $T'^{\mu\nu} = \zeta\Delta^{\mu\nu}\partial_{\lambda}u^{\lambda}$,
$$T'^{\mu}_{\mu} = \zeta \Delta^{\mu}_{\mu}\partial_{\lambda}u^{\lambda}$$
Our above computations can lead us to the fact that $\Delta^{\mu}_{\mu} = 3$ and thus
$$T'^{\mu}_{\mu} = 3\zeta\partial_{\mu}u^{\mu}$$
Where $\partial_{\mu}u^{\mu}$ is the four-divergence of the four-velocity.
