Confusion about derivation for isotropic fluids

[Dazed&Confused]

In Woodhouse's 'General Relativity' he finds an expression for the energy-momentum tensor of an isotropic fluid. If $W^a$ is the rest-velocity of the fluid and $\rho$ is the rest density then the tensor can be written as $$T^{ab} = \rho W^aW^b - p(g^{ab} -W^aW^b)$$
for a scalar field $p$. The conservation law $\nabla_a T^{ab}$ is then written as $$W^a\nabla_a \rho + (\rho +p )\nabla_a W^a=0$$
and
$$(\rho +p )W^a\nabla_aW^b = (g^{ab} -W^aW^b)\nabla_a p.$$
The bit that confuses me is why these equations are written seperately. I would have thought that the conservation law would give the first (multiplied by $W^a$) added with the second so it would be
$$W^bW^a\nabla_a \rho + W^b(\rho +p )\nabla_a W^a + (\rho +p )W^a\nabla_aW^b - (g^{ab} -W^aW^b)\nabla_a p=0.$$

Can someone explain why we split them up?

 

[vanhees71]

You have
$$T^{ab}=\rho W^a W^b - p (g^{ab}-W^a W^b),$$
where
$$W_a W^a=1$$. Obviously $W^a$ is the fluid four-velocity in the sense of the Landau frame (i.e., the rest frame of the fluid elements is given by energy flow).

Now you have local energy-momentum conservation, which reads
$$\nabla_a T^{ab}=0.$$
You can use this to derive local energy-momentum conservation for an observer comoving with the fluid. He uses $W^a$ as his time-like basis vector. The energy density in this reference frame is
$$T^{ab} W_a W_b=\rho$$
and the momentum density is given by
$$\Pi_c=W_{a} T^{ab} (g_{bc}-W_b W_c)=0,$$
as it must be.

Now to get the two balance equations for proper energy and momentum just write
$$T^{ab}=(\rho+p) W^a W^b-pg^{ab}.$$
Now write out the covariant divergence
$$\nabla_a T^{ab}=W^a W^b \nabla_a (\rho+p) + (\rho+p) W^b \nabla_a W^a + (\rho+p) W^a \nabla_a W^b - \nabla^b p=0. \qquad (*)$$
Now project to the time component of this equation by contracting with $W_b$:
$$W^a \nabla_a (\rho+p) + (\rho+p) \nabla_a W^a - W_b \nabla^b p=W_a \nabla_a \rho +(\rho+p) \nabla_a W^a=0,$$
which is your first equation. Here I have used
$$W_b W^b=1 \; \Rightarrow \; \nabla_a (W^b W_b)=0=2 W^b \nabla_a W_b$$
and
$$\nabla_c g^{ab}=0.$$
For the 2nd equation (the relativistic analogue of the Euler equation of an ideal fluid btw!) just contract (*) with the projector to the space-like piece, i.e.,
$$\nabla_a T^{ab}(g_{bc}-W_b W_c)=0 \; \Rightarrow \; (p+\rho) W^a \nabla_a W_c-(g_{ac}-W_a W_c) \nabla^a p=0$$
Now contract this with $g^{bc}$ which leads to
$$(p+\rho) W^a \nabla_a W^b - (g^{ab}-W^a W^b) \nabla_a p=0.$$
Putting the 2nd term on the other side you get your equation.

 

[Dazed&Confused]

Hi thanks for you answer it is quite helpful. In your last step when you contract the second term, how did that work?

Edit: thanks I figured it out.

 

[robphy]

The short answer to why they are split up is that the first equation is a scalar equation in the sense that each term is a scalar, whereas the second equation is a vector equation. So, in tensor calculus, it wouldn't be sensible to add them.

 

[Dazed&Confused]

So the temporal component is the first equation and the spatial the second?

And yes the author does show that this reduces to the Euler and continuity equation for speeds much less than that of light.

 

[vanhees71]

Yes. These are equations of motion expressed in covariant form. You project to the time-like components of vectors (and more generally tensors) in the local restframe of the fluid with the projector
$$P_{\parallel}^{\mu \nu}=W^{\mu} W^{\nu}$$
and to the space-like part with its Minkowski-orthogonal complement,
$$P_{\perp}^{\mu \nu} = g^{\mu \nu}-W^{\mu} W^{\nu}.$$
In this way you also get the energy-momentum tensor of the isotropic fluid. Using the Eckart definition of the local restframe (where by definition the four-velocity is determined by the energy flow), it must be
$$T^{\mu \nu} = \rho P_{\parallel}^{\mu \nu} - p P_{\perp}^{\mu \nu},$$
which defines the energy density and pressure as scalars (!), i.e., measured in the local restframe of the fluid cells.