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On page 62 of his book “General Relativity”, Robert Wald writes the stress-energy tensor $T_{ab}$ for a perfect fluid in the form

$$T_{ab} = \rho u_au_b + P(\eta_{ab} + u_au_b), \tag{1}$$

and then says that the equation of motion, when the fluid is not subject to external forces, is

$$\partial^a T_{ab}=0. \tag{2}$$

He projects this equation parallel and perpendicular to $u_b$ to find

$$u^a\partial_a\rho + (\rho + P)\partial^au_a = 0, \tag{3}$$ $$(P +\rho) u^a\partial_a u_b + (\eta_{ab} + u_au_b)\partial^a P = 0. \tag{4}$$

I see that if I multiply equation $(3)$ by $u_b$ and add it to equation $(4)$, I obtain equation $(2)$, but why is the term $u_au_b \partial^a P$ in equation $(4)$ (the one perpendicular to $u_b$) and not in equation $(3)$ as $u_a\partial^aP$?

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I think the issue you’re having is that you didn’t use the contraction $u_bu^b=-1$ along the way. Anyway, below is the full answer.

In general, if you’re given a unit timelike vector (i.e a vector $u$ such that $g(u,u)=-1$), then the orthogonal projection of a vector $\xi$ onto the subspace spanned by $u$ is $P_{\parallel}(\xi)=-g(\xi,u)u$. A similar statement holds for covectors.

In index notation, if $u_au^b=-1$, then the projection of $\xi^a$ onto $u^b$ is $-(\xi_au^a)u^b$, where index raising and lowering is done with $g_{ab}$. The covector version is of course that the projection of $\xi_a$ onto $u_b$ is $-(\xi_au^a)u_b$.

Ok, so now coming to the stress energy tensor, let’s write it as $T_{ab}=(\rho+P)u_au_b+Pg_{ab}$, and let’s define $\xi$ to be the divergence: \begin{align} \xi_b&=\nabla^a(T_{ab})\\ &=\nabla^a(\rho+P)\,u_au_b+(\rho+P)(\nabla^au_a)u_b+(\rho+P)u_a(\nabla^au_b)+(\nabla^aP)g_{ab}+\underbrace{P\nabla^ag_{ab}}_{=0} \end{align} where I have used metric compatibility to say the last term vanishes. Now, as mentioned above, the projection onto $u$ is $-(\xi_au^a)u_b$, but since $\xi_b=0$ (the stress tensor is divergence free) this (co)vector equation is equivalent to the scalar equation $\xi_bu^b=0$. We can easily evaluate $\xi_bu^b$ using the above formula and the normalization condition $u_bu^b=-1$: \begin{align} 0&=\nabla^a(\rho+P)\,u_a\cdot (-1)+(\rho+P)(\nabla^au_a)\cdot (-1)+(\rho+P)u_a(\nabla^au_b)u^b+\underbrace{(\nabla^aP)g_{ab}u^b}_{=(\nabla^aP)u_a}\\ &=-(\nabla^a\rho)u_a-(\rho+P)(\nabla^au_a)+(\rho+P)u_a(\nabla^au_b)u^b, \end{align} where I have cancelled the term $(\nabla^aP)u_a$ which appears twice with opposite sign. Lastly, this third term vanishes by differentiating the normalization condition $u_bu^b=-1$ (in index-free notation, $g(u,u)=-1$ so applying $\nabla_u$ to both sides and using metric compatibility, and symmetry of $g$, we get $2g(\nabla_uu,u)=0$). Thus, we end up with the equation \begin{align} 0&=-\xi_bu^b=(\nabla^a\rho)u_a+(\rho+P)(\nabla^au_a), \end{align} which is your equation (3).

To get the perpendicular component, you simply look at $\xi_b+(\xi_au^a)u_b$, which simplifies to \begin{align} 0&=\xi_b+(\xi_au^a)u_b\\ &=\xi_b-\left[(\nabla^a\rho)u_a+(\rho+P)(\nabla^au_a)\right]u_b\\ &=(\nabla^aP)u_au_b+(\rho+P)u_a(\nabla^au_b)+(\nabla^aP)g_{ab}\\ &=(\rho+P)u_a(\nabla^au_b)+(g_{ab}+u_au_b)\nabla^aP, \end{align} which gives you equation (4).

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  • $\begingroup$ if you want to, you can replace $g_{ab}$ with $\eta_{ab}$ and all covariant derivatives with partial derivatives, but since Wald introduces the necessary differential geometry early on in the book, I figured why not just prove the more general version, since it’s not much more effort anyway. $\endgroup$
    – peek-a-boo
    Jun 12, 2023 at 23:58

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