# How do I project the equation $\partial^a T_{ab} = 0$ parallel and perpendicular to $u_b$?

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 $$u_a\partial^aP$$?

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).
• 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. Jun 12, 2023 at 23:58