# Four-velocity of a static, spherically-symmetric star

Let us assume the static, spherically-symmetric space time given by $$ds^2 = -e^{2\Phi(r)}dt^2+e^{2\Lambda(r)}dr^2+r^2d\theta^2+r^2\sin^2\theta\,d\phi^2.$$ For a static star, in which case the fluid has no motion, the only nonzero component of the four velocity field is $U^0$. According to Schutz's A First Course in General Relativity, the normalization condition $\mathbf{U}\cdot\mathbf{U} = -1$ demands $$U^0 =e^{-\Phi}\,\,\,\mathrm{and}\,\,\,U_0=-e^\Phi.$$

However, it seems to me that the normalization condition just as easily allows $$U^0 =-e^{-\Phi}\,\,\,\mathrm{and}\,\,\,U_0=e^\Phi.$$ Is it true that both are acceptable? Is Schutz's statement just a convention to keep $U^0$ positive?

Short answer: you want the velocity vector to be future directed, which means that $U^0$ must indeed be positive.
We postulate that time is not completely symmetric, in the sense that there's such a thing as future and past. If you have that and you choose the $t$ coordinate so that it increases with time, then that forces $U^0$ to be positive. If you wanted to be difficult you could instead define $t$ so that it increases into the past, and then $U^0$ would be negative; I don't know why you'd do that, though.
I say this because the notion of future/past directed is only well defined in a time oriented spacetime; otherwise, you could take your "future" directed vector $\mathbf{U}$, parallel transport it on a closed loop and end up with a "past" directed vector. Thankfully, this doesn't happen in this example, or indeed in any other example you're likely to find.