Confusion regarding 4-Velocity Derivative Identity (for conservation of energy momentum tensor) in Carroll's Spacetime and Geometry

Question

During Carroll's discussion of the energy-momentum tensor for a perfect fluid (page 36), he writes out that its divergence should be zero. He then expands this as follows:

$$\partial_\mu T^{\mu\nu} = \partial_\mu(\rho + p ) U^\mu U^\nu + (\rho + p)(U^\nu\partial_mu U^\mu + U^\mu \partial_\mu U^\nu )+\partial^\nu p$$

Next he writes:

To analyze what this equation means, it is helpful to consider separately what happens when we project it into pieces along and orthogonal to the four-velocity field $U^\mu$. We first note that the normalization $U_\nu U^\nu = -1$ implies the useful identity $$U_\nu \partial_\mu U^\nu = \frac12 \partial_\mu (U_\nu U^\nu)=0.$$

This is where I am confused. Could someone explain this implication? I am having trouble understanding what this equality means and how we can derive this identity.

(Apologies if the question is more lengthy than it should be, I wanted to give context to the equation by including the build up).

Answer 1

Consider an analogue that you may have seen in introductory mechanics. If a particle travels with constant speed, then the acceleration vector is perpendicular to the velocity vector.

To prove this, you want to show that the time-derivative of the magnitude of the velocity vector is zero. But the magnitude is sometimes awkward to deal with... so instead show that the time-derivative of the square-magnitude (i.e. the velocity vector dotted with itself) is zero.
What happens when you carry that out?

Now consider the statement about the 4-velocity being a unit vector.