Orthogonal Projection of Perfect Fluid Energy Momentum

[GL_Black_Hole]

Homework Statement

Derive the relativistic Euler equation by contracting the conservation law $$\partial _\mu {T^{\mu \nu}} =0$$ with the projection tensor $${P^{\sigma}}_\nu = {\delta^{\sigma}}_\nu + U^{\sigma} U_{\nu}$$ for a perfect fluid.

Homework Equations

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

The Attempt at a Solution

Going term by term:
$$P^{\sigma}_\nu \partial_\mu (\rho +p) U^{\mu} U^{\nu} = \partial_\mu (\rho +p) U^{\sigma} U_{\mu} + U^{\sigma} U_{\nu} \partial_\mu (\rho +p)U^{\mu} U^{\nu}$$ but $$U^{\nu} U_{\nu} =-1$$ so this term is zero.

Next,
$$P^{\sigma}_\nu (\rho +p) U^{\nu} \partial_\mu U^{\mu} = 0$$
Then:
$$P^{\sigma}_\nu (\rho +p) U^{\mu} \partial_\mu U^{\nu} = (\rho +p)U^{\mu} \partial_\mu U^{\sigma}$$,
and finally:
$$P^{\sigma}_\nu \partial ^{\nu} p = \partial ^{\sigma} p$$.
But if I compare my answer to the result Carroll states on pg. 36 on his GR text I am off by a term:
$$U^{\sigma} U^{\mu} \partial_\mu p$$.

Where did I lose this term?

 

[TSny]

$$P^{\sigma}_\nu \partial ^{\nu} p = \partial ^{\sigma} p$$

On the right hand side, where's the contribution from the second term of the projection tensor?