# Derivation of radial momentum equation in Kerr geometry

• A
abby11
I am trying to derive the radial momentum equation in the equatorial Kerr geometry obtained from the equation $$(P+\rho)u^\nu u^r_{;\nu}+(g^{r\nu}+u^ru^\nu)P_{,r}=0 \qquad$$. Expressing the first term in the equation as $$(P+\rho)u^\nu u^r_{;\nu}=(P+\rho)u^r u^r_{;r}$$ I obtained the equation $$uu'+\dfrac{u^2}{r}+\dfrac{1}{P+\rho}\left(u^2+\dfrac{\Delta}{r^2}\right)P'=0$$ where primes refer to derivative w.r.t. the coordinate r. But the correct equation should be $$uu'+\frac{1}{r\Delta}\left(a^2-r-\frac{A\gamma^2K}{r^3}\right)u^2-\frac{A\gamma^2K}{r^6}+\frac{1}{P+\rho}\left(\frac{\Delta}{r^2}+u^2\right)P'=0 \qquad$$ It seems that I am missing some terms in expanding the first term ## (P+\rho)u^\nu u^r_{;\nu} ## where I had summed up ##\nu## to only r since the equation only involves radial derivatives. Can someone please point out what I am missing?

## Answers and Replies

Mentor
I am trying to derive the radial momentum equation in the equatorial Kerr geometry obtained from the equation
$$(P+\rho)u^\nu u^r_{;\nu}+(g^{r\nu}+u^ru^\nu)P_{,r}=0 \qquad$$

Where are you getting this equation from? It doesn't look anything like a "radial momentum equation" in Kerr geometry, since Kerr geometry is a vacuum geometry so ##\rho = P = 0## everywhere.

since the equation only involves radial derivatives

The equation has a covariant derivative, which includes terms in the connection coefficients that involve other values of ##\nu## besides ##r##.