# A Derivation of radial momentum equation in Kerr geometry

#### 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?

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#### PeterDonis

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 $r$.

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