# Energy of a circular orbit in the Schwarzschild metric?

I am looking at circular orbits in the Schwarzschild metric and the energy associated with them. In places I look the energy is given as: $$\newcommand{\p}[2]{\frac{\partial #1}{\partial #2}} \newcommand{\f}[2]{\frac{ #1}{ #2}} \newcommand{\l}[0]{\left(} \newcommand{\r}[0]{\right)} \newcommand{\mean}[1]{\langle #1 \rangle}\newcommand{\e}[0]{\varepsilon} \newcommand{\ket}[1]{\left|#1\right>} E=\l 1-\f{r_s}{r}\r\f{dt}{d\tau}=\f{r-2GM}{r(1-3GM)}$$ examples of where this is done is Moore, 2013 pg 126 and Baumgarte & Shapiro, 2010 pg10. But is the energy not simply given by the zeroth component of the 4-momentum i.e.: $$E=\f{dt}{d\tau}=\f{r-2GM}{r(1-3GM)(1-\f{r_s}{r})}$$ where deos the extra factor of $1-\f{r_s}{r}$ come from and which is the correct expression?

• Isn't this a very standard thing in any GR/black hole textbook? e.g. p.3-9 in Taylor & Wheeler; p.342 Shapiro & Teukolsky. Commented May 2, 2017 at 14:05

This is a consequence of the standard result that if the metric is independent of a coordinate $x^a$, then the corresponding covariant momentum $p_a = g_{a\mu}p^\mu \propto g_{a\mu} \dot{x}^\mu$ is conserved along geodesics. (This can be generalized a little but let's not get into that.) The easiest way to prove this is to notice that the Lagrangian
$$L = \frac12 g_{\mu\nu}\dot{x}^\mu \dot{x}^\nu$$
(with an affine parameter) has the geodesic equation as its E-L equation, and the covariant momentum $p_\mu$ is just the canonical momentum $\partial L / \partial \dot{x}^\mu$.
Applying this to a stationary spacetime, we find that $p_t$ is conserved, not $p^t$, so it deserves to be called energy.