The Schwartzschild metric in standard coordinates with signature $(1,-1,-1,-1)$ is given by $$ds^2=(1-\frac{r_s}{r})\ dt^2 - (1-\frac{r_s}{r})^{-1}\ dr^2 - r^2(d\theta^2+\sin^2\theta\ d\phi^2).$$ As the Schwartzschild metric is independent of time then it has a global time-displacement symmetry described by a Killing vector vector field $\xi^\mu$ given by $$\xi^\mu = (1,0,0,0).$$ This implies that a particle free-falling on a geodesic path with four-velocity $V^\mu$ has a constant of motion $\epsilon$ given by $$\epsilon=\xi_\mu V^\mu.$$
My question is: can such an expression be consistently interpreted in General Relativity as simply the energy per unit mass?
For example one could assert that a stationary observer at radial coordinate $r$ with four-velocity $U^\mu=((1-r_s/r)^{-1/2},0,0,0)$ has an energy per unit mass, $\epsilon_{obs}$, given by \begin{eqnarray} \epsilon_{obs} &=& \xi_\mu U^\mu \\ &=& (1-\frac{r_s}{r})(1-\frac{r_s}{r})^{-1/2} \\ &=& \sqrt{1-\frac{r_s}{r}}. \end{eqnarray} When the observer measures the energy per unit mass of the particle he compares it with his own energy scale. Therefore he perceives the ratio of these energies $\epsilon/\epsilon_{obs}$ given by $$E_{obs}=\frac{\epsilon}{\epsilon_{obs}}=\frac{\epsilon}{\sqrt{1-\frac{r_s}{r}}}.$$ This result agrees with the standard calculation procedure where one interprets $\epsilon$ as a constant of motion, uses it to calculate $V^0$, and then derives the energy per unit mass measured by the stationary observer using $$E_{obs}=U_\mu V^\mu.$$
