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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 time-displacement symmetry described by a Killing vector $\xi^\mu$ given by $$\xi^\mu = (1,0,0,0).$$ This implies that a particle free-falling on a geodesic path with four-velocity $P^\mu$ has a constant of motion $\epsilon$ given by $$\epsilon=\xi_\mu P^\mu.$$ I understand that $\epsilon$ can be interpreted as the particle energy measured by a stationary observer far from the origin, where the metric is flat, with four-velocity $U^\mu=\xi^\mu$.

Can one also interpret $\epsilon$ as the particle energy measured by a local observer who is free-falling with the particle?

I presume one must transform $\epsilon=\xi_\mu P^\mu$ to the local coordinates of the free-falling observer.

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 time-displacement symmetry described by a Killing vector $\xi^\mu$ given by $$\xi^\mu = (1,0,0,0).$$ This implies that a particle free-falling on a geodesic path with four-velocity $P^\mu$ has a constant of motion $\epsilon$ given by $$\epsilon=\xi_\mu P^\mu.$$ I understand that $\epsilon$ can be interpreted as the particle energy measured by a stationary observer far from the origin, where the metric is flat, with four-velocity $U^\mu=\xi^\mu$.

Can one also interpret $\epsilon$ as the particle energy measured by a local observer who is free-falling with the particle?

I presume one must somehow transform $\epsilon=\xi_\mu P^\mu$ to the local coordinates of the free-falling observer.

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 time-displacement symmetry described by a Killing vector $\xi^\mu$ given by $$\xi^\mu = (1,0,0,0).$$ This implies that a particle free-falling on a geodesic path with four-velocity $P^\mu$ has a constant of motion $\epsilon$ given by $$\epsilon=\xi_\mu P^\mu.$$ I understand that $\epsilon$ can be interpreted as the particle energy measured by a stationary observer far from the origin, where the metric is flat, with four-velocity $U^\mu=\xi^\mu$.

Can one also interpret $\epsilon$ as the particle energy measured by a local observer who is free-falling with the particle? Perhaps one can take the observer and particle to be instantaneously at rest so that one can say that the observer's four-velocity is instantaneously given by $U^\mu=\xi^\mu$?

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 time-displacement symmetry described by a Killing vector $\xi^\mu$ given by $$\xi^\mu = (1,0,0,0).$$ This implies that a particle free-falling on a geodesic path with four-velocity $P^\mu$ has a constant of motion $\epsilon$ given by $$\epsilon=\xi_\mu P^\mu.$$ I understand that $\epsilon$ can be interpreted as the particle energy measured by a stationary observer far from the origin, where the metric is flat, with four-velocity $U^\mu=\xi^\mu$.

Can one also interpret $\epsilon$ as the particle energy measured by a local observer who is free-falling with the particle?

I presume one must transform $\epsilon=\xi_\mu P^\mu$ to the local coordinates of the free-falling observer.