I’m studying General Relativity on Schutz’s book. On Chapter 7 he talks about conserved quantities along geodesics, with the equation
\begin{equation}
m\frac{dp_{\beta}}{d\lambda}=\frac{1}{2}g_{\nu\alpha,\beta}p^\nu p^\alpha
\end{equation}
and he concludes that “if all the components of the metric are independent of $x^\beta$ for some index $\beta$ then $p_\beta$ is a constant along any particle’s trajectory”.
For example, in the well-known Schwarzschild metric, $p_0$ is conserved as the metric is independent of $t$. But I could perform a coordinate change to make the metric “time dependent”. Does this mean that this concept of conserved quantities along geodesics is coordinate-dependent? Is there a preferred reference frame in which this quantity is conserved?
Edit
I'm probably confused about the concept of reference frame and coordinate system. I'll try to state the source of my confusion. Schutz says that, with the Schwarzschild metric
\begin{equation}
ds^2=-e^{2\Phi(r)}dt^2+e^{2\Lambda(r)}dr^2+r^2d\Omega^2
\end{equation}
"since the metric is independent of $t$ any particle that follows a geodesic has constant momentum component $p_0\equiv -E$". Then he states that "a local inertial observer at rest (momentarily) at any radius $r$ of the spacetime measures a different energy, namely $E^*=Ee^{-\Phi}$".
What does this imply? Does this imply that when an observer which is at rest at some point of the spacetime measures a quantity I should use a locally Minkowskian coordinate system (tangent space of the point P of the observer on the Manifold) and in that coordinate system the metric is not independent of time, since he sees that this quantity changes according to the point of spacetime he measures the quantity from? (Indeed, $\Phi$ is a function of $r$). Will any observer ever see this quantity conserved when he measures it or is it just a mathematical construct?
Best Answer
I haven't read Schutz, but from reading your question it sounds like his presentation of this topic has some deficiencies, and your confusion may be natural given those deficiencies. He's discussing this in terms of the (non-covariant) derivative of the metric with respect to a coordinate, which immediately creates some serious problems. That quantity simply isn't measurable. If you want to measure the metric or its derivatives, you end up with the following restrictions:
A local observer can't measure the metric. (This is for the same reason that you can't measure an absolute potential energy. The metric plays a role in GR analogous to that of the potential in Newtonian gravity.)
A local observer can't measure the derivative of the metric. (That derivative would basically be the gravitational field, which is not measurable because of the equivalence principle.)
A local observer can measure the second derivative of the metric, which is essentially a measure of tidal stresses.
So Schutz's presentation makes use of a derivative that has no physical interpretation.
Yes, any time any observer measures a vector quantity (such as the energy-momentum vector), they are implicitly doing so in some local Minkowski frame. (They could, for example, measure the vector's inner product with some other vector, but then they are effectively using this other vector as a coordinate axis of some Minkowski frame.)
To verify this conservation law, the observer needs to have global information, not just local information. Basically they need to measure the quantity $E^*$ in a local static frame, then use their global knowledge (of the metric and of their position in the spacetime) to determine $E$. They can then verify that $E$ is conserved.
The inability to determine such a conservation law based on purely local information is baked in to the structure of GR. Energy-momentum is a vector, and you can't compare vectors at different points in spacetime except by parallel transport. You can certainly verify that a test particle's energy-momentum vector is preserved under parallel transport along its own geodesic of motion, but you end up with a triviality, which is essentially that the test particle had the same free-fall motion that you did. This is just a test of the equivalence principle, and it holds even in a spacetime that does not have any symmetry.
To resolve the issues you're talking about in a more satisfactory way than in Schutz's presentation, you really need to use the notion of a Killing vector.