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In deriving the equations of motion for geodesics in spherically symmetric spacetimes through Hamiltonian formalism, we can find some constants of motion, namely, $E$ and $L$, the energy per unit of rest mass and angular momentum measured by a Schwarzschild observer. Nevertheless, it's not clear to me the real difference between the energy $E$ (related to timelike killing field) and the Hamiltonian $\mathscr{H}$, once the Hamiltonian is usually interpreted as the total energy of the system?

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  • $\begingroup$ Total energy of what system? With or without the energy of spacetime itself? $\endgroup$ – Qmechanic Jan 8 at 7:33
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The Hamiltonian $\mathscr{H}$ of a geodesic is equal to the norm of the 4-velocity/4-momentum (depending on what affine parameter is being used). Conservation of the Hamiltonian simply expresses that the norm of the 4-velocity or 4-momentum (i.e. the particle mass squared) is conserved along the geodesic. This is true in any spacetime.

The energy of a geodesic is found by contracting 4-momentum with a timelike Killing vector. Hence it exists only for stationary spacetimes (and in particular in Schwarzschild).

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