Let us assume that a spacetime has a global time-displacement symmetry described by a timelike Killing vector field ##\xi^\mu=(1,0,0,0)##.(adsbygoogle = window.adsbygoogle || []).push({});

Further assume that we have a particle with four-momentum ##P^\mu=m\ V^\mu##.

It is well-known that a stationary observer with four-velocity ##U^\mu## measures the energy of the particle, ##E_{obs}##, relative to himself as

$$E_{obs}=U_\mu V^\mu.$$

My question: Is the absolute energy of the particle given by ##E=\xi_\mu V^\mu##?

I use the word "absolute" as ##\xi^\mu## is a global symmetry of the spacetime so that I would have thought that ##E## should be the energy of the particle relative to the spacetime itself.

PS I'm sorry if I've asked similar questions in the past. I feel that I have formulated my question more succinctly in this post.

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# I Absolute Energy in General Relativity?

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