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DrGreg

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**In Special Relativity**(flat spacetime) we can define a contravariant 4-momentum vector as

[tex]P^{\alpha} = m\frac{dX^{\alpha}}{d\tau}[/tex]

for a particle of mass

*m*where [itex]X^{\alpha}[/itex] is the "coordinate vector" of a particle. Or as

[tex]\textbf{P} = (h\nu, \hbar \textbf{k})[/tex]

for a photon (in Minkowski coordinates).

Better still, we can define a covariant 4-momentum covector as

[tex]P_{\alpha} = g_{\alpha\beta}P^{\beta}[/tex]

We can then define energy relative to a frame as the time component (rescaled into the correct units) of the vector, and momentum relative to a frame as the space component (similarly rescaled) of the vector. In Minkowski coordinates, it doesn't really matter whether we use contravariant or covariant 4-momentum; after rescaling you get the same answer either way.

However, in non-Minkowski coordinates, the two methods do give different answers. Does it matter which choice we make? And if so, why?

If we choose Rindler coordinates (coordinates in which a Born-rigid accelerating rocket is stationary), does the 4-momentum vector or covector naturally give rise to something resembling Newtonian potential energy?

Finally, can any of the above be adapted to make sense in

**general relativity**(curved spacetime). I'm aware of the standard Physics FAQ (Is Energy Conserved in General Relativity?) which says that, in general, there is no global conservation of energy, only local. But it also says

"In certain special cases, energy conservation works out with fewer caveats. The two main examples are static spacetimes and asymptotically flat spacetimes."

..."The Schwarzschild metric is both static and asymptotically flat, and energy conservation holds without major pitfalls."

..."The Schwarzschild metric is both static and asymptotically flat, and energy conservation holds without major pitfalls."

So, for example, does it make any sense to talk about potential energy in Schwarzschild coordinates?