Four Momentum in General Relativity

[MrBillyShears]

Alright, I'm rather new to General Relativity, and I'm getting confused with four momentum. Back in SR, $p^α=mU^α$, but, this relationship doesn't hold in curved space, does it? Because, now I'm seeing that four momentum is somehow a covector in GR, and $p_0=-E$, so the time component of the contravariant component can't be E, since the 00 component of the metric isn't always -1. So, is four momentum now redefined in GR, and if the SR relationship between four velocity isn't valid, what is the new relationship to U? Is four momentum naturally a covariant thing?

 

[Dale]

Back in SR, $p^α=mU^α$, but, this relationship doesn't hold in curved space, does it? Because, now I'm seeing that four momentum is somehow a covector in GR,

I am not sure if this is your concern, but you would probably call both $p^α=mU^α$ and $p_α=mg_{\mu α}U^{\mu}$ "the four momentum". If you need to distinguish between the two verbally you would call one "the four momentum vector" and the other "the four momentum covector".

and $p_0=-E$, so the time component of the contravariant component can't be E, since the 00 component of the metric isn't always -1.

$p_0=-E$ would only hold in metrics where the 00 component is -1. In GR that won't always be the case. The 00 component may not be the time-time component. In fact, there may not even be a time-time component. Regardless, the definition above will hold, even if the components don't have such a clear interpretation.

 

[Dirac271]

That's ironic. I'm confused now too. In the introductory book I'm reading on GR, the author defines $p_0=-E$ when discussing trajectories in the Schwarzschild metric, and then he goes to define $p^0=g^{00}p_0=(1-{2M}/r)^{-1}E$. Why does he do this then? Is this a matter of convention, defining the vector momentum from the covector momentum?

 

[Dale]

Sorry, I was indeed being a little ambiguous (actually a little wrong). What I was thinking and what I should have said is that $p_0=\pm p^0=\pm E$ only in metrics where the 00 component is $\pm 1$, and even then only if the 0i components are 0 too.

The vector and covector momentum are always related as $p_{\mu}=g_{\mu\nu}p^{\nu}$, regardless of the metric. The vector and the covector are in some ways fundamentally different quantities, but nevertheless it is a very strong convention to think of them as the same quantity just with raised or lowered indexes.

 

[atyy]

See also post #8 by snoopies622 and George Jones's reply in #12 at www.physicsforums.com/showthread.php?t=263012.