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Four Momentum in General Relativity

  1. Sep 15, 2014 #1


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    Alright, I'm rather new to General Relativity, and I'm getting confused with four momentum. Back in SR, [itex]p^α=mU^α[/itex], 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 [itex]p_0=-E[/itex], 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?
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  3. Sep 15, 2014 #2


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    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".

    ##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.
  4. Sep 15, 2014 #3
    That's ironic. I'm confused now too. In the introductory book I'm reading on GR, the author defines [itex]p_0=-E[/itex] when discussing trajectories in the Schwarzschild metric, and then he goes to define [itex]p^0=g^{00}p_0=(1-{2M}/r)^{-1}E[/itex]. Why does he do this then? Is this a matter of convention, defining the vector momentum from the covector momentum?
  5. Sep 19, 2014 #4


    Staff: Mentor

    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.
    Last edited: Sep 19, 2014
  6. Sep 19, 2014 #5


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