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I know that in SR, the 4-velovity $$ u^\mu = (d t/d \tau,d x/d \tau,d y/d \tau,d z/d \tau ) $$ and $$p^\mu = m u^\mu.$$ How do these generalize to GR? I imagine there are new complications, particularly by what we let $p^0$ be. And are these only defined for geodesics, or in general?

Mikkel Rev
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    You would just have general curved metric, that will change your proper time. – Kosm Apr 12 '17 at 18:55
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    In exactly the same way, in fact. What you have written down are also sensible for massive particles in GR. – John Apr 12 '17 at 19:02

2 Answers2

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Are you mentioning $p^0$ because you think of it as the energy?

If that is where your question is coming from, then perhaps this answer may be of use. Energy is something that an observer measures about an object. Let me explain.

The observer has a 4-velocity, $u^\mu_{\text{obs}}$, and the object has a 4-momentum, $p^\mu_{\text{obj}}$. The energy that the observer measures for the object is the projection of the object's 4-momentum on the observer's 4-velocity: $$E = p^\text{obj}_\mu u^\mu_\text{obs}=p^\mu_\text{obj}g_{\mu\nu}u^\nu_\text{obs}$$ Here we see the metric tensor coming into play, irregardless if we are in a flat spacetime, or in a curved one.

If we think of ourself as the observer, then ${\bf u}=(1,0,0,0)$. The energy that we measure an object to have will be $E = g_{00}p^0_\text{obj}+g_{i0}p^i_\text{obj}$. If the time-space components of the metric vanish locally, the energy is $E = g_{00}p^0_\text{obj}$. In flat spacetime $g_{00} = 1$, and we get back to the initial idea that energy is the time component of the 4-momentum.

Hope that helps.

Damian Sowinski
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  • Are you saying that in GR $p^0$ is not necessarily the energy? – Mikkel Rev Apr 12 '17 at 20:19
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    Yes, but in SR as well. Energy depends on both the object being observed and the observer. Just think of a photon. Sure it has an energy $E=hf$ if you measure it, but that frequency depends on your motion relative to the source of the photon. Different observers will measure different frequencies due to Doppler shift. – Damian Sowinski Apr 12 '17 at 20:40
  • And do these 4-vectors exist even if the four position does not parameterize a geodesic? – Mikkel Rev Apr 12 '17 at 20:41
  • Not exactly sure what you mean there. A non-null path in spacetime can be parametrized by its arclength. Null paths can also be parametrized, but you have to use an affine parameter. The derivative (with respect to the parameter) gives the 4-velocity of the curve at each point. So I want to say, Yes, 4-velocities can always be constructed from curves. That's just mathematics. Now whether or not an object can move along that curve is a physics question. Perhaps a good place to start is looking into energy conditions – Damian Sowinski Apr 12 '17 at 21:00
  • @DamianSowinski yes, because $p^0$ is different in different frames of references. Don't quite get your point: why isn't $p^0$ equal to the energy in SR? – Prof. Legolasov Apr 13 '17 at 05:00
  • The energy that the observer measures the object to have is given by the projection of the object's 4-momentum onto the observer's 4-velocity. If both the observer and the object are moving relative to the stationary frame, then the energy measured by the observer will, in general, not be equal to simply $p^0$, but will also depend on the $p^i$. I'd suggest the Caltech Lecture Notes for further reading. – Damian Sowinski Apr 13 '17 at 06:06
  • I have small query here. You say that we measure energy by the eqn p.U, but we also have an expression for energy by the conserved quantity coming from the timelike Killing vector field. So which one is the true energy – Shashaank Apr 29 '21 at 19:27
  • @Shashaank see https://physics.stackexchange.com/questions/218121/conservation-of-energy-and-killing-field – Damian Sowinski Aug 18 '21 at 20:05
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On a general Lorentzian manifold with a time-positive metric, $d\tau^2 = g_{\mu\nu}dx^{\mu}dx^{\nu}$. In SR $g_{\mu\nu} = \eta_{\mu\nu}$ which gives the familiar relation:

$$ d\tau^2 = dt^2 - dx_idx^i $$

In GR the Einstein Field Equations tell you what $g_{\mu\nu}$ is so you can compute the altered line element.

Otherwise, the form of the 4-velocity and 4-momentum is the same in GR for a massive particle.

gautampk
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