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This follows Carroll's Gravity book (page 110).

An observer with four-velocity $U^{\mu}$ (such that $g_{\mu\nu}U^{\mu}U^{\mu}=-1$) measures the energy of a particle along a geodesic $$p^{\mu} = \frac{dx^{\mu}}{d\lambda}\tag{3.62}$$ to be $$ E = - p_{\mu}U^{\mu} \ , \tag{3.63} $$ where the particle is on either a timelike or null geodesic.

Can someone give some more intuition as to where this comes from? I see that it is an invariant and so will be the same in all frames, sot hat makes sense.

However what's confusing me is for generic $U^{\mu} = ( U^0, \mathbf{U} )$ and $p^{\mu} = (p^0, \mathbf{P})$ this has a nontrivial form, and I don't understand why this is the energy of the particle. Should this be simply understood as a definition of energy?

Qmechanic
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QuantumEyedea
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2 Answers2

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Take a local orthonormal frame $\{\mathbf{e}_\mu\}$, located at just one spacetime event, in which the observer is at rest. Orthonormal means that $\mathbf{e}_\mu \cdot \mathbf{e}_\nu = \eta_{\mu\nu}$, so that in a small neighborhood of the event, coordinates work just like in special relativity. And the observer being at rest means that $\mathbf{e}_0 = \mathbf{U}$, because then the components of $\mathbf{U}$ in this frame are $(1, 0, 0, 0)$.

Well, we know from special relativity that the energy is just $p^0$. In our local frame, the dot product $-p_\mu U^\mu$ equals $-p_0 = p^0$, because $\mathbf{U} = \mathbf{e}_0$ and because the metric is, at this point, the Minkowski metric. But the expression $-p_\mu U^\mu$ is a scalar, so it will give the same result no matter what coordinates you use. So you can calculate $-p_\mu U^\mu$ in any frame and you'll get the same result, which is the zero-th component of the four-momentum in the rest frame of the observer.

Or, if you prefer a more geometrical explanation: Since at this point the frame $\{\mathbf{e}_\mu\}$ is orthonormal, to get the components of the four-momentum we just project:

$$p^0 = - \mathbf{p}\cdot \mathbf{e}_0, \quad p^1 = \mathbf{p}\cdot \mathbf{e}_1,\quad \dots$$

(where the minus sign comes from the Minkowski metric).

Javier
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  • I know this question is old, but I don't really get how that minus sign comes from the Minkowski metric. Could you explain a bit further? – Jorge Casajus Mar 18 '21 at 14:21
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    @JorgeCasajus We have $\mathbf{p} = p^0 \mathbf{e}_0 + p^i \mathbf{e}_i$, so $\mathbf{p} \cdot \mathbf{e}_0 = p^0 (\mathbf{e}_0 \cdot \mathbf{e}_0) = -p^0$, since $\mathbf{e}_0 \cdot\mathbf{e}_0 = -1$. – Javier Mar 18 '21 at 14:38
  • Does this only work for an observer at point $P$ on the geodesic, who is trying to measure the momentum of the particle that is passing through point $P$? Carroll was not very specific about where the observer is on the spacetime manifold. – TaeNyFan Mar 24 '21 at 12:19
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    @TaeNyFan: yes, to measure the momentum of a particle in general relativity you have to be at the same spacetime event as the particle (or close enough). You can only take dot products between vectors at the same point. – Javier Mar 24 '21 at 14:06
  • @Javier Makes sense, thanks! – TaeNyFan Mar 24 '21 at 14:33
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    But I was wondering, that Energy measured depends on the frame. Like if I move in the direction of a particle I will see it as less energetic than if I move opposite to it. So energy should change with the frame used i.e the 4 velocity used. But here energy is a scalar quantity. Am I missing something – Shashaank Apr 29 '21 at 14:29
  • @Shashaank This scalar quantity is not just "the energy"; it's the energy measured by an observer with four-velocity $U^\mu$. All observers will agree on its value, though they will use different coordinate expressions. – Javier May 03 '21 at 14:07
  • Linked: https://physics.stackexchange.com/a/118614/50207 – Rexcirus Aug 17 '22 at 15:12
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Energy is the time-component of 4-momentum, and the time axis of an observer is given by its 4-velocity. The contraction $p_\mu u^\mu$ is just the projection of momentum onto that axis.

For example in flat spacetime, we have $$ u^\mu = (c,\vec 0) \qquad p_\mu = m\gamma (-c,\vec v) $$ in the rest frame of our observer and hence $$ -p_\mu u^\mu = \gamma mc^2 = E $$

Christoph
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