8

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

9

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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2

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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