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MTW (chapter 5) and others state that $-T^a_b v^b $ should be interpreted as the four-momentum density in the reference frame of an observer with four-velocity $v^a$, where $T^a_b$ is the stress-energy tensor. This makes sense as a useful machine we'd really like to have.

Okay, very first/simplest test case: Take the stress energy of dust in coordinates where it is at rest, so that $T^{ab} = \rho u^a u^b $ with $u^a = (1,0,0,0)$. Now here comes an observer with four-velocity $v^a = (\gamma, 0,0,v \gamma) $. Since $u_a v^a = - \gamma$, the four-momentum density in the observer's frame is supposed to be $(\gamma \rho,0,0,0)$? What happened to the momentum of the dust in the observer's frame? He should see dust particles zipping by, right?

I must be missing something extremely basic for this to be bugging me so much. But what?

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You are calculating the momentum in the dust rest frame. In the observer rest frame we have $v^a = (1,0,0,0)$ and $u^a = (\gamma,0,0,-v\gamma)$ so that $-T^a{}_bv^a = \rho\gamma^2(1,0,0,-v)$. Note that this, of course is the same expression we would get if we transformed your result to the observer restframe.

In other words, the correct statement is that $-T^a{}_bv^b$ is the 4-momentum observed by the observer, expressed in whatever frame we use for our calculations.

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  • $\begingroup$ Thank you, it all makes perfect sense now. Somehow I got the idea that the answer was automatically going to be expressed in the reference frame of the observer (possibly due to MTW's phrase "as measured in observer's Lorentz frame" ), but now I can see that's obviously impossible, due to the way $T^{ab}$ and $v^a$ transform. $\endgroup$ – Chronos Oct 20 '17 at 21:53

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