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In Roger Penrose's The Road to Reality, relativistic angular momentum $M^{ab}$ at $x$ is defined on page 437 to be the ‘tensor quantity’ $$M^{ab}=x^a p^b - x^b p^a,$$

where $x^a$ is ‘the $4$-vector of the point on the particle's world line at the time that its angular momentum is being considered’ and $p^a$ is the energy-momentum $4$-vector.

My question is this: How am I to actually make sense of this? As far as I understand, $x$ is simply the coordinate (in some coordinate patch) on our Lorentzian $4$-manifold (which has a flat connection in the context of special relativity and possibly one with curvature in that of general relativity resulting from the presence of mass; although I suppose this is irrelevant, I think Penrose is talking about the former here). So, there is no notion of tensoring $x$ with anything.

There is the possibility that he is talking about the standard real $4$-space with a $(+,-,-,-)$-metric, in which case $x$ may be seen as a $4$-vector, but I think this should be seen as an accidental feature and should not be fundamental for the definition. Obviously this would be similar to Newtonian mechanics – even then there are ambiguities resulting from non-specified identifications. In any case, what is the definition in the general case (on a Lorentzian $4$-manifold)?

Seeing as we wish to tensor $x$ with $p$, we may, for example, identify the tangent space with a small coordinate neighbourhood via the exponential map and push $x$ up in this way. I don't know if this is the standard procedure. Also, please note that I have not yet read further into the book.

P.S. I have a mathematical background, so it is entirely possible that I didn't get some key points.

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  • $\begingroup$ Just a slightly different point of view: Your angular momentum components here are precisely the infinitesimal generators of the Lorentz group (rotations boosts). In this regard, you can think of the finite angular momentum tensor (exponentiation of the Lie algebra) as a local Lorentz transformation of some body) $\endgroup$ – R. Rankin Jan 8 '19 at 5:16
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There are two things going on here that need to be cleared up.

As far as I understand, $x$ is simply the coordinate (in some coordinate patch) on our Lorentzian 44-manifold [...]

Not true. The notation in the equation $M^{ab}=x^ap^b-x^bp^a$ looks like concrete index notation, which would be coordinate-dependent. It isn't. The modern convention is that Latin indices are abstract indices, and only Greek indices are concrete and tied to a particular coordinate system. Penrose was the one who basically invented and popularized this notation. That means that $x^a$ is to be interpreted not as one component of a vector but as an entire vector, and $a$ does not take on a numerical value like 2 or 3.

Seeing as we wish to tensor $x$ with $p$, we may, for example, identify the tangent space with a small coordinate neighbourhood via the exponential map and push $x$ up in this way. I don't know if this is the standard procedure.

In a flat spacetime this issue doesn't even arise. You can simply pick an arbitrary point as the origin, and then you can refer to a position as a vector. No coordinates are implied or required.

For the generalization to a curved spacetime, I think you have the right idea when you talk about a neighborhood, because this definition clearly only works in a neighborhood. However, you don't need to go about this by picking some random set of coordinates and subjecting them to manipulations. For a more explicit sketch of the approximation procedures, see Hawking and Ellis, The large scale structure of space-time, pp. 62-63. They invoke local orthonormal coordinates, but that's just a convenience.

The larger and unavoidable issue in a curved spacetime is that you don't get integral conservation laws, even for scalars. This is discussed in Misner, Thorne, and Wheeler, Gravitation, p. 457. For example, energy isn't conserved on cosmological scales. The point of introducing a quantity like $M^{ab}$ is that it's going to satisfy a differential conservation law.

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  • $\begingroup$ Thanks for the clarifications: Penrose does explicate how he distinguishes between abstract and concrete indices earlier in the book, I had forgotten about this; as for your second clarification, what I suggested was indeed concerned with the general non-flat case. It would've been nice of Penrose to go into a bit of conceptual detail (perhaps in a footnote) like you did. Very interesting points. $\endgroup$ – L'Air de Panache Aug 6 '17 at 16:27

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