Undergrad What was the four-momentum meant to include?

[dsaun777]

Hello,
was the four-momentum of relativity, P^ν, supposed to include all mass and energy contributions from every field i.e. electromagnetic, strong, gravitational...
Or is it just the momentum of what was known in Einstein's time?

 

[Ibix]

Four momentum only works for point particles, or things you can approximate as point particles, but includes every contribution to their energy and momentum. For example, most of the mass of a proton is due to the binding energy of the quarks, so there's an awful lot of strong force contributing to the $m$ in a "ball of mass $m$" that you would treat as a point particle.

For fields and the like (when you can't lump them in to a point particle) you'd need a stress-energy tensor.

 

[dsaun777]

Four momentum only works for point particles, or things you can approximate as point particles, but includes every contribution to their energy and momentum. For example, most of the mass of a proton is due to the binding energy of the quarks, so there's an awful lot of strong force contributing to the $m$ in a "ball of mass $m$" that you would treat as a point particle.

For fields and the like (when you can't lump them in to a point particle) you'd need a stress-energy tensor.

Can you then contract and integrate the stress-energy tensor to arrive at some four-momentum? I suppose it depends on what kind of spacetime you are working in right?

 

[Ibix]

If I have this straight, if you have a family of observers following timelike paths that form a congruence $u^a$ then the energy momentum density they measure at an event is $u_aT^{ba}$. You integrate over some finite region of a spacelike 3-surface (formally, an achronal one) that encloses your "point particle" and you get its four momentum.

I may not have that quite right - sure others will correct me if so.

 

[PeterDonis]

If I have this straight, if you have a family of observers following timelike paths that form a congruence $u^a$ then the energy momentum density they measure at an event is $u_aT^{ba}$. You integrate over some finite region of a spacelike 3-surface (formally, an achronal one) that encloses your "point particle" and you get its four momentum.

I may not have that quite right - sure others will correct me if so.

This is pretty much correct. The only clarification I would make is that the congruence $u^a$ describes the worldlines of pieces of the matter whose energy-momentum density you want to obtain, not "observers". Assuming that these worldlines occupy a suitably small "world tube", surrounded by enough vacuum to treat the matter as an isolated region, then, if one is OK with modeling the matter as a point particle, one would do the integral you describe over the intersection of the world tube with an achronal 3-surface to obtain the energy-momentum density 4-vector for the matter at the "point" that represents that intersection.

One other caution here is that, if the congruence $u^a$ is not hypersurface orthogonal (meaning it is impossible to find an achronal 3-surface that is everywhere orthogonal to $u^a$, which is what we would naturally want to support an interpretation as "the matter at some instant of time"), modeling the matter by a simple energy-momentum 4-vector will not be enough. Heuristically, the "point particle" will have spin as well as 4-momentum, and it will take some additional geometric object besides the 4-momentum density vector to describe the spin.

 

[Ibix]

This is pretty much correct. The only clarification I would make is that the congruence $u^a$ describes the worldlines of pieces of the matter whose energy-momentum density you want to obtain, not "observers".

Ah, right - that makes more sense. It was bothering me how the $u^a$ would disappear in the integral (which it would need to do if it really were some arbitrary family of observers and the result had to be an invariant). But if it's part of the specification of the material then of course I don't expect it to vanish. Thanks.