I Mass/energy in general relativity

1. Jun 22, 2017

Umaxo

Hi,

i got little confused after this conversation

There was also this conversation:

So i want to clear the confusion.

As was pointed out to me, in STR mass is defined as m2=E2 - p2, which can be done for all objects. But I was taught, that in GR there are, in general, some problems.

I know it can be done by computation/measurement of 4-momentum in asymptoticly flat region and then using STR definition of mass, but not all solutions of einstein equations are asymptoticly flat.

So are there really no dificulties with mass in GR as is Mister T saying? He says, that "the mass of a macroscopic object it includes the energies of the particles that make up that object", but again i was taught there is problem with this approach, because there is no gravitational energy in GR, so there is no contribution from gravity to overall mass if one tries to compute it this way. However, gravitation has its inprint on the mass of the object.

And about the question wheter there is indeed only one mass in modern physics - one could imagine observer who doesnt have acces to informations from asymptoticly flat region and has only acces to information near some object where there is "wild" curvature. But he still wants to somehow measure properties of the object, so different definition of mass would be needed for him. So are there some different kinds of masses in GR (i guess they should reduce to same number if pu can be appropriately defined), or indeed there is only definition m2=pupu?

Thanks for all the future replies:)

2. Jun 22, 2017

Vitro

@Umaxo, I don't think Mister T's comment was in the context of GR for an object of large dimensions in a curved spacetime where there is significant tidal gravity across the object, or an object that's both large and very massive and itself causes significant spacetime curvature on its various parts. I don't know if to such an object one could assign a naive scalar mass, I would think not.

3. Jun 22, 2017

phinds

I'm no expert on this stuff but I believe that is exactly correct. Mass is affected by gravity, but gravity does not contribute to mass. Binding energy in atoms/molecules does contribute to mass.

4. Jun 22, 2017

Mister T

What I'm saying is that no difficulty arises from referring to the rest mass as simply the mass.

That is something that someone with a knowledge of GR would have to answer for us. But calling the mass the rest mass won't remove any difficulty in the physics. In fact, my point is that it introduces difficulty for people trying to learn physics.

5. Jun 22, 2017

pervect

Staff Emeritus
What sort of mass are we talking about? The ADM, Bondi, and Komar masses are conserved quantities in an asymptotically flat (for ADM and Bondi masses) or stationary (for the Komar mass) spacetime. And they behave as much as Umaxo describes (small details may vary, I haven't read the thread that closely).

Energy density is one of the components of the stress energy tensor. There is a conservation law for the stress energy tensor as a whole, $\nabla_a T^{ab} = 0$, but the energy density itself can and does vary with time. These seem to behave as the other posters are talking about.

So I think there is some talking past each other here.

6. Jun 22, 2017

PAllen

Bondi mass, by design, is not necessarily conserved. It excludes radiation reaching (null) infinity, so it decreases over time for a radiating system. In fact, a definition of the energy of radiation including GW is the difference between ADM mass (which is conserved) and Bondi mass.

7. Jun 23, 2017

pervect

Staff Emeritus
That's a good point, though the correct view then gets rather vague. I suppose the old sci.physics.faq does one of the best jobs, http://math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html

"Is energy conserved in General Relativity" is the question and title of the FAQ, and the answer starts out as "In special cases, yes. In general — it depends on what you mean by "energy", and what you mean by "conserved".

This has the advantage of actually being correct, but has the disadvantage of having little if any predictive power without more explanation.

8. Jun 23, 2017

Umaxo

I need to check the definitions of these, but due to my exams i dont have enaugh time right now, so please be patient with me:)

But i want ask this, since i think it could be answered without knowing much about the above masses:

Is there anyone who can verify this?

- In this case, mass could be always defined simply from integrating energy-momentum tensor, right? Then the above masses, as i understood, are defined only because we dont know much about details of matter distribution and interactions inside the objects in question, so we need to come to conlusion by indirect methods, like the inprit matter leaves on spacetime geometry. And the reason why there are at least those 3 masses mentioned is because it is not that easy to analyze the inprints. Is that correct?

Last edited: Jun 23, 2017
9. Jun 23, 2017

PAllen

All 3 most common special cases of system mass/energy in GR (Komar, ADM, and Bondi) are affected by gravitational potential energy. The same components each locally measured, at the same temperature, will have lower total mass if they are closer together. The reflects the decrease in potential energy.

So, generally, I would disagree with that statement. What I am guessing it refers to is that the stress-energy tensor T does not include any explicit gravitational contribution. Thus, gravity does not explicitly appear in the source term of the GR field equations. However, the non-linearity of the GR field equations is considered by many to imply 'gravity gravitates' to some extent.

10. Jun 23, 2017

Umaxo

So this is wrong then:
right? Since, in general we dont have gravitational potential energy.

11. Jun 23, 2017

jbriggs444

I do not know how you see a conflict in those two passages. @PAllen points out that mass lowers as potential energy decreases. @Mister T points out that mass is not additive because energy contributes to it.

12. Jun 23, 2017

Umaxo

I know there is problem with defining conserved energy on macroscopic scales, because it cannot be made invariant.

Since in general energy is not conserved, i believe this implies you cannot in general define potential energy. Thus you cannot always compute mass as sum of its consituents energies (which i now realize @Mister T didnt wrote). Is this right? Or problems with invariance of conserved macroscopic energy doesnt mean that you cannot make computations?

So, I agree the way @Mister T formulated it (and i guess meant) doesnt contradict what @PAllen said. I guess I just read more than was written.

13. Jun 23, 2017

jbriggs444

The problem with not computing mass as the sum of the masses of the constituents is far more basic. No mumbling about curved space or global definitions is required.

Take two steel balls. One kilogram each. Individually. This is an invariant measure. No matter how fast or how slow these balls are moving according to any frame of reference you choose to adopt, their individual masses are always 1 kg each.

Now accelerate the balls to some significant fraction of the speed of light relative to one another. Consider the two balls as one combined system. What is the mass of the two balls considered together? It is greater than 2 kg. No matter what frame of reference you choose, at least one of the two balls has kinetic energy that contributes significantly to the combined mass. It turns out that the mass of the combined system computed this way by including the kinetic energy is always the same no matter what frame of reference you use to evaluate it. That is to say that it is "invariant".

Mathematically, $m^2 = E^2 - p^2$. If you have the two kilogram masses and a frame of reference in which their center of mass is at rest (p=0) then $E^2$ is greater than 2 kg because it includes 2 kg from the balls' rest energies plus the contribution from their kinetic energies in this frame. So the combined mass is greater than 2 kg.

Take two protons, two neutrons and two electrons. Weigh them. Put them together into a helium atom and weigh them again. The mass is not the same. The atom weighs less. Mass is not an additive quantity. [Of course, you had to allow a good quantity of energy to radiate away when you assembled that atom -- that's what balances the energy books]

14. Jun 23, 2017

Mister T

Conserved means amount stays the same. Invariant means the same in all reference frames. They are two different things.

This conversation started with a discussion of rest mass. Simply calling it the mass implies nothing about the physics. It's equivalent to rest energy regardless of what you call it. It's a quantity that's invariant but not conserved. Total energy is not invariant but it is conserved.

15. Jun 23, 2017

Staff: Mentor

No. That definition is the Komar mass; the other two PAllen mentioned (ADM mass and Bondi mass) do not require any knowledge of the stress-energy tensor, only the asymptotic spacetime geometry.

16. Jun 24, 2017

Umaxo

Yes, i wrote nonsense.

I need to think about it a little:) I will get back after my exams in tuesday.

17. Jun 24, 2017

Umaxo

I didnt try to suggest that mass can be computed from masses of its constituens.

I was asking, wheter this can be done in general in GR - to compute mass of composite object through energies (not masses) of its constituents. For two particles, the computation in STR should go as $m_{total}^2=p_{total}^2= (p_u+p´_u)(p^u+p´^u)=m^2+m´^2+2p_u p´^u$ in natural units, where m is mass, p is 4momentum. But you cannot directly do this in GR, because every 4-momentum lives on its own tangent space and you cannot simply add them up for macroscopic objects.
You could perhaps parallel transport them to same location and add them up there and compute mass from it. But parallel transport gives different results for different paths, so in this way you wouldnt get invariant mass.

18. Jun 24, 2017

Umaxo

But the fact that ADM mass and Bondi mass do not require knowledge of stress-energy tensor doesnt by itself imply that mass cannot be computed from some kind of integral of stress-energy tensor.

I guess i went wrong by talking about stress-energy tensor and integrals. See #17, I think i made it little clearer what my question is.

Or at least if someone could answer this weaker question:

Is it in general possible to define invariant mass, if we know all the information of interrior of macroscopic object (i.e. all the fields, state equations, geometry etc. inside) without referring to asymptotic behaviour, i.e. just by some computation inside the star? - all the masses mentioned require special spacetimes, but if answer to this question is yes, then one could in principle compute invariant mass for any spacetime one can imagine.

19. Jun 24, 2017

Staff: Mentor

Yes, it does, because it means that "mass", without specifying what definition of mass you are using, is not a well-defined term. On one definition of "mass"--Komar mass--you do compute it from an integral of the stress-energy tensor. On other definitions--ADM mass or Bondi mass--you don't. So your question as it stands is not well-defined, since you didn't specify what definition of mass you are using.

And, btw, you also cannot fall back on a general statement that the different definitions will all give the same answer, because in general that is not the case. The different definitions apply in different domains: Komar mass applies to any stationary spacetime, while ADM and Bondi mass apply to any asymptotically flat spacetime. But there are asymptotically flat spacetimes which are not stationary, and there are stationary spacetimes which are not asymptotically flat.

In fact, even in cases where both definitions apply--spacetimes which are both stationary and asymptotically flat--the different definitions don't always give the same answer. The simplest counterexample is Schwarzschild spacetime. It is a vacuum solution, so the stress-energy tensor is zero everywhere, which means the Komar mass is zero. But the ADM mass (and Bondi mass, which for any stationary spacetime is equal to the ADM mass) is $M$, the "mass" that appears in the metric.

Unfortunately, no, it isn't.

20. Jun 24, 2017

Umaxo

Thanks, that actually answered all of my questions:)

and i also want to thank for all of the replies, they were very helpfull.