Mass of a body

1. Oct 7, 2004

blue_sky

Let consider a body of finite dimentions and known density. If I ask: which is the total mass of the body in GR
a) make the question sense?
b) if yes, how I can calculate the total mass?

blue

2. Oct 7, 2004

pmb_phy

Yes. The question makes sense. But the answer is the same in GR as it is in SR. The total (rest) mass of a body is found by, in the rest frame, intetgrating the energy density over the volume of the body and then dividing by c2. This will be observer dependant. But when most people speak of the rest mass of such a body in GR they are evaluating this integral in a specific frame, i.e. as measured by a far-away observer. The mass then obtained is unique.

Pete

3. Oct 7, 2004

Stingray

Pmb, your definition is not really useful in full GR. There's no local energy density for you to use. You can define it to be the stress-energy tensor doubly contracted with an observer's 4-velocity, but this is defined only along a single line. You can define a surface "orthogonal" to the observer in a particular sense, but it cannot be extended to (spacelike) infinity in most cases. You therefore cannot use an observer "at infinity" in this way to get a unique answer.

That said, there does exist a formalism (by WG Dixon) which deals with extended bodies in full GR. In it, it is possible to define an observer-independent "rest mass" for a compact body. The prescription is vaguely similar to what you have described, but it makes no use of observers at infinity to get uniqueness. It is also not a constant, and nobody really knows how it relates to things like the ADM mass. So don't take the label of "mass" too seriously.

There are some other quasilocal mass definitions in GR as well, but all of them are quite complicated. It is also not clear exactly how they should be interpreted. To be honest, I think everyone would be a lot happier if they stopped trying to force particular concepts into theories which do not naturally admit them.

4. Oct 12, 2004

pmb_phy

re "This will be observer dependant" - That was incorrect. I should have said "observer independant".
I'm going by the definition as given in MTW.
Sure there is. T00. I believe that when you define mass in this way the gravitational energy which contributes to this mass is built into it.
That's (relativistic) mass density and not rest mass. I made it clear that I was refering to rest mass of the system.
Why? This is exactly how MTW state that it is to be measured. I don't have the text at hand but will try to remember the page number next time I'm near the internet.

Pmb

5. Oct 12, 2004

Stingray

Which is just Tab va vb. This is the energy density measured by a set of observers moving with 4-velocities va(x)=ga0(x). You then go on to say that Tab ua ub for an observer with 4-velocity ua is not what you were talking about:

You're contradicting yourself. (I meant to imply an integration over the density to obtain a mass if that's what you were complaining about.)

Also, one usually wants something called rest mass to be a coordinate invariant. T00 does not have this property. Even neglecting the problems with choosing an appropriate surface of integration, it is different in every coordinate system. Nobody uses gauge-dependent expressions like this except sometimes in the context of a particular approximation scheme.

It makes more sense if you define a reference frame by a uniquely defined pointlike observer. This is often possible locally, but a suitable surface of integration cannot in general be extended over large distances.

You're taking their statements out of context. Energies can be defined for observers at infinity when the spacetime possesses a Killing vector or when you are measuring the total energy of the spacetime. This thread is not about total energy, and most spacetimes do not have Killing vectors.

6. Oct 12, 2004

pervect

Staff Emeritus
If you also assume that the geometry of space-time is asymptotically flat at infinity, you can determine the total mass.

I've already presented the basic formula for this in another thread (I think it's several other threads by now). This formula can be found in Wald's general relativity (where I got it).

$$M = 2 \int_{\Sigma} (T_{ab} - \frac{1}{2} T g_{ab}) n^a \xi^b dV$$

Here Tab is the stress-energy tensor, gab is the metric tensor, na is the unit future normal to , and is the Killing vector representing the time translation symmetry of the static system. T is the trace of Tab.

Note that Pete's answer is not correct, there is more to determining the mass of a body than integrating the energy density.

To compute T_ab, assuming it's a perfect fluid, you'll need both the density (which will set T_00), and the pressure. If you assume a real fluid, you'll need an "equation of state" - the density of the fluid vs. it's pressure.

7. Oct 13, 2004

pmb_phy

Please go back and reread what I was saying and this time pay close attention to it. To be specific I stated
That quite litterally means that T00 is the rest energy density and not the quantity T(u,u)/c2 which is different. T(u,u) is the (relativistic) mass density and the two are most definitely not the same. I fail to see what part of that is unclear or how you could possibly have misunderstood what I stated.
Nope. Sorry. Absolutely not. I said to evaluate T00[/sub]2 in the rest frame and that is, by definition, the rest mass density (but not the proper mass density). It has a unique value and as such it is not observer dependant.
So you say.
The topic of this thread is, and I quote, the "total mass of the body in GR".

Pmb

8. Oct 13, 2004

blue_sky

Thanks; it's the Killing vector always existing in full GR?

9. Oct 13, 2004

pervect

Staff Emeritus
A timelike Killing vector, which is what's needed for the particular formula I posted, exists only in static space-times. If your body is not imploding or exploding, it should represent a static space-time, so that shouldn't be a problem.

Your system will still have a well defined energy even if you don't have a static space-time as long as it's an asymptotically flat space-time. However, you can't use the formula above directly for a non-static space-time, at least not without a little more work to extend the concept of a Killing vector slightly to represent an asymptotic time-translation symmetry. When you do this extension, you'll get a formula for the mass of your body, which is known as the Bondi mass.
[end edit]

I should probably point out that if your body is spherically symmetrical, the exterior metric (and thus the exterior gravitational field) of your body should be given by the same formulas that apply for a black hole of the same total mass as your body - a Schwarzschild metric if your body is not rotating, and a Kerr-Newmann metric if your body is rotating.

Since your body is not a black hole (or, perhaps I should say assuming your body isn't a black hole), the size of the body will be larger than the event horizon of the black hole of the same mass. This means that the region outside the body where the black-hole formulas work will be outside the event horizion of the black hole.

Last edited: Oct 13, 2004
10. Oct 13, 2004

blue_sky

Static... in full GR... looks difficult to accept.
If the body is a planet orbitating you can't then speak of total mass in full GR?

blue

11. Oct 13, 2004

Stingray

pmb,
I'm sorry I apparently did your statements a little too quickly. I agree with the basic idea of your definition of rest mass. Just as a practical note, it is difficult to define a unique rest frame (but possible), and doing the integral is very difficult in general (it requires bitensors). Also, this mass is not conserved, and is not equal to the mass that pervect presented.

I was thrown off by your statement about observers at infinity, which is incorrect. There are notions of energy defined by observers at infinity, but they are not the same as what you were talking about. A common, very general definition is called the ADM energy. It is the generalization that pervect mentions of the formula he gave above, and is a constant.

By the way, I said that this thread was not about total energy because usually we don't think of the universe as consisting of only one object (or at least I don't). In GR, it is not a simple matter to generalize things. The ADM energy gives the total energy of the spacetime. It cannot make any local statements. So it can't be used to find individual masses in a binary system -- just the total mass.

12. Oct 14, 2004

pervect

Staff Emeritus
I'm not sure why - static means just that the metric coefficients are constant. I don't see what's difficult to accept, what your problem is. I apologize if this is getting too technical, but you did ask for a formula...

If you have a planet orbiting a star, you can talk about the total mass of the planet-star system in GR, but strictly speaking you can NOT talk about the mass of the planet.

Actually, though, the binding energy of the planet to the star is usually so small it can be neglected. So you can get an approximate mass of the planet that's good enough for practical purposes. This only works if gravity is weak enough, though - it's probably fine for a planet orbiting a star, but it probably would have problems for two black holes closely orbiting each other.

13. Oct 14, 2004

pervect

Staff Emeritus
I was thinking, and I realized there's a conceptually rather simple explanation for the general form of the formula I gave, taking the simplest case (spherical symmetry, along with the previously assumed static spacetime and asymptotic flatness). We can break down the intergal into two parts:

1) What I will term he total "bare" energy the system, given by the intergal of the energy density T00 over the volume. This is the usual weak field approximation to the total mass.

2) The binding energy of the system, given by the intergal of (T11+T22+T33) * g00 over the volume. This is just the intergal of the locally measured pressure * the volume * g00, which corrects the pressure as mesured by a local observer to the pressure seen by an observer at infnity, times a scaling constant (I make the scaling constant out to be a factor of 3 - it's origin isn't totally clear to me yet).

We get the total energy by taking the "bare" energy minus the binding energy.

And here's a little more rationale as to why we can separate the energy out in this fashion. Suppose we have a large cloud of diffuse gas. If the cloud is large enough, the gravity is weak, and we can use the weak-field approximation of integrating T00 over the volume.

Now we allow the system of diffuse gas to collapse gravitationally, to a state where gravity is strong. The total mass must remain the same during the collapse (assuming that this occurs in an asymptotically flat space-time where energy is conserved). The intergal of T00 increases due to the energy gained during collapse, but that increase is due to the binding energy of the system which we subtract out of the integral. The result is that energy is conserved.

Last edited: Oct 14, 2004
14. Oct 14, 2004

Garth

There is a measurement problem.
If we define mass as the integral of density, then how do we measure density? By measuring mass andtaking the volume differential?

15. Oct 14, 2004

blue_sky

Pervect, maybe I'm wrong but I dont understand if it is possible, in general and w/o simplification, to define the total mass of a body in full GR.
A metric with coefficients that are costant looks to me a semplification. If you look at the universe, can we use a static metric to describe it? If not, the simplification used to define the totall mass looks to me too strong.

blue

Last edited: Oct 14, 2004
16. Oct 14, 2004

pervect

Staff Emeritus
One needs special conditions to define energy in General Relativity. Totally arbitrary space-times do not conserve energy in General Relativity. The metric with constant coefficients, a static space-time, is one way of keeping energy conservation that's particularly simple, but it's not the only way. The more general way is asymptotic flatness of the space-time, this is a boundary condition to the solutions of the field equations. This is the usual approach to energy conservation in GR.

The math behind defining the energy of a body even in a space-time that's both static AND asymptotically flat is messy enough that it's all I'd want to attempt to cover in a discussion - if you want more detail as to how it's derived for non-static (but asymptoticall flat) space-times you'll need to go to a textbook, if my discussion is not enough.

With asymptotic flatness as a boundary condition, one can define the energy of an isolated system. Without it, one cannot (except for the special case where the space-time is static - and even here, one runs into a problem of setting the proper scale factor).

You might find the sci.physics.faq "Is energy conserved in Genral Relativity" helpful

http://math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html

17. Oct 14, 2004

blue_sky

Thanks.
What happens if we consider the space (not time) closed; still valid the energy conservation?

18. Oct 15, 2004

pervect

Staff Emeritus
I think the answer depends on the details of how space is closed. Unless I'm mistaken (unfortunately, this is all too possible), the Einstein static uiverse is asymptotically flat, so it would be an example of a spatially closed universe that does meet your conditions.

A more typical Friedmann cosmology (one where the density of matter exceeded the critical limit, one that would be expanding and then later contracting) would not (IMO again) conserve energy.

You might try posting this question to sci.physics.research to see if you can get someone else to give an answer.

19. Oct 15, 2004

Garth

The Einstein static universe is cylindrical, it is space-like closed and time-like static and eternal. However, it would also be unstable w.r.t. small fluctuations and was the basis of Lemaitre's 'coasting' universe. You may be thinking of the Einstein-de Sitter model which was flat.

However if there are no local, or peculiar, motions, with zero pressure (a dust universe), so the universe simply expands with Hubble flow, then energy is conserved in a Friedmann universe as it is equal to energy-momentum.
- Garth

Last edited: Oct 15, 2004
20. Oct 15, 2004

pervect

Staff Emeritus
When Wald defines asymptotic flatness, he does it by conformally mapping a flat space-time to a cylindrical Einstein universe. But there are a bunch of other conditions on the mapping that get very involved. Still, I'd be surprised if a trivial mapping didn't satisfy them (these conditions on the mapping). So that's why I picked an Einstein universe as being liikely to conserve enregy. But I haven't worked out all the details, and I might be missing something major. It's a somewhat odd argument, the usual idea behind asymptotic flatness is making infninity a place via a conformal transform.

I don't think the standard closed Friedmann universe will conserve energy, due to cosmological redshift - and it's not even spatially flat.