How Is Total Mass Calculated in General Relativity for a Finite Body?

[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

 

[pmb_phy]

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

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

 

[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.

 

[pmb_phy]

re "This will be observer dependant" - That was incorrect. I should have said "observer independant".

Pmb, your definition is not really useful in full GR.

I'm going by the definition as given in MTW.

There's no local energy density for you to use.

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.

You can define it to be the stress-energy tensor doubly contracted with an observer's 4-velocity,..

That's (relativistic) mass density and not rest mass. I made it clear that I was referring to rest mass of the system.

You therefore cannot use an observer "at infinity" in this way to get a unique answer.

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

 

[Stingray]

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.

Which is just T^ab v_a v_b. This is the energy density measured by a set of observers moving with 4-velocities v^a(x)=g^a0(x). You then go on to say that T^ab u_a u_b for an observer with 4-velocity u^a is not what you were talking about:

That's (relativistic) mass density and not rest mass. I made it clear that I was referring to rest mass of the system.

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.

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.

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.

 

[pervect]

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

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).

https://www.physicsforums.com/showthread.php?t=30020&page=7&pp=15


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

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.

 

[pmb_phy]

Which is just T^ab v_a v_b. This is the energy density measured by a set of observers moving with 4-velocities v^a(x)=g^a0(x). You then go on to say that T^ab u_a u_b for an observer with 4-velocity u^a is not what you were talking about:

Please go back and reread what I was saying and this time pay close attention to it. To be specific I stated

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.

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.

You're contradicting yourself.

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.

You're taking their statements out of context.

So you say.

This thread is not about total energy, ..

The topic of this thread is, and I quote, the "total mass of the body in GR".

Pmb

 

[blue_sky]

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).

https://www.physicsforums.com/showthread.php?t=30020&page=7&pp=15


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

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.

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

 

[pervect]

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.

[edit]
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.

 

[blue_sky]

A timelike Killing vector, which is what's needed for the particular formula I posted, exists only in static space-times.

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

 

[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.

 

[pervect]

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

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.

 

[pervect]

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.

adding...

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.

 

[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?

 

[blue_sky]

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...

Pervect, maybe I'm wrong but I don't 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

 

[pervect]

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

 

[blue_sky]

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

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

 

[pervect]

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

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.

 

[Garth]

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.

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

 

[pervect]

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

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.

 

[Garth]

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..

Yes the Einstein cylindrical universe is conformally flat.

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

It depends on how you define the energy to be conserved. My example was taking the energy/mass content of a co-expanding and co-moving typical volume of a Friedmann universe. You are correct about red shift, as my example was a dust universe there was no radiation in it. It was just a 'thought example'.
Garth

 

[Garth]

We are cross posting with the thread "Energy not defined in GR?" Of course mass and energy are closely related concepts so this is not surprising.

My quote from that thread,

Energy and momentum are frame dependent concepts; therefore it is necessary to define a frame of reference, a preferred frame in order to restore the principle of the conservation of energy.

The need to choose a particular frame, a 'preferred' frame' in order to conserve energy, and therefore in order to be able to define it consistently, disappears when space-time is flat. Therefore Wald's definition is entirety consistent with my point.

However there are frames in which it is possible to define a sort of energy in GR even in the presence of curvature that is when the metric is static. Metrics can only be static in the presence of mass and curvature in the frame of reference of the Centre of Mass of the system in question. This is my 'preferred frame', preferred not by arbitrary choice, but by the need to define a consistent and conserved energy. It is also that frame picked out by Mach's Principle.

As a matter of interest it this insight that is the foundation of the SCC theory of gravitation.

Garth

 

[blue_sky]

We are cross posting with the thread "Energy not defined in GR?" Of course mass and energy are closely related concepts so this is not surprising.

My quote from that thread,

The need to choose a particular frame, a 'preferred' frame' in order to conserve energy, and therefore in order to be able to define it consistently, disappears when space-time is flat. Therefore Wald's definition is entirety consistent with my point.

However there are frames in which it is possible to define a sort of energy in GR even in the presence of curvature that is when the metric is static. Metrics can only be static in the presence of mass and curvature in the frame of reference of the Centre of Mass of the system in question. This is my 'preferred frame', preferred not by arbitrary choice, but by the need to define a consistent and conserved energy. It is also that frame picked out by Mach's Principle.

As a matter of interest it this insight that is the foundation of the SCC theory of gravitation.

Garth

Thanks garth... where can I read more on that subject?

blue

 

[Garth]

Thanks garth... where can I read more on that subject?

blue

If the subject you are referring to is Self Creation Cosmology the relevant SCC papers are:
i. [The original paper, Barber, G.A. : 1982, Gen Relativ Gravit. 14, 117. 'On Two Self Creation Cosmologies'.]
ii "'A New Self Creation Cosmology, a 'semi-metric' theory of gravitation'," http://www.kluweronline.com/oasis.htm/5092775, Astrophysics and Space Science 282: 683–730, (2002).
but the new theory can be recovered in five electronic eprints that followed;

iii "The Principles of Self Creation Cosmology and its Comparison with General Relativity",
http://arxiv.org/abs/gr-qc/0212111

iv "Experimental tests of the New Self Creation Cosmology and a heterodox prediction for Gravity Probe B", http://arxiv.org/abs/gr-qc/0302026 .

v. 'The derivation of the coupling constant in the new Self Creation Cosmology',
http://arxiv.org/abs/gr-qc/0302088 .

vi "The Self Creation challenge to the cosmological concordance model"
http://arxiv.org/abs/astro-ph/0401136 .

and

v "Self Creation Cosmology - An Alternative Gravitational Theory"
http://arxiv.org/abs/gr-qc/0405094 to be published in "Progress in General Relativity and Quantum Cosmology. " Nova Science Publishers, Inc. New York..


You could start with the last first - if you see what I mean!
- Garth

 

[pervect]

OK, there are some very interesting additional equaitons in MTW for a star with constant proper density rho, and a schwarzschild radius R, and mass M

These are also presented in page 15 of a web reference below

http://www.pma.caltech.edu/Courses/ph136/yr2002/chap25/0225.1.pdf

The total mass can be written simply as $$\frac{4}{3}\pi \rho R^3$$

Pressure as a function of depth is given by eq 25.44 in the URL above. 2M/R must be less than 8/9, or the pressure becomes infinite at the center.

This simple result might deceive one at first glance into thinking that one can just integrate T₀₀ over the volume. This is not at all the case. The density, rho, is constant in proper coordinates, but it is not constant in Schwarzschild coordinates. In Schwarzschild coordinates

$$T^{00} = \rho e^{-2 \Phi}$$

where $$\Phi$$ is given by eq. 25.45 in the URL above.

Similarly, the proper volume dV for a spherical shell of radius r and thickness dr (in a local coordinate system where the density is the constant rho) is:

$$dV = 4 \pi r^2 e^\Lambda dr$$

where $$e^\Lambda$$ is given by 25.36 in the URL

I haven't checked out the formulas from Wald for mass I gave using this particular metric yet, but they should work.

[edit]
I've done some more checking, and it turns out that v^a = [1,0,0,0] is not a Killing vector inside the star for the constant density case. You have to make the time component a function of r. So the Killing vector adds an additional scale factor to the intergal I presented, it will not just pick out the first component of T₀₀.
[end edit]

Incidentally, it turns out that 3 * the volume intergal of the pressure _in Newtonian theory_ is equal to the gravitational binding energy - MTW, pg 607, exercise 23.7 - so it's logical that the binding energy in the general case is proportional to 3x the intergal of the pressure, as this is the Newtonian formula for binding energy.

[edit-add]

Some numerical results from MTW are interesting:

For the criitical case, where 2M/R = 8/9, the maximum possible mass

the total normalized mass M is .838, the binding energy is .536, the mass that you'd get if you diassembled the star or integrated the energy-density over the actual volume of the star is 1.374

These are in rather strange units of sqrt(3/32 * pi * rho) (geoemtric units)

 

[pmb_phy]

Also, this mass is not conserved, and is not equal to the mass that pervect presented.

Its conserved in the sense that Tab[/sub]_; = 0.

I was thrown off by your statement about observers at infinity, which is incorrect.

I made no statement about observers at infinity.

There are notions of energy defined by observers at infinity, but they are not the same as what you were talking about.

I was talking about mass, not energy. I don't consider them to be the same thing since in general they are not e.g. see Rindler's SR text on this point in section on relativistic mechanics of continuum).

Pmb

 

[blue_sky]

This simple result might deceive one at first glance into thinking that one can just integrate T₀₀ over the volume. This is not at all the case. The density, rho, is constant in proper coordinates, but it is not constant in Schwarzschild coordinates. In Schwarzschild coordinates

$$T^{00} = \rho e^{-2 \Phi}$$

where $$\Phi$$ is given by eq. 25.45 in the URL above.

Similarly, the proper volume dV for a spherical shell of radius r and thickness dr (in a local coordinate system where the density is the constant rho) is:

$$dV = 4 \pi r^2 e^\Lambda dr$$

I haven't checked out the formulas from Wald for mass I gave using this particular metric yet, but they should work.

Pervect, my original question was also rised because also using the "simple" Schwarzschild solution you must use
$$dV = 4 \pi r^2 dr$$
Instead of the right formula for $$dV$$ that u mentioned before to obtain the total mass.

blue

 

[pervect]

Any one of the several embedding diagrams on the WWW will show that the area of an equatorial slice through a massive star is greater than 4*pi*R^2, where R is the radius of the star in Schwarzschild coordinates

for instance :

wikipedia

This implies that the volume inside the Schwarzshild radius is greater than 4/3 pi r^3, just as the area is greater than 4*pi*r^2.

It is indeed interesting that we get the right answer for total mass when we multiply the density by the non-physical "volume" expressed in Schwarzschild coordinates (and ignore binding energy). It even works when the density is a function of radius.

I'm afraid I don't find MTW's explanation of why this occurs particularly illuminating, I'll quote some of it later on.

For me, associating the energy with a time-like symmetry, and applying Noether's theorem, is a much more convincing argument for talking about the existence and distribution of energy inside a spherical star than MTW's argument. The downside is that the math to compute the energy distribution via the Killing vector approach doesn't seem to be easy (GRTII/maple so far hasn't found a solution I want to deal with).

Here is MTW's argument, as written. Note that in a snipped section, they mention that this argument only works for spherically symmetrical space-times, a very specialized case and different from any other of the specialized cases I have been considering so far.

In Schwarzschild coordinates it [the mass-energy] is defined by

"total mass energy inside radius r" = $$\int_0^r 4 \pi r^2 \rho dr$$

The fully convincing argument for this definition is is found only by considering a generalization of it to the time-dependent spherically symmetric stars. [some references to other parts of the text trimmed]. For them one finds that the mass-energy m associated with a given ball of matter (fixed baryon number) can change in time only to the extent that locally measurable energy fluxes can be detected at the boundry of the ball. [Such energy fluxes could be the power expended by pressure forces against the moving boundry surface, or heat fluxes, or radiation (photon or neutrino fluxes). But since spherically symmetric gravitational waves do not exist (chap. 35 and 36), neither physical intuition or Einstein's equation require that the problem of localizing gravitational wave energy be faced.] Thus the energy m is localized, not by a mathematical convention, but by the circumstance that transfer of energy (with this defintion of m) is detectable by local measurements.

Anyway, that's the quote from my textbook, but I can't say I find it all that convincing personally.

I am convinced that MTW's defintion for m(r) gives the right answer outside the star, because that can be checked by looking at the far-field outside the star.

 

[pmb_phy]

This simple result might deceive one at first glance into thinking that one can just integrate T₀₀ over the volume.

Nothing is simple in GR.


I made this mistake in this thread. I incorrectly stated that the mass of a body (e.g. mass of a star) was obtained by integraring T₀₀ over the volume. I had seen this in MTW and didn't pay too much attention to it.

However I see that what I read in MTW was not quite what I assumed it was.

Since all forms of energy have mass and thus contribute to the gravitational field it follows that gravitational energy must also contribute to the total mass of a body. The gravitational field contributes to the mass-energy through its effects on the metric. Recall that the energy of a particle does not include simply kinetic energy and rest energy. It also includes gravitational energy which makes its effects know through the "gravitational potentials" g_ab. Just because the total energy is not a linear sum of proper energy, kinetic energy and gravitational energy it would be incorrect to think that its not there at all. "Not well defined .." should never be mistaken for "does not exist/is not there". Life exists. However the term "life" is not well defined.

In the above formula for the mass of a body one has to consider that the particles which make up the fluid (that is used as a source in the derivation that pervect refers to) are emersed in a g-field and that this g-field affects their gravitational mass.

If you integrate $\rho(r)$ over the coordinate volume of a spherically symmetric mass distribution then you'll get the total mass, M, of the star. If you integrate $\rho(r)$ over the proper volume then you'll get the total proper mass, M_p. The difference between these masses is the gravitational binding energy.

This is not at all the case. The density, rho, is constant in proper coordinates, but it is not constant in Schwarzschild coordinates.

[itex]rho, proper mass density, is a scalar and is thus independant of the coordinate system used. The quantity [itex]T_{00}[/tex] is the rest mass-energy density. I'm not 100% sure but I believe that the term you used, i.e. [itex]T^{00}[/tex], shouldn't be thought of as the energy density. The energy density of a fluid is the sum of the energies of the individual particles which make up the fluid. The energies of these particles is not P⁰, its P₀. This sum must be equal to [itex]T^{00}[/tex]. But let me get back to you on this point.

I haven't checked out the formulas from Wald for mass I gave using this particular metric yet, but they should work.

Its in Wald on page 126 and in Schutz on page 259 (or there about).

Pete

 

[blue_sky]

I'am startint to think thaf in full GR the integral over a finite volume don't have any phisical meaning; what is describing the full GR world are only the local equations.
So in full GR is a non sense to define the mass of a body with finite dimentions.
Any of you support this?

blue

 

[Garth]

In GR you need to have invariant particle (rest) mass in order to make measurements of mass, length and time. If an atom increases in mass, for example, it will shrink and 'vibrate' more quickly. Rulers will likewise shrink and clocks speed up relative to the original standards of measurement.

Garth

 

[blue_sky]

In GR you need to have invariant particle (rest) mass in order to make measurements of mass, length and time. If an atom increases in mass, for example, it will shrink and 'vibrate' more quickly. Rulers will likewise shrink and clocks speed up relative to the original standards of measurement.

Garth

Actually my thoght is that an integral over a finite 3 volume in full GR doen't make sense because c is finite and there is no rule which say how to relate 2 different point of the 3 volume (the 3 volume integral assume to make the integral at the same time for all the points... but this can't be done in full GR due to the finite speed of light, at least can be done only in particular cases)... so in general doen's make sense the total mass of a body.
What you think about that?

 

[Garth]

What I think is that it is GR that needs to be modified, but that is a personal (and how!) opinion!
There are two issues here for GR, or any physical theory, the first is how to apply the theory's equations consistently, the question you raise above, i.e. 'doing the mathematics', and the second is how to relate the terms and definitions in those equations to physical objects and measurements. In order to have any correlation between the GR theory world and the real world the theory demands that particle (rest) masses have to be invariant, otherwise we would be in an 'Alice Through The Looking Glass' world in which the mathematical terms used would lose, or at least change, their meaning.

If we now consider an extended gravitating body with particle number conservation then in GR the total rest mass of all those particles has to be constant. The question is then how to add on the mass equivalent of all the energy fields present, especially the binding energy of the gravitational field itself. It is here that the going gets tough!
- Garth

 

[pervect]

Actually my thoght is that an integral over a finite 3 volume in full GR doen't make sense because c is finite and there is no rule which say how to relate 2 different point of the 3 volume (the 3 volume integral assume to make the integral at the same time for all the points... but this can't be done in full GR due to the finite speed of light, at least can be done only in particular cases)... so in general doen's make sense the total mass of a body.
What you think about that?

It sounds like you've read the sci.physics.faq on energy in GR :-) Of course there are some important special cases where this issue can be resolved.

 

[Garth]

In relating the total mass of a body to the equations used in GR we insert a mass M into the gravitational potentials of the metric describing the gravitational field around a spherical body, the Schwarzschild solution. But where does this M come from, what value do we attribute to it? It is actually defined by the fitting of the GR solution to the Newtonian, the normalisation of the Robertson parameter alpha to be unity. Hence M is defined to be the mass of the body as calculated from the orbit of a distant satellite (where space-time is 'nearly' flat) using Kepler's laws.

It is also necessary, as you have obviously been thinking, to make the rest of the theory consistent, hence the need to define by different routes the mass of a gravitating body in a consistent way. Its not easy away from that asymptotic null infinity!

Garth