The Schwartzschild solution - why no Stress Tensor?

[TerryW]

I've been working my way through GR using mainly D'Inverno's Introducing Einstein's General Relativity, but with MTW as well and a couple of other books. I began to get a sense of what was going to happen and got a big surprise when I reached the Schwartzschild solution. The process goes something like this - we solve the equations for a spherically symmetric vacuum and lo and behold, the solution tells us that there is a mass at the middle! But it is a mass which doesn't generate a stress tensor. I've looked through all my books and the only book that even mentions this apparent illogicality is Penrose's Road to Reality, and he just makes reference to another book by Pais. Why does no one want to talk about this? Can anyone explain?

 

[stevenb]

Why does no one want to talk about this? Can anyone explain?

My guess is that this is considered so trivial by experts that it does not need to be mentioned, or stressed. I'm not sure why they would feel this way because many people (including myself) ask this question when we first learn the subject. Trust me, once you get it, you'll slap your head and say "of course, so simple".

The Schwartzschild solution is the solution outside the region where there is a Stress Energy Tensor. For example, the exterior of a star is a region where there is no mass/energy, (i.e. vacuum is assumed). The full exact solution (including the star interior) would need to be modified, and the boundary condition would need to match at the surface of the star. However, as long as certain conditions are met (spherical symmetry being an important one), the exterior solution is not dependent on the structural details of the interior stress energy tensor.

I think one thing that trips us up in the early stages of learning (you may have passed this stage however) is that we somehow think that there can be no spacetime curvature (at a point) if there is no stress energy tensor (at that point), but this is false of course. Einstein's equations are in terms of the Ricci tensor, not the full Riemann curvature tensor. The Ricci tensor is a partially contracted version of the Riemann tensor, and the Ricci scalar is a fully contracted Ricci tensor. Hence, even though the Ricci tensor is zero in vaccum, there can still be spacetime curvature in regions where there is no stress energy tensor (that is, in vacuum). Gravity waves is another example where there is curvature (i.e. nonzero Riemann tensor), but the Ricci tensor is zero. This (initially) unintuitive result would not happen in a hypothetical 3 dimensional world, but it does happen in our real 4 dimensional spacetime.EDIT: By the way, just as a point of interest, I'm sure you haven't missed the direct similarity with Newton's theory of gravity, in which the exterior gravitational field is independent of the actual structure of a spherically symmetric mass. You also remember that Newton's theory shows the gravitational field inside a spherically symmetric outer shell of mass to be zero. Similarly, a corollary of Birkoff's theorem is that the vacuum metric inside a spherically symmetric stress energy tensor is the Minkowski metric. So, a full solution of a problem with vacuum both outside and inside a nonzero spherically symmetric stress energy tensor region would require matching the boundary conditions with the outer Shwartzschild solution and the inner Minkowski solution, with a more complicated solution (have fun solving that one! ) in the region between.

 

[PeterDonis]

The Schwartzschild solution is the solution outside the region where there is a Stress Energy Tensor. For example, the exterior of a star is a region where there is no mass/energy, (i.e. vacuum is assumed). The full exact solution (including the star interior) would need to be modified, and the boundary condition would need to match at the surface of the star. However, as long as certain conditions are met (spherical symmetry being an important one), the exterior solution is not dependent on the structural details of the interior stress energy tensor.

I agree with this as far as it goes, but there's one point it doesn't cover: what about black holes? A black hole does not have an "interior" where there is a nonzero stress-energy tensor: the stress-energy tensor is zero right down to the singularity at r = 0.

Mathematically, the "eternal black hole" spacetime (the full Schwarzschild spacetime geometry, including the singularity, which is completely static in time, so it lasts forever in both "time directions"--from t = negative infinity to t = positive infinity) is a perfectly valid solution to the Einstein Field Equation in vacuum (zero stress-energy tensor); the fact that it appears to have a "mass" present that can be externally measured, for example by putting objects in orbit about the hole and plugging their observed orbital radii and periods into Kepler's Third Law, is an "illusion", so to speak, caused by the curvature of the spacetime.

Physically, however, this picture is not very satisfactory, because we would like an account of how this "eternal" black hole came into existence, and there isn't one--the mathematical solution just says "it's always been there", which isn't physically reasonable. Instead, we expect, physically, that an actual black hole spacetime will have come into being by the collapse of a sufficiently massive object such as a star. This works basically the way stevebd described--you match the "exterior" solution, which is Schwarzschild, with an "interior" solution that describes the star, the simplest one being a perfect fluid--except that it isn't static: the star decreases in size with time, until finally it is small enough that a black hole horizon forms around it (and at that point, the "exterior" vacuum Schwarzschild geometry must include a portion inside the horizon). Soon after that ("soon" as it would be seen by an observer riding on the surface of the collapsing star), the matter in the star collapses to zero radius and forms the r = 0 singularity; but that's hidden behind the horizon so it isn't visible externally. All that remains outside the horizon (and inside it, down to the singularity) is the vacuum spacetime geometry that now bears the "imprint" of the mass of the star that collapsed. This is possible because, as stevebd said, there can be spacetime curvature (usually called "Weyl curvature") even where there is no stress-energy; but again, physically, we expect that this curvature is there because, at some time in the past, some nonzero stress-energy was somewhere in the vicinity and left its "imprint" on the spacetime geometry as it passed.

 

[bcrowell]

Mathematically, the "eternal black hole" spacetime (the full Schwarzschild spacetime geometry, including the singularity, which is completely static in time, so it lasts forever in both "time directions"--from t = negative infinity to t = positive infinity) is a perfectly valid solution to the Einstein Field Equation in vacuum (zero stress-energy tensor); the fact that it appears to have a "mass" present that can be externally measured, for example by putting objects in orbit about the hole and plugging their observed orbital radii and periods into Kepler's Third Law, is an "illusion", so to speak, caused by the curvature of the spacetime.

IMO the "reality" versus "illusion" distinction isn't usually useful in and of itself. However, you could certainly object the "point" where this mass lives isn't actually a point on the manifold.

Physically, however, this picture is not very satisfactory, because we would like an account of how this "eternal" black hole came into existence, and there isn't one--the mathematical solution just says "it's always been there", which isn't physically reasonable.

I could make the same complaint about an electron.

 

[PeterDonis]

IMO the "reality" versus "illusion" distinction isn't usually useful in and of itself.

I don't necessarily disagree, which is why I put "illusion" in scare-quotes. I was referring to the fact that, as you say:

the "point" where this mass lives isn't actually a point on the manifold.

So technically, there isn't a "real" mass there. But physically, we would say that GR as a theory breaks down at the singularity, and some new physics (presumably a quantum gravity theory) has to take over.

I could make the same complaint about an electron.

True (although a single electron is not the same as an entire spacetime). However, I probably should have clarified that I wasn't thinking of the general question of why anything exists at all, or whether it's possible for something to come into existence out of nothing. I was thinking of an issue that's specific to the "eternal" black hole spacetime. The issue has two parts: (1) the fact that the spacetime is asymptotically flat, and (2) the fact that it contains a "white hole" portion as well as the "black hole" portion.

(1) means that the black hole spacetime is not really self-contained: in order to explain how one could come into existence, you have to explain where the asymptotically flat region comes from. For a black hole that forms from a collapsed star, that's easy: the asymptotically flat region doesn't actually extend to infinity, it just "merges" into the overall background spacetime of the universe as a whole. But that kind of black hole spacetime doesn't include the white hole portion. For an "eternal" black hole that (2) does include the white hole portion, I'm not sure this way of viewing the asymptotically flat region will work, because the white hole has a "past horizon" associated with it--particles can come out of it, but not go into it. I'm not sure there's a way to merge that kind of spacetime structure into an overall background spacetime like the ones we know of in cosmology.

 

[atyy]

The Schwarzschild solution does not say there is a mass at the middle.

The M in the Schwarzschild solution is a "mass parameter", but the pure Schwarzschild solution is a vacuum solution with an (unphysical?) singularity.

In modelling the solar system the pure Schwarzschild solution is not used. The Schwarzschild solution models only the outside of the sun. The inside of the sun is modeled with another solution of Einstein's equation in which matter does have a stress-tensor (one such solution is called the Schwarzschild interior solution, not to be confused with any part of the pure Schwarzschild vacuum solution)

 

[bcrowell]

The Schwarzschild solution does not say there is a mass at the middle.

The M in the Schwarzschild solution is a "mass parameter", but the pure Schwarzschild solution is a vacuum solution with an (unphysical?) singularity.

In modelling the solar system the pure Schwarzschild solution is not used. The Schwarzschild solution models only the outside of the sun. The inside of the sun is modeled with another solution of Einstein's equation in which matter does have a stress-tensor (one such solution is called the Schwarzschild interior solution, not to be confused with any part of the pure Schwarzschild vacuum solution)

I totally disagree. The Schwarzschild solution does say there is a mass at the middle. It's not just a parameter. It's a mass. If you're dealing with the sun, it's the mass of the sun. If you're dealing with a black hole, it's the mass of the black hole.

 

[atyy]

I totally disagree. The Schwarzschild solution does say there is a mass at the middle. It's not just a parameter. It's a mass. If you're dealing with the sun, it's the mass of the sun. If you're dealing with a black hole, it's the mass of the black hole.

What's your definition of mass?

 

[FunkyDwarf]

One could define it as the mass observed at infinity.

EDIT: Sorry, yes defining something with itself is bad. Make that gravitational potential at infinity.

 

[PeterDonis]

What's your definition of mass?

It's what you measure by putting an object in an elliptical orbit about the black hole, and measuring the orbital period and the semi-major axis, and plugging into Kepler's Third Law. The orbit will be the same regardless of whether the central object is a star or a black hole of the same mass. So the term "mass" is justified for the black hole as well as the star, even though the black hole is a vacuum solution.

 

[atyy]

One could define it as the mass observed at infinity.

EDIT: Sorry, yes defining something with itself is bad. Make that gravitational potential at infinity.

It's what you measure by putting an object in an elliptical orbit about the black hole, and measuring the orbital period and the semi-major axis, and plugging into Kepler's Third Law. The orbit will be the same regardless of whether the central object is a star or a black hole of the same mass. So the term "mass" is justified for the black hole as well as the star, even though the black hole is a vacuum solution.

So mass does not exist unless infinity does.

By this definition, does mass exist in our universe?

 

[PeterDonis]

So mass does not exist unless infinity does.

By this definition, does mass exist in our universe?

The elliptical orbit we use to measure the central mass will have a finite semi-major axis and a finite orbital period. So no infinity is required to do the actual measurement.

If you mean that the values we plug into Kepler's Third Law need to be "at infinity" for the law to be correct, technically that's true if you use the straightforward Newtonian definition of the law. But that's an idealization; actual measurements are always made at finite times and distances, and an *exact* calculation would not use the bare Newtonian form of Kepler's Third Law, it would have to include relativistic corrections because spacetime at a finite radius is not flat. But for many real-life problems those corrections are too small to measure.

And yes, there is plenty of mass by this definition in the universe--even though the example that prompted the definition wasn't about the universe, it was specifically about a black hole. In fact, this definition is part of what makes cosmologists believe there is dark matter in the universe: we measure the Keplerian orbits of objects in galaxies and find that there is more mass required to account for the orbital parameters than we can actually see in those galaxies. So we deduce that there must be additional mass there that isn't visible. (I say "part of" because it's not only the Keplerian orbits of individual objects, but how the orbits vary with distance from the center of the galaxies, that doesn't match with the visible mass; that variation requires adding something like a continuous mass distribution of dark matter to the model, not just additional dark matter "at the center". See next paragraph.)

There is a more general definition of "mass" that can be applied to situations like the universe as a whole. Mass, or energy, is the 00 component of the stress-energy tensor. Even more generally, the stress-energy tensor itself can be thought of as playing a role analogous to "mass" in Newtonian gravity, since all of its components contribute to "gravity" through the Einstein Field Equation. Which, of course, takes us back to the original question. So the really most general answer is, the precise definition of "mass" depends on the specific model you're using, and what you're using it for.

 

[pervect]

The stress-energy tensor does define a notion of mass density. However, integrating the 00 component of the stress-energy tensor (density per unit volume) in general doesn't even define a covariant quantity (one that is independent of coordinates), so the stress energy tensor does NOT directly provide a suitable definition of "mass". This shouldn't be surprising - taking just one component out of a tensor and integrating it isn't a good way to get a covariant quantity in general.

In the case of the Schwarzschild metric (one with a interior solution that has no singularities), one can make some additional restrictions that allow the notion of integrating the mass-energy over the volume to provide a covariant quantity - basically by defining a specific notion of volume that's orthogonal to the time-symmetry of the Schwarzschild metric.

However, the quantity thus computed will not be the mass of the system. It can, however, be thought of the mass "before assembly", and the difference between the mass as defined by the mass parameter and the mass "before assembly" defined in this manner can be explained by "gravitational binding energy".

There's a section in MTW's textbook that performs this calculation if a reference is wanted or needed.

However, there is in general no way to account properly for "gravitational binding energy" in the more general case - it's a special feature of the static geometry that allows it to be defined. In fact, there are strong arguments (presented in most textbooks, including MTW) that show that there's no way of defining a covariant notion that defines "where" gravitational binding energy is stored.

This interesting feature was realized rather early, in fact, and led to Noether's theorem, when Hilbert asked Emily Noether for help with the "poor behavior" of energy in General Relativity.

Note that when I talk about "gravitational binding energy", I include the scare quotes. In this case, the reason for the "scare quotes" is that this notion is defined in Newtonian mechanics, but is not in general defined in General Relativity (except in certain special cases, such as the static geometry being discussed).

 

[atyy]

The mass parameter is not the "physical" integral of the mass-energy (it is a "look-alike" integral).

 

[TerryW]

Thanks for your reply, stevenb, but I am still a long way from slapping my head! I wonder if this really is so trivial that the experts don't feel the need to mention it, or maybe they are a bit embarrassed about it!. Looking through the derivation of the Schwartzschild solution in D'Inverno, the factor -2m/r is introduced as a 'let's suppose the constant of integration is ... -2m/r". Well why not suppose it is +2m/r, or something else? You say that there can be spacetime curvature if there is no stress energy tensor, but how does this curvature come about. Where in the universe is there a curved vacuum spacetime without the presence of a mass somewhere? In MTW, there is a quote along the lines of 'matter tells spacetime how to curve and spacetime curvature tells matter how to move'. This seems to be pretty clear - no matter, no curvature. In Feynman's Lectures on Gravitiy, he says (p141) 'We cannot calculate the T ^UV for a system consisting of the Earth and the moon…' So it does look like there is a bit of a problem here.

 

[atyy]

In this post, by mass I mean a field that has localizable stress-momentum-energy.

In any realistic situation in which a part of the vacuum Schwazschild solution is used, it is "joined" to another solution of the Einstein equation in which mass is present. In such a situation, the mass parameter of the vacuum Schwarzschild equation is determined by the stress-momentum-energy in the non-vacuum part of the spacetime. (However the relationship between the mass parameter and the stress-momentum-energy is not completely straightforward.)

 

[bcrowell]

So mass does not exist unless infinity does.

By this definition, does mass exist in our universe?

This is actually correct. There is no definition of a scalar mass-energy that applies to all spacetimes in general relativity. There are only special-purpose definitions that work in certain types of spacetimes. One of those types is an asymptotically flat spacetime.

 

[bcrowell]

Thanks for your reply, stevenb, but I am still a long way from slapping my head! I wonder if this really is so trivial that the experts don't feel the need to mention it, or maybe they are a bit embarrassed about it!. Looking through the derivation of the Schwartzschild solution in D'Inverno, the factor -2m/r is introduced as a 'let's suppose the constant of integration is ... -2m/r". Well why not suppose it is +2m/r, or something else?

The t-t component of the metric can be interpreted in the weak-field limit as twice the gravitational potential. It has to match up with the Newtonian result in the weak-field limit.

You say that there can be spacetime curvature if there is no stress energy tensor, but how does this curvature come about. Where in the universe is there a curved vacuum spacetime without the presence of a mass somewhere?

What you're objecting to is a well-known feature of relativity. Since mass exists in the universe, we obviously can't point you to a region of space that has never experienced any nonzero effect from mass.

 

[stevenb]

You say that there can be spacetime curvature if there is no stress energy tensor, but how does this curvature come about. Where in the universe is there a curved vacuum spacetime without the presence of a mass somewhere?

Well yes, you do make a good point here. There does need to be a presence of mass (or more accurately T) somewhere. We can draw an analogy with electromagnetics. We know that there needs to be a source somewhere (the 4 vector J which includes charge density and current), but still we often study the wave equation without sources. In other words EM waves are allowed solutions in pure vacuum with no sources. This basically means that the waves may be generated by sources, and then the wave can travel far away to a pure vacuum region. In other words, we approximate a universe with sources with one without sources. Similarly, gravity waves can travel far from the mass source, and it is meaningful to talk about gravitational waves as vacuum solutions. There is one key difference with gravity however, since gravity waves themselves carry energy and momentum, and can in principle be a source in themselves.

More to the point in the case of a star, yes there is the mass source that curves space in the external vacuum, but the curvature can be far from the mass at a point where there is no stress energy tensor. That's all I meant in saying there need not be a source at the point where the curvature is. The solution far from the star needs to match the Newtonian limit, and it turns out that (just as in the case of Newtonian gravity) the vacuum solution is not dependent on the structure of the source, provided that the geometry is spherically symmetric.

 

[bcrowell]

As a somewhat more exotic example of what stevenb is talking about, you can have a cosmological model in which the initial big bang singularity is highly nonisotropic and inhomogeneous (sort of like a more chaotic version of the Misner mixmaster universe), but there is no matter whatsoever. Thermodynamically, there is actually an argument to be made that out of all possible big bangs, this is the most probable. In such a universe, spacetime curvature is all over the place in the form of gravitational waves, but it was never caused by any matter, because matter has never existed and never will.

Our own universe's cosmology is vacuum-dominated right now, so it's actually quite a good approximation for many purposes to model it as not containing any matter fields. Nevertheless it has curvature that is evolving over time.

 

[atyy]

Again, in this post, I mean mass as fields with localizable momentum-stress-energy, as opposed to all the various valid nolocal notions of "mass" that bcrowell has in mind, some of which are discussed in http://relativity.livingreviews.org/Articles/lrr-2009-4/ . So the Schwarzschild mass parameter is not a mass by definition.

As TerryW points out in his OP, strictly speaking, MTW has it wrong.

But, as stevenb and bcrowell point out, no more wrong than saying EM fields are generated by charges.

In the EM case, it should be EM fields are generated by charges and boundary conditions.

Same with the gravitational field, except that additionally the vacuum equations are nonlinear, because gravity is a "source" of its own curvature. This is mainly heuristic, but can be made somewhat precise as in pervect's example in post #13.

However, most of these are irrelevant for realistic applications of the vacuum Schwazrschild solution, where it is joined to a solution describing the inside of the sun or Earth which do have mass.

 

[TerryW]

Thanks again stevenb, but still no slapping of forehead! I could go along with the analogy if we were taking the vacuum field equations and producing a solution which implies the existence of gravity waves in the way that Maxwell's equations can be solved to produce EM waves, but this isn't the case. My view is that this is more akin to the derivation of the static field associated with a point charge, but it is clear that no one has come up with a stress tensor associated with a remote mass, so there isn't an equation to solve to get the curvature from the stress tensor. I will have to see if I can get hold of a copy of Pais's book, referred to in Penrose's Road to Reality to see what he has to say.

 

[PeterDonis]

My view is that this is more akin to the derivation of the static field associated with a point charge, but it is clear that no one has come up with a stress tensor associated with a remote mass, so there isn't an equation to solve to get the curvature from the stress tensor.

What do you mean by "stress tensor associated with a remote mass"? If the mass is remote--i.e., it's somewhere else besides where you are--then the stress-energy tensor where you are is zero; it's only nonzero elsewhere, where the mass is. This is the type of scenario the Schwarzschild solution models--elsewhere, where the mass is, you join the "exterior" Schwarzschild solution that gives the external field of the mass to an "interior" solution that describes the mass itself, but if the mass is "remote", all you need to worry about is the exterior portion.

 

[TerryW]

What do you mean by "stress tensor associated with a remote mass"? If the mass is remote--i.e., it's somewhere else besides where you are--then the stress-energy tensor where you are is zero; it's only nonzero elsewhere, where the mass is. This is the type of scenario the Schwarzschild solution models--elsewhere, where the mass is, you join the "exterior" Schwarzschild solution that gives the external field of the mass to an "interior" solution that describes the mass itself, but if the mass is "remote", all you need to worry about is the exterior portion.

Hi PeterDonis,

Well, you can have a stress tensor associated with an EM field where the charges are in some other place so why not a stress tensor associated with a gravitational field where the mass is remote? Is it possible perhaps to find a solution to Einstein's equations for a spherical dust cloud? I could go along with an argument that ran along the lines that a central mass was once a spherical dust cloud and as it condensed, it left an imprint on spacetime which it then drags around with itself.

 

[Altabeh]

The mass parameter is not the "physical" integral of the mass-energy (it is a "look-alike" integral).

Well, if you think that M is just an unphysical parameter, you are going the wrong way! We don't play with words like "parameter" in a dangerous way as you do. If M was to be taken as a parameter only, then there would be no meaning to the Schwartzschild radius because we just identify it by the help of M and nothing else! It is here indeed the same old classical mass that we used to deal with in Newtonian mechanics. More generally, it is said that if $$T_{00}=\rho(\vec{x})$$, then $$M=\int_0^R \rho d^3\vec{x}$$ and R is the radius of the gravitating body!

AB

 

[stevenb]

Well, if you think that M is just an unphysical parameter, you are going the wrong way! We don't play with words like "parameter" in a dangerous way as you do. If M was to be taken as a parameter only, then there would be no meaning to the Schwartzschild radius because we just identify it by the help of M and nothing else! It is here indeed the same old classical mass that we used to deal with in Newtonian mechanics. More generally, it is said that if $$T_{00}=\rho(\vec{x})$$, then $$M=\int_0^R \rho d^3\vec{x}$$ and R is the radius of the gravitating body!

AB

What about the potential energy of pressure or the kinetic energy of pulsations? It is still possible to maintain spherical symmetry and have mass-energy contributions beyond the integration of the mass density part of the T₀₀ component of the Stress Tensor. Practically, these other components are small, and usually utterly insignificant, but they are there in the physics of GR.

Perhaps a better way to say it is the M is what a distant Newtonian physicist would interpret as the integration of mass density over the gravitational source. If he did gravitational force measurements taking data only in a region where the Newtonian limit is valid, then he would have not have the measurement fidelity to distinguish the acutal internal stucture of the source.

Actually, that is an understatement. For, even measurements near the spherically symmetric source, but exterior to it, can't distinguish the actual internal structure, as proved by Birkoff's Theorem. One needs to probe the interior of the source itself to know what's happening in there.

I understand TerryW's concerns. Sometimes the whole thing seems like magic. But that's the price we sometimes pay for elegant solutions. They are so amazing and powerful that they seem impossibly simple and even somewhat magical.

 

[Altabeh]

The process goes something like this - we solve the equations for a spherically symmetric vacuum and lo and behold, the solution tells us that there is a mass at the middle! But it is a mass which doesn't generate a stress tensor. I've looked through all my books and the only book that even mentions this apparent illogicality is Penrose's Road to Reality, and he just makes reference to another book by Pais. Why does no one want to talk about this? Can anyone explain?

Yes there is a vanishing stress tensor because there is no mass where you're calculating those equations. If there was, then you'de probably hit another metric, for instance, Florides metric which deals with the metric of the interior of a of star with $$m=2MG/c^2$$ where M is the mass of gravitating body and of course here the components of stress tensor is no longer taken to be vanishing.

See http://members.iinet.net.au/~housewrk/Papers/new%20SH%20sol.pdf

AB

 

[Altabeh]

What about the potential energy of pressure or the kinetic energy of pulsations? It is still possible to maintain spherical symmetry and have mass-energy contributions beyond the integration of the mass density part of the T₀₀ component of the Stress Tensor. Practically, these other components are small, and usually utterly insignificant, but they are there in the physics of GR.

In fact you answer your own question. Above I just supposed T_00 was the only non-zero element. If there is principal radial stress and or the principal stress tangential to any surface of radius r<=R is non-zero and positive (for the sake of physicality), we would still get a completely spherically symmetric metric of the form known as "interior" solution or Florides solution to the EEEs where definitely some other components of stress tensor turn out to be non-zero!

Perhaps a better way to say it is the M is what a distant Newtonian physicist would interpret as the integration of mass density over the gravitational source. If he did gravitational force measurements taking data only in a region where the Newtonian limit is valid, then he would have not have the measurement fidelity to distinguish the acutal internal stucture of the source.

I don't see any problem in using the mass-density integral relation to get mass even if there are stresses inside a gravitating body! Yet, this seems to be a problem at first glance because as you said that is a "distant" physicist who measures M through that integral in the precense of (even small) pressures; but the more you look into the EEEs and their structure, the better you get the idea that here the pressure components of stress tensor must be evaluated through EEEs not by a piori. If you keep doing this the other way around, then there is no longer going to be a guaranteed spherical symmetry as can be seen from Florides' work!

Actually, that is an understatement. For, even measurements near the spherically symmetric source, but exterior to it, can't distinguish the actual internal structure as proved by Birkoff's Theorem. One needs to probe the interior of the source itself to know what's happening in there.

And this is what they have done after Schwartzschild!

AB

 

[pervect]

Well, if you think that M is just an unphysical parameter, you are going the wrong way! We don't play with words like "parameter" in a dangerous way as you do. If M was to be taken as a parameter only, then there would be no meaning to the Schwartzschild radius because we just identify it by the help of M and nothing else! It is here indeed the same old classical mass that we used to deal with in Newtonian mechanics. More generally, it is said that if $$T_{00}=\rho(\vec{x})$$, then $$M=\int_0^R \rho d^3\vec{x}$$ and R is the radius of the gravitating body!

AB

This is not a correct formulation. As I mentioned in a previous post, MTW computes the integral of $\rho \, dV$. Last time, I didn't provide a specific reference. You can find this calculation on pg 604 of "Gravitation", Misner, Thorne, Wheeler for the case of a star of uniform density. The result is that the integral of [tex]\rho \, dV[/itex] is different - larger - than the M that appears in the Schwarzschild metric.<br /> <br /> The reason for the difference, as I explained earlier, is that the integral of density * volume, represents the mass of the system "before assembly". The second number is the mass "after assembly". The difference between the two can be ascribed to "Gravitational binding energy". The total energy of the system before assembly is different than the total energy after assembly, because of the gravitational binding energy. MTW goes on to show that for a small enough star, the formula thus obtained for gravitational binding energy is equal to the Newtonian number, as one would expect.<br /> <br /> I'll mention something else that might not be too important now but could be later. The integral of rho*dV is not a frame (really coordinate) independent quantity. I'll say "frame" because it might make the meaning clearer - if you think of SR, a volume element depends on one's velocity, so it depends on the choice of "frame". <br /> <br /> However, because there is a meaningful notion of a "static observer" in the Schwarzschild case, this can be worked around. In more general cases, the integral of rho*dV might not even be well-defined in GR.<br /> <br /> I think it bears repeating, that here isn't anyone single concept of mass in general relativity as there is in Newtonian mechanics. This surprises a lot of people. There are, however, at least three concepts in common use in GR, and probably more - though they are all very closely related.<br /> <br /> I'm pretty sure , for instance, that three of the commonly used masses (the Komar, Bondi, and ADM masses) exist and are equal to each other for the Schwarzschild metric. However, I'm not 100% positive as I have only seen the former (the Komar mass) calcuated for that case. So, in some sense, you are probably right that the parameter M has a lot of significance, but you are not right when you assume that it's the same significance as it has in Newtonian mechanics. <br /> <br /> Let me urge readers to *please* do a little bit of research into the topic, and not just fire off posts based on Newtonian physics, it seems to me that we've been going around trying to squash the same missapplications of Newtonian physics again and again and again in this thread.[/tex]

 

[atyy]

An additional reference is p259 of http://books.google.com/books?id=qhDFuWbLlgQC&dq=schutz+bernard&source=gbs_navlinks_s