Dear Bernard Lavenda,

Please forgive my lack of familiarity with your terminology. I'm not able to answer most of your question. But I have given thought to your point about the Schwarzschild solution hardly being empty. The solution indeed seems unsatisfactory. The Schwarzschild metric, for better or worse, applies only to the vacuum surrounding a central mass M, and not to the space where M resides. It seems disappointing, however, that the mass M is put in by hand as a constant of integration to meet the boundary condition of Newtonian gravity, and does not appear in the energy-momentum tensor, which is zero. Why, after all, should general relativity depend on Newtonian gravity and not be self-contained? If the mass must be put in by hand, are not Einstein's field equations lacking? The answer I have reached is: to correctly solve Einstein's equations for a central mass M, the 00 component of energy-momentum tensor should not be zero, but should contain a density distribution rho(r) that represents the central mass. If this is done, Einstein's field equations automatically give 1-2M/r, where M is not a constant of integration, but is in fact determined by Einstein's constant kappa=-8PI G/c^2.. This to me is far more satisfactory. I am finishing a paper now solving Einstein's equations for a central thin-shell density distribution modeled by a broadened spherical delta function. The exterior metric is identical to Schwarzschild's, but space is not a vacuum, since the mass is present in the energy-momentum tensor. I think it is a shame we are taught only the Schwarzschild metric, which is trivial and must be artificially tailored, rather than the true central mass solution.

Dear Kathleen Rosser,

I couldn't agree with you more. It is an even greater shame that we are not taught the inner solution of the Schwarzschild metric! I believe this is an artifact of people playing around with coordinate transformations in the '60's. If we had stuck to the original works, things like black holes would never have crossed our minds as a result of prolonging the outer solution into a domain where the metric no longer exists. It's like prolonging the Poincare disc model beyond the rim of the disc!

The switch from the outer to inner solution consists of replacing constant mass by constant mass density. In the former case, the tidal forces are not zero but add up to zero (i.e. vanishing of the eigenvalues of the Ricci tensor) and determine a dynamic state of equilibrium resulting in a prolate ellipsoid instead of a spherical mass. (Two semi-axes equal while the third is larger.) However, in the latter case, the tidal forces are all equal to the mass density, indicating a metric of constant (negative) curvature.

There is an interesting parallelism here (I believe) with the difference between the Laplace and Poisson equations. A scalar potential, -GM/r, obeys the Laplace equation, yet the potential -(4/3)pi G rho r^2 obeys the Poisson equation! From Newton's proposition 71, we know that the potential at any distance from where the source is located is the same as if all the mass were located at that point. This allows us to go from -GM/r->-(4/3)pi G rho r^2. But the metrics are entirely different. The Laplace equation is analogous to the vanishing of the eigenvalues of the Ricci components, while the Poisson equation pertains to the individual tidal force components being equal. Yet, it is difficult to imagine to go from a source free to a source solution merely by replacing M by rho.

I think that all this leads to questioning Einstein's field equations. The conservation of G resides in the Bianchi identities, while T requires physical considerations. The former holds always while the latter is conditional. The 00 component should be proportional to the density, but the remaining components are highly questionable. For instance, in the Friedmann model, the mass density is equal to the negative of the hydrostatic pressure. Apart from being implicated in the inflationary scenario, a negative pressure will never lead to expansion and highlights the inappropriateness of the Einstein equations. In other words, there is no Poisson equation where the hydrostatic pressure is the source. For hydrostatic equilibrium the gradients of the hydrostatic pressure and gravitational potential must sum to zero which is very different from what the Einstein equations have to say.

It is well known that the pressure must be zero over any free surface of an astronomical body, so the energy balance equation between hydrostatic pressure and the gravitational and rotational potentials fixes the value of the pressure at every point inside a body. It is therefore hardly conceivable that the hydrostatic pressure be determined by a component of the Ricci tensor and the scalar potential, or equivalently, G.

This brings me to one other point. The idea of solving the Schwarzschild problem is to find a Ricci flat metric on R^2xS^2. Mathematicians love to do this, like in the book Riemannian Geometry by Peterson Sec. 2.5. After going through the solution, he concludes that "The metric, however, lives on S^2xR^3 rather than R^2xS^2, and is the standard flat metric on this space." The equation that lets him to conclude this has dimension d=3 and not d=4, as I would have expected. So in which space is the Schwarzschild metric Ricci flat?

Dear Bernard,

Dear Stefano,

Why not there would be only matter pressure via vacuum, no gravitational interactions in themselves at all?

Dear Bernard Lavenda,

First taught GR by James B. Hartle, I was told that the Schwarzschild metric can extend to inner regions of black holes because the event horizon is only a coordinate singularity, as can be proven by a tronsformation to Kruskall coordinates. However Dirac, in his concise textbook General Theory of Relativity (1975, p36,) notes that black holes would take forever to form. I take this to mean that unless they are primordial, black holes do not exist.

About questioning Einstein's field equations (EFE), I have published here on ResearchGate a review article "Current Conflicts in General Relativity: Is Einstein's theory incomplete?" which gives eight reasons why EFE are currently being questioned in mainstream journals, particularly Phys Rev D. Of the eight, I conjecture that the most crucial flaw in EFE is the need for pressure P in the diagonal elements of Tmunu. P must be determined by an equation of state, which is often ad hoc, and in many cases contains more information than EFE themselves. To wit, for static spherical mass distributions, the g00 metric component, and thus time dilation and redshift, depends critically on P. And so by manipulating the equation of state, one can produce any redshift desired, regardless of the mass distribution. This seems unsatisfactory.

I completely agree that the pressure components are highly questionable, and particularly so for the FRW metric as applied in LambdaCDM cosmology, with equations of state driving the expansion rate. And of course dark energy requires, according to recent observations, P < - rho, which corresponds to so-called phantom matter.

The problem is not so much that general relativity and its applications are wrong, which they seem to be, but that apparently no one has succeeded in coming up with a better solution. Hundreds of modified gravity theories are currently being studied, many reported in Phys Rev D, with minimal success. What is the answer?

--------

Dear Steffano Quatrini,

Thanks for mentioning the Opennheimer and Volkof solution. I will look it up. It may pertain to the paper I'm currently finishing on continuous thin shell solutions to Einstein's field equations without the need for junction conditions.

Dear Bernard Lavenda: The Scharzschild solution is vacuum, because it is valid in a spacetime region with zero energy-momentum tensor T_{\mu\nu}.

Interior solutions to the Einstein field equations (with T_{\mu\nu} \neq 0) also exist, mostly by some approximating methods or idealizations, like e.g. by solving the Tolman--Oppenheimer--Volkov equation, as somebody else already mentioned.

The fact that there is mass parameter M in the Schwarzschild solution does not contradict the whole scenario at all, it is a misunderstanding that the Schwarzschild equation requires zero T_{\mu\nu} in the whole spacetime. It just requires T_{\mu\nu} in the domain of validity. Therefore, if one approximates the gravitational field around a star (like the Sun, neglecting rotation and everything else around like planets), spacetime is Schwarzschild. If you consider a rotating star, spacetime will be Kerr. There is no necessity of a horizon or whatsoever -- the Schwarzschild or Kerr solution is just not valid in the interior region, that's all!

And of course, as the non-relativistic limit should be reached in the case of low mass density, the mass parameter M must fit to the Newtonian gravitational field.

Therefore, there is absolutely no inconsistency with general relativity or vacuum vs matter at all.

Best regards

Oliver

Hi Oliver,

The inner solution was solved by Schwarzschild himself in the 1916 paper. Unlike the outer solution, the metric is one of constant, negative curvature.

If the energy-stress tensor vanishes, as you say, in "the domain of validity", so too must the mass. Be it constant or otherwise, it would cause gravitational attraction, and this is what is found. (It is claimed that there is even gravitational "repulsion"!)

Best regards,

Bernard

" If the energy-stress tensor vanishes, as you say, in "the domain of validity", so too must the mass." What mass should vanish? Surely not the one caused by the central mass which might be or might not be behind a horizon! Otherwise spacetime would be flat!

It is just the completely comparable scenario in Newtonian gravity: of course, the Newtonian gravitational field around a mass M does include the parameter M, even though the vacuum solution goes like the typical 1/r law, whereas the interior solution is a different one.

Again: treated relativistically (which is unnecessary overkill of course), the gravitational field / spacetime around the sun (but not inside) is Schwarzschild.

I don't understand: General relativity can't treat the two-body problem. There is only one mass, the central one, and, perhaps, a "test particle"--whatever that is.

I also don't understand "vacuum" and "interior" to what?

According to Newton's proposition 71, the force of gravity due to a spherical shell on any point of its interior is zero. Hence, the potential is constant in the interior of the shell, which is the same if all the mass were concentrated at a single point. This is definitely not Schwarzschild, which has an unsurpassable boundary at r=2M, contrary to the affirmation of Kruskal's transformation which swaps "space" and "time". At r=2M, causal geodesics determinate; since a nonlinear transformation destroys the geodesic property what is being extended?

Dear Bernard,

interior means "inside the star, inside its matter". It does not mean "beyond the horizon"! To find the interior solution, you have to make a sophisticated, or maybe rather idealized ansatz for the equation of state of the matter.

"Outer" means: outside the matter distribution, in vacuum.

The Schwarzschild solution does not imply that in all radial symmetric situations, we must have a horizon and a black hole. The outer solution around the sun is Schwarzschild which to a good degree can be approximated by Newton, of course.

Eventually it is all a question of the parameter M, which -- by means of r_S= 2GM/c^2 -- defines the Schwarzschild radius, and thus sets a scale for the radial coordinate r. If for r

Dear Oliver,

"To find the interior solution, you have to make a sophisticated, or maybe rather idealized ansatz for the equation of state of the matter. "

No, it is one of constant density and whose space component is conformally equivalent to the Poincare disc model, or the spatial component of the Einstein, de Sitter metrics. If you don't like my book, "A New Perspective on Relativity", see Moller's "The Theory of Relativity" Eq (103) on p. 330. It should have been obvious from these models that there is something very fishy about the energy stress tensor, which likewise appears in the Friedmann metric: the sum of the rest energy and hydrostatic pressure is zero, meaning one is the negative of the other.

Specifying the pressure via Einstein's equation makes it over redundant. The pressure must be zero over any free surface of an astronomical body, and the conservation of energy fixes it at every point inside the body (cf. Jeans, "Astronomy and Cosmogony" Sec. 188. Moreover, it can't contradict the condition for hydrostatic equilibrium, which it does in all these solutions.

" Eventually it is all a question of the parameter M, which -- by means of r_S= 2GM/c^2 -- defines the Schwarzschild radius, and thus sets a scale for the radial coordinate r." No, r=2m defines the boundary of the model. There is no extension beyond 2m. Beyond that, it's a whole new ball game and a new model with a new metric is required. That's why Schwarzschild did what he did: Berl Ber. p. 424 (1916). It's a good question why this solution has been completely neglected in all texts on general relativity for it would certainly deter one from introducing Kruskal coordinates, black holes, closed surfaces, etc, etc, etc.

" If for r

Bernard, I think we either talk past each other, or misunderstand each other, or both. In any case I think you miss my point, or I miss yours.

Your starting question above actually was crystal-clear. You asked whether the vanishing of the Ricci tensor means emptiness/vacuum, and how would this be consistent with a Schwarzschild solution and a nonvanishing M. And I have explained to you that everything is fine, and it would be a misunderstanding on your side. Yes, it means vacuum, yes it is all consistent.

What I am saying is that the Schwarzschild solution is a vacuum solution, therefore it is only valid if T_{\mu\nu}=0. It does not matter if the mass M comes from a singularity "mass" at r=0 (in which case the Schwarzschild solution is valid also for r

Dear Oliver,

Could you explain what you mean by gravity not always being attractive in GR? What are you referring to?

See, for example, C H McGruder II, "Gravitational repulsion in the Schwarzschild field" Phys Rev D25 (1982) 3191.

Because all known black hole solutions are vacuum solutions, with T_{\mu\nu}=0. You do realize that T_{\mu\nu} denotes the energy-momentum tensor? I am just asking because I get the feeling you are mixing it up with temperature.

How can you have a stress tensor zero when you have angular momentum, charge, mass, etc as the "first law" of black hole thermodynamics implies? No, I am not confusing T with temperature. And if a bh has temperature and entropy, how can the energy-stress tensor ever be zero? (I don't agree at all with what is commonly referred to as black hole entropy, since it is not an entropy at all!)

Of course, it is true that Schwarzschild himself found an interior solution (which is not called "Schwarzschild solution", in order to avoid confusion), but of course this is only one of many possible ones. There are infinitely many others, all depending on the matter model under consideration. For a neutron star, a dust model is surely not sufficient. Also, the interior solution found by Schwarzschild has no natural correspondence with the (vacuum, outer) Schwarzschild solution. In especial, it is not preferred in any way.

It is referred to as the interior solution, and there are matching conditions with the outer solution. See Moller, p. 331 Eq (107). The inner solution was derived for a "perfect" fluid which exhausts the energy-stress tensor. It's spatial part appears in the Friedmann metric, so I wouldn't say it is without cosmological interest.

What I am saying is that the Schwarzschild solution is a vacuum solution, therefore it is only valid if T_{\mu\nu}=0. It does not matter if the mass M comes from a singularity "mass" at r=0 (in which case the Schwarzschild solution is valid also for r

See, for example, C H McGruder II, "Gravitational repulsion in the Schwarzschild field" Phys Rev D25 (1982) 3191.

OK, thank you, I will read it and see what the author means.

How can you have a stress tensor zero when you have angular momentum, charge, mass, etc as the "first law" of black hole thermodynamics implies? No, I am not confusing T with temperature. And if a bh has temperature and entropy, how can the energy-stress tensor ever be zero?

Because where there is no matter, the stress-energy tensor by definition is zero. Therefore it's called a vacuum solution. It's actually quite simple: Just have a look at how the Schwarzschild (or Kerr) solution is derived from the Einstein equations, and you'll see.

To be more precise: of course, non-gravitational fields like the electromagnetic field also contribute to the stress-energy tensor. Nevertheless it is a perfectly valid approach to neglect this for the model's simplicity's sake.

(I don't agree at all with what is commonly referred to as black hole entropy, since it is not an entropy at all!)

Of course I disagree with you, but that is another (off-)topic, and a discussion by itself.

It is referred to as the interior solution, and there are matching conditions with the outer solution. See Moller, p. 331 Eq (107). The inner solution was derived for a "perfect" fluid which exhausts the energy-stress tensor. It's spatial part appears in the Friedmann metric, so I wouldn't say it is without cosmological interest.

It is "a" interior solution, not "the" interior solution. Every other interior solution needs to match exactly the same conditions to the outer solution -- of which there is one and only one, and this is the Schwarzschild solution

And you say it yourself: it corresponds to a perfect fluid (dust) that is of no or very little astrophysical significance. Modelling a star as a perfect fluid does rarely make sense. In cosmology, however, dust is a perfectly valid approximation to matter distribution, but that is not what we are talking about.

Here is where we differ radically. The Schwarzschild "outer" solution is not valid for r0, after extension to Eddington-Finkelstein coordinates.

It's just the general relativists that ignore this and make things up to suit themselves without any physical or mathematical justification for it.

That is a harsh statement, which in turn means you would consider yourself more knowledgeable than all the other quite well-renowned relativists who have studied their topics for decades, written books, given lectures, maybe won Nobel prizes, but are all quite in line with what I have written above? (Nota bene: I would not consider myself someone anywhere near such people...)

ADDENDUM: I recommend your reading of http://www.blau.itp.unibe.ch/newlecturesGR.pdf, chap 23.7. as an example. The Tolman--Oppenheimer--Volkov equation is explained, which describes astrophysically relevant interior solutions for a radially symmetric star, which the interior solution found by Schwarzschild is not.

Alternatively: every other GR text.

There is very little you can criticize about Schwarzschild's inner solution: the space component is conformally equivalent to the Beltrami metric, which is the classic example of a non-Euclidean geometry with negative, constant curvature.

The interior solution found by Schwarzschild is perfectly valid from a mathematical point of view, and represents a highly idealized matter model.

It is just not of great relevance to astrophysics as modelling a star's interior by dust is not overly sensible.

Therefore the TOV equation for the interior solution delivers a broader ansatz for matter models that reflect additional features found in stars like pressure, interaction, radiation and so forth.

Regarding the TOV equation. The pressure just happens to be defined twice in general relativity: once as the rr component of the Einstein's equation and once from the vanishing of the divergence of the energy-stress tensor. Regarding the former, the tt- and rr-components of the Ricci tensor involve the essentially same terms with opposite signs. The extra terms vanish when the metric coefficients satisfy a=1/b as they do for both the outer and inner Schwarzschild solutions. If a=(1-2m/r) all Ricci components vanish while if a=1-rho r^2, they are all equal to 3 rho.

Now, from Einstein's equations you get rho=-p, an "unfortunate" result. Yet, because you are considering a perfect fluid, you get a relativistic generalization of Euler's equation [Eq (126.7) in Landau & Lifshitz, "Fluid Mechanics." But, there they consider the "projection" of the divergence of the energy stress tensor along the 4-velocity. Also note that the density has now become the energy density which when added to the pressure becomes the enthalpy density. But, the rest mass density is NOT the internal energy density.]

Moller complicates things a bit when he attempts to fit inner and outer solutions at r=r_1 which he claims correspond to zero pressure at the surface of the sphere. The fact that rho=-p, follows from his equations (96a,b) on page 329 when a=1/b is introduced. Thus, if rho is constant so too is p, and his eqn (94) says that their sum vanishes. This leaves his A and B constants in the tt component of the metric fixed at r=r_1, the surface of the sphere. His choice does not make the pressure vanish in eq(102) because there is a lambda term, which he subsequently "neglects". But in comparison to zero, lambda cannot be neglected. All that is required is that the pressure be constant on the surface and it is because the sum of the constant density and pressure vanishes, his eqn (94).

This "unfortunate" result is repeated in the Friedmann metric which led Robertson to conclude that all cosmological models "require a negative pressure equal to the energy density!" This requires that "both rho and p vanish." [Robertson, Noonan, "Relativity and Cosmology" page 374 Eqn (17.7).]

The moral of the story is that you don't need the Einstein equations to determine pressure when you get it through the divergence of the energy-stress tensor for a perfect fluid.

It was also incorrect of Robertson to draw the analogy of the vanishing of the energy-stress tensor with the first law of thermodynamics since the energy density is not the rest energy density The latter is not "the average proper density of all forms of energy", as Robertson claims. That wouldn't be the internal energy in any case. This shows that formal similarities can be very deceiving, and I myself fell into that trap.

Dear Bernard,

although I need some time to fully reproduce everything you said and look it up:

"The moral of the story is that you don't need the Einstein equations to determine pressure when you get it through the divergence of the energy-stress tensor for a perfect fluid."

That is correct! And even more: the usual ansatz for finding interior solutions is to put in a pressure, or, more generally, to define an energy-stress tensor on the rhs of the Einstein equations, in order to solve for the metric. Therefore, modelling the stellar interior from equations of state that which come from outside the theory of GR (plasma physics, astrophysics, whatever) is the first step, and then deduce the metric and all entities derived from it (Ricci, Riemann etc.) is the second.

It might also well be that for a given certain parameter set, there is no solution to the Einstein equations, resp. the TOV equation at all any more. If this is the case, it is a strong sign that no stable star exists, and either a collapse to another, stable parameter set occurs, or, complete gravitational collapse and BH formation follows. This is how the upper limits for masses of white dwarfs or neutron stars are deduced.

Best regards

Oliver

Dear Oliver,

My point is that when you take the divergence of the stress tensor and set it equal to zero you should come out with the condition of hydrostatic equilibrium. Barring the error in interpreting the mass density as an internal energy density and summing it with the pressure to obtain the enthalpy density ( a la Landau and Liftshitz) that is basically what you get. You don't need another equation which defines what you already know because it may lead to a contradiction (and it does!).

When you set the pressure equal to the rr component of the Einstein tensor, you get it equal to the negative of the mass density (times c^2). This is because the tt component of the Ricci tensor is the negative of the rr component apart from two additional terms that cancel for ab=1 in the metric ds^2=-adt^2+dr^2/b+... This has been verified both the inner solution of Schwarzschild, the Einstein-de Sitter models and the Friedmann models.

Best regards,

Bernard