# Classical Gravity versus GR

## Main Question or Discussion Point

I have a question about classical gravity versus GR. If we use Newtonian gravity for a sphere, gravity is zero in the center due to vector addition. So if we were to plot the gravity force versus distance, from far space, to the center of a spherical planet, it starts near zero, climbs until we reach the surface of the planet, and then decays back to zero.

If we use this gravity profile to explain a GR space-time well, in the fabric of space-time, is there a peak in the center of the well? The center point has the same gravity number as distant space so should it have the same space-time fabric height? It would look like a mountain in a hole, with the peak height the same as distant space.

I never see GR explained with a peak in the center, so is this Newtonian gravity peak virtual and the cause of the GR affect? Here one possible explanation, using the particle-wave nature of matter. If we just assume there were only particles without waves, for the sake of argument, the center point would see the most potent particle exchange due to the lowest distance summation. If we assume only waves, without particles, this is more consistent with the zero gravity in the center, due to wave addition. Due to the particle-wave nature both need to be consistent, yet each acting separately lead to two different results. To make these consistent, the particles that should appear in center can not appear there. But due to the conservation of energy, they will still need to appear elsewhere where they are consistent with their wave nature. The result is GR. The contraction of space-time allows nature to get the potential energy of the inconsistent particles into compacted space-time to compensate for the wave-particle inconsistency in the center of gravity.

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Mentz114
Gold Member
The interior solution for a sphere of matter in GR predicts flat space at the center, in agreement with the Newtonian result. This applies to matter with finite pressure throughout and with a regular energy-momentum tensor.

I don't think one can derive GR in the way you suggest.

The interior solution for a sphere of matter in GR predicts flat space at the center, in agreement with the Newtonian result. This applies to matter with finite pressure throughout and with a regular energy-momentum tensor.

I don't think one can derive GR in the way you suggest.
Suppose there is a small cavity at the center, what the space in it would be? "flat" or "Not flat"? Suppose next, if there is a clock in the cavity, would the clock run slow compared with a clock at infinity?

Suppose there is a small cavity at the center, what the space in it would be? "flat" or "Not flat"? Suppose next, if there is a clock in the cavity, would the clock run slow compared with a clock at infinity?

The clock in the cavity would run slower than a clock at infinity and slower than a clock on the surface of the massive body. The force of gravity goes to zero as you descend from the surface to the centre. The gravitational potential at the surface is lower than at infinty and continues to get more negative as you descend to the centre. It is the potential rather than the force of gravity that influences clock rates.

A.T.
If we use this gravity profile to explain a GR space-time well, in the fabric of space-time, is there a peak in the center of the well? The center point has the same gravity number as distant space so should it have the same space-time fabric height? It would look like a mountain in a hole, with the peak height the same as distant space.
No. you misunderstand how gravity is modeled in GR. There is no "space-time fabric height". The distortion of space-time has an maximum in the center of a sphere, causing a maximal time dilation there. Free fallers are always drawn towards areas of higher time dilatation. But at a point of locally maximal time dilatation, there is nowhere else to go.

Try this visualization:
If you set "initial position" to zero, the free falling object is placed at the center of the sphere. And it stays there because it's geodesic world line doesn't deviate in any space direction. The object moves only trough time by moving straight ahead trough space-time.

A.T.
The interior solution for a sphere of matter in GR predicts flat space at the center, in agreement with the Newtonian result.
Are you sure, that the interior Schwarzschild metric is flat in the center? The only thing, that is required from it, to agree with Newton, is that the worldline of an object resting in the center is a geodesic in space-time. To satisfy this, space-time doesn't need to be flat there. You can have a geodesic with a constant space-coordinate in curved space-time too. The spacial part of the interior solution has a hypersphere-like non-zero curvature at every point, also at the center of the mass. And if space alone is curved there, space-time can hardly be flat.

If the sphere mass had a small cavity at the center, I would of course agree that space-time is flat within the cavity. But the interior Schwarzschild solution considers a filled spherical mass.

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Going back to the OP, I think there's a little confusion about two things. First, there's a difference between the gravitational force as a function of r, which is what goes to zero at the center and at r=infinity, and the gravitational potential, which is what is sometimes modeled as a curved sheet with balls rolling around on it.

Second, it wasn't clear to me whether the OP meant a spherical mass shell or a solid sphere - I assumed it was the latter. In that case, you have to invoke the very thin tunnel from the surface to the center to discuss the force at points below the surface of the sphere. Contrary to what I think Mentz114 was suggesting, the force is not zero in this region, but varies as a linear function of r (like a spring) from zero at the center to the value at the surface given by the usual inverse square law, which defines the force everywhere else.

This gives rise to a potential function that has a minimum at r=0, rises parabolically to the radius of the sphere, and linearly out to a max value at infinity. There's no peak at the center, as far as the potential is concerned; that would imply an outward force, which is not the case.

Mentz114
Gold Member
A.T.:
Are you sure, that the interior Schwarzschild metric is flat in the center?
I'm looking at a metric which when r = 0 goes to ds^2 = dr^2 - c^2dt^2. I interpret this as 'flat', but I could be wrong.

A.T.:

I'm looking at a metric which when r = 0 goes to ds^2 = dr^2 - c^2dt^2. I interpret this as 'flat', but I could be wrong.
The metric ds^2 = dr^2 - c^2dt^2 is flat, it's the same with metric at infinity, but it can't be the metric in the cavity. Since a clock in the cavity run slower than a clock at infinity

A.T.:

I'm looking at a metric which when r = 0 goes to ds^2 = dr^2 - c^2dt^2. I interpret this as 'flat', but I could be wrong.
Wikipedia says "The Schwarzschild metric is a solution of Einstein's field equations in empty space, meaning that it is valid only outside the gravitating body. That is, for a spherical body of radius R the solution is valid for r > R. To describe the gravitational field both inside and outside the gravitating body the Schwarzschild solution must be matched with some suitable interior solution at r = R."

It is obvious that an obsever moving down a tunnel to the centre of the Earth will (besides getting very hot) pass a point where the radius is less than the Shwarzchild radius of the Earth's mass but there will be no event horizon at that point. At the centre of the Earth the radius is zero but there is no singularity. This is because the mass within the Shwarzchild radius is less than the Shwarzchild mass.

If we consider only the enclosed mass as the observer descends down the tunnel then the gravitational gamma factor:

$$\sqrt{\left(1-\frac{2GM}{c^2R}\right)}$$

becomes (assuming even density distribution):

$$\sqrt{\left(1-\frac{2GM_ER^2}{c^2R_E^3}\right)}$$

where $R_E$ and $M_E$ are the radius and mass of the Earth.

The modified formaula contains no singularites as would reasonably be expected within the Earth but it also shows that the gravitational gamma factor goes to unity (ie is the same as the gamma factor at infinity) at the centre of the Earth, contradicting what I said in the my last post :(

A.T.
If we consider only the enclosed mass as the observer descends down the tunnel then the gravitational gamma factor:
I doubt if you can do this, and simply use the outer Schwarzschild solution. I think you have to use the inner Schwarzschild solution (sorry German only):
http://de.wikipedia.org/wiki/Schwarzschild-Metrik#Innere_L.C3.B6sung

A.T.
A.T.:
I'm looking at a metric which when r = 0 goes to ds^2 = dr^2 - c^2dt^2. I interpret this as 'flat', but I could be wrong.
The inner solution for r=0 has a non vanishing factor at dt that accounts for the time dilation:
http://de.wikipedia.org/wiki/Schwarzschild-Metrik#Innere_L.C3.B6sung
But as far as I know, you have to consider the second derivates of a metric, to tell if it has intrinsic curvature. Curvature is not defined for a dimensionless point, but for an infinitesimal path enclosing an area > 0:
http://en.wikipedia.org/wiki/Introduction_to_mathematics_of_general_relativity#Curvature_tensor

Mentz114
Gold Member
A.T:
The interior solution I've referred to is in Stephani, page 124. He calls it the interior Schwarzschild solution. No doubt different metrics may be found with different assumptions about the energy-momentum tensor.

A.T.
A.T:
The interior solution I've referred to is in Stephani, page 124. He calls it the interior Schwarzschild solution. No doubt different metrics may be found with different assumptions about the energy-momentum tensor.
Sure, but as Xeinstein already noted, the space-time line element in the center of a mass>0 cannot be the same as far away in flat space-time. It has to contain the factor of gravitational time dilation.

Aside from that, for the question of space-time curvature at the mass-center: I think that you have to consider how the line element given by the metric changes around that point, by taking the derivates.

A.T.:

I'm looking at a metric which when r = 0 goes to ds^2 = dr^2 - c^2dt^2. I interpret this as 'flat', but I could be wrong.
I agree. I thought you meant "flat everywhere for r < radius of the sphere, not just at the center. It certainly must be flat there, just by symmetry.

Mentz114
Gold Member
I must apologise, I misread the print. There is a residual g_tt at r=0. Its value is -

$$-\frac{9}{4} (1-Ar_0^2) , A=\frac{1}{3}\kappa\mu c^2$$

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I must apologise, I misread the print. There is a residual g_tt at r=0. Its value is -

$$-\frac{9}{4} (1-Ar_0^2) , A=\frac{1}{3}\kappa\mu c^2$$
Looking at the inner Shwarzchild metric in the link given by A.T. http://de.wikipedia.org/wiki/Schwarzschild-Metrik#Innere_L.C3.B6sung and in this paper http://dbserv.ihep.su/~pubs/prep2005/ps/2005-29e.pdf I get the residual with R=0 as

$$dS^2 = \frac{9}{4}\left(\frac{1}{3}-\sqrt{1-\frac{2GM_o}{c^2R_o}}\right)^2c^2dt^2$$

Where $M_o$ and $R_o$ are the mass and radius of the gravitational body and $R_o$ is greater than the Shwarzchild radius of the body.

Does that seem about right? It looks similar to, but not exactly the same as the Mentz formula.

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pervect
Staff Emeritus
I have a question about classical gravity versus GR. If we use Newtonian gravity for a sphere, gravity is zero in the center due to vector addition. So if we were to plot the gravity force versus distance, from far space, to the center of a spherical planet, it starts near zero, climbs until we reach the surface of the planet, and then decays back to zero.

If we use this gravity profile to explain a GR space-time well, in the fabric of space-time, is there a peak in the center of the well? The center point has the same gravity number as distant space so should it have the same space-time fabric height? It would look like a mountain in a hole, with the peak height the same as distant space.

I never see GR explained with a peak in the center, so is this Newtonian gravity peak virtual and the cause of the GR affect?
You never see a "gravity profile" in GR at all - you have taken some analogies meant to help illustrate the theory a little too seriously. So your conceptual model of GR is basically wrong. Unfortunately, it's hard to correctly describe a correct model of GR without using a lot of math.

For example, gravity in GR is not really a force. You might try reading the downloads of the first few chaptors of Taylor's book "Exploring Black Holes" at

to get some better idea of what GR is actually aobut.

I noticed something interesting about the inner Shwarzchild solution for a solid gravitational body.

By ignoring radial and rotational motion the metric simplifies to:

(Eq 1) $$dS^2 = \left({3 \over 2}\sqrt{1-{R_s \over R_o}}-{1\over 2}\sqrt{1-{R_s R^2 \over R_o^3}}\right)^2 c^2dt^2$$

Taking the square root of both sides this becomes:

(Eq 2) $$dS = \left({3 \over 2}\sqrt{1-{R_s \over R_o}}-{1\over 2}\sqrt{1-{R_s R^2 \over R_o^3}}\right) c dt$$

At the surface of the body $R = R_o$ so the inner metric becomes:

(Eq 3) $$dS = \left({3 \over 2}\sqrt{1-{R_s \over R}}-{1\over 2}\sqrt{1-{R_s \over R}}\right) c dt$$

=>

(Eq 4) $$dS = \left(\sqrt{1-{R_s \over R}}\right) c dt$$

..which is the same as the external metric (as it should be).

Now if we set R to zero in (Eq 2) the inner metric becomes:

(Eq 5) $$dS = \left({3 \over 2}\sqrt{1-{R_s \over R_o}}-{1\over 2}\right) c dt$$

It is fairly easy to see that if the the radius of the massive body is 9/8 times the Shwarzchild radius then the proper time rate (dS) of a clock at the centre of the body goes to zero and this is for a solid body that has not become a black hole! For radius less than 9/8 the Shwarzchild radius (but still greater than the Shwarzchild radius) the proper clock rate at the centre of the massive body becomes negative suggesting time reversal. The authors of the second paper suggest the time reversal represents a negative form of gravity that prevents a fully fledged black hole from forming.

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Mentz114
Gold Member
Further correction

Looking at the inner Shwarzchild metric in the link given by A.T. http://de.wikipedia.org/wiki/Schwarzschild-Metrik#Innere_L.C3.B6sung and in this paper http://dbserv.ihep.su/~pubs/prep2005/ps/2005-29e.pdf I get the residual with R=0 as

$$dS^2 = \frac{9}{4}\left(\frac{1}{3}-\sqrt{1-\frac{2GM_o}{c^2R_o}}\right)^2c^2dt^2$$

Where $M_o$ and $R_o$ are the mass and radius of the gravitational body and $R_o$ is greater than the Shwarzchild radius of the body.

Does that seem about right? It looks similar to, but not exactly the same as the Mentz formula.
I was half-blind with tiredness when I made my earlier post, and I have to make another correction -

$$-\left(\frac{1}{2} - \frac{3}{2}\sqrt{1-Ar_0^2}\right)^2 , A=\frac{1}{3}\kappa\mu c^2$$

This is what kev gets from the other source. Phew, sorry for the blunders.

Mentz114
Gold Member
kev:
It is fairly easy to see that if the the radius of the massive body is 9/8 times the Shwarzchild radius then the proper time rate (dS) of a clock at the centre of the body goes to zero and this is for a solid body that has not become a black hole! For radius less than 9/8 the Shwarzchild radius (but still greater than the Shwarzchild radius) the proper clock rate at the centre of the massive body becomes negative suggesting time reversal. The authors of the second paper suggest the time reversal represents a negative form of gravity that prevents a fully fledged black hole from forming.
I should point out that the 'second paper' claims to use results outside GR, from Field theory gravity ( FTG).

Stephani shows that if the radius falls below the critical value, which you define above, the interior solution is unstable because pressure diverges, and rules out the time reversal as a physical effect. Fascinating stuff.

If we look at classical gravity, the gravitational force is function of mass and distance. With GR the same phenomena is now a function of space-time. This works well for large mass. But if I threw a rock of mass M at a window, is the GR distortion in space-time from that little rock sufficient to explain the potential affect that overcomes the strong the EM forces that are binding the glass together? Is GR a Newtonian approximation if we assume high mass, but does it break down at low mass?

Another way to look at it, can space-time distortions gain momentum if acted upon by a force to create enhanced space-time affects at low velocity? I could have used a small iron magnetic for the GR space-time, and used EM force for momentum.

kev:

I should point out that the 'second paper' claims to use results outside GR, from Field theory gravity ( FTG).

Stephani shows that if the radius falls below the critical value, which you define above, the interior solution is unstable because pressure diverges, and rules out the time reversal as a physical effect. Fascinating stuff.
The second paper http://dbserv.ihep.su/~pubs/prep2005/ps/2005-29e.pdf uses the classic interior Schwarzchild solution as can be found in other papers. The part that is outside GR is concluding that gravity reversal acts to prevent a body with regions of negative time being a stable configuration. The interior solution is based on an assumption of constant density distribution within the body. Stephani concludes that the divergence of pressure acts to redistribute the density gradient within the body to prevent time reversal. In a way, both authors are in agreement that that negative time is unstable and undesirable. The authors of this paper http://odarragh.astro.utoronto.ca/Schwarzschild.pdf [Broken] reach a similar conclusion. This is the concluding paragraph of the paper:

"The pressure vanishes at r0, and is positive for smaller r, but to remain physically reasonable, we must require that p remain finite as r approaches zero. This requires that the denominator in the above expression remain positive, from which one can derive the condition

$$r_0 > { 9 \over 8} 2m$$

Thus we cannot expect to have stable configurations of uniform density if the radius is less than 9/8 of the Schwarzschild radius. While this result has been derived for a uniform
density sphere, such a limit exists for configurations with density increasing monotonically
toward the centre of the sphere."

My guess is that whatever mechanism acts to prevent negative time also acts to prevent the formation a classic black hole with a singularity of infinite density at the centre.

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Mentz114
Gold Member
Kev:

My guess is that whatever mechanism acts to prevent negative time also acts to prevent the formation a classic black hole with a singularity of infinite density at the centre.
I'll go along with that.

As an aside, in the Russian paper on page 3 the authors write "We have to emphasize that inequalities (3) and (5) are not a consequence of GR." But (5) looks just like the critical radius defined in GR, so I don't understand that. Perhaps they didn't know ?

If we look at classical gravity, the gravitational force is function of mass and distance. With GR the same phenomena is now a function of space-time. This works well for large mass. But if I threw a rock of mass M at a window, is the GR distortion in space-time from that little rock sufficient to explain the potential affect that overcomes the strong the EM forces that are binding the glass together? Is GR a Newtonian approximation if we assume high mass, but does it break down at low mass?

Another way to look at it, can space-time distortions gain momentum if acted upon by a force to create enhanced space-time affects at low velocity? I could have used a small iron magnetic for the GR space-time, and used EM force for momentum.
Sorry for quoting myself, but this low mass affect is important. With velocities out of the range of relativistic I can create momentum affects from the mass that are beyond what the GR and SR math suggest. I can throw a small solid H2 rock at a large hollow sphere of solid H2, that has more GR space-time affect, and still break it up.

The math puts us in the ozone layer of thinking where practical reality is different.