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## Classical Gravity versus GR

 Quote by siphon 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.
 Blog Entries: 6 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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 Quote by kev Looking at the inner Shwarzchild metric in the link given by A.T. http://de.wikipedia.org/wiki/Schwarz...re_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.

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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.

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 Quote by Mentz114 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 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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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.

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 Quote by siphon 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.
Are you referring to the fact that in GR there are no forces, just trajectories ?

 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?
The momentum of the rock breaks the glass. Gravity doesn't come into it.

 Is GR a Newtonian approximation if we assume high mass, but does it break down at low mass?
It's the other way round - Newtonian theory is a weak-field approximation of GR.

 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.

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 Sorry for quoting myself, but this low mass affect is important.
What effect is that ? You just invented it. Please explain.

 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.
Forgive me for being extremely sceptical - show us your equations.

 Quote by siphon 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?
Why are you invoking gravity, either Newtonian or GR, to explain a rock breaking a piece of glass? The electromagnetic interactions between the molecules of the rock and of the glass are what give rise to the forces that eventually break the glass.
 The overall affect has to do with the momentum of the mass. The final transfer of this momentum will involve EM forces. The inertia of the mass makes this extreme enough to break the glass. Is this classical mass observation equal to GR space-time momentum being converted to the stresses placed on the EM force? At the lower ends of mass, GR arguments appear to break down relative to observation. If we took a planet at velocity V and doubled that to 2V with V being small, the SR or GR changes would be negligible. But the classical momentum doubles. The difference for GR and SR will be a very small, but finite difference. The newtonian assumption will be closer to the final affect, in reality, which is a double strength collision.

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Siphon:
 At the lower ends of mass, GR arguments appear to break down relative to observation.
What do you mean by this ? There's no mass limit of applicability in GR. Anyway, as far as rock throwing on earth, Newton's laws of motion will suffice.

 If we took a planet at velocity V and doubled that to 2V with V being small, the SR or GR changes would be negligible. But the classical momentum doubles. The difference for GR and SR will be a very small, but finite difference. The newtonian assumption will be closer to the final affect, in reality, which is a double strength collision.
What difference? You're not expressing yourself very well.

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 Quote by siphon The overall affect has to do with the momentum of the mass. The final transfer of this momentum will involve EM forces. The inertia of the mass makes this extreme enough to break the glass. Is this classical mass observation equal to GR space-time momentum being converted to the stresses placed on the EM force? At the lower ends of mass, GR arguments appear to break down relative to observation. If we took a planet at velocity V and doubled that to 2V with V being small, the SR or GR changes would be negligible. But the classical momentum doubles. The difference for GR and SR will be a very small, but finite difference. The newtonian assumption will be closer to the final affect, in reality, which is a double strength collision.
Just about any text on relativity shows that the equations of relativity very closely aproximate the classical Newtonian equations for momentum, kinetic energy etc at low speeds. See http://hyperphysics.phy-astr.gsu.edu...releng.html#c6

 Quote by siphon The overall affect has to do with the momentum of the mass. The final transfer of this momentum will involve EM forces. The inertia of the mass makes this extreme enough to break the glass. Is this classical mass observation equal to GR space-time momentum being converted to the stresses placed on the EM force? At the lower ends of mass, GR arguments appear to break down relative to observation.
Can you explain what you mean by this?

Assuming we're not in the neighborhood of any large masses (as I believe you are), space-time is effectively flat here, so the rock follows a space-time geodesic that is essentially a straight line, just as with Newtonian gravity (which we expect, since GR reproduces Newton for flat space-time). Since both the glass and the rock want to follow straight line geodesics, but they instead run into each other, they undergo acceleration, and force etc.

How are you seeing the description of this event in GR?
 Blog Entries: 6 Hi, I noticed another interecting aspect of the equation for internal gravitational time dilation of a solid body : $$\frac{dS}{dt} = \left({3 \over 2}\sqrt{1-{R_s \over R_o}}-{1\over 2}\sqrt{1-{R_s R^2 \over R_o^3}}\right)$$ where $$R_x$$ and $$R_o$$ are the Schwarzchild radius and surface radius of the body, R is the radius within the body where the measurement is being made, dS is time rate of a clock at R as measured by an obsever at infinity and dt is the time rate of a clock at infinity. Trying out various numerical tests it turns out that the equation can be re-written as $$\frac{dS}{dt} = {3 \over 2}(P_M)-{1\over 2}(P_E)$$ where $$P_M$$ is the gravitational time dilation factor due to the total mass of the body and $$P_E$$ is the gravitational time dilation factor due to the enclosed mass with radius R. This makes it easy to work out the factor for bodies that do not have a uniform density such as as a sphere with a dense core or even a hollow cavity. It also makes it clear that the time rate is constant everywhere within a centered cavity (but slower than the time rate at the external surface of the body). If we had a body that has all its mass within a very thin shell just outside the Shwarschild radius we would have a body that looked and behaved in every way like a black hole externally, except for a small quantity of extremely red shifted radiation escaping from its surface. Internally, the time dilation factor everywhere within the hollow cavity below the Sharzchild radius would be zero.