# Classical Gravity versus GR

1. Mar 23, 2008

### 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? 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.

2. Mar 23, 2008

### Mentz114

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.

3. Mar 23, 2008

### Xeinstein

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?

4. Mar 23, 2008

### yuiop

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.

5. Mar 23, 2008

### A.T.

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.

6. Mar 23, 2008

### A.T.

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.

Last edited: Mar 23, 2008
7. Mar 23, 2008

### belliott4488

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.

8. Mar 23, 2008

### Mentz114

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.

9. Mar 23, 2008

### Xeinstein

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

10. Mar 23, 2008

### yuiop

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 :(

11. Mar 23, 2008

### A.T.

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

12. Mar 23, 2008

### A.T.

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

13. Mar 23, 2008

### Mentz114

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.

14. Mar 23, 2008

### A.T.

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.

15. Mar 23, 2008

### belliott4488

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.

16. Mar 23, 2008

### Mentz114

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

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

Last edited: Mar 23, 2008
17. Mar 23, 2008

### yuiop

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.

Last edited: Mar 23, 2008
18. Mar 23, 2008

### pervect

Staff Emeritus
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.

19. Mar 23, 2008

### yuiop

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.

Last edited: Mar 23, 2008
20. Mar 24, 2008

### Mentz114

Further correction

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.