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Classical Gravity versus GR 
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#1
Mar2308, 09:14 AM

P: 20

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 spacetime well, in the fabric of spacetime, 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 spacetime 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 particlewave 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 particlewave 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 spacetime allows nature to get the potential energy of the inconsistent particles into compacted spacetime to compensate for the waveparticle inconsistency in the center of gravity. 


#2
Mar2308, 09:50 AM

PF Gold
P: 4,087

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


#3
Mar2308, 10:10 AM

P: 91




#4
Mar2308, 10:22 AM

P: 3,967

Classical Gravity versus GR
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
Mar2308, 10:50 AM

P: 4,034

Try this visualization: http://www.adamtoons.de/physics/gravitation.swf 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 spacetime. 


#6
Mar2308, 11:25 AM

P: 4,034

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


#7
Mar2308, 11:39 AM

P: 666

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
Mar2308, 11:46 AM

PF Gold
P: 4,087

A.T.:



#9
Mar2308, 01:09 PM

P: 91




#10
Mar2308, 01:11 PM

P: 3,967

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: [tex] \sqrt{\left(1\frac{2GM}{c^2R}\right)}[/tex] becomes (assuming even density distribution): [tex] \sqrt{\left(1\frac{2GM_ER^2}{c^2R_E^3}\right)}[/tex] where [itex]R_E[/itex] and [itex] M_E[/itex] 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
Mar2308, 01:26 PM

P: 4,034

http://de.wikipedia.org/wiki/Schwarz...re_L.C3.B6sung 


#12
Mar2308, 01:49 PM

P: 4,034

http://de.wikipedia.org/wiki/Schwarz...re_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/Introdu...rvature_tensor 


#13
Mar2308, 02:15 PM

PF Gold
P: 4,087

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 energymomentum tensor. 


#14
Mar2308, 02:27 PM

P: 4,034

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


#15
Mar2308, 02:36 PM

P: 666




#16
Mar2308, 04:04 PM

PF Gold
P: 4,087

I must apologise, I misread the print. There is a residual g_tt at r=0. Its value is 
[tex]\frac{9}{4} (1Ar_0^2) , A=\frac{1}{3}\kappa\mu c^2[/tex] 


#17
Mar2308, 04:50 PM

P: 3,967

[tex] dS^2 = \frac{9}{4}\left(\frac{1}{3}\sqrt{1\frac{2GM_o}{c^2R_o}}\right)^2c^2dt^2[/tex] Where [itex]M_o[/itex] and [itex]R_o[/itex] are the mass and radius of the gravitational body and [itex]R_o[/itex] 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. 


#18
Mar2308, 06:39 PM

Emeritus
Sci Advisor
P: 7,628

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 http://www.eftaylor.com/download.htm...ral_relativity to get some better idea of what GR is actually aobut. 


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