
#19
Mar2308, 06:43 PM

P: 3,966

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) [tex] 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[/tex] Taking the square root of both sides this becomes: (Eq 2) [tex] 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[/tex] At the surface of the body [itex] R = R_o [/itex] so the inner metric becomes: (Eq 3) [tex] dS = \left({3 \over 2}\sqrt{1{R_s \over R}}{1\over 2}\sqrt{1{R_s \over R}}\right) c dt[/tex] => (Eq 4) [tex] dS = \left(\sqrt{1{R_s \over R}}\right) c dt[/tex] ..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) [tex] dS = \left({3 \over 2}\sqrt{1{R_s \over R_o}}{1\over 2}\right) c dt[/tex] 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. 



#20
Mar2408, 04:38 AM

PF Gold
P: 4,081

[tex]\left(\frac{1}{2}  \frac{3}{2}\sqrt{1Ar_0^2}\right)^2 , A=\frac{1}{3}\kappa\mu c^2[/tex] This is what kev gets from the other source. Phew, sorry for the blunders. 



#21
Mar2408, 05:25 AM

PF Gold
P: 4,081

kev:
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. 



#22
Mar2408, 10:02 AM

P: 20

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 spacetime. This works well for large mass. But if I threw a rock of mass M at a window, is the GR distortion in spacetime 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 spacetime distortions gain momentum if acted upon by a force to create enhanced spacetime affects at low velocity? I could have used a small iron magnetic for the GR spacetime, and used EM force for momentum. 



#23
Mar2408, 10:05 AM

P: 3,966

"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 [tex] r_0 > { 9 \over 8} 2m[/tex] 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. 



#24
Mar2408, 10:14 AM

PF Gold
P: 4,081

Kev:
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 ? 



#25
Mar2408, 10:17 AM

P: 20

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



#26
Mar2408, 10:20 AM

PF Gold
P: 4,081





#27
Mar2408, 10:32 AM

PF Gold
P: 4,081





#28
Mar2408, 01:53 PM

P: 662





#29
Mar2408, 02:20 PM

P: 20

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 spacetime 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. 



#30
Mar2408, 03:14 PM

PF Gold
P: 4,081

Siphon:




#31
Mar2408, 03:24 PM

P: 3,966





#32
Mar2408, 04:28 PM

P: 662

Assuming we're not in the neighborhood of any large masses (as I believe you are), spacetime is effectively flat here, so the rock follows a spacetime geodesic that is essentially a straight line, just as with Newtonian gravity (which we expect, since GR reproduces Newton for flat spacetime). 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? 



#33
Mar2708, 09:31 PM

P: 3,966

Hi,
I noticed another interecting aspect of the equation for internal gravitational time dilation of a solid body : [tex] \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) [/tex] where [tex] R_x[/tex] and [tex] R_o[/tex] 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 rewritten as [tex] \frac{dS}{dt} = {3 \over 2}(P_M){1\over 2}(P_E) [/tex] where [tex]P_M[/tex] is the gravitational time dilation factor due to the total mass of the body and [tex]P_E[/tex] 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. 


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