Is there a peak in the center of a GR space-time well?

[yuiop]

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 approximate the classical Newtonian equations for momentum, kinetic energy etc at low speeds. See http://hyperphysics.phy-astr.gsu.edu/Hbase/relativ/releng.html#c6

 

[belliott4488]

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?

 

[yuiop]

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 Schwarzschild 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.