Dimensions of Contraction due to Gravity

In summary, the gravitational length contraction is only in the radial direction. There is no contraction in all 3 dimensions.
  • #1
DC0
11
0
I would like some information concerning GR and the dimensional contraction of a standard rod due to gravity. As the potential energy per unit mass decreases as one approaches the center of a planet, time dilates and therefore, space needs to contract, relative to one in deep space. I’ve seen it written that there in no contraction and also that there is contraction in only the radial direction. It seems the contraction should be in all 3 dimensions in order to maintain a constant speed of light in all 3 dimensions. What is the real story? Thanks
 
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  • #2
Hi DC0, welcome to PF!

Unfortunately, I am not aware of any standard definition of gravitational length contraction, as there is for gravitational time dilation.
 
  • #3
DC0 said:
I would like some information concerning GR and the dimensional contraction of a standard rod due to gravity. As the potential energy per unit mass decreases as one approaches the center of a planet, time dilates and therefore, space needs to contract, relative to one in deep space. I’ve seen it written that there in no contraction and also that there is contraction in only the radial direction. It seems the contraction should be in all 3 dimensions in order to maintain a constant speed of light in all 3 dimensions. What is the real story? Thanks

This is a matter of the choice of coordinate system or "chart".

Locally, everything still works as in SR, with the speed of light being the usual constant. However, to compare distances with distances at other locations as if in a flat space, one has to choose how to map the curved space-time to a conventional flat coordinate system. The choice is a matter of convention, not of physics, and is analogous to mapping some area on the Earth using a flat piece of paper.

For theoretical purposes, Schwarzschild coordinates are often used, as these simplify the GR calculations. However, the coordinate speed of light in the radial and tangential directions is then different, and these coordinates have some very odd properties as they cross the surface of a very massive object.

For practical purposes, such as NASA calculations for orbits and paths of space probes within the solar system, it is often easier to use isotropic coordinates, where the scale factor is assumed to be the same in both the radial and tangential direction, so the coordinate value of the speed of the light is the same in all directions but differs slightly from the local speed of light.This also makes it easier to compare with Newtonian gravity. There is a simple mathematical formula for conversion between Schwarzschild and isotropic coordinates (e.g. see MTW exercise 31.7).

Also, when there are multiple gravitational sources, the effects of those sources on time and space (in the weak approximation) as described for individual sources in isotropic coordinates can be added together in the same way as Newtonian potentials (see equation 7.59 in Carroll "Spacetime and Geometry").
 
  • #4
During a Super Nova, pressure and gravity causes matter to compress to about the density of a neutron, and as mass is added and the neutron star grows, time t on the surface dilates and slows relative to a point out in remote space T. This will happen even to a greater extent down toward the center of the core. The relative rate of the slowing of time is expressed by:

t / T = √ (1+2 U / C^2 ) where U = (1/2)( G M r2 / R3 ) - (3/2)( G M / R) , where M is the mass of the neutron star, R is the radius, and r is a radius within the star.
https://www.physicsforums.com/showthread.php?t=40391

Here I used the comparison of the rate of clock ticks rather than the time between ticks. As matter (using no contraction) (at a density of 8x10^17 Kg/m^3) is added, it will cross a transition point (at 2.6 solar masses) where the time rate at the center will come to a stand still (t/T = 0).

If nothing else happens, as more matter is added, the radius r at which time freezes increase: Setting t/T to 0 and solving for r we get:

r (where t/T=0) = √(3 R^2 – R^3 C^2 / (G M) )

This equation shows that as the mass increases to 4.8 solar masses, where t/T = 0 at the surface, the condition for a black hole

Reh=2GM/C^2

is met. This is where the escape velocity is equal to the speed of light.
But something else may be happening and this is where my initial question comes into play.

I divided a neutron star into 500 shells of equal mass and density. As the rate of time
slows for each shell with thickness dr at radius rn, the radial dimensions of space contract by the same factor. With this contraction of space, relative to a point out in remote space, the thickness of each shell will contract and the radius at each shell rn will have to be re-summed from the center. The change in potential energy per unit mass across each shell Udrn, can be calculated by

Udrn = m G dr / rn^2

Where m is the total mass below rn, G is the gravitational constant, and dr is the thickness of the shell.

The potential energy per unit mass Un at shell n can now be summed from the surface down to rn and added to the potential energy at the surface. The relative rate of time t/T at each layer can now be re-calculated.

t / T = √(1+2 Un / C^2 )

Un is a negative value. If (1+2 U / C^2 ) becomes less than 0, t/T is set to 0. After the contraction, t/T will be seen to be closer to if not equal to 0 for each shell rn. Sense time freezes at r where t/T=0, imaginary results are set to 0 where matter was frozen This is like one approaching the speed of light, t/T will never quite get to 0.

These calculations may need to be iterated several times. If the condition t/T = 0 for each successive shell works its way to the surface, you will have a black hole

Using this model where space is allowed to contract in the radial direction, a 2 solar mass neutron star at an average density of 1x10^17 Kg/m^3 will contract down to a black hole. The matter in this model is not crushed by gravity to a singularity but space is contracted to a singularity. Relative to the matter, it is still at the same density, but relative to remote space, the event horizon of the black hole has a radius of 6 Km and an average density of 1.3x10^18 Kg/m^3.

In this model, a 2 solar mass neutron star with a radius of 21 Km reached to the point where time from the center to the surface came to a stand still (t/T = 0). The radial dimensions of space contracted and a black hole with a 6 Km radius was formed.
 
  • #5
DC0 said:
The relative rate of the slowing of time is expressed by:

t / T = √ (1+2 U / C^2 ) where U = (1/2)( G M r2 / R3 ) - (3/2)( G M / R) , where M is the mass of the neutron star, R is the radius, and r is a radius within the star.
https://www.physicsforums.com/showthread.php?t=40391

Nobody appears to have brought it up in that thread, but this formula is not for a real neutron star (or other massive object); it's for a highly idealized, unrealistic scenario where the object has a constant density throughout. However, qualitatively the formula does give a reasonable idea of the behavior of t/T inside a real object--but that is by no means the only important factor in determining how much mass it can have before it forms a black hole. See below.

DC0 said:
Here I used the comparison of the rate of clock ticks rather than the time between ticks. As matter (using no contraction) (at a density of 8x10^17 Kg/m^3) is added, it will cross a transition point (at 2.6 solar masses) where the time rate at the center will come to a stand still (t/T = 0).

No, that isn't what will happen. What will happen is that the object will become unstable and collapse into a black hole before this "transition point" is reached.

One way of seeing this is to look at what happens to the pressure at the center of the object as you add more mass to it. Mathematically, at the "transition point" where t/T = 0 at the center, the pressure at the center is infinite. So this state isn't really physically possible; something else must happen before this point is reached (and the only possible something is collapse to a black hole).

DC0 said:
If nothing else happens, as more matter is added, the radius r at which time freezes increase: Setting t/T to 0 and solving for r we get:

r (where t/T=0) = √(3 R^2 – R^3 C^2 / (G M) )

This equation shows that as the mass increases to 4.8 solar masses, where t/T = 0 at the surface, the condition for a black hole

Reh=2GM/C^2

is met.

That isn't the condition for a black hole to form; a black hole will form well before this point. See above. (It is true that a black hole will have t/T = 0 at its horizon, but that's very different from saying an ordinary massive object has to have t/T = 0 at its surface before it will form a black hole.)

DC0 said:
I divided a neutron star into 500 shells of equal mass and density. As the rate of time
slows for each shell with thickness dr at radius rn, the radial dimensions of space contract by the same factor.

Were you trying to use the actual equations of GR for this, or is this just your own personal model? I suspect it's the latter, because what you are describing is not at all the way the actual equations of GR (as well as quantum mechanics, which is crucial to this problem because neutron degeneracy pressure is by far the main source of pressure in a neutron star) are used to estimate the maximum possible mass of a neutron star. See this Wikipedia article for a start:

http://en.wikipedia.org/wiki/Tolman–Oppenheimer–Volkoff_limit
 
  • #6
I’ve made the assumption that if the radius of a star R with mass M was equal to or smaller than

Reh = 2GM/C^2

then you would have a black hole. Also using the equations for 1/ (time dilation)

t / T = (1+2 U / C2 )^1/2 where U = (1/2)( G M r^2 / R^3 ) - (3/2)( G M / R)

and setting the conditions at the surface to r = R and t/T = 0 and then solving for R, we get

R = 2GM/C^2.

Sense this is the same equation it should also meet the condition of a black hole.

To allow the density to vary, I divided the star up into 500 shells of equal mass. I know that the density of a neutron star varies with depth, but I didn’t know this relationship. If I had this info I could have applied it to each shell. A uniform density gave me a starting point and the shells allowed the density to vary as the star contracted. This allowed me to calculate the potential energy per unit mass at each radius even though the density of each shell was changing. t/T could then be calculated using

t / T = (1+2 U / C2 )^1/2

This contraction is not one that puts an additional pressure on matter as in neutron degeneracy pressure. The radial dimensions of space contracts. Relative to the mass it self this does not cause an increase in pressure. For example with this method a black hole was created with 2.1 solar masses at a density of 1x10^17. This is 1/3 the density of a neutron. When the surface radius becomes equal to or less than Reh, then we have a black hole.

You are right, this in probably my own personal model of GR. The shells were only used to calculate the potential energy per unit mass U in order to come up with new values of t/T. This gets back to my original question, How does space contract as you get close to t/T = 0 ? I have a ways to go before I understand the equations of GR. I assumed here that the contraction to be similar to that in SR and that as you approach the speed of light the contraction of space is equal to t/T.
 
  • #7
DC0 said:
I’ve made the assumption that if the radius of a star R with mass M was equal to or smaller than

Reh = 2GM/C^2

then you would have a black hole.

That's true, but the star won't be static if this is the case, and all the equations you are using for "time dilation" and so forth are only valid if the star is static, so you can't use them the way you're using them.

DC0 said:
Sense this is the same equation it should also meet the condition of a black hole.

No, because this equation is only valid for a static star; see above.

DC0 said:
This contraction is not one that puts an additional pressure on matter as in neutron degeneracy pressure. The radial dimensions of space contracts.

This doesn't look valid to me; there is nothing in GR that I'm aware of that works like this.

DC0 said:
You are right, this in probably my own personal model of GR.

I'm not sure it's even a "model of GR"; see above.

DC0 said:
This gets back to my original question, How does space contract as you get close to t/T = 0 ?

What do you mean by "space contracting"? That is, what physical observable corresponds to "space contracting"?

DC0 said:
I assumed here that the contraction to be similar to that in SR and that as you approach the speed of light the contraction of space is equal to t/T.

This is an incorrect assumption; the two cases are not analogous.
 

FAQ: Dimensions of Contraction due to Gravity

What is "Dimensions of Contraction due to Gravity"?

"Dimensions of Contraction due to Gravity" refers to the phenomenon where an object's dimensions (length, width, and height) appear to decrease when it is subjected to a strong gravitational force.

How does gravity cause contraction?

Gravity causes contraction by pulling on the particles that make up an object, compressing them closer together and reducing the object's overall size.

Is the amount of contraction due to gravity the same for all objects?

No, the amount of contraction due to gravity depends on the mass and density of the object. Objects with a higher mass and density will experience more contraction than lighter and less dense objects.

Does the strength of gravity affect the amount of contraction?

Yes, the strength of gravity directly affects the amount of contraction an object experiences. The stronger the gravitational force, the more significant the contraction will be.

Can we observe dimensions of contraction due to gravity in everyday life?

Yes, dimensions of contraction due to gravity can be observed in everyday life, especially in objects that are subjected to strong gravitational forces, such as planets and stars. It can also be seen in smaller scales, such as in the compression of objects on the Earth's surface.

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