[jeffreyH (OP)]

Does the mass-energy density affect gravity at higher densities? Does any amplification of field strength occur? How could we test if this were true?

[Kryptid]

The gravitational force experienced between two bodies is related to the mass and distance between them. In some sense, density doesn't matter. If you replaced the Sun with a high-density object like a black hole of equal mass, the planets would retain their normal orbits because the mass/distance factor has remained unchanged. The difference with a black hole is that, since all of the mass is concentrated in the center, you can get very close to it (thus experiencing a much higher attractive force). If you got the same distance from the center of the Sun, the gravitational attraction would not be nearly so strong because much of the Sun's mass would be above you as well as below you (effectively cancelling out some of the attractive force).

[jeffreyH (OP)]

Can this ever be confirmed experimentally? The mathematics may point in this direction but does the experimental situation agree.

[jeffreyH (OP)]

In this situation if we have a mass M1 with radius R1 and a smaller mass M2 with radius R2 then the gravitational field strengths at the respective surfaces would be G1 and G2.
If we compress M1 to the same radius as R2 then length contraction and time dilation should increase. However by the reasoning above the change in the dimension of R1 should alter G1 at the surface (if we wish to preserve orbits around the mass) even though the mass has not increased and is simply compressed. Therefore G1 must increase with compression if the above statements are true. Meaning that the same mass exhibits higher field strength at its surface under compression. I.E. Gravitational amplification.

[jeffreyH (OP)]

I am working on an equation for mass-energy density for gravity. I will post a graph of the results when done.

[Kryptid]

In this situation if we have a mass M1 with radius R1 and a smaller mass M2 with radius R2 then the gravitational field strengths at the respective surfaces would be G1 and G2.
If we compress M1 to the same radius as R2 then length contraction and time dilation should increase. However by the reasoning above the change in the dimension of R1 should alter G1 at the surface (if we wish to preserve orbits around the mass) even though the mass has not increased and is simply compressed. Therefore G1 must increase with compression if the above statements are true. Meaning that the same mass exhibits higher field strength at its surface under compression. I.E. Gravitational amplification.

The field strength at the surface has increased because you have decreased the distance to the center. Shell theorem predicts that the strength of gravity at the surface of a sphere is dependent only upon the mass of the sphere and its radius, not its density: http://en.wikipedia.org/wiki/Shell_theorem

For example, imagine if the Earth was a giant, hollow shell only one foot thick, but that this shell was super-dense such that it had the same mass as the real Earth. Shell theorem predicts that the gravitational force at the surface of this "shell Earth" is the same as that of "real Earth". You could do the same by positing that the Earth is hollow and 99% of its mass is tied up inside of a black hole at its center. This "black hole Earth" still exhibits the 1G force at its surface. The density and arrangement of mass within a sphere does not affect its surface gravity (assuming that the mass is distributed in a spherically-symmetrical manner).

[jeffreyH (OP)]

In this situation if we have a mass M1 with radius R1 and a smaller mass M2 with radius R2 then the gravitational field strengths at the respective surfaces would be G1 and G2.
If we compress M1 to the same radius as R2 then length contraction and time dilation should increase. However by the reasoning above the change in the dimension of R1 should alter G1 at the surface (if we wish to preserve orbits around the mass) even though the mass has not increased and is simply compressed. Therefore G1 must increase with compression if the above statements are true. Meaning that the same mass exhibits higher field strength at its surface under compression. I.E. Gravitational amplification.

The field strength at the surface has increased because you have decreased the distance to the center. Shell theorem predicts that the strength of gravity at the surface of a sphere is dependent only upon the mass of the sphere and its radius, not its density: http://en.wikipedia.org/wiki/Shell_theorem

For example, imagine if the Earth was a giant, hollow shell only one foot thick, but that this shell was super-dense such that it had the same mass as the real Earth. Shell theorem predicts that the gravitational force at the surface of this "shell Earth" is the same as that of "real Earth". You could do the same by positing that the Earth is hollow and 99% of its mass is tied up inside of a black hole at its center. This "black hole Earth" still exhibits the 1G force at its surface. The density and arrangement of mass within a sphere does not affect its surface gravity (assuming that the mass is distributed in a spherically-symmetrical manner).

It is not simply having decreased the distance to the centre. I can decrease the distance to the centre be digging down into the earth to the same point. However the gravitational effect would decrease and not increase. The radius makes no sense inside the mass itself as it becomes a fraction. In the inverse square nature of gravitation this will not explain compression without a functional adjustment of mass to radius.

[jeffreyH (OP)]

Here is a plot I have done as density increases/radius decreases. Does anyone know at what percentage of the radius of a mass that we get the Schwarzschild radius? I would like to see if it falls correctly on this graph. The y-axis is gravity strength and the x-axis percentage of the uncompressed radius.

[jeffreyH (OP)]

Oops. That wasn't the density plot.

[Pmb]

Does the mass-energy density affect gravity at higher densities?

Yes.

Does any amplification of field strength occur?

I don’t understand what this means so I’ll just say no.

How could we test if this were true?

When you tell me what amplification of field strength occur[/I] means I’ll let you know.

[Kryptid]

It is not simply having decreased the distance to the centre. I can decrease the distance to the centre be digging down into the earth to the same point. However the gravitational effect would decrease and not increase.

I explained this before with the Sun example. The reason that the gravity decreases when you dig into the Earth is because some of the mass of the planet is now above your head and therefore pulling on you in a direction away from the center. At the center of the Earth, gravity is at its weakest because all of the mass is around you, not below you. It pulls on you in all directions roughly equally, cancelling out the attractive force. In a black hole, the center is where gravity is the strongest because that's where all of its mass is concentrated.

[jeffreyH (OP)]

Does the mass-energy density affect gravity at higher densities?

Yes.

Does any amplification of field strength occur?

I don’t understand what this means so I’ll just say no.

How could we test if this were true?

When you tell me what amplification of field strength occur[/I] means I’ll let you know.

The way I am looking at it the compression of mass produces tighter gravitational flux lines. The reason why photons get trapped is that they are outnumbered by a denser graviton field. It is like a laser beam for photons except this is an intensification of gravitons much like a laser beam. That is the 'amplification' I am exploring.

[Kryptid]

The way I am looking at it the compression of mass produces tighter gravitational flux lines. The reason why photons get trapped is that they are outnumbered by a denser graviton field. It is like a laser beam for photons except this is an intensification of gravitons much like a laser beam. That is the 'amplification' I am exploring.

The more proper analogy would be to compare a gravitational field with an electromagnetic field, not a laser. Although both laser beams and EM fields are made up of photons, they have rather different properties (virtual vs. real photons). The best equivalent to a laser would be a uniform beam of gravitational waves.

[jeffreyH (OP)]

It is not simply having decreased the distance to the centre. I can decrease the distance to the centre be digging down into the earth to the same point. However the gravitational effect would decrease and not increase.

I explained this before with the Sun example. The reason that the gravity decreases when you dig into the Earth is because some of the mass of the planet is now above your head and therefore pulling on you in a direction away from the center. At the center of the Earth, gravity is at its weakest because all of the mass is around you, not below you. It pulls on you in all directions roughly equally, cancelling out the attractive force. In a black hole, the center is where gravity is the strongest because that's where all of its mass is concentrated.

You appear to be talking about a mass at normal density. I am dealing with collapsing masses. The radius here is related to an isolated mass and not a two mass interaction.

[jeffreyH (OP)]

The way I am looking at it the compression of mass produces tighter gravitational flux lines. The reason why photons get trapped is that they are outnumbered by a denser graviton field. It is like a laser beam for photons except this is an intensification of gravitons much like a laser beam. That is the 'amplification' I am exploring.

The more proper analogy would be to compare a gravitational field with an electromagnetic field, not a laser. Although both laser beams and EM fields are made up of photons, they have rather different properties (virtual vs. real photons). The best equivalent to a laser would be a uniform beam of gravitational waves.

A uniform beam works for me. Bearing in mind that like light the waves are not parallel and disperse with distance. If we could ever produce a directed graviton field we may be able to create a graviton uniform beam that would act laser like..

[jeffreyH (OP)]

Interestingly with this sort of technology it would be possible to tow quite large object in space.

[jeffreyH (OP)]

A clarification on the two graphs above. They are both in natural units. Graph 1 is the increase in gravitational strength as the radius contracts. Graph 2 is the mass density increase as the radius contracts. There is a proportionality as would be expected but what ties then together? That is my next mission.

[Kryptid]

You appear to be talking about a mass at normal density. I am dealing with collapsing masses. The radius here is related to an isolated mass and not a two mass interaction.

What's the fundamental difference between "normal" densities and collapsing masses? It's just a matter of degree, really. Besides, if a higher density object has even slightly more gravity than an object of similar mass with a lower density, then that would mean that the Sun (hypothetically) collapsing into a black hole would affect the orbits in the Solar System. This is in contradiction to current physics knowledge. Look what NASA has to say on the subject: http://spaceplace.nasa.gov/review/dr-marc-sun/black-hole-sun.html

If the Sun were somehow compressed enough to become a black hole, it would be less than 6 kilometers (well under 4 miles) across. It would exert no more gravitational force on Earth or the other planets in the solar system than it does now. Why? Because it would contain no more matter than it does now and it would be no closer to the planets than it is now.

[jeffreyH (OP)]

You appear to be talking about a mass at normal density. I am dealing with collapsing masses. The radius here is related to an isolated mass and not a two mass interaction.

What's the fundamental difference between "normal" densities and collapsing masses? It's just a matter of degree, really. Besides, if a higher density object has even slightly more gravity than an object of similar mass with a lower density, then that would mean that the Sun (hypothetically) collapsing into a black hole would affect the orbits in the Solar System. This is in contradiction to current physics knowledge. Look what NASA has to say on the subject: http://spaceplace.nasa.gov/review/dr-marc-sun/black-hole-sun.html

If the Sun were somehow compressed enough to become a black hole, it would be less than 6 kilometers (well under 4 miles) across. It would exert no more gravitational force on Earth or the other planets in the solar system than it does now. Why? Because it would contain no more matter than it does now and it would be no closer to the planets than it is now.

You misunderstand. I agree with you. Yes the orbits would be consistent outside of the collapse. Within the collapsing region density prevails. Working out towards the original radius you arrive at the same values as you had before collapse.

[jeffreyH (OP)]

I have attached a plot of standard gravity plot against density. Where these cross at a value of 1 represents normal solid matter at an ideal density. To the right series 1 represents densities above melting point, first liquid and then gas. To the left is solid matter compression which will ultimately lead to a singularity. Series 2 is the inverse square plot for gravity without reference to density. I am thinking through some conclusions at the moment.