Gravitons and Push Gravity question.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by rousejohnny</i>
<br />A gravity wave is produced spherically outward from or towards the center of gravity.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Even in mainstream physics, this is not good terminology. "Gravity waves" are an atmospheric phenomenon used in meteorology and hydrodynamics. You mean "gravitational waves" as used in relativity, the existence of which is still just a theory. These hypothetical entities are so weak that, assuming they exist, no effect from them has ever been detected in our solar system. So any discussion of their possible effects is even more hypothetical.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">You have a consistent pull from (some prefer a push towards) a center of gravity. In order for the arch to maintain the circle and conserve energy, its magnitude must be maintain resulting in an increase in gravitational "density" per any given area as you move towards the center.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">This seems to introduce a confusion of a different type. These hypothetical "gravitational waves" have nothing to do with ordinary gravitational force. (If they did, they would have been detected long ago because gravitation and changes therein are the dominant forces observed daily in our solar system.)

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">It is the speed, angle and mass of a body that enters an area of greater gravitational potential that determines the geometry of its path.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">This continues the previous confusion. Although gravitational force and gravitational potential have a mathematical relationship, they are in principle physically separate things. A close analogy here is acceleration and velocity, which have the same type of derivative relationship, yet each can take on any value, and the causal connection works only one way: accelerations cause changes in velocity, and gravitational forces cause gradients in gravitational potentials.

In the latter case, if we start with a large, hollow, spherical shell and place a test body anywhere inside, the gravitational potential depends on the mass and radius of the shell, but there is no gravitational force acting on the body. The body's path is linear and its speed is uniform (until it bounces off the shell) regardless of its own speed, angle, and mass. Moreover, we can add additional uniform, spherical shells outside the first one, and the same statements will remain true about no gravitational force on the body, even though the interior gravitational potential can increase through this process essentially without limit.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">The gravity acts linearly and any "curvature" of spacetime is the result of the linear forces influence and its sphereical propagation at some degree less than instantanious upon the mass entering the system.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Curvature of space paths should not be mixed up with curvature of "spacetime" paths in GR. Think of "spacetime" as proper time and you will avoid such confusions. See also our article "Does space curve?" at metaresearch.org/cosmology/gravity/spacetime.asp If you insist on using now-falsified geometric GR to understand the motions of bodies, my advice is to lower your expectations for ever understanding the physics behind the math because no one has yet shown a way to do that without invoking miracles such as spontaneous creation of new momentum.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Frame-dragging from the rotation and less than instantaneous speed of a gravity wave would combine to produce curvature.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">This combines all the previous confusions. I doubt you could find any physicist who would follow you to this conclusion.

For a greatly improved understanding of gravitation, I recommend our CD "Gravity", which contains 30 articles (many from past issues of the Meta Research Bulletin), a dozen colorful presentations, and some animations, to give the most complete understanding of the nature and origin of gravity presently available. For more information about the CD, see metaresearch.org/publications/CDs/GravityContents.asp -|Tom|-

Can the shell you have described above be explored in more detail? Say the mass of the shell is a few meters thick and few kilograms per square meter and the diameter is 100 AU or whatever size is easy to work with. Would the gravity of the shell have any effect on space in the interior? This detail has been left out of the topic since the time of Newton or maybe I'm wrong about this. Anyway, Newton did say the effect inside is different than outside the mass in that anything outside would be effected as though the mass was centered at the center of the sphere. He never said what effect would be encountered inside the sphere. Has anyone kicked this detail around since Newton?

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Jim</i>
<br />Has anyone kicked this detail around since Newton?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">It's a homework exercise for physics students. There is zero gravitational force everywhere inside a uniform spherical shell. Nonetheless, the gravitational potential inside still slows atomic clocks. -|Tom|-

I certainly did hack this up pretty well. I must learn to express myself more clearly. The issue I asked initially about the person in a pocket within the earth and why that person would weigh more in the pocket than they do on the surface is what I am trying to confront. Let's use gravitons that are pushing towards the center of gravity. As they push towards the center of the earth they maintain their quantity, what changes is the area in which they occupy is decreasing and the gravitons are being compressed following the inverse square law all the way to the bottom. The whole circle thing above was obviously a poor conceptual tool for the mindset of this site. The point is that there is no blocking of gravitons from the other direction that generates the increase in gravity, it is the compression of the same number of gravitons in a much smaller area. The number of gravitons attracted to the center of gravity is based on the size of the mass attracting them.

The circle merely represented magnitude of gravity. If a hula hoop had its circumference reduced while maintaining its basic geometric shape, by the laws of conservation the plastic would have to become much more dense. Likewise with gravitons, as they move towards a center point they too become compressed.

I hope that is more clear?

If you had a spherical shell, where the walls were say 8000 miles thick, with a diameter of say one light year. One would expect to be attracted to the shell wall you are closest to. Should you be touching the shell wall, one would expect to be attracted to that shell wall close to the equivalent of Earths gravity. Tom would say that you wouldn,t have an attraction to the shell wall, no matter where you are within that shell. Is this correct?

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Skarp</i>
<br />Tom would say that you wouldn't have an attraction to the shell wall, no matter where you are within that shell. Is this correct?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">That is correct. And it is not just Tom that says so. Every physics teacher covering gravity and every student who hopes to pass the course had better say so too, and be able to at least demonstrate why it must be so. The second year students would be expected to prove mathematically that it is rigorously true for anywhere in the interior of all uniform spherical shells.

In a short nutshell, consider an arbitrary point inside the shell. Make it close to one wall, as in your example. Draw a cone with apex at the given point, expanding toward any part of the nearby wall. That cone will intercept a certain area of shell having a certain mass m at a certain distance d. The gravitational force it would produce on the point will be proportional to m/d^2.

Now extend the same cone in the exact opposite direction. It will eventually intercept a much larger area of the far wall having a mass M at a distance D and producing a force on the point of interest proportional to M/D^2.

However, the area of the base of a cone is proportional to the cone height squared. Because mass intercepted is propotional to area on the shell, it must be the case that M/m = D^2/d^2. Every force toward a near wall with a small mass is exactly compensated by an equal and opposite force toward the far wall with a large mass. The larger mass is (D/d)^2 times bigger than the small mass, but is also weaker because it is farther away by the same factor.

So at every point inside a uniform spherical shell, there is zero net force of gravity from the shell. -|Tom|-