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 />With the concept of graviton driven pushing gravity, why aren't we crushed? Certainly, if we were on a neutron star we would be, so where do these graviton cancel out in this phenomena on earth or in space that does not crush us from both sides?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Gravitons do not crush things; they accelerate things. For example, astronauts and their spacecraft in orbit are free to accelerate in response to all gravitational forces, so the astronauts feel nothing and are in fact weightless. When they are back on the ground, they would still be weightless if the ground did not push back and resist the acceleration of gravity.

So on Earth, and more so on a neutron star, it is not the gravity that crushes but the surface of the planet or neutron star pushing back.

Another example is the inside of a uniform spherical shell. The gravitational potential inside can become indefinitely large if the mass of the shell is large enough. But anyone or anything inside still feels no gravitational force whatever. Gravity does not crush; it just pushes things around if the gravitons are unbalanced, such as when they are partially blocked by a source mass in some direction.

On Earth, we feel just 1g of acceleration because we are hit by 100% of gravitons raining on us from above, but only 99.999...% of the gravitons raining on us from below because Earth blocks 0.000...1% of those gravitons. But that imbalance is enough to push us down and press us against the Earth, which pushes back firmly and makes us feel weight.

On a neutron star, the blockage of gravitons from below is orders of magnitude more efficient, so the imbalance is orders of magnitude greater, resulting in a very strong downward push and a very strong back-push by the neutron star's surface, crushing us. To prevent that, if only for a moment, we might jump into a crevice on the neutron star. While we are in free fall, we fell no crushing force whatever, showing that gravity is not the culprit. It's the rapidly approaching floor of the crevice that we must fear. [:0]


A different question than the one you asked is: "Do matter ingredients inside material bodies feel a crushing force from gravitons?" And the answer is "yes". That kind of force has nothing to do with graviton imbalances or accelerations. A matter ingredient (MI) is defined as the largest entity that cannot be penetrated by gravitons. MIs are orders of magnitude smaller than typical quantum particles. An MI in isolated space would absorb a lot of gravitons from all directions and would feel a crushing force. But it has very high density and very large mass compared to gravitons, so it is easily able to resist the crushing force. Of greater concern to an MI is energy balance because gravitons are continually depositing new energy.

Understanding the thermodynamics of gravitons is a larger subject, addressed in detail in <i>Pushing Gravity</i> and our new "Gravity" CD. But to offer a brief analogy, the situation is like that for hot and cold macroscopic bodies in isolated space, which are nonetheless always exposed to "background" microwave radiation. So warm bodies normally radiate enough heat into space to compensate for what they absorb from the background radiation, which keeps them in equilibrium. And very cold bodies do the reverse -- absorb at least as much heat from background radiation as they emit, which keeps them in or near equilibrium. It is similar for MIs bathed in background graviton radiation, except that elysium is an intermediary that aids the heat balance process. -|Tom|-

So these MIs are not observable and yet they are more massive than is the graviton. Wouldn't it be more likely that the MIs would be located before the graviton? I wonder if the MIs share common traits such as the baryonic particles and electrons do, ie. balanced charge?

<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 />So these MIs are not observable...<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">They are observable in principle, of course. But humans have not yet invented instrumentation with that much resolution.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">...and yet they are more massive than is the graviton.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Yes, an MI is to a graviton as the Earth is to a large meteoroid.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Wouldn't it be more likely that the MIs would be located before the graviton?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Certainly.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">I wonder if the MIs share common traits such as the baryonic particles and electrons do, ie. balanced charge?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">No, charge is a force, and forces are caused by specific particles or specific mediums. In the case of charge, elysons (elysium constituents) are responsible.

Protons are so dense that they have strong surface gravity (for their scale), and hold thick elysium atmospheres gravitationally bound around them. When two protons try to approach, their elysium atmospheres compress and, in spring-like fashion, produce a repulsive force that results in a rebound rather than a collision. Protons and all particles are immersed in an elysium "sea", so their own elysium atmospheres are merely denser elysium comcentrations than their surroundings. Particles with nagative charges emit more gravitons than they absorb and therefore have a zone of lesser elysium surrounding them.

There's a lot more to this model. Your questions merely scratch the surface. To get a sense of the "Meta Cycle", see the graphic at the end of metaresearch.org/cosmology/gravity/meta_cycle.asp To get a real appreciation of the whole model, I recommend the two resources I mentioned. To stay up-to-date on the most recent developments, I also recommend the Meta Research Bulletin. -|Tom|-

Tom,

One last question. Say I was in a pocket somewhere deep down in the earth, shouldn't my weight be the same there as it is on the surface according to pushing gravity?

<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 />Say I was in a pocket somewhere deep down in the earth, shouldn't my weight be the same there as it is on the surface according to pushing gravity?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Each matter ingredient (MI) blocks some gravitons from striking each other matter ingredient, resulting in a force tending to push the two together. This predicted force of pushing gravity is identical to the pulling force predicted by Newtonian gravity. Until one gets to ultra-high matter densities or distances large compared with a kiloparsec, there cannot be a difference between pushing gravity and pulling gravity because, particle by particle, they both predict identical forces.

Even in pulling gravity, the question of what happens below the surface is non-trivial. Some of Earth's mass (the part above the observer) produces an upward force, lowering the observer's weight. However, the observer is also closer to the Earth's denser interior, producing a stronger downward force. Which of the two dominates depends on the density profile of the Earth and the depth of the observer. -|Tom|-

Here is a concept that I have been playing around with. It is some twisting of the inverse spare law thinking in terms of circumferance and mass and a linear pull or pushing gravity. Well here is a brief summary.

A gravity wave is produced spherically outward from or towards the center of gravity. Any two lines from the center point outward that intersects the gravitational sphere (single wave of gravity) perpendicularily would generate an arch. The curvature of this arch would generate a circle, depending on the shape of the mass and distance from the center, the perfection of this "circle" would vary.

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.

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. 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. Frame dragging from the rotaion and less than instantanious speed of a gavity wave would combine to produce curvature.