Gravitons and Push Gravity question.

The connection with prior posts is clear to me. The two bodies will effect the third body in very simple ways if you keep angular momentum at zero when starting the third body in motion. The way it has been done makes little sense unless you are using pen and paper to calculate all the tiny bits. A good overall perspective can be made from two starting points 90 degrees apart much like is done when tidal charts are computed. Call the two starting points the spring and neap point. Starting at the spring point the third body falls into the nearest star mass. Starting at the neap point the third body falls between and through the two star masses and continues on as you have posted above. Starting from other positions some combination of these two extremes happens. How does the math work out in this model? Thats what I wonder.

<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 />How does the math work out in this model? Thats what I wonder.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">As I said, hairy second-order differential equations. Intuition is a very poor guide in such cases, for reasons I explained in detail in chapter six, "Orbits", of <i>Dark Matter, Missing Planets and New Comets</i>. -|Tom|-

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by tvanflandern</i>
<br /><blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by north</i>
<br />is the lead(Pb) dense enough not to allow gravitons to have an effect?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">No ordinary matter is dense enough to produce significant gravitational shielding. All ordinary matter is as transparent to gravitons as a chain-link fence is to the wind.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">i would think that lead is denser than the Earths crust.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Yes, lead is denser. Earth's crust averages only about 3 g/cc. Why is that relevant?

If a body was dense enough for graviton shielding (say, matter from a neutron star, for which a teaspoonful would weigh tons), it would not fall as fast as ordinary matter because its inertial mass would exceed its gravitational mass. But to other outward appearances, the super-dense body would act normal. -|Tom|-
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Tom

through all this though, i think there is the assumption that the density of gravitons, no matter the gravitons density , is enough to stop 10 tons of lead? or put another way, since the density of gravitons is less at the center of a mass than at the crust and as well, in a three dimensional picture, there is less gravitons within this hole than that which would surround a planets outer crust, what makes you think that there is enough gravitons to stop the mass of lead enough to make it "bob" or oscillate? is there not a critical density of gravitons to density of mass ratio? i would suggest there is.

so we have gravitons coming in Jim's hole from the north as well as from the south,which would balance out, as well from any other angle within this sphere, which balances out, there is no gravitons left in which to stop the mass of lead.

the gravitons at this point would be balanced, there is as much gravitons coming from the north angles, as there is from the south angles and all angles in between.

but from the the north where the drop originates, the momentum of the lead(Pb) mass is increasing its acceleration, it has three things here, 1) the density of the lead its self, 2) the leads(Pb)momentum, 3) density of gravitons.

on the south side we have only the density of gravitons. which this density of gravitons must be able to overcome,1) the momentum of the lead over time and 2) the density of the gravitons from the north, which is ceaseless. since the density of gravitons is equal at either the north or south of this hole in a sphere in the begining and since the south part of the sphere has no mass momentum only the gravitons momentum , it seems to me that the south gravitons must not only over come the density of the lead and the leads momentum but also the the gravitons that are still coming from the north.

therefore i see as yet no reason that the south gravitons have the ability to overcome the leads(Pb) density, momentum(acceleration) and gravitons coming from the north, there is an imbalance. which leads to the conclusion that this mass of lead has nothing strong enough to stop from going into space at the south side of this hole in this sphere.

therefore i see no reason for any oscillation at this spheres core.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by north</i>
<br />there is the assumption that the density of gravitons, no matter the gravitons density, is enough to stop 10 tons of lead?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">You have missed a basic point about gravitons, essential to understanding nature. Gravitons are so small that all ordinary matter is highly transparent to them. The small percentage of gravitons that actually hit something push on each tiny matter ingredient (MI) independently. Whatever the graviton "wind" does to one MI, it does to all MIs. So it doesn't matter how many of them are in a body, or how dense that body is. One MI, one atom of any kind, a huge sphere of iridium (much denser than lead), or the Moon -- all bodies fall at the same rate in the Earth's field.

You may have been misled by your intuition, based on experience with mechanical forces, where larger and denser masses are harder to move. But that is not true for gravity. A lead ball and a feather move and accelerate with equal ease in a gravitational field. 10 tons of lead is nothing special to gravitons. Neither is 1 million tons of lead.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">since the density of gravitons is less at the center of a mass than at the crust<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">That difference is insignificant. As an approximation, one in every 100 million gravitons might be missing at the center, as compared with the surface. But that small imbalance is all it takes to create the force of gravity as we know it.

But because you assumed the difference in graviton streaming density between center and surface was significant when it is not, most of what you deduced from that incorrect assumption does not follow.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">what makes you think that there is enough gravitons to stop the mass of lead enough to make it "bob" or oscillate?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">The "stopping" does not occur until the mass of lead is rising, moving upwards from the center. What exactly do you expect gravity to do to a rising mass? Why shouldn't it slow, then stop, then start falling again, just as for a ball thrown straight up? I don't know where your intuition derailed here because the predicted behavior is so familiar from everyday experience.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">is there not a critical density of gravitons to density of mass ratio? i would suggest there is.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Not for ordinary matter. Now if you want to talk neutron stars or Mitchell stars, then what you say might have some meaning.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">we have gravitons coming in Jim's hole from the north as well as from the south, which would balance out, as well from any other angle within this sphere, which balances out, there is no gravitons left in which to stop the mass of lead.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Of course not. That is why there is zero force at the center. The slowing does not begin until the lead mass has sailed past the center and starts rising up the hole on the other side. The mass is not stopped until the lead gets to the top of the other hole. Then it starts to fall back and repeat the process.

Naturally, the speed of the falling body increases as it falls because gravity is pulling it downward. And that speed slows as the body rises because gravity is still pulling it downward. At any distance r from the center on one side, the speed will be exactly the same at distance r on the other side of the hole. So the speed reaches a maximum at the center, and is zero at both surfaces.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">i see as yet no reason that the south gravitons have the ability to overcome ... gravitons coming from the north<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Gravitons are far too small to collide with each other. On average, a graviton must travel about 1000-2000 parsecs before hitting any other graviton.

And in any case, the gravitons in the hole are too few to matter compared to the gravitons flying through the solid Earth from all directions. And those <i>are</i> imbalanced at places away from the center because there is more graviton-absorbing Earth in some directions than in their opposite directions. -|Tom|-

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by tvanflandern</i><br />
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.
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Hi, Tom. I'm curious as to your interpretation of gravitons and black holes. Do black holes block them or emit them? To be fair, I don't believe in gravitons, but I like to understand the competing theories.

James

Ok, who put "Uninflamable" signs on gasoline tanker trucks?

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Youjaes</i>
<br />I'm curious as to your interpretation of gravitons and black holes.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">In the 18th century, Mitchell was the first to describe stars with gravity so strong that light cannot escape. These are still called "Mitchell stars". That is what strong gravity fields produce. The modern concept of event horizons that reverse space and time is a mathematical model with little physical plausibility. Einstein himself proved it was a physical impossibility in a 1939 paper while he was at Princeton. Over a decade later, one of his students, Wheeler, revived the concept. But Wheeler never even addressed, much less overcame, the objections Einstein raised.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Do black holes block them or emit them?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Invent any answer you like. Black holes are a purely mathematical concept, not something that exists in physical reality. So you can make them graviton-compatible or not, as you please.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">To be fair, I don't believe in gravitons, but I like to understand the competing theories.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">That's okay, because I side with Einstein and argue that "black holes" don't exist. [}] -|Tom|-