The inverse square law

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by EBTX</i>
<br />If you say that a mass of 1 unit produces shadowing effect of 1 unit ... then won't two masses produce 2 units of shadowing effect? Is this not correct by your theory?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Doubling the mass doubles the shadow effect. If the two masses are separate, then like two shadows of objects cast by two flashlights, the shadows will sometimes add and sometimes not, depending on location.

A spot equi-distant between two masses feels zero force because the two separate forces cancel.

In general, each mass must be considered separately and the effects of each separate mass then added vectorially (i.e., the sum depends on direction and distance of each). -|Tom|-

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Doubling the mass doubles the shadow effect<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
Now, if the object mass that the shadow falls on is doubled without changing its area presented to the shadow, will there be any increase in the shadowing effect? Can changing the mass of the shadowed object cause more CGs to hit it from the rear than what is represented by the shadow?

You see, if we double the mass that the shadow falls on, we need to have the same shadowing effect on that added mass as was on the original mass ... (i.e. we need to double the CGs hitting the side away from the shadow of the now doubled object) ... or the shadowing theory breaks down.

Let's say that the shadow represents a shortfall of 42 CGs on the "windward" side of the object it falls on. Now if we double the objects mass that the shadow is falling on, that same shadow must now represent a shortfall of 84 CGs on the now doubled object. How could this be so?

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by EBTX</i>
<br />if the object mass that the shadow falls on is doubled without changing its area presented to the shadow, will there be any increase in the shadowing effect?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">That depends on your definition of "the shadow effect". I'll elaborate below.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Can changing the mass of the shadowed object cause more CGs to hit it from the rear than what is represented by the shadow?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">A body's own mass does not have any influence on its own acceleration in the graviton shadow of another body. That is beacuse a body may be visualized as a loose collection of individual matter ingredients (MIs) with each MI acted upon independently by gravitons (CGs) from all other MIs. So it doesn't matter how many MIs there are.

But a body's own mass is very much a factor in how strong a shadow it casts on other bodies, which determines how much those other bodies will accelerate.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">You see, if we double the mass that the shadow falls on, we need to have the same shadowing effect on that added mass as was on the original mass ... (i.e. we need to double the CGs hitting the side away from the shadow of the now doubled object) ... or the shadowing theory breaks down.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Let's say the normal flux is 100 gravitons per second hitting each side (front and back) of each MI inside the target body. Then let's say that the shadow of another mass reduces that flux to 98 gravitons per second on the facing side only. Then each MI in the target body will start to accelerate toward the shadow by the net momentum from 2 extra gravitons per second. And that remains true no matter how many MIs are in the target body.

What does change is how strong a shadow the target body casts back on the source mass, and therefore how much the source mass is induced to accelerate.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Let's say that the shadow represents a shortfall of 42 CGs on the "windward" side of the object it falls on. Now if we double the objects mass that the shadow is falling on, that same shadow must now represent a shortfall of 84 CGs on the now doubled object. How could this be so?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">In my example, if the target body originally had 21 MIs, then the shortfall would indeed be 42 CGs per second on the shadowed side of the target body. (That would be its leeward side, not its windward side, by definition of "shortfall".) Then if we doubled the number of MIs, we would double the shortfall because each MI feels the same 2 CGs/sec shortfall. -|Tom|-

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Then if we doubled the number of MIs, we would double the shortfall because each MI feels the same 2 CGs/sec shortfall. <hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
So, we're not talking about actual quantities per second of actual CGs as real entities? I am beginning to see your point of view but it still seems inconsistent with your other statements.

Perhaps this will clarify the issue I have.
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There is a hoola hoop in space some distance from a planet. It's in a plane perpendicular to a vertical line to the center of the planet. Is it your position that ...

" a finite number of CGs (X) normally go through the hoola hoop from each direction ... but ... from the planet side which casts a shadow ... that number is less (X-N). Therefore, the excess CGs coming through the side away from the planet is then just 'N'. If we double the planet mass this excess is increased to '2N' "

If so, we have exactly 'N' CGs available to push whatever is the size of the hoola hoop at that same distance. So, we put a mass there and record its acceleration. Then we double this mass (keeping it at exactly the same volume as the original) and get 1/2 the acceleration since we have the same quantity of propellant ('N') pushing a double mass. Now, double the planet mass too ... the acceleration produced must now again be equal to that of the first mass since we now have twice the propellant ('2N' CGs). The "forces" on these bodies must then be
<center>
F = 1M x 1A (on the single mass single planet)
F = 1M x (1/2)A (on the double mass single planet)
F = 2M x 1A (on the duble mass double planet)
</center>

[Please remember that I am here talking about CGs as a "physical propellant" (a pusher) as you suggest they are. I realize that a doubled mass single planet should produce the same acceleration as the single mass, single planet ... but ... by the stated theory of "pushing", it simply cannot do so unless modified by some sort of miraculous multiplying effect. You say that the CGs have a 'propelling' effect which I take to be analogous to momentum transfer as is the case with other types of mechanical interactions. Hence, they are logically equivalent to ... and should match the properties of ... other mechanical propellants. If there are 'N' propellants they cannot do the same job on two masses as they can on one mass. This would make no physical sense. I see 'N' as a fixed, real number of particles per unit time ... however large they may be ... so that once they hit and rebound, they have done their job and can provide no more accelerations. By saying that each particle feels the shortfall (rather than the aggregate body), you are in effect increasing that 'N' shortfall to greater than 'N' per unit time without increasing the source body's mass which is the cause of the 'N' shortfall. I am saying that if you double the object mass, each particle in that mass will "feel" only 1/2 the excess CGs that they previously felt because 'N' is a finite number distributed over twice as many object particles.]

Now, we go back to the single mass single planet and move the object mass to 1/2 the distance. It's excess CG count must then increase fourfold by the inverse square law.

<center>F = 1M x 4A (on on a single mass 1/2 the distance)</center>
Thus, the inconsistency ...

<font size="4">2M x 1A</font id="size4"> (on the double mass, double planet)
<font size="4">does not =</font id="size4">
<font size="4">1M x 4A </font id="size4">(on a single mass, single planet, 1/2 the distance)

which is a required equality by Newton's gravitational equation.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by EBTX</i>
<br />There is a hoola hoop in space some distance from a planet. It's in a plane perpendicular to a vertical line to the center of the planet.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Why not a sphere? Shape plays no role in either Newtonian gravity (NG) or in pushing gravity (PG).

But okay, we have a hoola hoop.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Is it your position that ... " a finite number of CGs (X) normally go through the hoola hoop from each direction ... but ... from the planet side which casts a shadow ... that number is less (X-N). Therefore, the excess CGs coming through the side away from the planet is then just 'N'. If we double the planet mass this excess is increased to '2N' " If so, we have exactly 'N' CGs available to push whatever is the size of the hoola hoop at that same distance.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">That all sounds good to me. We then have exactly 'N' CGs available to push each and every same-mass hoola hoop at that same distance on that side of the planet.

Just adding a detail to your correct description, suppose our standard-size hoola hoop contains N matter ingredients (MIs). Then by your premise, each matter ingredient gets pushed by exactly one graviton (CG), totaling N CGs available to push the entire hoola hoop. If we had five hoola hoops, we would have 5N matter ingredients being pushed by 5N gravitons. It doesn't matter how many MIs or hoola hoops we have or how big the hoola hoops are. Each individual MI gets pushed by the same imbalanced CD flux as every other MI at that distance.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">So, we put a mass there and record its acceleration. Then we double this mass (keeping it at exactly the same volume as the original) and get 1/2 the acceleration since we have the same quantity of propellant ('N') pushing a double mass.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">It should now come as no surprise that this is incorrect. If we double the mass, that doubles the number of MIs, and is the equivalent of adding a second hoola hoop. It's the same net flux for each matter ingredient, so the net flux for a doubled mass must be doubled.

That is why the acceleration of target bodies does not depend on the body's own mass. Each MI is accelerated independently from all others, so it doesn't matter how many there are.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">I realize that a doubled mass single planet should produce the same acceleration as the single mass, single planet<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">?? Why did you say that? Doubling the source mass doubles the acceleration of target bodies in its field. In PG, that is because the source mass blocks twice as many gravitons, making the shadow or net flux twice as strong for each and every MI in a target body. -|Tom|-

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">It should now come as no surprise that this is incorrect. If we double the mass, that doubles the number of MIs, and is the equivalent of adding a second hoola hoop. It's the same net flux for each matter ingredient, so the net flux for a doubled mass must be doubled.
<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
I didn't mean to talk of "hoola hoops" ;o) I meant to speak of the flux through an area per unit time. The hoola hoop just delimits an area. Anyway, what I have trouble understanding is ... a finite number of CGs in a finite time (N/t) comes through a finite volume containing a finite number of MIs (M). So that we have ...

<center><font size="4">N/tM </font id="size4">= CGs per unit time per MI </center>

Then we double the mass (in the same volume) and have 2M. Now we have ...

<center><font size="4"> N/t2M </font id="size4">= CGs per unit time per MI </center>

Yet your answer implies that ...

<center><font size="4">N/tM =N/t2M </font id="size4"> </center>

which does not make sense mathematically or physically unless you are implying that MIs feel the excess flux of CGs as a percentage rather than as a constant number N.

Hmmmm ... wait a minute ...

Is this what you mean ... ?

The excess flux is a real number like "523" ... then you put one mass in the way and it gets hit by "36" of the 523. Then you put another mass there as well (doubling) and it absorbs another "36" of the remaining (523-36=)"487" ...?

The light dawns ... I concede the point.

Everything is AOK as long as the CGs are incredibly abundant which is necessarily the case with MM (even in the Standard Model the number of gravitons must number in the "gazillions" per unit time per unit volume). Then, the only problem with the inverse square law is its algebraic inconsistency about which we can agree to disagree, i.e.

<center><font size="4">(M1 x M2) / r^2 = M1/r^2 x M2/r^2 (!)</font id="size4"> </center>

as opposed to what I believe is true ...

<center><font size="4">(M1 x M2)/ r^2 = M1/r x M2/r</font id="size4"> </center>
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