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The Slabinski Article: Cross sectional area propor
20 years 11 months ago #8063
by PhilJ
Replied by PhilJ on topic Reply from Philip Janes
Okay, Tom; I'll post this here instead of starting a new thread. My previous criticisms of the Slabinski article were of little, if any, consequence. Now, I get down to the heart of the article. Sorry if this is a bit long, but I think it's important.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Slabinski: <i>Pushing Gravity </i>(<i>PG</i>), page 125:
"We have the surprising result that scattering of gravitons by the mass particle causes no decrease in the rate at which gravitons reach <i>A</i>test!"<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
Well, duh! That result was intuitively obvious to me from the start. It means that, against a uniform background of CG's, the CG's deflected by the mass particle toward Atest exactly match those deflected elsewhere, which had been moving toward Atest before being scattered by the mass particle. This is analogous to the fact that an idealized 100% reflective silver globe suspended in a uniformly illuminated white room will seem to disappear (if not for the reflection of the observer). By extension, two silver globes in a white room will reflect only the whiteness of the walls, and each globe's reflection in the other will likewise be the same pure white as the walls.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Slabinski: <i>PG</i>, page 126:
"At m1, there is a deficiency of gravitons coming from the m2 direction [due to absorption by m2]. The momentum transfer to m1 due to this excess depends both on absorption of part of the excess and on scattering of part of the excess [by m1]." <hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
I won't call myself a complete idiot, but I do admit I am just now beginning to see how this is supposed to work. The graviton shadow, cast by particle 2's absorption of CG's, is disbursed by particle 1's scattering of CG's, resulting in a net momentum of particle 1 toward particle 2--and vise versa. Slabinski concludes (on page 128, formula 29) that this mechanism accounts for about 99.999,999,999,999,999,999,999,999,999,999,999,3% of gravitational force.
Recalling the analogy of the silver globes in the white room: Suppose each globe absorbs a tiny fraction of the incident light (as any real silver globe will actually do). Then the reflected image of each globe, seen on the face of the other, will be darker than the walls. Therefore, the light reflected from the side of each globe facing the other will impart less momentum to the globe than the light from the opposite side of the globe. That imbalance will result in a very slight force of attraction between the two globes. In the case of gravity, that force is much stronger. The greater the ratio of reflection to absorption, the greater the ratio of momentum to heat. (Am I slow, or what?)
Now that I think I get what Slabinski is saying, a new question comes to mind: What effect does the scattering of particle 2's graviton shadow by particle 1 have on a new particle 3? Particle 3 should see a reflected or refracted image of particle 2's gravity in the direction of particle 1 and an image of particle 1's gravity in the direction of particle 2. When one of the particles is between the others, this scattering is similar to the concept of gravitational shielding (due to absorption of CG's), but probably many orders of magnitude more intense. By the way, the scattering effect should attenuate the absorption shielding effect. Like sunlight passing thru a cloud of opaque white smoke particles, CG's don't need a straight unobstructed path.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Slabinski: <i>PG</i>, page 124, formula 5:
"All we know about <i>f</i> is that the fraction of particles scattered through angles between Theta and (<i>Theta <i>+</i>d Theta</i>) is <i>f</i>(<i>Theta</i>)(2 <i>pi</i> sin <i>Theta d Theta</i>), so
Integral from 0 to <i>pi</i> {<i>f</i>(<i>Theta</i>)2 <i>pi</i> sin <i>Theta d Theta</i>} = 1 <hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
[Sure would be nice to have an editor that allows more fonts and special characters.]
The distribution of graviton scattering is crucial to predicting the effect. I haven’t figured out yet, whether Slabinski assumed a function, <i>f</i>, that gives the maximum or minimum possible ratio when he gets to formula 29 at the end of the article. If MI's and CG's act like perfectly elastic spheres, with MI's being much more massive than CG's, then the CG's should be reflected, on average, more than 90 degrees–back toward their direction of origin. (My brain is objecting to any suggestion that it even attempt to calculate the precise average angle.) On the other hand, if CG's are scattered in the manner of photons passing thru a spherical elysium atmosphere, the average scattering angle might be very small. (I won’t even hint that my poor brain attempt that calculation.) The average scattering angle could be anywhere between zero and 180 degrees.
Extending the model to macro-size objects, consider, the Sun, Earth and an Earth satellite:
Suppose Earth scatters the Sun's graviton shadow, on average, more than 90 deg., i.e., back toward the Sun. Then, the back-scattered solar graviton shadow (always toward Earth) will be more intense on the sunny side than on the eclipsed side. The satellite's orbital radius should then be shorter when it is on the sunny side of Earth.
Now, suppose Earth scatters the Sun's graviton shadow, on average, less than 90 deg., i.e., away from the Sun. Then, the forward-scattered solar graviton shadow (always toward Earth) will be more intense on the eclipsed side of Earth. The satellite's orbital radius should then be shorter when it is on the eclipsed side of Earth.
I am pretty sure the effect on the satellite of Earth's gravitational shadow, scattered by the Sun, is much less than the other way around.
Now, another problem comes to mind: Macro objects should not scatter CG's in the same way MI's do; scattering gravitons only once. What are the chances that a CG will penetrate to the center of Earth and be scattered by only one MI before returning to outer space?
Let us suppose that a graviton must pass thru the Earth a trillion times before being absorbed by an MI. If scattering occurs 2.9 x 10^29 times more often than absorption (per Slabinski's formula 29), then a graviton will penetrate, on average, only 4 x 10^-18 meter into the Earth before being scattered. After being scattered a certain number of times (depending on the average angle of each scattering), that CG's course will be completely random, and it is likely to be headed back toward the surface before penetrating to any appreciable depth. Even our atmosphere might be dense enough to shield Earth from the Sun's gravity. For Slabinski's solution to be viable, scattering must occur far less often, and the number of CG's passing thru Earth must be something like 10^40 times the number that are absorbed by Earth–give or take ten or twenty orders of magnitude. We need to detect and measure the effect of graviton scattering before we can narrow that number down to anything meaningful; and we might as well forget about looking for any gravity shielding due to absorption.
Tom, you may need to rethink your estimates of the size, spacing, speed, momentum and mean distance between collisions of CG's.
Just a side note: If a CG is scattered many times on its way thru an object, it's path may be many times longer than the thickness of the object. Therefore, the speed of gravity might be much slower within any dense object than it is in open space.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Slabinski: <i>Pushing Gravity </i>(<i>PG</i>), page 125:
"We have the surprising result that scattering of gravitons by the mass particle causes no decrease in the rate at which gravitons reach <i>A</i>test!"<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
Well, duh! That result was intuitively obvious to me from the start. It means that, against a uniform background of CG's, the CG's deflected by the mass particle toward Atest exactly match those deflected elsewhere, which had been moving toward Atest before being scattered by the mass particle. This is analogous to the fact that an idealized 100% reflective silver globe suspended in a uniformly illuminated white room will seem to disappear (if not for the reflection of the observer). By extension, two silver globes in a white room will reflect only the whiteness of the walls, and each globe's reflection in the other will likewise be the same pure white as the walls.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Slabinski: <i>PG</i>, page 126:
"At m1, there is a deficiency of gravitons coming from the m2 direction [due to absorption by m2]. The momentum transfer to m1 due to this excess depends both on absorption of part of the excess and on scattering of part of the excess [by m1]." <hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
I won't call myself a complete idiot, but I do admit I am just now beginning to see how this is supposed to work. The graviton shadow, cast by particle 2's absorption of CG's, is disbursed by particle 1's scattering of CG's, resulting in a net momentum of particle 1 toward particle 2--and vise versa. Slabinski concludes (on page 128, formula 29) that this mechanism accounts for about 99.999,999,999,999,999,999,999,999,999,999,999,3% of gravitational force.
Recalling the analogy of the silver globes in the white room: Suppose each globe absorbs a tiny fraction of the incident light (as any real silver globe will actually do). Then the reflected image of each globe, seen on the face of the other, will be darker than the walls. Therefore, the light reflected from the side of each globe facing the other will impart less momentum to the globe than the light from the opposite side of the globe. That imbalance will result in a very slight force of attraction between the two globes. In the case of gravity, that force is much stronger. The greater the ratio of reflection to absorption, the greater the ratio of momentum to heat. (Am I slow, or what?)
Now that I think I get what Slabinski is saying, a new question comes to mind: What effect does the scattering of particle 2's graviton shadow by particle 1 have on a new particle 3? Particle 3 should see a reflected or refracted image of particle 2's gravity in the direction of particle 1 and an image of particle 1's gravity in the direction of particle 2. When one of the particles is between the others, this scattering is similar to the concept of gravitational shielding (due to absorption of CG's), but probably many orders of magnitude more intense. By the way, the scattering effect should attenuate the absorption shielding effect. Like sunlight passing thru a cloud of opaque white smoke particles, CG's don't need a straight unobstructed path.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Slabinski: <i>PG</i>, page 124, formula 5:
"All we know about <i>f</i> is that the fraction of particles scattered through angles between Theta and (<i>Theta <i>+</i>d Theta</i>) is <i>f</i>(<i>Theta</i>)(2 <i>pi</i> sin <i>Theta d Theta</i>), so
Integral from 0 to <i>pi</i> {<i>f</i>(<i>Theta</i>)2 <i>pi</i> sin <i>Theta d Theta</i>} = 1 <hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
[Sure would be nice to have an editor that allows more fonts and special characters.]
The distribution of graviton scattering is crucial to predicting the effect. I haven’t figured out yet, whether Slabinski assumed a function, <i>f</i>, that gives the maximum or minimum possible ratio when he gets to formula 29 at the end of the article. If MI's and CG's act like perfectly elastic spheres, with MI's being much more massive than CG's, then the CG's should be reflected, on average, more than 90 degrees–back toward their direction of origin. (My brain is objecting to any suggestion that it even attempt to calculate the precise average angle.) On the other hand, if CG's are scattered in the manner of photons passing thru a spherical elysium atmosphere, the average scattering angle might be very small. (I won’t even hint that my poor brain attempt that calculation.) The average scattering angle could be anywhere between zero and 180 degrees.
Extending the model to macro-size objects, consider, the Sun, Earth and an Earth satellite:
Suppose Earth scatters the Sun's graviton shadow, on average, more than 90 deg., i.e., back toward the Sun. Then, the back-scattered solar graviton shadow (always toward Earth) will be more intense on the sunny side than on the eclipsed side. The satellite's orbital radius should then be shorter when it is on the sunny side of Earth.
Now, suppose Earth scatters the Sun's graviton shadow, on average, less than 90 deg., i.e., away from the Sun. Then, the forward-scattered solar graviton shadow (always toward Earth) will be more intense on the eclipsed side of Earth. The satellite's orbital radius should then be shorter when it is on the eclipsed side of Earth.
I am pretty sure the effect on the satellite of Earth's gravitational shadow, scattered by the Sun, is much less than the other way around.
Now, another problem comes to mind: Macro objects should not scatter CG's in the same way MI's do; scattering gravitons only once. What are the chances that a CG will penetrate to the center of Earth and be scattered by only one MI before returning to outer space?
Let us suppose that a graviton must pass thru the Earth a trillion times before being absorbed by an MI. If scattering occurs 2.9 x 10^29 times more often than absorption (per Slabinski's formula 29), then a graviton will penetrate, on average, only 4 x 10^-18 meter into the Earth before being scattered. After being scattered a certain number of times (depending on the average angle of each scattering), that CG's course will be completely random, and it is likely to be headed back toward the surface before penetrating to any appreciable depth. Even our atmosphere might be dense enough to shield Earth from the Sun's gravity. For Slabinski's solution to be viable, scattering must occur far less often, and the number of CG's passing thru Earth must be something like 10^40 times the number that are absorbed by Earth–give or take ten or twenty orders of magnitude. We need to detect and measure the effect of graviton scattering before we can narrow that number down to anything meaningful; and we might as well forget about looking for any gravity shielding due to absorption.
Tom, you may need to rethink your estimates of the size, spacing, speed, momentum and mean distance between collisions of CG's.
Just a side note: If a CG is scattered many times on its way thru an object, it's path may be many times longer than the thickness of the object. Therefore, the speed of gravity might be much slower within any dense object than it is in open space.
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20 years 11 months ago #7971
by tvanflandern
Replied by tvanflandern on topic Reply from Tom Van Flandern
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by PhilJ</i>
<br />Well, duh! That result was intuitively obvious to me from the start.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">I tip my hat to your superior intellect. It wasn't obvious to me, and I was very familiar with the uniformly illuminated, perfectly reflecting sphere argument. So it was one of a (fortunately) small number of outright reasoning mistakes I made in <i>Dark Matter...</i>. Slabinski originally wrote his article as a critique of MM as developed in my book.
But in behavior typical of a good model and unheard of for bad models, MM actually made more sense, not less, when the error was corrected. No ad hoc fixes were needed.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Am I slow, or what?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Everybody before Slabinski in the last 250 years, including Kelvin and Maxwell who debated these points in the late 19th century, missed that point. So you have nothing to beat yourself up over. Slabinski found it, not because he was looking for it, but because he was thorough in his development.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Now that I think I get what Slabinski is saying, a new question comes to mind: What effect does the scattering of particle 2's graviton shadow by particle 1 have on a new particle 3?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Most of the time, the answer is nothing. On the rare occasions when there is an effect, it decreases the net force produced. This is called "gravitational shielding". So far, it is too small an effect to be detectable in any but the most critical applications.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Let us suppose that a graviton must pass thru the Earth a trillion times before being absorbed by an MI. If scattering occurs 2.9 x 10^29 times more often than absorption (per Slabinski's formula 29), then a graviton will penetrate, on average, only 4 x 10^-18 meter into the Earth before being scattered.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote"> No, that is not what "scattered" means in this context. I suggested one meaning in my chapter in PG right before Slabinski's: the scattering occurs entirely in the elysium surrounding every matter ingredient. Another possibility is that every single graviton impact is mostly prefectly elastic but leaves behind that extremely tiny fraction of its momentum.
That change of perspective about the meaning of "scattering" eliminates the remainder of your objection as you wrote it. -|Tom|-
<br />Well, duh! That result was intuitively obvious to me from the start.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">I tip my hat to your superior intellect. It wasn't obvious to me, and I was very familiar with the uniformly illuminated, perfectly reflecting sphere argument. So it was one of a (fortunately) small number of outright reasoning mistakes I made in <i>Dark Matter...</i>. Slabinski originally wrote his article as a critique of MM as developed in my book.
But in behavior typical of a good model and unheard of for bad models, MM actually made more sense, not less, when the error was corrected. No ad hoc fixes were needed.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Am I slow, or what?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Everybody before Slabinski in the last 250 years, including Kelvin and Maxwell who debated these points in the late 19th century, missed that point. So you have nothing to beat yourself up over. Slabinski found it, not because he was looking for it, but because he was thorough in his development.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Now that I think I get what Slabinski is saying, a new question comes to mind: What effect does the scattering of particle 2's graviton shadow by particle 1 have on a new particle 3?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Most of the time, the answer is nothing. On the rare occasions when there is an effect, it decreases the net force produced. This is called "gravitational shielding". So far, it is too small an effect to be detectable in any but the most critical applications.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Let us suppose that a graviton must pass thru the Earth a trillion times before being absorbed by an MI. If scattering occurs 2.9 x 10^29 times more often than absorption (per Slabinski's formula 29), then a graviton will penetrate, on average, only 4 x 10^-18 meter into the Earth before being scattered.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote"> No, that is not what "scattered" means in this context. I suggested one meaning in my chapter in PG right before Slabinski's: the scattering occurs entirely in the elysium surrounding every matter ingredient. Another possibility is that every single graviton impact is mostly prefectly elastic but leaves behind that extremely tiny fraction of its momentum.
That change of perspective about the meaning of "scattering" eliminates the remainder of your objection as you wrote it. -|Tom|-
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20 years 11 months ago #7994
by PhilJ
Replied by PhilJ on topic Reply from Philip Janes
<center><b>K<font size="1">abs</font id="size1"> can’t be that high!</b> </center>
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Slabinski, <i>PG</i>, page 128,
K<font size="1">abs</font id="size1"> <1.7 x 10^-8 cm^2 / g             (28)
<i>K<font size="1">scat</font id="size1"> </i>/ <i>K<font size="1">abs</font id="size1"></i> > 2.9 x 10^29                 (29)<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
If you multiply <i>K<font size="1">abs</font id="size1"></i> or <i>K<font size="1">scat</font id="size1"> </i> by a density, you get the inverse of the average distance traveled by a CG thru the given density of matter before it is absorbed or scattered. If <i>K<font size="1">abs</font id="size1"></i> = 1.7 x 10^-8 cm^2 / g, then a CG travels, on average, 590 km thru water before being absorbed. The Earth has a density of 5.52 and a diameter of 12,760 km. Therefore, a CG, on average, would make it about 107 km, or 0.84% of the way thru Earth, before being absorbed. That, alone, would result in practically 100% gravitational shielding.
Taking the maximum value of <i>K<font size="1">abs</font id="size1"></i> and the minimum of <i>K<font size="1">scat</font id="size1"> </i> / <i>K<font size="1">abs</font id="size1"></i>, it appears that a CG would be scattered approximately 5 x 10^21 times per centimeter in water. Passing thru Earth (along a nearly straight zig zag path) a CG would be scattered about 3.3 x 10^31 times. Most CG’s will not keep scattering back toward their original course, so if the average angle of scattering is more than a tiny fraction of an arcsecond, most CG’s would lose their way and come out of the Earth in random directions, resulting in nearly 100% gravity shielding. If the average scatter angle is that small, you may need to drastically increase your estimate of <i>K<font size="1">scat</font id="size1"> </i> / <i>K<font size="1">abs</font id="size1"></i>? My pocket calculator can’t compute cos <i>Theta</i> for angles much smaller than 0.01 degree. If a CG is scattered 0.01 degree, the momentum imparted to the MI is only (1-cos 0.01degree) <i>m<font size="1">g</font id="size1"> v<font size="1">g</font id="size1"></i> = 1.5 x 10^-8 <i>m<font size="1">g</font id="size1"> v<font size="1">g</font id="size1"></i>.
I’m not sure if this invalidates anything. It may simply mean that your minimum value for <i>K<font size="1">abs</font id="size1"></i> can be lowered by another thirty or fourty orders of magnitude. I’m not sure what that would that do to <i>K<font size="1">scat</font id="size1"> </i> / <i>K<font size="1">abs</font id="size1"></i>.
<center><b>Approaching the problem from a different angle</b>:</center>
CG scattering is presumed to be some symmetrical statistical distribution function. Slabinski refers to that function as <i>f</i>(<i>Theta</i>), and says we know nothing about <i>f</i> -- other than the fact that every CG which is scattered is scattered at some angle <i>Theta</i> relative to its previous direction of travel.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Slabinski: <i>PG</i>, page 124, formula 5:
Integral from 0 to <i>pi</i> {<i>f</i>(<i>Theta</i>)2 <i>pi</i> sin <i>Theta d Theta</i>} = 1<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
I think we can surmise more than that about <i>f</i>. (As I have stated before, my math skills are pretty rusty, so don’t be surprised if I get this next part wrong.)
<i>f</i>(<i>Theta</i>) is the probability that a CG will be scattered by an angle between <i>Theta</i> and <i>Theta</i> + <i>dTheta</i>.
So the mean value of <i>f</i>(<i>Theta</i>) is:
Mean(<i>f</i>(<i>Theta</i>)) = integral from 0 to <i>pi</i> {<i>Theta</i>(<i>f</i>(<i>Theta</i>) 2 <i>pi</i> sin <i>Theta dTheta</i>)}
The mean value of the change in direction of a CG after n successive scatterings is:
Mean(<i>f</i>^<i>n</i>(<i>Theta</i>)) = integral from 0 to <i>pi</i> {Theta(<i>f</i>^<i>n</i>)(<i>Theta</i>) 2 <i>pi</i> sin <i>Theta</i> <i>d</i>^<i>n</i>(<i>Theta</i>))}
My intuition tells me that, for any <i>f</i> (except special cases in which the CG is only scattered by certain rational multiples of <i>pi</i>), as the number <i>n</i> of successive scatterings becomes sufficiently large, (<i>f</i>^<i>n</i>(<i>Theta</i>) will approach the constant, <i>dTheta</i> / <i>pi</i>, the mean value of the change in direction of CG’s will approach 90 degrees, and the force of gravity will be effectively shielded.
Since we have yet to observe any measurable gravity shielding, Mean(<i>f</i>^<i>n</i>(<i>Theta</i>)) must be quite small, and Mean(<i>f</i>(<i>Theta</i>)) must be smaller still. However, the closer Mean(<i>f</i>(<i>Theta</i>)) is to zero degrees, the less momentum we get from each scattering event. (I still don’t know what happened to <i>Theta</i> on page 128; it should play a key role in “Part 4. Numerical Limits”. Perhaps the answer is hidden in Slabinski’s references to Tom’s “The Speed of Gravity”, 1998a; and “The Meta Cycle” 1998b.) My intuition tells me the smaller Mean(<i>f</i>(<i>Theta</i>)) is, the greater the ratio of <i>K<font size="1">scat</font id="size1"> </i> / <i>K<font size="1">abs</font id="size1"></i> in formula 29.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Slabinski, <i>PG</i>, page 128,
K<font size="1">abs</font id="size1"> <1.7 x 10^-8 cm^2 / g             (28)
<i>K<font size="1">scat</font id="size1"> </i>/ <i>K<font size="1">abs</font id="size1"></i> > 2.9 x 10^29                 (29)<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
If you multiply <i>K<font size="1">abs</font id="size1"></i> or <i>K<font size="1">scat</font id="size1"> </i> by a density, you get the inverse of the average distance traveled by a CG thru the given density of matter before it is absorbed or scattered. If <i>K<font size="1">abs</font id="size1"></i> = 1.7 x 10^-8 cm^2 / g, then a CG travels, on average, 590 km thru water before being absorbed. The Earth has a density of 5.52 and a diameter of 12,760 km. Therefore, a CG, on average, would make it about 107 km, or 0.84% of the way thru Earth, before being absorbed. That, alone, would result in practically 100% gravitational shielding.
Taking the maximum value of <i>K<font size="1">abs</font id="size1"></i> and the minimum of <i>K<font size="1">scat</font id="size1"> </i> / <i>K<font size="1">abs</font id="size1"></i>, it appears that a CG would be scattered approximately 5 x 10^21 times per centimeter in water. Passing thru Earth (along a nearly straight zig zag path) a CG would be scattered about 3.3 x 10^31 times. Most CG’s will not keep scattering back toward their original course, so if the average angle of scattering is more than a tiny fraction of an arcsecond, most CG’s would lose their way and come out of the Earth in random directions, resulting in nearly 100% gravity shielding. If the average scatter angle is that small, you may need to drastically increase your estimate of <i>K<font size="1">scat</font id="size1"> </i> / <i>K<font size="1">abs</font id="size1"></i>? My pocket calculator can’t compute cos <i>Theta</i> for angles much smaller than 0.01 degree. If a CG is scattered 0.01 degree, the momentum imparted to the MI is only (1-cos 0.01degree) <i>m<font size="1">g</font id="size1"> v<font size="1">g</font id="size1"></i> = 1.5 x 10^-8 <i>m<font size="1">g</font id="size1"> v<font size="1">g</font id="size1"></i>.
I’m not sure if this invalidates anything. It may simply mean that your minimum value for <i>K<font size="1">abs</font id="size1"></i> can be lowered by another thirty or fourty orders of magnitude. I’m not sure what that would that do to <i>K<font size="1">scat</font id="size1"> </i> / <i>K<font size="1">abs</font id="size1"></i>.
<center><b>Approaching the problem from a different angle</b>:</center>
CG scattering is presumed to be some symmetrical statistical distribution function. Slabinski refers to that function as <i>f</i>(<i>Theta</i>), and says we know nothing about <i>f</i> -- other than the fact that every CG which is scattered is scattered at some angle <i>Theta</i> relative to its previous direction of travel.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Slabinski: <i>PG</i>, page 124, formula 5:
Integral from 0 to <i>pi</i> {<i>f</i>(<i>Theta</i>)2 <i>pi</i> sin <i>Theta d Theta</i>} = 1<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
I think we can surmise more than that about <i>f</i>. (As I have stated before, my math skills are pretty rusty, so don’t be surprised if I get this next part wrong.)
<i>f</i>(<i>Theta</i>) is the probability that a CG will be scattered by an angle between <i>Theta</i> and <i>Theta</i> + <i>dTheta</i>.
So the mean value of <i>f</i>(<i>Theta</i>) is:
Mean(<i>f</i>(<i>Theta</i>)) = integral from 0 to <i>pi</i> {<i>Theta</i>(<i>f</i>(<i>Theta</i>) 2 <i>pi</i> sin <i>Theta dTheta</i>)}
The mean value of the change in direction of a CG after n successive scatterings is:
Mean(<i>f</i>^<i>n</i>(<i>Theta</i>)) = integral from 0 to <i>pi</i> {Theta(<i>f</i>^<i>n</i>)(<i>Theta</i>) 2 <i>pi</i> sin <i>Theta</i> <i>d</i>^<i>n</i>(<i>Theta</i>))}
My intuition tells me that, for any <i>f</i> (except special cases in which the CG is only scattered by certain rational multiples of <i>pi</i>), as the number <i>n</i> of successive scatterings becomes sufficiently large, (<i>f</i>^<i>n</i>(<i>Theta</i>) will approach the constant, <i>dTheta</i> / <i>pi</i>, the mean value of the change in direction of CG’s will approach 90 degrees, and the force of gravity will be effectively shielded.
Since we have yet to observe any measurable gravity shielding, Mean(<i>f</i>^<i>n</i>(<i>Theta</i>)) must be quite small, and Mean(<i>f</i>(<i>Theta</i>)) must be smaller still. However, the closer Mean(<i>f</i>(<i>Theta</i>)) is to zero degrees, the less momentum we get from each scattering event. (I still don’t know what happened to <i>Theta</i> on page 128; it should play a key role in “Part 4. Numerical Limits”. Perhaps the answer is hidden in Slabinski’s references to Tom’s “The Speed of Gravity”, 1998a; and “The Meta Cycle” 1998b.) My intuition tells me the smaller Mean(<i>f</i>(<i>Theta</i>)) is, the greater the ratio of <i>K<font size="1">scat</font id="size1"> </i> / <i>K<font size="1">abs</font id="size1"></i> in formula 29.
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20 years 11 months ago #7997
by tvanflandern
Replied by tvanflandern on topic Reply from Tom Van Flandern
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by PhilJ</i>
<br />K<font size="1">abs</font id="size1"> can’t be that high!<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Yes, it can. You have the wrong meaning of the factor. It is the the cross-sectional-area-to-mass ratio for absorption of a graviton by a matter ingredient. Because the mass of a matter ingredient can be indefinitely small, the collisional cross-section can be indefinitely small also, meaning that any graviton could fly through an entire planet with little chance of being absorbed.
My suggestion in my previous message, which you did not pick up on either, was this: Instead of considering scattering and absorption as separate attributes, consider that each single graviton that encounters a matter ingredient is mostly scattered but slightly absorbed by that 30-orders-of-magnitude ratio. In that way, there is no difference between scattered and absorbed gravitons. They are each just slightly inelastic.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Perhaps the answer is hidden in Slabinski’s references to Tom’s “The Speed of Gravity”, 1998a; and “The Meta Cycle” 1998b.)<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">I will try to get the latter posted to the web site in the near future so it is available. The former reference is already there. -|Tom|-
<br />K<font size="1">abs</font id="size1"> can’t be that high!<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Yes, it can. You have the wrong meaning of the factor. It is the the cross-sectional-area-to-mass ratio for absorption of a graviton by a matter ingredient. Because the mass of a matter ingredient can be indefinitely small, the collisional cross-section can be indefinitely small also, meaning that any graviton could fly through an entire planet with little chance of being absorbed.
My suggestion in my previous message, which you did not pick up on either, was this: Instead of considering scattering and absorption as separate attributes, consider that each single graviton that encounters a matter ingredient is mostly scattered but slightly absorbed by that 30-orders-of-magnitude ratio. In that way, there is no difference between scattered and absorbed gravitons. They are each just slightly inelastic.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Perhaps the answer is hidden in Slabinski’s references to Tom’s “The Speed of Gravity”, 1998a; and “The Meta Cycle” 1998b.)<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">I will try to get the latter posted to the web site in the near future so it is available. The former reference is already there. -|Tom|-
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20 years 11 months ago #8000
by PhilJ
Replied by PhilJ on topic Reply from Philip Janes
A search of your site for "meta cycle" yields only a mention in your resume, with no link. I browsed the site for a while and couldn't find any link to the article.
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20 years 11 months ago #8001
by tvanflandern
Replied by tvanflandern on topic Reply from Tom Van Flandern
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by PhilJ</i>
<br />A search of your site for "meta cycle" yields only a mention in your resume, with no link. I browsed the site for a while and couldn't find any link to the article.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">I will need time to get the article to one of our site volunteers, who must convert it to ASP and integrate it into the web site with the same page template. While this might happen sooner, I'd give it a week. -|Tom|-
<br />A search of your site for "meta cycle" yields only a mention in your resume, with no link. I browsed the site for a while and couldn't find any link to the article.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">I will need time to get the article to one of our site volunteers, who must convert it to ASP and integrate it into the web site with the same page template. While this might happen sooner, I'd give it a week. -|Tom|-
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