The Slabinski Article: Cross sectional area propor

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20 years 9 months ago #8074 by PhilJ
Replied by PhilJ on topic Reply from Philip Janes
First off, Tom, I have been a bad boy. I’ve been a little hasty about posting my calculations -- discovering errors a day or two later. Let me admit to a rather large error of scale. I said I thought your <i>K<font size="1">abs</font id="size1"></i> was 30 or 40 orders of magnitude too high; actually the gravity shielding is that much too high, which is equivalent to <i>K<font size="1">abs</font id="size1"></i> being too high by a factor of ln(30 or 40 orders of magnitude). Ln(10^30) = 69.08; ln(10^40) = 92.10. My latest calculations, below, indicate that your minimum <i>K<font size="1">abs</font id="size1"></i> may be lowered by a factor of ln(10^80), which equals 184. My revised estimate, therefore, is <i>K<font size="1">abs</font id="size1"></i> &lt; 10^-10 cm^2 / g.

<hr noshade size="1">
<center><font color="orange"><i><b>Ooops!</b></i> <font face="Comic Sans MS"> It came to me in a nap. Right the first time? Told you my math was crap! </font id="Comic Sans MS"></font id="orange">
(This time I’ll admit to a feeble attempt at poetry.) </center>

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Me, this thread, my first post of 1/17/04:
[given <i>K<font size="1">abs</font id="size1"></i> = 1 cm^2 / g, and <i>rho</i> = 1 g / cm^3] “... the average number of balloons in the path of a given photon is one balloon per centimeter of depth.” <hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
Up to that point, I was doing okay. So, in one hundred centimeters, a photon will encounter an average of 100 balloons, and the average distance between balloons is one centimeter.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Me, this thread, my first post of 1/17/04:
“Therefore, 1 / (2 <i>K<font size="1">abs</font id="size1"> rho</i>) = average distance traveled by a CG before being absorbed.” <hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
That should be:
“Therefore, 1 / (<i>K<font size="1">abs</font id="size1"> rho</i> ) = average distance traveled by a CG before being absorbed.”

I must have been thinking of the probability that a photon will penetrate to a depth of at least one centimeter; I thought that was one half the average distance traveled between balloons, but again I was wrong; any statistician can tell you without thinking, it is one eth. I, on the other hand, had to derive it as follows:

In the analogy of little black balloons: Suppose <i>b</i> dots are randomly scattered on an area <i>a</i> of the screen, and each dot has area <i>c</i> such that <i>c</i> &lt;&lt; <i>a</i>. What fraction of a random area of the screen is not covered by dots (due to overlapping dots)? The probability that a randomly selected point is not within a first dot is (1 - <i>c</i>/<i>a</i>). The probability that the point is not within the first <i>n</i> dots is (1 - <i>c</i>/<i>a</i>)^<i>n</i>. For <i>n</i> = <i>b</i>, that probability becomes <i>p</i> = (1 - <i>c</i>/<i>a</i>)^<i>b</i>. As <i>c</i>/<i>a</i> approaches zero, (1 - <i>c</i>/<i>a</i>)^<i>b</i> converges to 0.36787944, which happens to be, not one half, but one <i>e</i>th (i.e., 1/<i>e</i>).

Note: Try not to confuse my <i>p</i> with Slabinski’s <i>rho</i>. I’m having a hard time thinking of a suitable name for this variable which does not start with the letter “p” -- probability, percent, portion, permeability, etc. By the way, my <i>rho</i> is density; his <i>rho</i> is momentum. It’s been way too long since I played with this notation. What are the traditional symbols for probability and density, anyway?


<hr noshade size="1">
The relevance of the above to the Meta Model is:
&nbsp &nbsp &nbsp &nbsp At depth <i>d</i><font size="1">1</font id="size1"> = 1/<i>K<font size="1">abs</font id="size1"></i> <i>rho</i>, &nbsp &nbsp &nbsp &nbsp <i>cb</i> = <i>a</i>, and&nbsp &nbsp &nbsp &nbsp <i>p</i> = 1/<i>e</i>.
&nbsp &nbsp &nbsp &nbsp At depth <i>d</i>, &nbsp &nbsp <i>p</i> = <i>e</i>^(-<i>d</i>/<i>d</i><font size="1">1</font id="size1">).
Therefore the probability that a photon will penetrate the cloud of balloons (or the analogous CG will penetrate matter) at least to depth <i>d</i> is:
&nbsp &nbsp &nbsp &nbsp <i>p</i> = <i>e</i>^(-<i>d</i> <i>K<font size="1">abs</font id="size1"></i> <i>rho</i>).

Staying with the CGS system, apparently favored by Slabinski: Earth’s average density is 5.52 g / cm^3. That is the density of any solid angle of Earth (a conical piece). I was wrong to use that density in my earlier posts. For a CG passing thru Earth, we should be using the average density of a thin cylindrical piece, which is somewhat higher because the denser compression-solid iron core is in the center. I don’t know where to look for estimates of the Earth’s density as a function of depth, so I’ll make a wild guess (Can anyone supply a better value?) that the average is:
<i>rho</i> = 8.0 g / cm^3.

<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:
“<i>K<font size="1">abs</font id="size1"></i> &lt; 1.7 x 10^-8 cm^2 / g &nbsp &nbsp &nbsp &nbsp (28)”<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
Using that value of <i>K<font size="1">abs</font id="size1"></i>, with the diameter of Earth = <i>d<font size="1">e</font id="size1"></i> = 1.267 x 10^9 cm, and <i>rho</i> = 8.0:
&nbsp &nbsp &nbsp &nbsp &nbsp <i>p</i> = <i>e</i>^(-de <i>K<font size="1">abs</font id="size1"></i> <i>rho</i>) = 1.33 x 10^-71.
In other words, no gravity from the Sun would reach the far side of Earth. Apparently, assuming that at least 99.999,999.9% of the Sun’s gravity does reach the far side, <i>K<font size="1">abs</font id="size1"></i> must be off by a factor of at least ln(10^80), which is roughly 184. (Can anyone reading this supply a better estimate of the minimum percent of the Sun’s gravity that penetrates Earth? What is the maximum percent of gravity shielding that could be, as yet, undiscovered?)

So I expect that
&nbsp &nbsp &nbsp &nbsp &nbsp <i>K<font size="1">abs</font id="size1"></i> &lt; (1.7 x 10^-8) / 184 cm^2 / g &lt; 9.24 x 10^-11.

For a nice, round number, let’s say that <i>K<font size="1">abs</font id="size1"></i> &lt; 10^-10 cm^2 / g.

p.s.: I am working on a slightly different logical path, which should lead to the same result, but doesn’t. Instead, it yields <i>K<font size="1">abs</font id="size1"></i> &lt; 2.445 x 10^-9 cm^/g. I’m pretty sure the error is in that different path, but I can’t absolutely rule out the possibility that the error is in the above calculations.


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20 years 9 months ago #8131 by PhilJ
Replied by PhilJ on topic Reply from Philip Janes
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Tom, this discussion, 1/13/04
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.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Tom, this discussion, 1/17/04
“This is not the meaning of Kabs.”<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
Okay; you alluded to a new model in your article in <i>PG</i> -- contemporary with Slabinski’s analysis of your earlier model. He defined the constant, <i>K<font size="1">abs</font id="size1"></i>, to deal with your earlier model; and that definition is the meaning of <i>K<font size="1">abs</font id="size1"></i> upon which his calculations are based. Any recalculation of <i>K<font size="1">abs</font id="size1"></i> using Slabinski’s formulas, but based on a different meaning of <i>K<font size="1">abs</font id="size1"></i>, would have no validity. You cannot now claim that his calculations support your new model unless you can show that the two models, though physically different, are mathematically equivalent -- as I suspect they are. If you do show that equivalence, then any conclusions that I may draw from Slabinski’s calculations should be applicable to your new model. So where’s the beef?

I therefore continue to pigheadedly pursue my critique of the Slabinski article as it appears in <i>PG</i>, and not as you seem to wish it had been written. I hope my attitude does not alienate you to the point of not paying attention to my arguments.

<hr noshade size="1">

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Me, yesterday:
“As <i>c</i>/<i>a</i> approaches zero, (1 - <i>c</i>/<i>a</i>)^<i>b</i> converges to 0.36787944, which happens to be, not one half, but one <i>e</i>th (i.e., 1/e).”<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">

This needs clarification. I forgot to mention that I was calculating (1-<i>c</i>/<i>a</i>)^<i>b</i> for the special case in which the total area of dots equals the area of screen; i.e., <i>bc</i> = <i>a</i>. In that case, as <i>c</i>/<i>a</i> approaches zero, <i>b</i> approaches infinity. For example: (1-0.1)^10 = 0.9^10 = .34867844; (1-0.01)^100 = 0.99^100 = .366032341, etc. That sequence converges to 1/<i>e</i>, which is the probability that a photon will penetrate the cloud of balloons to the depth d1.

Yes, I’m still a bad boy! I’m afraid the math corner of my brain has atrophied to the point that I have to correct my earlier posts before adding anything new. A simple arithmetic error:

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Me, yesterday:
“Using that value of <i>K<font size="1">abs</font id="size1"></i> [= 1.7 x 10^-8 cm^2 / g] , with <i>d<font size="1">e </font id="size1"></i>= 1.267 x 10^9 cm, and <i>rho</i> = 8.0 g/cm^3:
&nbsp &nbsp &nbsp &nbsp <i>p</i> = e^(-<i>d<font size="1">e</font id="size1"></i> <i>K<font size="1">abs</font id="size1"></i> <i>rho</i>) = 1.33 x 10^-71.”<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">


WRONG! That should have been <i>p</i> = 1.465 x 10^-75.
<i>p</i> = <i>e</i>^(-<i>de</i> <i>K<font size="1">abs</font id="size1"></i> <i>rho</i>)
= <i>e</i>^(-1.267 x 10^9)(1.7 x 10^-8)(8.0)
= <i>e</i>^-172.3 = 1.465 x 10^-75

I need another day to recalculate my new estimate of <i>K<font size="1">abs</font id="size1"></i>. Right now, I think the new <i>K<font size="1">abs</font id="size1"></i> will be &lt; 5.804 x 10^-12 cm^2 / g. I haven’t triple checked that result, yet, so don’t bank on it.

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20 years 9 months ago #8192 by tvanflandern
<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 />Okay; you alluded to a new model in your article in <i>PG</i> -- contemporary with Slabinski’s analysis of your earlier model. ... So where’s the beef?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">There is no "new model". I simply pointed out that you are using an inappropriate meaning of "scattered" as it relates to "absorbed". There are many ways to partition all graviton encounters into these two categories. In one of them, "scattered" gravitons involve no actual contact with the MI. In another of them, the momentum from a single graviton can be partly scattered and partly absorbed.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">I therefore continue to pigheadedly pursue my critique of the Slabinski article as it appears in <i>PG</i>, and not as you seem to wish it had been written. I hope my attitude does not alienate you to the point of not paying attention to my arguments.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Pressures on my time vary over a great range, but are presently getting tight again. When you did not understand my point, I had no choice but to bow out. Sometimes I have to do that even when your point is right on the mark. -|Tom|-

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15 years 1 month ago #23059 by buffoil
Replied by buffoil on topic Reply from
Apologies for the thread necromancy. A site search showed that the exact topic of interest to me had already been brought up here:

<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 />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...<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">

For some time I've been curious to see the actual mathematical demonstration that, within the basic Lesage scenario, perfectly elastic graviton particle scattering off of two matter ingredient particles results in zero net gravitational force. Le Sage, Maxwell, Kelvin, Darwin, Poincare and so on all stated that the return reflections of elastic gravitons from each of the gravitationally interacting bodies exactly cancels the erstwhile gravitational impulse of the Le Sage mechanism. The conclusion then is that for a Le Sage type mechanism to work, some form of inelastic interaction must take place, i.e. absorption of some graviton particles or some slowing down of them.

However -- and my calculus is years rusty -- it seems to me that the integration Slabinski uses on pg. 124 in <i>Pushing Gravity</i> includes the <i>test area</i> A as both sink and <i>source</i> of gravitons. But in a real Le Sage type situation the test area under a total reflection regime would actually be comprised of another matter ingredient, thus blocking out all gravitions from that direction. Relating this to PhilJ's analogy with the two silver globes in a white room, I'm thinking that in fact there is no possible source of reflected photons at the points on the globes' reflecting surfaces on a line drawn between their centers. If you could peek out through a partially silvered section of one globe looking directly at the other globe, I'm thinking that you would in all likelihood see a black dot!

Anybody up enough on Slabinski's integration to point out where I'm going wrong on this?

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15 years 2 weeks ago #23073 by PhilJ
Replied by PhilJ on topic Reply from Philip Janes
Hello, Buffoil. I just dropped in after several months absence to see if any rational beings were once again frequenting the site.

Before addressing your question, I should point out that, since I last posted on this thread, I have abandoned the LeSage model entirely. My own Fractal Foam Model of Universes is based on chaotic interplay of ethereal shear waves, which propagate at light speed, and ethereal pressure waves, which are at least 20 billion times faster. (I guess I should start a thread devoted to my model.)

I am the wrong person to ask for mathematical proof, but I'll dust off my copy of the Slabinski article and try to answer you on an intuitive level. I stand by my assertion that perfectly reflective silver globes in a perfectly uniform white room would be invisible to one another. Imagine an observer in the room wearing a perfectly reflective white suit and peering out thru a peep hole. He would see the reflection of his peep hole as a black dot in each globe. He would also see much smaller relections of that black dot following less direct paths of reflection off both spheres and back to his eye.

I think I understand your concern. You are wondering how light can reach the center line between the two spheres. It does so by multiple reflections. Looking out thru your half-silvered window, you see the other sphere; in it you see a reflection of your own sphere; in the reflection of your own sphere, you see a reflection of the other sphere. Each successive reflection is much smaller than the previous one. The limit of the nth reflection's diameter as n approaches infinity is zero. So the size of the black dot that you are expecting to see is zero.




Fractal Foam Model of Universes: Creator

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15 years 2 weeks ago #23764 by buffoil
Replied by buffoil on topic Reply from
<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 />Hello, Buffoil. I just dropped in after several months absence to see if any rational beings were once again frequenting the site.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">

Well, it'd been a half decade since my last visit here! I went searching to see whether my particular question had ever been discussed, and came across your posts. Since I was here, and since you hadn't quite touched on my exact question, I figured no harm in bringing it up. This forum is very quiet compared to what I remember.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">...I have abandoned the LeSage model entirely...<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">

I've never been comfortable with some of the complexities piled on top of the basic Le Sage model by its supporters thus to accomodate the problem of heating of ordinary matter raised by Kelvin, Maxwell, Poincare and so on. It's understandable that many people could be turned off by these seemingly strained accomodations -- planetary heating, exploding planets, mass accretion, and such -- seeing them as something akin to adding epicycles to epicycles. Is it because of these "entities multiplied beyond necessity" that you abandoned Le Sage?

In fact, it's precisely because a Le Sage regime involving only perfectly elastic reflections would do away with all these subsequent deferents and epicycles that I'm curious to understand why perfectly elastic collisions have been by an apparent near-unanimous consensus ruled out. I'm willing to be convinced that the physical and mathematical models rule out Le Sage under the aegis of a perfectly elastic interaction regime, but I'm not yet apprised of the evidence.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">You are wondering how light can reach the center line between the two spheres. It does so by multiple reflections.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">

But how? Consider the reflective spherical segments facing one another along the center line between the two spheres as infinitesimal planar reflecting surfaces. Better yet, to invoke the crude classic model, consider them as points on directly opposite sides of a pool table. If a pool ball ("photon") follows the nominal rule of incident deflection it would seem that there is no way for it to "get inside" the line connecting those two points unless a player ("light source") intentionally places a ball there and then shoots it so as to bounce back and forth along that line. What am I missing?

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