The Slabinski Article: Cross sectional area propor

<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 />It is fairly obvious where this assumption comes from; gravity-mass = inertia-mass.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">"Gravitational mass" has no meaning for entities this small. There is no assumption of equality of mass types. Rather, it is taken for granted that only mass in the sense of "quantity of matter" exists.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">If MI's and CG's behave anything like spherical objects, and their mass is proportional to their volume, we should expect that A(sub)scat = K(sub)scat <i>m</i>(super)2/3 and A(sub)abs = K(sub)abs <i>m</i>(super)2/3.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Your attention to detail is hiding the bigger picture. The power of m is irrelevant to the subsequent discussion. Because the gravitons are all considered equivalent, and the matter ingredients are equivalent too, the body radius for each is a constant and the r-dependence is absorbed into the coefficient K. The only unstated assumption here is that all masses can be reduced to the sum of lots of individual (identical) matter ingredients (MIs). So what is true for one MI is true for any larger body.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">On the other hand, if all MI's and CG's are created equal (implying that no known subatomic particle consists of a single MI), it doesn't really matter what power is connected to the inertia-mass-at least until we are able to nail down their mass and velocity within an order of magnitude.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">That is the key point that answers your question. -|Tom|-

Dear Philj and Tom,

Is Slabinsky's article downloadable as a PDF somewhere?

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Jan</i>
<br />Is Slabinsky's article downloadable as a PDF somewhere?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Not to my knowledge. There is still considerably more knowledge in books than on the Internet, although a trend to narrow that gap obviously exists. -|Tom|-

Perhaps I did not make myself clear.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">quote: me: start of this discussion:
“gravity-mass = inertia-mass”<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
By gravity-mass, I mean“m” in formula #17 (on page 126 of PG)
F1 = (G m(sub)1 m(sub)2) / r^2.
By “inertia-mass”, I mean “m” in the formula rho(sub)g = m (sub)g v(sub)g [which is implicit in the derivation of formula #12 on page 125 of PG]. Slabinski assumes that these two masses are identical.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">me: same paragraph:
“We do know that the weight of any macro-size object is proportional to its momentum, and (I think) we know that electrons proportionally add both weight and momentum to an atom.” <hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
My mistake: I meant “inertia”, rather than “momentum”.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">me: next paragraph:
“If MI's and CG's behave anything like spherical objects, and their mass is proportional to their volume, we should expect that A(sub)scat = K(sub)scat m(super)2/3 and A(sub)abs = K(sub)abs m(super)2/3.” <hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
I, too, am assuming too much. I am thinking of an MI as being spherical, with a constant density. In that case its mass would be proportional to r^3, while the area of its profile would be proportional to r^2. If an MI’s profile area (and therefore its weight) is proportional to its inertia-mass, as Slabinski assumes, then that mass is probably concentrated in a spherical shell.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Tom: 2nd message in this thread:
"Gravitational mass" has no meaning for entities this small.”<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
You may be right, but the gravity-mass of a “mass particle” is key to Slabinski’s article, as is the inertia-mass. Anyway, your comment, above, anticipates my next criticism of the article, i.e., “It’s better to talk about momentum and leave the mass of a CG out of the formulas.” I have started in a separate discussion [url] metaresearch.org/msgboard/topic.asp?TOPIC_ID=442 [/url] on that, so please don’t respond here.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Tom: next paragraph:
“...the gravitons are all considered equivalent, and the matter ingredients are equivalent too, the body radius for each is a constant...” <hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
It is clear that Slabinski treats all gravitons as being equal (though, I intuitively suspect any “normal” random distribution of masses and velocities might yield the same result). His use of “m(sub)1" and “m(sub)2" for the two “mass particles” does appear to allow different size masses. He does not explicitly state that he is talking about MI’s (That’s your nomenclature, not his.) However, his absorption and scattering radii would be meaningless for any macro-sized particle, and probably for any known boson, meson, etc., as well.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Tom: 2nd same sentence:
“...and the r-dependence is absorbed into the coefficient K.”<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
It’s moot, now (if the MI’s mass is concentrated in a spherical shell, as I just stated), but that 2/3 power is connected to m. If I were right about the MI having a uniform density, Slabinski’s ultimate ratio (formula 29) would be off by about ten orders of magnitude.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Jan: 3rd message of this thread.
“Is Slabinsky's article downloadable as a PDF somewhere?”<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
You can order Pushing Gravity (PG) from Tom on this site. Are you poor, like me? I don’t know how libraries in the Netherlands work, but in the USA, we have an interlibrary loan system. It took about two weeks for me to get PG at the tiny library in this out-of-the-way town of 2000 people in western Washington State. The copy I got came from the University of Puget Sound, Washington; one of the articles in it was written by a professor there.

<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 />“gravity-mass = inertia-mass”. By gravity-mass, I mean“m” in formula #17 (on page 126 of PG) F1 = (G m(sub)1 m(sub)2) / r^2. By “inertia-mass”, I mean “m” in the formula rho(sub)g = m (sub)g v(sub)g [which is implicit in the derivation of formula #12 on page 125 of PG]. Slabinski assumes that these two masses are identical.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">The concept of inertia in connection with gravitons was covered in last year's MRB article "Does gravity have inertia", to which the answer was "no". Formula #16 has the same form as formula #17 (Newton's law of gravitation), which allows an identification of terms for macroscopic bodies where gravitational shielding is not significant. However, formula #17 is not used in Slabinski's mathematical modeling of the Meta Model. It just helps to set numerical constraints on some parameters of the model.

To repeat and expand: At this very-small-scale level, the concepts of "gravitational mass" and "inertial mass" do not exist. Mass is a measure of the quantity of matter. Likewise, energy is undefined, but momentum is the product of mass times velocity.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">I am thinking of an MI as being spherical, with a constant density.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">My response assumed the same thing.

To get a feel for what's happening physically, one must think in terms of masses, velocities, collisions, and physical principles. Any attempt to be guided by math, rather than the other way around, is guaranteed to produce great confusion.

You are attemption to use induction to relate macroscopic concepts to MM parameters. That cannot be done uniquely even in principle. One must use deduction from first principles to define the model. Then, if and only if following a unique deductive path leads to equations or concepts similar to macroscopic ones can we learn what the macroscopic equations/concepts really mean. There is no unique pathway from macroscopic equations/concepts to MM equations/concepts.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">the gravity-mass of a “mass particle” is key to Slabinski’s article, as is the inertia-mass.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">That is a case on point. Worked out from a deductive, physical perspective, no such associations can be drawn.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">I have started in a separate discussion ..., so please don’t respond here.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">My available time for discussions fluctuates, but is never high. Please confine anything you want my comments on to a single thread. (If you are opening group discussions, the number of separate threads doesn't matter. Responses will determine which threads live and which die.) I chose this thread to answer only because you said the most things here. and these generalizations apply to the other threads too. -|Tom|-

Philj,
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Are you poor, like me? <hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">

No, but not rich either. I have so many other books that I wish to purchase; however, I saw PG on www.amazon.co.uk for only £12.28, which is probably the cheapest option for me if we take shipment into account.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">I don’t know how libraries in the Netherlands work, but in the USA, we have an interlibrary loan system. <hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
Too my knowledge, we do have an interlibrary loan service, but this only works for university libraries in the Netherlands . However, there is a catch, only students or staff can become members of the university library. This counts me out then.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">It took about two weeks for me to get PG at the tiny library in this out-of-the-way town of 2000 people in western Washington State.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
It is impossible to ask for certain academic books at a small public library. This must sound old fashioned to you, I know. []