- Thank you received: 0
New concept: CG Fragments
- tvanflandern
- Topic Author
- Offline
- Platinum Member
Less
More
20 years 11 months ago #6860
by tvanflandern
Reply from Tom Van Flandern was created 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 />Your Meta Model looks very promising to me, but I'm scratching my head over the type(s) of collisions that might cast shadows in a uniform medium.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
Collisions don't cast shadows, bodies do. Or more exactly, "matter ingredients" do. MIs are defined as the largest particle that cannot be penetrated by gravitons. A larger body is just a collection of MIs, each acting independently.
All of your questions are covered in <i>Pushing Gravity</i> (PG), the bible on this subject. I can only give short responses here.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">If CG's are absorbed by a particle (in capture colisions), the particle will cast a shadow, but what happens to the energy?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
In absorption, the energy is absorbed, and is probably responsible for the excess heat flow from large masses. In scattering collisions, no energy is lost as heat. There is just an exchange of momentum. The heat "problem" in Le Sage-type models was debated by Maxwell and Lord Kelvin in the 19th century. See PG for numerical estimates and a comparison with observations.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">If a shadow in the medium can accelerate a massive particle appreciably, then CG's must have appreciable momentum. Energy is momentum times velocity; if a CG's velocity is many times the speed of light, then CG's must be extremely energetic. The particles should become extremely hot--whatever "hot" means in this context--unless energy is not conserved.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
Look at the formulas in PG. Graviton velocity enters the gravitational constant formula with a higher power than it enters the heat absorption formula. So it follows that there must exist a size small enough to keep the heat down, combined with a velocity large enough to transfer significant momentum.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">I believe perfectly elastic collisions (like really bouncy billiard balls) between spherical CG's and spherical particles would cast no shadow.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
True. And perfectly absorbed collisions would generate too much heat. But the right ratio of scattered gravitons to absorbed gravitons solves this problem because scattered gravitons do cast shadows provided that absorption of other gravitons creates an asymmetry. See Slabinski's paper in PG. Slabinski was the first to solve this problem. -|Tom|-
<br />Your Meta Model looks very promising to me, but I'm scratching my head over the type(s) of collisions that might cast shadows in a uniform medium.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
Collisions don't cast shadows, bodies do. Or more exactly, "matter ingredients" do. MIs are defined as the largest particle that cannot be penetrated by gravitons. A larger body is just a collection of MIs, each acting independently.
All of your questions are covered in <i>Pushing Gravity</i> (PG), the bible on this subject. I can only give short responses here.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">If CG's are absorbed by a particle (in capture colisions), the particle will cast a shadow, but what happens to the energy?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
In absorption, the energy is absorbed, and is probably responsible for the excess heat flow from large masses. In scattering collisions, no energy is lost as heat. There is just an exchange of momentum. The heat "problem" in Le Sage-type models was debated by Maxwell and Lord Kelvin in the 19th century. See PG for numerical estimates and a comparison with observations.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">If a shadow in the medium can accelerate a massive particle appreciably, then CG's must have appreciable momentum. Energy is momentum times velocity; if a CG's velocity is many times the speed of light, then CG's must be extremely energetic. The particles should become extremely hot--whatever "hot" means in this context--unless energy is not conserved.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
Look at the formulas in PG. Graviton velocity enters the gravitational constant formula with a higher power than it enters the heat absorption formula. So it follows that there must exist a size small enough to keep the heat down, combined with a velocity large enough to transfer significant momentum.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">I believe perfectly elastic collisions (like really bouncy billiard balls) between spherical CG's and spherical particles would cast no shadow.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
True. And perfectly absorbed collisions would generate too much heat. But the right ratio of scattered gravitons to absorbed gravitons solves this problem because scattered gravitons do cast shadows provided that absorption of other gravitons creates an asymmetry. See Slabinski's paper in PG. Slabinski was the first to solve this problem. -|Tom|-
Please Log in or Create an account to join the conversation.
20 years 10 months ago #6954
by PhilJ
Replied by PhilJ on topic Reply from Philip Janes
Obviously, I'm not quite up to speed on this forum, having read only the 1993 version of <i>DMMP&NC</i>. (Does that have a shorter abbreviation?) I shall have to get a copy of <i>PM</i>.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Graviton velocity enters the gravitational constant formula with a higher power than it enters the heat absorption formula. <hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
Without knowing those formulas and how they were derived, I can only ask, "Does that mean gravity does not conserve momentum and energy?" The simplest calculations are usually the easiest to debug, and it looks to me like a simple and straightforeward problem to convert the momentum imparted by gravitons into the equivalent energy. You only have to assume a graviton's average speed (let's go with 10 billion c) and a proportion of gravitons that are absorbed by the Earth (let's guess 1 per million). Someone please tell me what's wrong with the following calculations.
Rounding off Earth's gravity, a 1 kilogram object weighs 10 Newton. Let's assume that object <u>absorbs</u> enough more CG's per second from above than from below to give it a weight of 10 Newton. To conserve momentum, the CG's that account for the imbalance must lose 10 Newton second of momentum each second. The remainder of CG's absorbed equally from all directions contribute the same energy per collision, but no net momentum.
<center>E = mv</center>
The energy of a particle = momentum times velocity. Since the CG's strike the object at various angles, the horizontal components of momentum vectors cancel, while the scalar energies do not. Therefore, the actual energy absorbed is significantly greater, but I'll be generous and ignore that. Taking the speed of CG's as 10 billion c, E = (10 Newton second/second) x (3.0 E18 meter/second) = 3.0 E19 Joule. Now if the Earth only absorbs one millionth of the CG's passing thru it, the object absorbs about 2 million times as many CG's as those accounting for the 10 Newton imbalance; so the total energy absorbed in one second is greater than 6.0 E25 Joule each second; that's 60 yottaWatt (per <i>Webster's</i>, yotta- = 10 E24) or roughly 1/2 trillion times Earth's total allotment of solar power.
(Earth diameter = 12,760 km; area of cross section = 128 billion meter^2; solar power = 1 kW/meter^2; total solar power = 1.28 E14 Watt. Incidentally, dividing that power by c gives the force exerted by sunlight on Earth = about 426,000 Newton.)
<center>E = mc^2</center>
Could that much energy be accounted for by increasing mass of the object? If E = mc^2, then m = E/c^2. Now, 60 yottaWatt = 60 yottaJoule/second; so every second you would need to convert that much energy to (6 E25 Joule) / (9 E16 meter^2/second^2) = 666,666,667 kilograms of matter. Consequently, the mass of that 1 kg object (and every other object in the universe) would double in about a nanosecond. That's a lotta mass to sweep under the elysium! And I'm pretty sure it's more than enough to support the exploding planet hypothesis, too. (Any accountants reading this? How much is 67% per nanosecond compounded continuously for three million years?)
So, are my assumptions totally wrong? Have I misstated the physics? Or are my calculations wrong?
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Graviton velocity enters the gravitational constant formula with a higher power than it enters the heat absorption formula. <hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
Without knowing those formulas and how they were derived, I can only ask, "Does that mean gravity does not conserve momentum and energy?" The simplest calculations are usually the easiest to debug, and it looks to me like a simple and straightforeward problem to convert the momentum imparted by gravitons into the equivalent energy. You only have to assume a graviton's average speed (let's go with 10 billion c) and a proportion of gravitons that are absorbed by the Earth (let's guess 1 per million). Someone please tell me what's wrong with the following calculations.
Rounding off Earth's gravity, a 1 kilogram object weighs 10 Newton. Let's assume that object <u>absorbs</u> enough more CG's per second from above than from below to give it a weight of 10 Newton. To conserve momentum, the CG's that account for the imbalance must lose 10 Newton second of momentum each second. The remainder of CG's absorbed equally from all directions contribute the same energy per collision, but no net momentum.
<center>E = mv</center>
The energy of a particle = momentum times velocity. Since the CG's strike the object at various angles, the horizontal components of momentum vectors cancel, while the scalar energies do not. Therefore, the actual energy absorbed is significantly greater, but I'll be generous and ignore that. Taking the speed of CG's as 10 billion c, E = (10 Newton second/second) x (3.0 E18 meter/second) = 3.0 E19 Joule. Now if the Earth only absorbs one millionth of the CG's passing thru it, the object absorbs about 2 million times as many CG's as those accounting for the 10 Newton imbalance; so the total energy absorbed in one second is greater than 6.0 E25 Joule each second; that's 60 yottaWatt (per <i>Webster's</i>, yotta- = 10 E24) or roughly 1/2 trillion times Earth's total allotment of solar power.
(Earth diameter = 12,760 km; area of cross section = 128 billion meter^2; solar power = 1 kW/meter^2; total solar power = 1.28 E14 Watt. Incidentally, dividing that power by c gives the force exerted by sunlight on Earth = about 426,000 Newton.)
<center>E = mc^2</center>
Could that much energy be accounted for by increasing mass of the object? If E = mc^2, then m = E/c^2. Now, 60 yottaWatt = 60 yottaJoule/second; so every second you would need to convert that much energy to (6 E25 Joule) / (9 E16 meter^2/second^2) = 666,666,667 kilograms of matter. Consequently, the mass of that 1 kg object (and every other object in the universe) would double in about a nanosecond. That's a lotta mass to sweep under the elysium! And I'm pretty sure it's more than enough to support the exploding planet hypothesis, too. (Any accountants reading this? How much is 67% per nanosecond compounded continuously for three million years?)
So, are my assumptions totally wrong? Have I misstated the physics? Or are my calculations wrong?
Please Log in or Create an account to join the conversation.
- tvanflandern
- Topic Author
- Offline
- Platinum Member
Less
More
- Thank you received: 0
20 years 10 months ago #7503
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 />Does that mean gravity does not conserve momentum and energy?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">It means that gravity is a dynamic force like a falling waterfall, losing small amounts of energy converted to heat.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">So, are my assumptions totally wrong? Have I misstated the physics? Or are my calculations wrong?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">I'll choose door #2. [] In Slabinski's article in <i>PG</i>, you will see derivations of the gravity and heat formulas from scratch. Slabinski drew three conclusions, the third of which was new and remarkable:
<ul><li>If all gravitons are scattered, there is no net force.</li><li>If all gravitons are absorbed, the heat generated would vaporize any mass in microseconds.</li><li>If some gravitons are absorbed and some are scattered, the absorption asymmetry creates a net force for the <i>scattered</i> gravitons, but heat only from the absorbed gravitons.</li></ul>So one only needs to set the scattering to absorption ratio high enough to get whatever force-to-heat ratio one wants. Some numerical estimates of this ratio appear in <i>PG</i>. -|Tom|-
<br />Does that mean gravity does not conserve momentum and energy?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">It means that gravity is a dynamic force like a falling waterfall, losing small amounts of energy converted to heat.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">So, are my assumptions totally wrong? Have I misstated the physics? Or are my calculations wrong?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">I'll choose door #2. [] In Slabinski's article in <i>PG</i>, you will see derivations of the gravity and heat formulas from scratch. Slabinski drew three conclusions, the third of which was new and remarkable:
<ul><li>If all gravitons are scattered, there is no net force.</li><li>If all gravitons are absorbed, the heat generated would vaporize any mass in microseconds.</li><li>If some gravitons are absorbed and some are scattered, the absorption asymmetry creates a net force for the <i>scattered</i> gravitons, but heat only from the absorbed gravitons.</li></ul>So one only needs to set the scattering to absorption ratio high enough to get whatever force-to-heat ratio one wants. Some numerical estimates of this ratio appear in <i>PG</i>. -|Tom|-
Please Log in or Create an account to join the conversation.
20 years 10 months ago #7132
by PhilJ
Replied by PhilJ on topic Reply from Philip Janes
Since "scattering" is not mentioned in <i>DMMP&N</i>C and I await an interlibrary loan of <i>PM</i>, I shall take a guess at what you mean by "scattering". I suppose you mean that one CG is absorbed and a subsequent CG, after a finite time, is scattered, making off with 99.99999999999999% of the energy left by the first CG. That would explain the disappearance of the energy, but it gives rise to a new mystery; how does the MI hold onto that energy and then transfer it. Perhaps your "scattering" isn't so different from my suggestion (in the original post) that the CG splits into fragments.
The CG's split, in retrospect, looks even simpler if the incoming CG is made up of loosely bound fragments, which can easily be seperated without altering the character or energy state of the MI. No mysterious energy storage and transfer mechanism required. I like Occam's razor, too; and this model looks a whole lot simpler to me than what I believe your "scattering" involves.
I hope you realize that my intent is not to shoot down the MM, but to rescue it from perceived flaws. []
The CG's split, in retrospect, looks even simpler if the incoming CG is made up of loosely bound fragments, which can easily be seperated without altering the character or energy state of the MI. No mysterious energy storage and transfer mechanism required. I like Occam's razor, too; and this model looks a whole lot simpler to me than what I believe your "scattering" involves.
I hope you realize that my intent is not to shoot down the MM, but to rescue it from perceived flaws. []
Please Log in or Create an account to join the conversation.
- tvanflandern
- Topic Author
- Offline
- Platinum Member
Less
More
- Thank you received: 0
20 years 10 months ago #7254
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 />Since "scattering" is not mentioned in <i>DMMP&N</i>C and I await an interlibrary loan of <i>PM</i>, I shall take a guess at what you mean by "scattering".<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">The following is from <i>PG</i>:
Imagine a vast, unlimited field of small meteoroids moving at random. Now add two relatively large comets to this field. If collisions between meteoroids and comets were perfectly elastic, each rebounding meteoroid on one side of a comet would exactly replace the missing continuation of a rebounding meteoroid on the opposite side. So the field of meteoroids would remain essentially the same as it would have been if the comets were not present. No net force on the comets occurs. However, if the comets completely absorb impacting meteoroids, then meteoroids are lost from the system. More meteoroids are absorbed on each comet’s outer surface than on its inner surface (the one facing the other comet) because some meteoroids absorbed by the other comet never make it to the first comet’s inner surface. In this case, the comets feel a net force pushing them toward one another.
This analogy illustrates the way gravitons (replacing meteoroids) interact with units of matter (replacing comets). To make the analogy better, we must introduce an extended halo of small asteroids orbiting the nucleus of each comet. These asteroids are the counterparts of elysium in the actual model, which acts as a sort of extended atmosphere around a matter unit. The new feature is that collisions with the comet nucleus or with any of its orbiting cloud of asteroid satellites by meteoroids will produce a net force on the entire comet cloud. If the total mass of the satellites is comparable to or greater than the mass of the comet nucleus, then all collisions of meteoroids with satellites will contribute more to the net force on the comet than collisions with the nucleus because the asteroid satellites have a greater cross-sectional area exposed to collisions. The greater cross-sectional area assures more total collisions on the satellites than if they were collected into a single mass. In general, the more finely the satellites are broken up without loss of total mass, the greater will be the excess of the resulting contribution to net force on the comet from the satellites.
The importance of this to our model is that meteoroid impacts on the comet nucleus, like graviton impacts on a unit of matter, leave a heat residue; whereas meteoroid impacts on asteroid satellites, like graviton impacts on the elysium cloud around a unit of matter, contribute to the net force on the nucleus without heating it. The elysium cloud then absorbs the main heat generated by graviton impacts. But that heat is harmlessly (and undetectably) carried away as heated elysium packed near a unit of matter is exchanged with cooler elysium from the general pool of elysium filling space.
Referring back to Slabinski’s article, we can now identify gravitons impacting a unit of matter with “absorbed gravitons”, and gravitons impacting the denser elysium cloud around a unit of matter (mostly not absorbed or only partially absorbed) with “scattered gravitons”. Both contribute to the gravitational constant. The latter are far more numerous than the former. But only the “absorbed gravitons” contribute to the heating of matter because the elysium heating is quickly dissipated into the general reservoir of elysium, which has already reached its own thermodynamic equilibrium with the gravitons. Just as our planetary electromagnetic radiation budgets were not complete until we could measure contributions from all wavelengths, the gravitational energy budgets will not be complete until we can measure the portions carried away by elysium.
So heat is deposited by gravitons, then is leisurely lost as the elysium circulates and freshens in separate activities that are not part of the graviton absorption/scattering process. This brings to mind the heat generated by a refrigerator. Most of it must be siphoned off and dumped to allow the important part of the process to operate. The net result is just what we need to make the LeSage graviton model work. The gravitational constant (Slabinski's eqn. 16) depends on the products of absorption and scattering coefficients, the latter being huge compared to the former. Meanwhile, the heat flow (Slabinski's eqn. 19) depends only on the absorption coefficient (the part of the heat absorbed by matter instead of by elysium), and is therefore miniscule in comparison. We simply neglect the presently undetectable part of the thermodynamic cycle representing elysium heat. ...
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">I suppose you mean that one CG is absorbed and a subsequent CG, after a finite time, is scattered, making off with 99.99999999999999% of the energy left by the first CG.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">There is no association of absorbed and scattered gravitons, and nothing makes off with any of the heat produced by absorbed gravitons. It remains present and is observed as the excess heat flow from large planets. Anyway, the ratio of scattered to absorbed gravitons is large enough that the energy percentage would not be nearly that far away from 100%. []
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">I hope you realize that my intent is not to shoot down the MM, but to rescue it from perceived flaws. []<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">I appreciate the good intentions. But you need to get up to speed on the model details and work already done before proposing "improvements".[] -|Tom|-
<br />Since "scattering" is not mentioned in <i>DMMP&N</i>C and I await an interlibrary loan of <i>PM</i>, I shall take a guess at what you mean by "scattering".<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">The following is from <i>PG</i>:
Imagine a vast, unlimited field of small meteoroids moving at random. Now add two relatively large comets to this field. If collisions between meteoroids and comets were perfectly elastic, each rebounding meteoroid on one side of a comet would exactly replace the missing continuation of a rebounding meteoroid on the opposite side. So the field of meteoroids would remain essentially the same as it would have been if the comets were not present. No net force on the comets occurs. However, if the comets completely absorb impacting meteoroids, then meteoroids are lost from the system. More meteoroids are absorbed on each comet’s outer surface than on its inner surface (the one facing the other comet) because some meteoroids absorbed by the other comet never make it to the first comet’s inner surface. In this case, the comets feel a net force pushing them toward one another.
This analogy illustrates the way gravitons (replacing meteoroids) interact with units of matter (replacing comets). To make the analogy better, we must introduce an extended halo of small asteroids orbiting the nucleus of each comet. These asteroids are the counterparts of elysium in the actual model, which acts as a sort of extended atmosphere around a matter unit. The new feature is that collisions with the comet nucleus or with any of its orbiting cloud of asteroid satellites by meteoroids will produce a net force on the entire comet cloud. If the total mass of the satellites is comparable to or greater than the mass of the comet nucleus, then all collisions of meteoroids with satellites will contribute more to the net force on the comet than collisions with the nucleus because the asteroid satellites have a greater cross-sectional area exposed to collisions. The greater cross-sectional area assures more total collisions on the satellites than if they were collected into a single mass. In general, the more finely the satellites are broken up without loss of total mass, the greater will be the excess of the resulting contribution to net force on the comet from the satellites.
The importance of this to our model is that meteoroid impacts on the comet nucleus, like graviton impacts on a unit of matter, leave a heat residue; whereas meteoroid impacts on asteroid satellites, like graviton impacts on the elysium cloud around a unit of matter, contribute to the net force on the nucleus without heating it. The elysium cloud then absorbs the main heat generated by graviton impacts. But that heat is harmlessly (and undetectably) carried away as heated elysium packed near a unit of matter is exchanged with cooler elysium from the general pool of elysium filling space.
Referring back to Slabinski’s article, we can now identify gravitons impacting a unit of matter with “absorbed gravitons”, and gravitons impacting the denser elysium cloud around a unit of matter (mostly not absorbed or only partially absorbed) with “scattered gravitons”. Both contribute to the gravitational constant. The latter are far more numerous than the former. But only the “absorbed gravitons” contribute to the heating of matter because the elysium heating is quickly dissipated into the general reservoir of elysium, which has already reached its own thermodynamic equilibrium with the gravitons. Just as our planetary electromagnetic radiation budgets were not complete until we could measure contributions from all wavelengths, the gravitational energy budgets will not be complete until we can measure the portions carried away by elysium.
So heat is deposited by gravitons, then is leisurely lost as the elysium circulates and freshens in separate activities that are not part of the graviton absorption/scattering process. This brings to mind the heat generated by a refrigerator. Most of it must be siphoned off and dumped to allow the important part of the process to operate. The net result is just what we need to make the LeSage graviton model work. The gravitational constant (Slabinski's eqn. 16) depends on the products of absorption and scattering coefficients, the latter being huge compared to the former. Meanwhile, the heat flow (Slabinski's eqn. 19) depends only on the absorption coefficient (the part of the heat absorbed by matter instead of by elysium), and is therefore miniscule in comparison. We simply neglect the presently undetectable part of the thermodynamic cycle representing elysium heat. ...
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">I suppose you mean that one CG is absorbed and a subsequent CG, after a finite time, is scattered, making off with 99.99999999999999% of the energy left by the first CG.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">There is no association of absorbed and scattered gravitons, and nothing makes off with any of the heat produced by absorbed gravitons. It remains present and is observed as the excess heat flow from large planets. Anyway, the ratio of scattered to absorbed gravitons is large enough that the energy percentage would not be nearly that far away from 100%. []
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">I hope you realize that my intent is not to shoot down the MM, but to rescue it from perceived flaws. []<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">I appreciate the good intentions. But you need to get up to speed on the model details and work already done before proposing "improvements".[] -|Tom|-
Please Log in or Create an account to join the conversation.
20 years 10 months ago #7150
by PhilJ
Replied by PhilJ on topic Reply from Philip Janes
I'll reserve further comments on this topic 'til I have studied <i>PM</i>.
Please Log in or Create an account to join the conversation.
Time to create page: 0.310 seconds