Debunking Setterfields Speed of Light Model

p. 29: "where k is Boltzmann’s constant (1.38 x 10-16 erg K-1)" I think Jellison failed to remember that his ergon also has a mass factor. His two zeta^-2 would then nicely cancel out instead of leaving one zeta to linearly multiply molecular velocity with the c-decay factor.That would kind of eliminate his 6th point in his summary and with that his only argument for the real physical impossiblity of Setterfield's ideas.I keep feeling I'm missing something.. but I can't find what...

I apologise, he does have a lot more arguments. I was merely referring to the points in his summary as I admittedly only skim read the rest. In his summary he mainly attacks Setterfield's methods, not the idea itself.Points 1, 2, 3 and 4 (in the summary) all point out errors in Setterfield's calculations, scientific method and interpretations, they do not pertain to the physical impossibility of Setterfield's ideas and so do not disprove them, they only discredit Setterfield as a good theoretical scientist.And ok, point number 5 is a bit beyond me atm, i'm still trying to figure out what he means exactly..So yes, my statement was premature and so, again, my apologies.Thank you for pointing it out. Edited by Khy, : No reason given.

True, I do not mean to defend Setterfield.. at all tbh. I only meant to point out that Jellison made an error in point 6, but I went on to say too much. xD

I mentioned it in an earlier post, but on page 29 where he uses the boltzmann equation to calculate the molecular velocity he makes a mistake with the boltzmann constant. He correctly multiplies 'm' (mass) by the factor (zeta)^-2 in the divisor. This would indeed lead to the function with a powerless zeta as a factor as he displayed in the second function on that page.He forgot one thing though. Mass is also a factor in the definition of energy (he used the ergon which is defined as 1 g·cm2/s2), if he had used the same factor on his boltzmann's constant, it would cancel out with the zeta^-2 in the divisor. Molecular speeds would then not increase with an increasing c-decay factor.I hope this is clear enough.. I am not too familiar with english physics/mathematics terminology.

Thank you very much, that will help a great deal Lets try that: translated into base quantities would be L is length/distance, T is time, is temperature and M is mass.As you can see, there is an M in both the dividend and the divisor under the root and so both should be multiplied by . As he said on page 6 in equation number two: .I'm assuming Jellison ended up with thus and finally That would indeed lead to an increasing molecular velocity with an increasing c-decay factor.This is what he should have done: and thus which is no different than the standard boltzmann equation, and it certainly does not lead to an increasing molecular velocity with an increasing c-decay factor because in the dividend cancels out with the on in the divisor.aka earth would not have lost it's atmosphere, and interstellar dusts and gases and stars would not have been as chaotic as they would have been if Jellison was correct.Edited by Khy, : No reason given.Edited by Khy, : added some (t)'sEdited by Khy, : No reason given.Edited by Khy, : No reason given.

Ok, this is a complex matter, but I'll try to get my head around it cuz its fun.1. A meter is defined as: "the length of the path travelled by light in vacuum in 1 299,792,458 of a second."2. The second is defined as: "the duration of 9,192,631,770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium 133 atom."If we want to retain any frame of reference we are forced to say that in Setterfield's universe the meter is defined as it is today, by today's lightspeed. In the same way we have to retain the present day definition of time. If we would apply the factor to time and distance, then speed, (distance/time) which uses both quantities would not change either as they would cancel out against eachother as well according to equation 1 and 5.About the gravitational constant:On page 6 Jellison deduces from the consequenses of on that G should be defined as: (equation 3)If we try to deduce G by diving into physical quantities, then that would be: and in Setterfield's c-decay world as: and that would indeed result in (equation 3): Consistency is a good thing, you cannot let one mass obey the c-decay factor, and another not. Then you're making the same wishful mistakes as Jellison did. You can also not apply the c-decay factor to every quantity because you would lose your frame of reference (which is completely dependent on present day's lightspeed, so it makes sense not to apply the factor to time and distance.)Edited by Khy, : No reason given.

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You don't apply the c-decay factor to units of energy etc, you apply them to the actual masses, lengths, units etc that are actually pertinent to the problem.

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Energy is defined by mass. If you use a constant, then that constant, and all terms in it's entire definition ARE pertinent to the problem.

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How is maintaining some arbitrary definition going to affect whether our atmosphere escapes. We should get the same answer to that question regardless of how we define a second or a meter.

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It is not, that is exactly what I'm saying and that's why I did not use the c-decay factor on time or distance quantities.

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I'm not making any error. I'm not saying Jellison is right. I'm saying that you aren't correct.

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I'm terribly sorry, I mixed up Setterfield and Jellison, I meant to point out that Jellison might be making similar mistakes as the ones he acuses Setterfield of.

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One erg of energy is expended when either a 1g or 10000000000000 g mass is accelerated across a friction free surface by a 1 dyne force over a distance of 1 cm.

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A dyne is define as: Are you saying that a 1 dyne force will accelerate 1g just as fast as it will accelerate 10000000000000g, because according to its definiton above a dyne will accelerate 1 gram at a rate of 1 cm per second squared. That means it would accelerate a mass of 10000000000000g at a rate of 1/10000000000000 cm per second squared.

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We should also get the same answer regardless of how we define the gram or kilogram, yet you do chose apply the c-decay factor to mass units.

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I did not chose to do that, Jellison did and I only pointed out that he was inconsistent and wrong that earth would lose its atmosphere, as you said: "We should also get the same answer regardless..."Edit:

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Not correct. Jellison does not deduce anything on page 6. The effect on G was given by Setterfield with the purpose of making sure that the force of gravity between attracting bodies remained the same while masses changed. In fact all of the relationships on page 6 are supplied by Setterfield.

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True, I read too quick! Those formulas are, however, all logical implications of changing the speed of light without throwing today's physical laws out the window.Edited by Khy, : No reason given.Edited by Khy, : No reason given.

Oh dear, yes indeed, it would appear I'm the idiot. I'm very sorry for the implication.You are indeed right in everything you're saying, I really should stop trying to argue for anything beyond my initial claim... I just find it strange to apply the c-decay factor to one mass, but ignore another mass in the same equation, namely the gram in the definition of the dyne in the erg in Boltzmann's constant.If the mass of an oxygen molecule decreases, then why wouldn't the mass of the IPK (File:CGKilogram.jpg - Wikipedia) and consequently the gram decrease? I'm quite sure less force would be needed to accelerate the lighter 'c-decay'-gram at 1cm/s^2.
Edited by Khy, : No reason given.

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But that would simply mean that using standard amounts of material could not be used to establish the gram or the kilogram (......) the definition of the gram and kilogram would have to change so that a 1 newton force always provided a 1 m/s acceleration to a 1 kg object.

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So the change in force needed to accelerate an S-world-kilogram would have to be proportional to the decrease in mass compared to a present day kilogram. That's the only way to keep the definition of a Newton.That decrease would be the same as the decrease in mass of an oxygen molecule. My initial claim is not just that there is a change in k (Nm/s^-2), but that that change is zeta^-2(t). As stated should be so for any mass in S-world.Edited by Khy, : No reason given.

I don't see the problem.If at some point in the past an oxygen molecule would have a mass 0.5 times that of what it is now, then any other unit of mass (including the kilogram) would also be 0.5 times the size of what it is now containing the same amount of molecules that all just have half of their present mass...The force needed to accelerate half a present day kg (which would be 1 S-world-kg) to 1 m/s/s is half a present day newton (1 S-world-N). I don't see where F=ma does not work out.If you use the above-mentioned 1 S-world-N in the definition of k, you get 0.5Nm/s^-2 as its definition.

A newton and a kilogram would indeed change their value over time, that is the whole idea of Mr Setterfield.

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Over some time scale, F=ma is not true because F= dp/dt = d(mv)/dt where m is changing so that we cannot simply use F= m(dv/dt) = ma.

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Thats true indeed... I can probably find a reason/solution for that that you'd disagree with xD, but I'm gonna call it a day... thanks for your knowledgeable corrections and the fun discussion... I enjoyed it.