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# Debunking Setterfields Speed of Light Model

Author Topic:   Debunking Setterfields Speed of Light Model
Khy
Junior Member (Idle past 2654 days)
Posts: 12
Joined: 03-09-2012

 Message 16 of 41 (655376) 03-09-2012 8:12 PM Reply to: Message 15 by NoNukes03-09-2012 6:05 PM

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 gÂ·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.

 This message is a reply to: Message 15 by NoNukes, posted 03-09-2012 6:05 PM NoNukes has responded

 Replies to this message: Message 17 by Panda, posted 03-09-2012 9:46 PM Khy has responded Message 19 by NoNukes, posted 03-10-2012 9:19 AM Khy has not yet responded Message 25 by Dr Adequate, posted 03-10-2012 3:29 PM Khy has not yet responded

Panda
Member (Idle past 1966 days)
Posts: 2688
From: UK
Joined: 10-04-2010

 Message 17 of 41 (655384) 03-09-2012 9:46 PM Reply to: Message 16 by Khy03-09-2012 8:12 PM

LaTeX
 Khy writes:(he used the ergon which is defined as 1 gÂ·cm2/s2)

FYI

$\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

produces:

It is a little unintuitive, but it might prove useful to you.
More details can be found here: 'Hypertext Help with LaTeX'

If I were you
And I wish that I were you
All the things I'd do
To make myself turn blue

 This message is a reply to: Message 16 by Khy, posted 03-09-2012 8:12 PM Khy has responded

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NoNukes
Inactive Member

 Message 18 of 41 (655430) 03-10-2012 9:05 AM Reply to: Message 8 by Perdition03-09-2012 10:54 AM

Re: How do you get off of inifinite speed?
 I don't know of any math that could get you off of infinite speed once you're on it. So, either he calculated a very high rate of speed that is being erroneously labelled "inifinite" or his math fails at this very simple level...or I'm missing something (an eminitely probable proposition).

I don't find this particular line of reasoning condemning of Setterfield's work. Extrapolating back to time zero is probably unjustified, and it really isn't necessary for the speed to be infinite at time zero, given that we cannot even see any objects for which light has traveled the full 13.7 billion years of the universe's existence.

Under the big bang theory, we don't extrapolate back to time t=0. Would it be a fair criticism of modern cosmology to argue that BBT does not accurately reflect what happens at t=0?

Under a government which imprisons any unjustly, the true place for a just man is also in prison. Thoreau: Civil Disobedience (1846)

The apathy of the people is enough to make every statue leap from its pedestal and hasten the resurrection of the dead. William Lloyd Garrison

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NoNukes
Inactive Member

 Message 19 of 41 (655431) 03-10-2012 9:19 AM Reply to: Message 16 by Khy03-09-2012 8:12 PM

 Mass is also a factor in the definition of energy (he used the ergon which is defined as 1 gÂ·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.

As I understand your point, you are suggesting that the Jellison did not include the effects on Boltzmann's constant that would result from the rest of the stuff Setterfield has postulated.

I don't think the unit analysis method you've put forward is sufficient to determine what the effect on k should be, but I think, a more generalized view of your position does have merit. After all if Plank's constant, isn't constant, lots of stuff is probably wrong. Perhaps even the values used for temperature need to be defended.

Perhaps there is a way to trace through the physics to determine the effect on Boltzmann's constant. I, personally, am not up to it, and I'm not sure it would be worth the effort. I'd probably just not use the escaping atmosphere argument.

Under a government which imprisons any unjustly, the true place for a just man is also in prison. Thoreau: Civil Disobedience (1846)

The apathy of the people is enough to make every statue leap from its pedestal and hasten the resurrection of the dead. William Lloyd Garrison

 This message is a reply to: Message 16 by Khy, posted 03-09-2012 8:12 PM Khy has not yet responded

Khy
Junior Member (Idle past 2654 days)
Posts: 12
Joined: 03-09-2012

 Message 20 of 41 (655435) 03-10-2012 9:56 AM Reply to: Message 17 by Panda03-09-2012 9:46 PM

Re: LaTeX
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)'s

Edited by Khy, : No reason given.

Edited by Khy, : No reason given.

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NoNukes
Inactive Member

 Message 21 of 41 (655440) 03-10-2012 11:55 AM Reply to: Message 20 by Khy03-10-2012 9:56 AM

Re: LaTeX
I'm pretty sure that this unit analysis method would not produce the proper dependency of k on mass except by the sheerest of coincidences.

Once you've reduced everything to base units, what does the L correspond to? Would we be able to say that since atoms are supposed to have gotten bigger (or smaller, I cannot remember what Setterfield finally said, and I'm not going to look) that we need to introduce a corresponding ratio everwhere that L appears?

Let's consider the simpler equation for gravity.

F = GMm/r^2

Would it be correct to delve into the base units for G to determine how the constant G varies with Setterfield's assumptions in addition to considering what happens to M and m?

Under a government which imprisons any unjustly, the true place for a just man is also in prison. Thoreau: Civil Disobedience (1846)

The apathy of the people is enough to make every statue leap from its pedestal and hasten the resurrection of the dead. William Lloyd Garrison

 This message is a reply to: Message 20 by Khy, posted 03-10-2012 9:56 AM Khy has responded

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Khy
Junior Member (Idle past 2654 days)
Posts: 12
Joined: 03-09-2012

 Message 22 of 41 (655455) 03-10-2012 12:57 PM Reply to: Message 21 by NoNukes03-10-2012 11:55 AM

Re: LaTeX
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 â„ 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.

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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NoNukes
Inactive Member

 Message 23 of 41 (655458) 03-10-2012 2:02 PM Reply to: Message 22 by Khy03-10-2012 12:57 PM

Re: LaTeX
 Consistency is a good thing, you cannot let one mass obey the c-decay factor, and another not.

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. In the end, we will want to express our answers in terms of the units that we know.

 1. A meter is defined as: "the length of the path travelled by light in vacuum in 1 â„ 299,792,458 of a second."

That's the way those units are defined now. The meter could also have been defined as the length of a platinum/irridium bar held in some measurement lab, and the second could be defined as 1/86400 of the time required for a full earth rotation. The size of those units would be essentially the same as the units we use now.

It's just plain silly to talk about maintaining the definitions of the units when the underlying properties change. 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.

 Then you're making the same wishful mistakes as Jellison did

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

If Jellison made a mistake it was in not evaluating the effect on the other physical constants he used. But if he tried to take them into account using your method, he would have been making yet another error.

Under a government which imprisons any unjustly, the true place for a just man is also in prison. Thoreau: Civil Disobedience (1846)

The apathy of the people is enough to make every statue leap from its pedestal and hasten the resurrection of the dead. William Lloyd Garrison

 This message is a reply to: Message 22 by Khy, posted 03-10-2012 12:57 PM Khy has responded

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Khy
Junior Member (Idle past 2654 days)
Posts: 12
Joined: 03-09-2012

 Message 24 of 41 (655461) 03-10-2012 3:00 PM Reply to: Message 23 by NoNukes03-10-2012 2:02 PM

Re: LaTeX
quote:
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.

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.

quote:
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.

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.

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

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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Member
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Joined: 07-20-2006

 Message 25 of 41 (655463) 03-10-2012 3:29 PM Reply to: Message 16 by Khy03-09-2012 8:12 PM

 Mass is also a factor in the definition of energy (he used the ergon which is defined as 1 gÂ·cm2/s2)

This has lost something in translation, I fear.

Ergon

Ergon may refer to:

Ergon, an alien chicken from TV's Doctor Who.
Ergon, a concept from Aristotle's Nicomachean Ethics that is most often translated as function, task, or work.
Ergon, Inc., a privately held petroleum company based in Jackson, Mississippi.
Ergon Energy, an electricity company owned by the Government of Queensland.

You must mean erg.

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Member
Posts: 16099
Joined: 07-20-2006

 Message 26 of 41 (655465) 03-10-2012 3:49 PM Reply to: Message 22 by Khy03-10-2012 12:57 PM

Re: LaTeX
 1. A meter is defined as: "the length of the path travelled by light in vacuum in 1 â„ 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.

No, that's exactly what we can't do. The measured quantities can change, but how can a unit change? What we could say is that if we use the speed of light as our standard for a meter, then a meter as measured by this technique is getting shorter, thus throwing off our efforts to measure distances; but if we let the units vary too (which is nonsensical) then we would have to say that a meter as measured by that standard always stayed the same length, namely a meter ... and we would also, now I come to think of it, have to say that the speed of light was constant, since it is defined in meters per second, no matter how much it actually changed.

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NoNukes
Inactive Member

 Message 27 of 41 (655466) 03-10-2012 3:53 PM Reply to: Message 24 by Khy03-10-2012 3:00 PM

Re: LaTeX
 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

A joule of energy is expended when any mass is accelerated across a friction free horizontal surface by a 1 N force over a distance of 1 meter.

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.

That's how we define our units of energy. We don't change that definition when the mass of proton changes.

 NoNukes writes: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.

 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.

The above logic is not right. 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.

Boltzmann's constant relates the energy of molecules to the bulk temperature. The constant is the same for the atoms or molecules of any ideal gas. I don't see, from mere unit analysis, a reason why the constant would change simply because those atoms have become heavier.

Under a government which imprisons any unjustly, the true place for a just man is also in prison. Thoreau: Civil Disobedience (1846)

The apathy of the people is enough to make every statue leap from its pedestal and hasten the resurrection of the dead. William Lloyd Garrison

 This message is a reply to: Message 24 by Khy, posted 03-10-2012 3:00 PM Khy has responded

 Replies to this message: Message 29 by Khy, posted 03-10-2012 4:25 PM NoNukes has responded

NoNukes
Inactive Member

 Message 28 of 41 (655467) 03-10-2012 4:08 PM Reply to: Message 24 by Khy03-10-2012 3:00 PM

Re: LaTeX
Duplicate removed

Edited by NoNukes, : No reason given.

Under a government which imprisons any unjustly, the true place for a just man is also in prison. Thoreau: Civil Disobedience (1846)

The apathy of the people is enough to make every statue leap from its pedestal and hasten the resurrection of the dead. William Lloyd Garrison

 This message is a reply to: Message 24 by Khy, posted 03-10-2012 3:00 PM Khy has not yet responded

Khy
Junior Member (Idle past 2654 days)
Posts: 12
Joined: 03-09-2012

 Message 29 of 41 (655469) 03-10-2012 4:25 PM Reply to: Message 27 by NoNukes03-10-2012 3:53 PM

Re: LaTeX
quote:
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.

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.

quote:
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.

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:

quote:
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.

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.

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NoNukes
Inactive Member

 Message 30 of 41 (655470) 03-10-2012 4:44 PM Reply to: Message 29 by Khy03-10-2012 4:25 PM

Re: LaTeX
 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 accelerated a mass of 10000000000000g at a rate of 1/10000000000000 cm per second squared.

Your math is correct. But now calculate the kinetic energy of each mass after the force is applied over the 1cm distance. In each case the answer will be one erg although the final velocities in each case will be quite different.

 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.

No those formula are not logical implications of changing the speed of light.

In particular, the change to G is done for the express purpose of keeping the earth in its orbit while the mass of the sun changes.
The effect on G is independent, but exactly counteractive from the changing of the speed of light and the change in mass, and is Sutterfield's ad hoc nonsense. If there is a logical link, I'd appreciate you pointing that out.

ABE:
And of course we do have evidence that the speed of light was the same as it is today 168,000 years ago, and that at least some decay rates used for aging have not changed over millions of years.
ABE: end

The exercise should illustrate the lengths at least one creationist will go to in order to deny that a single word of Genesis is not literally true.

Edited by NoNukes, : No reason given.

Under a government which imprisons any unjustly, the true place for a just man is also in prison. Thoreau: Civil Disobedience (1846)

The apathy of the people is enough to make every statue leap from its pedestal and hasten the resurrection of the dead. William Lloyd Garrison

 This message is a reply to: Message 29 by Khy, posted 03-10-2012 4:25 PM Khy has responded

 Replies to this message: Message 32 by Khy, posted 03-10-2012 5:30 PM NoNukes has responded

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