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Author Topic:   Debunking Setterfields Speed of Light Model
NoNukes
Member
Posts: 9320
From: Central NC USA
Joined: 08-13-2010
Member Rating: 2.6


Message 31 of 41 (655472)
03-10-2012 5:04 PM
Reply to: Message 29 by Khy
03-10-2012 4:25 PM


Re: LaTeX
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..."

"We should get the same answer regardless..." means regardless of what units of measure we use, and not "regardless of whether the mass of a proton changes."

You, and you alone attempted to predict the effect on k using unit analysis. I have acknowledged that there may be some change to k that follows from Setterfield's assumptions, and perhaps we can detect the effect using one or more of the postulates on page 6. But attempting analyze that effect by merely looking at the definition of gram or kilogram is wrong.

I've also explained why the technique appears to work with G. What have I omitted?


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:
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Khy
Junior Member (Idle past 1750 days)
Posts: 12
Joined: 03-09-2012


Message 32 of 41 (655474)
03-10-2012 5:30 PM
Reply to: Message 30 by NoNukes
03-10-2012 4:44 PM


Re: LaTeX
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 (http://en.wikipedia.org/wiki/File:CGKilogram.jpg) 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.


This message is a reply to:
 Message 30 by NoNukes, posted 03-10-2012 4:44 PM NoNukes has responded

Replies to this message:
 Message 33 by NoNukes, posted 03-10-2012 5:48 PM Khy has responded

    
NoNukes
Member
Posts: 9320
From: Central NC USA
Joined: 08-13-2010
Member Rating: 2.6


Message 33 of 41 (655477)
03-10-2012 5:48 PM
Reply to: Message 32 by Khy
03-10-2012 5:30 PM


Re: LaTeX
Khy writes:

If the mass of an oxygen molecule decreases, then why wouldn't the mass of the IPK and subsequently the gram decrease

The mass of the IPK would decrease. Although it would be attracted to the earth by exactly the same force (according to Mr. S), the mass would become easier to accelerate through non gravitational means.

But that would simply mean that using standard amounts of material could not be used to establish the gram or the kilogram. If Newton's three laws are to remain correct, 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.

So what's your initial claim?


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 32 by Khy, posted 03-10-2012 5:30 PM Khy has responded

Replies to this message:
 Message 34 by Khy, posted 03-10-2012 6:21 PM NoNukes has responded

    
Khy
Junior Member (Idle past 1750 days)
Posts: 12
Joined: 03-09-2012


Message 34 of 41 (655479)
03-10-2012 6:21 PM
Reply to: Message 33 by NoNukes
03-10-2012 5:48 PM


Re: LaTeX
quote:
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.

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.


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Replies to this message:
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NoNukes
Member
Posts: 9320
From: Central NC USA
Joined: 08-13-2010
Member Rating: 2.6


Message 35 of 41 (655480)
03-10-2012 6:49 PM
Reply to: Message 34 by Khy
03-10-2012 6:21 PM


Re: LaTeX
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.

No. An S-world kilogram would always require 1 Newton of force to accelerate it to the 1 m/s/s. But the number of atoms in an S-world kilogram would have to change over time.

Otherwise, we'd have to let go of F=ma (at least when appreciable period of time were involved).

My initial claim is not just that there is a change in k (Nm/s^-2)

In my opinion you have yet to show why k should change. Maybe it would.


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 34 by Khy, posted 03-10-2012 6:21 PM Khy has responded

Replies to this message:
 Message 36 by Khy, posted 03-10-2012 7:12 PM NoNukes has responded

    
Khy
Junior Member (Idle past 1750 days)
Posts: 12
Joined: 03-09-2012


Message 36 of 41 (655485)
03-10-2012 7:12 PM
Reply to: Message 35 by NoNukes
03-10-2012 6:49 PM


Re: LaTeX
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.


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 Message 35 by NoNukes, posted 03-10-2012 6:49 PM NoNukes has responded

Replies to this message:
 Message 37 by NoNukes, posted 03-10-2012 7:34 PM Khy has responded

    
NoNukes
Member
Posts: 9320
From: Central NC USA
Joined: 08-13-2010
Member Rating: 2.6


Message 37 of 41 (655490)
03-10-2012 7:34 PM
Reply to: Message 36 by Khy
03-10-2012 7:12 PM


Re: LaTeX
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

What problem are you looking for? Using your units, a newton and a kg change their value over time. If that doesn't bother you, then so be it.

If we put Avagardo's number of oxygen atoms on a scale, they would always weight 16 grams, but over time that collection of atoms could be accelerated to increasingly higher velocities by adding the same amount of kinetic energy.

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.


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 36 by Khy, posted 03-10-2012 7:12 PM Khy has responded

Replies to this message:
 Message 38 by Khy, posted 03-10-2012 8:13 PM NoNukes has responded

    
Khy
Junior Member (Idle past 1750 days)
Posts: 12
Joined: 03-09-2012


Message 38 of 41 (655496)
03-10-2012 8:13 PM
Reply to: Message 37 by NoNukes
03-10-2012 7:34 PM


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

quote:
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.

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.


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Replies to this message:
 Message 39 by NoNukes, posted 03-10-2012 11:28 PM Khy has not yet responded

    
NoNukes
Member
Posts: 9320
From: Central NC USA
Joined: 08-13-2010
Member Rating: 2.6


Message 39 of 41 (655506)
03-10-2012 11:28 PM
Reply to: Message 38 by Khy
03-10-2012 8:13 PM


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

You continue to make the same mistake. And that mistake is in part the basis for your mistake regarding Dr. J's critique.

The size of our measuring units should never change. 1kg in universe Einstein should be equal to 1kg in universe Setterfield. The problem is that the inertial on similar objects is different in the two universes. That's why it is possible for oxygen to escape in one universe but be held in the atmosphere in another.

ABE:

On reflection, I suppose you could develop a system of units based on what you can measure in universe S. But if you do so, you should be prepared to develop the laws of physics in universe S as well. Given that a mass in S-universe units behaves differently over time to all forces except gravitational forces, I suspect that coming up with a proper statement of the laws of physics using S universe units won't be easy. In order to evaluate whether oxygen can escape the atmosphere, you'll have to deal with both gravitational forces, and with the energy/temperature relationships that determine the velocities of oxygen molecules. Good luck with that.

Edited by NoNukes, : No reason given.

Edited by NoNukes, : Rethinking the previous answer...


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 38 by Khy, posted 03-10-2012 8:13 PM Khy has not yet responded

Replies to this message:
 Message 40 by AdminModulous, posted 03-11-2012 4:29 AM NoNukes has acknowledged this reply

    
AdminModulous
Administrator
Posts: 885
Joined: 03-02-2006


Message 40 of 41 (655519)
03-11-2012 4:29 AM
Reply to: Message 39 by NoNukes
03-10-2012 11:28 PM


LaTeX is a pretty poor subtitle
Please can we consider using a subtitle that reflects the present state of the discussion?
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 Message 39 by NoNukes, posted 03-10-2012 11:28 PM NoNukes has acknowledged this reply

    
Perdition
Member (Idle past 587 days)
Posts: 1593
From: Wisconsin
Joined: 05-15-2003


Message 41 of 41 (655643)
03-12-2012 10:47 AM
Reply to: Message 18 by NoNukes
03-10-2012 9:05 AM


Re: How do you get off of inifinite speed?
I don't find this particular line of reasoning condemning of Setterfield's work.

As I said, I haven't read Setterfield's work, I was merely confused on the point I raised, namely, how do you decrease from infinite speed?

If it's really just an artifact of extrapolating back to time zero with a graphical asymptote, then I guess I can let it go.


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 Message 18 by NoNukes, posted 03-10-2012 9:05 AM NoNukes has acknowledged this reply

    
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