Naked Science Forum
On the Lighter Side => New Theories => Topic started by: jeffreyH on 05/04/2014 05:02:27
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I now have an independent calculation of the Planck mass and G. The method is not circular but at the moment I need to double and triple check my work. These values are 1.01069E-08 for the Planck mass and 6.74259E-11 for G. At this point it is highly likely that these are in error. The hour is late and I couldn't sleep. Not the best circumstances for doing this sort of work.
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I have attached a PDF detailing the non-circular calculation method for G. Comments welcome.
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It's not an independent check.
It's a circular argument.
it uses the value for the mass of the Earth
"We now multiply Mp by the mass of the earth"
But nobody has ever measured that directly. It is calculated from the measured value of the local acceleration due to (the Earth's) gravity, the Earth's radius, and the Universal gravitational constant G.
So, you are using G to calculate G
Sorry, but this method is dead.
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It's not an independent check.
It's a circular argument.
it uses the value for the mass of the Earth
"We now multiply Mp by the mass of the earth"
But nobody has ever measured that directly. It is calculated from the measured value of the local acceleration due to (the Earth's) gravity, the Earth's radius, and the Universal gravitational constant G.
So, you are using G to calculate G
Sorry, but this method is dead.
The point is you don't have to use the earth. A manufactured mass with an exactly known mass, radius and value for g is all you need. g can be measured just by recording the speed at which the mass attracts an object over a well defined period of time. These measurements need to be taken outside the earth's gravitational field. Far enough so that it's gravitational effect is negligible. It wouldn't be easy or cheap but that doesn't make the method dead. If the adjusted value of G is used as in the method you would arrive at probably the most accurate value for G.
The other point is that an awful lot of money is spent on various experiments that may or may not work. Wouldn't it be better to find one that does work and spend on that once instead of over and over again in hopefulness of an answer.
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Try (c/c^2) * (2 * 10^-2). Elegant and simple.
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c/c^2 = 3.33564E-09 More on this later
(c/c^2) * (2 * 10^-2) = 6.67128E-11 The factor of 2 balances 2GM/C^2
If we now do 3.33564E-09/C^2 we get 3.7114E-26 with a magnitude of 10^-26
So let's use that with 3.33564E-09. 3.33564E-09 * 10 ^ - 26 = 3.33564E-35
Divide this by 2 = 1.66782E-35
Current Planck length 1.61619926E-35
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"A manufactured mass with an exactly known mass, radius and value for g is all you need. g can be measured just by recording the speed at which the mass attracts an object over a well defined period of time. "
Or you can do it the easy way- as Cavendish did.
The point remains that you have calculated G by using a given value of G.
So you have not really achieved anything.
" If the adjusted value of G is used as in the method you would arrive at probably the most accurate value for G."
No it would not.
It would give you exactly the same value of G as you measured.
Do you not understand that you are using G to calculate G so, if you assumed that it was 42 it would come out to exactly 42?
Also c/(c^2) is a pointlessly complicated way of saying 1/c
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"A manufactured mass with an exactly known mass, radius and value for g is all you need. g can be measured just by recording the speed at which the mass attracts an object over a well defined period of time. "
Or you can do it the easy way- as Cavendish did.
The point remains that you have calculated G by using a given value of G.
So you have not really achieved anything.
" If the adjusted value of G is used as in the method you would arrive at probably the most accurate value for G."
No it would not.
It would give you exactly the same value of G as you measured.
Do you not understand that you are using G to calculate G so, if you assumed that it was 42 it would come out to exactly 42?
Also c/(c^2) is a pointlessly complicated way of saying 1/c
G was not used at all it is all derived from c. OK yes it's 1/c. I was lazy and just used the c^2 from the original equation. Where was G used in (c/c^2) * (2 * 10^-2) or (1/c) * (2 * 10^-2) ? The result is G. No other values were used. Are you saying we do not have an accurate value of c? Are you questioning the use of (2 * 10^-2) to scale the value?
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Or Rs of earth using the modified form 0.04*M/c^3
(0.04*5.97E+24)/c^3 = 0.008866077
And that eliminates G altogether. The factor 0.04 is derived from a couple of models on arxiv. One for quark confinement and the other for black hole models.
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You seem to have missed a subtle point here.
The thread is called
"An attempt to calculate G without circularity"
And you have not actually calculated G- except by using the mass of the Earth- which isn't directly measured, but calculated from G.
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Try (c/c^2) * (2 * 10^-2). Elegant and simple.
Elegant, simple and wrong
It gives 6.67128 E-11 seconds per metre
The right value is
6.67384 × 10-11 m3 kg-1 s-2
So, the number is clearly wrong (By about 0.04% which is more than the experimental error) and the units are completely screwed.
Why did you not notice that?
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I made no mistake on the units. Seconds per metre are the units you would need behind the event horizon as space is then one dimensional and limiting on angular momentum. It is also interesting that the error is 0.04% don't you think?
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And returning from the Twilight Zone....
If we treat GM as a single entity we may gain more insight. Starting with Rs = 2GM/c^2 we can rearrange as GM/dRs = c^2/2 where we adjust the delta value of Rs. Taking the values GM/1 = c^2/2, GM/2 = c^2/2 and GM/3 = c^2/2 we can show the relationship 1/2*c^2, 2/2*c^2 and 3/2*c^2 etc. This will give us the proportion of the original mass used to calculate Rs. We can then plot this. More on this later. This halving of the radius is because of the relationship between the Planck mass and a 2 Planck length Rs. All the above equations are in units of metres.
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Taking Rs as 1 we are starting from unity. The factors of c^2 are reminiscent of particle spin.
1/2*c^2 = electron
2/2*c^2 = c^2 = photon
3/2*c^2 = gravitino (c^2 + 1/2*c^2) like photon + electron
4/2*c^2 = graviton (c^2 + c^2) like two copy gluon
Since the photon and gluon have the same spin.
In the case of the gravitino this is intriguing.
From wikipedia.
http://en.wikipedia.org/wiki/Gravitino
"The other option is that the gravitino is unstable. Thus the gravitinos mentioned above would decay and will not contribute to the observed dark matter density. However, since they decay only through gravitational interactions, their lifetime would be very long, of the order of Mpl2 ∕ m3 in natural units, where Mpl is the Planck mass and m is the mass of a gravitino. For a gravitino mass of the order of TeV this would be 105 s, much later than the era of nucleosynthesis. At least one possible channel of decay must include either a photon, a charged lepton or a meson, each of which would be energetic enough to destroy a nucleus if it strikes one. One can show that enough such energetic particles will be created in the decay as to destroy almost all the nuclei created in the era of nucleosynthesis, in contrast with observations. In fact, in such a case the universe would have been made of hydrogen alone, and star formation would probably be impossible."
If the gravitino was a combination of a photon-like and an electron-like particle what would this mean? Alternatively any spin 1 + spin 1/2. I know it is usually 3 particle composites but this being considered a fermion would make it a possible composite.
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In the above scheme could
3/2*c^2 = gravitino (c^2 + 1/2*c^2) like photon + electron
actually be
3/2*c^2 = gravitino (c^2 + 1/2*c^2) like gluon + electron.
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The number is still wrong.
The units are still wrong.
You are still wrong.
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Well I have found some interesting relationships that would not have been apparent using standard methods. Of course I realize this is way off base. I am not arguing that any of this is standard physics or even valid in its own right. I am looking at relationships between values and blatantly ignoring unit conversions. Not really standard practice. On that point I don't care as long as I end up with an understanding of the relationships involved.
In physics much use is made of imaginary numbers as a means to an end. I am doing the equivalent with units. Let's call them imaginary units.
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"In physics much use is made of imaginary numbers as a means to an end. I am doing the equivalent with units. Let's call them imaginary units.
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No.
Lets call them the wrong units.
The mathematicians use imaginary numbers to make equations balance out.
Your ignoring the importance of units means that your "equations" don't balance, and they never will.
The "interesting relationships" you have found are just meaningless coincidences and you ought to accept that.