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Planck's Electric Charge ---- Factual One.


Kramer

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HUGE CONTROVERSY: PLANCK’S ELECTRIC CHARGE ---- FACTUAL ONE.
( Lay-man’s motto: Trouble the waters for clarification. )

Comparing the Planck’s constants derived with base “h” (historically) after with “h tire” and the constants with base “ electric charge – e” we see that they differ from each other. by the square root of ‘constant of fine structure”.
For example:

Planck’s mass = ( h-tire * C / G ) ^ 0.5 = 2.176646195*10^-8 Kg.
Electric mass = ( e / ( 4 * pi* ε0 * G ) ^ 0.5 ) = 1. 8593899458*10^-9 Kg.
Planck’s mass = electric mass / sqrt α.

Planck’s length = ( h-tire * G / C^3 ) = 1.61624*10^-35 m.
Electric radius = R = ( e / ( (4 * pi * ε 0 / G ) ^ 0.5) * C^2) = 1.3805438 * 10^-36 m.
Planck’s length = Electric radius / sqrt. α

Planck’s time = ( h-tire * G / C^5 ) ^ 0.5 = 5.39121 8 10 6 –44 sec.
Electric period = ( h / (M * C ^2) = 3 965007875* 10^-42 sec.
or ( ( 2* pi / α ) / R ) / C = 3.9650078* 10^-42 sec.
Planck’s time / electric period / (2*pi / sqrt
α )
etc…..
Evident that this discrepancy came from :
qpl / (1/ alpha) = e .
What is the physic’s meaning of Planck charge?
Are real the breaks of symmetry, or are a misconception?


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NO CONTROVERSY

 

That you don't understand something is not in the least bit controversial.

 

What are h tire, electric mass, electric radius and electric period? (other than nonsense terms) Is " h tire" h bar? (h/2pi)? Is electric actually electron? In any event, what period is this?

 

Planck units are derived, and depend fundamental constants (some of which comprise alpha), in various combinations that give you the units you want. Taking ratios of some might leave you with a number related to alpha.

 

The Planck charge is a convenient unit of charge when using Planck units. There is no physical basis for it, i.e there is no known particle that has this charge.

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The Planck charge is a convenient unit of charge when using Planck units. There is no physical basis for it, i.e there is no known particle that has this charge.

Absolutly right. Really we have no more than a useful system of units. At best the physics is just giving us handwaving scales at which we would expect new physics.

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Swansont

NOT CONTROVERSY

That you don't understand something is not in the least bit controversial.
----- I don’t think so. There is a controversy and a big one. Is it by the nature of matter : mass and mass-less particles, or by the schism of quantum from basic classic , an opera of humans,--- that it is exact, what I want to understand.
By the way about the “lack of my understanding” — isn’t this the same as “we speak about what we don’t understand’?. I simple intend to understand.

Let see:
Qpl = (4*pi*ε0*h-bar*C)^0.5
It derives from the physic’s equations: Qpl^2 / (4*pi* ε0* R) = h * ( C / (2*pi*R) ---I
Have you any objection? Simple is Equivalence of energies in classic.
Why not, then factual charge: e^2 / (4*pi*ε0*R) = h * ( C / (2*pi*R) ?----II

Simple, because system II isn’t true. There is not equivalence’
There appears controversy: energy of two factual electric charges differs from quantum energy. How to correct it? Change factual “e” in Qpl.
Why? You tell me.

What are h tire, electric mass, electric radius and electric period? (other than nonsense terms) Is " h tire" h bar? (h/2pi)? Is electric actually electron? In any event, what period is this?

----- “h tire” my mistake for “h bar”
Electric mass :

Electric radius :
Frequency :
Derived from EQUIVALENCE of all kind of energies in an elementary particle.
The same as are derived Planck Constants . Only Planck Constants are “CORRECTED” in conformity with factual electric charge.
From equations of energies :
E = e^2 / (4*pi*ε0*R) = G*M^2 / R = h*(C / ( ( 2 * pi / α )* R) = M*C^2 = ( C^4 ) / G ) * R etc…

Planck units are derived, and depend fundamental constants (some of which comprise alpha), in various combinations that give you the units you want. Taking ratios of some might leave you with a number related to alpha.
---I think that plank constant are extracted by all kind of physic’s laws making equivalence with Planck energy E = h*f. And when there appears discrepancies were infused non dimensional constants as in case with Coulomb law, or Boltzmann law .
There is nothing wrong with non-dimensional constants. But to change entity as electric charge, a physic’s one, which is measured many times and it is not fictive, I think is a huge controversy.

 

ajb
Absolutly right. Really we have no more than a useful system of units. At best the physics is just giving us handwaving scales at which we would expect new physics
.
----- Sorry that I don’t understand perfect English.

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Units are arbitrary and chosen by convenience. In many cases, using the charge on an electron or proton as the magnitude of the unit is the most convenient, but not always. The thing is, we do this all the time, and it's not "controversial". We measure distances in meters or kilometers. But we also have light-years and parsecs, and some people measure distances in feet and miles.

 

So yes, there are different units of charge. We also have Coulombs and StatCoulombs. No controversy.

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ajb
Are you asking why charge comes in units of the electric charge (fundamental charge) and not the Planck charge?
------ Do you think --- Planck constant charge is a unit? Different from what we use (Coulomb)? And others stay not changed ? “ Mpl “ is not in Kg ?
I really don’t understand, why you deviate from something simple:
Why Ee = (e^2) / (4*pi*ε0*R) = me * C^2 is not equal with
Ee = (h*C) / ( 2*pi*R) ?
Why Planck to equalize the equation, changed “e” with “ QPl “ ?
Why he insisted in equalization of energies for different laws?
Because for the law of conservation of energy.
Once and for ever: Change of one unit in physic attire change of all units in correspondence, if you don’t want to make a mess with conservation of energy.
And here we have the controversy:
Ee = (e^2) / (4*pi*ε0*Re) = Em = me * C^2 is not equal with
Eh= (h*C) / ( 2*pi*Re) ?
What to do? We can’t change “e” which is fine established, when instead “Qpl” is only a “convenient unit” without any measure support. We can’t change “h” because is well established in Physic, and especially in modern physic (as it is a “prima dona” in opera.)
We can’t change “C”. It is rock established.
Why don’t try with others participant in above formulas?
In Coulomb potential law of energy “ e “ is electric charge (Forget electron). Re is a distance between them. We can’t change it, it is a distance and fine.
But in “Eh.” equation this distance “Re” is linked with the oscillations, with frequency.
What is a frequency:
1- is a periodic movement of electric charge in a segment of space, and f = C / Re
II – is a periodic movement of electric charge in a circle of space and f = C / (2*pi*Re)
III-Is a periodic movement of electric charge in spherical trajectory. A spherical trajectory is a number of circle trajectories with a common center which pursue each other and fill a sphere. And this is one Hz.
In spherical oscillations fsph. = C / ( (2*pi /α) * Re) . If we adopt this kind of oscillations the controversy of Planck charge is solved.

swansont
nits are arbitrary and chosen by convenience. In many cases, using the charge on an electron or proton as the magnitude of the unit is the most convenient, but not always. The thing is, we do this all the time, and it's not "controversial". We measure distances in
meters or kilometers. But we also have light-years and parsecs, and some people measure distances in feet and miles.

So yes, there are different units of charge. We also have Coulombs and StatCoulombs. No controversy.
---- I suppose that the answer for “ajb” is in the same line with your rebut.

 



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Swansont

I don't understand your answer. Using different systems of units doesn't violate any conservation laws.

----- I don’t say this! At all --- I say the opposite. In Planck constants is changed only “one” constant “e” with “qpl”. And this I am trying to understand: Is it a Planck flaw or a law of nature that :

Qpl^2 / ( 4*pi*ε0*R ) is equal with h * C / (2*pi*R) in Planck costants
when
e^2 / (4*pi*ε0*R) is different with h * C / (2*pi*R) in fact? Or the fact is wrong?
Please don’t dodge my question.

If you have two things that are not equal, perhaps it's because nobody is claiming they are
equal.

------ With this you say that e^2 / (4*pi*ε0*R) is different with h * C / (2*pi*R) ?
That means we have not equivalence of energies.
Well.
In so called annihilation we have:

2 * me * C^2 = 2*e^2 / (4*pi*ε0*Re) different from 2 * h * C / (2 * pi* Re)
where is lost the difference of energies? Where is the conservation of energies?

Endyo816

I think OP is mixing quantum mechanic, relativistic and classical equations.
If there could be some clarification on what each "E" represents it would help to pin down the problem.

---- Right. In Planck area, I may say too in Einstein area, all kinds of energies have the same value. The kinds are expressed with different laws of physics: Coulomb. Einstein, Newton , Boltzmann, Planck ….This is the law of conservation of energy in whatever form it is appears.
And my thread is about this.
I dare to say that if we speculate ( in lay—mans manner), in some concepts about mass, about structure of matter, we my have for elementary particles:

E = ECoulomb = E Ejnstein = Eplanck = E Boltzmann = ENewton…..
and a different idea about breaking of symmetry.

 


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Swansont

I don't understand your answer. Using different systems of units doesn't violate any conservation laws.

----- I don’t say this! At all --- I say the opposite. In Planck constants is changed only “one” constant “e” with “qpl”. And this I am trying to understand: Is it a Planck flaw or a law of nature that :

Qpl^2 / ( 4*pi*ε0*R ) is equal with h * C / (2*pi*R) in Planck costants

when

e^2 / (4*pi*ε0*R) is different with h * C / (2*pi*R) in fact? Or the fact is wrong?

Please don’t dodge my question.

I'm not dodging anything. I told you I didn't understand what your answer was. You are writing down equations and not saying what they represent, or what the variables are.

 

It looks here like you are equating the planck charge and the fundamental charge. They are not equal. Nobody has claimed they are equal.

 

 

If you have two things that are not equal, perhaps it's because nobody is claiming they are

equal.

------ With this you say that e^2 / (4*pi*ε0*R) is different with h * C / (2*pi*R) ?

That means we have not equivalence of energies.

Well.

In so called annihilation we have:

2 * me * C^2 = 2*e^2 / (4*pi*ε0*Re) different from 2 * h * C / (2 * pi* Re)

where is lost the difference of energies? Where is the conservation of energies?

I don't recognize some of these equations and I don't know why you are setting them equal to each other.

 

mc^2 is the Einstein rest frame mass-energy equivalence.

2e^2 / (4*pi*ε0*Re) looks like the electric potential energy for two electrons separated by Re/2, or twice the energy if separated by Re

 

Is Re the classical electron radius?

 

Why should these be equal?

 

Where did you get hc/(2*pi*R)? What does that represent?

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The "E's" don't represent the same thing. All energy, but different forms in different situations.

 

As a quick example:

 

0c02f6f38c911dc8fea96feecd95eb31.png

 

[latex]\tau = 2\pi [/latex]

 

Tau [latex]\tau[/latex] is referring to two different things, torsion and Pi doubled.

 

I can write:

 

[latex]-n\cdot b' = 2\pi [/latex]

 

but it won't make much sense.

 

Hopefully this helps.

Edited by Endy0816
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Swanson

I'm not dodging anything. I told you I didn't understand what your answer was. You are writing down equations and not saying what they represent, or what the variables are.
----- Yes, you are. Or you see the thread boring and treat it with careless.
What is here that you didn’t understand?

Qpl^2 / ( 4*pi*ε0*R ) is equal with h * C / (2*pi*R) with Planck constant for charge.
This is equalization of two kind of energies: Coulomb and his own (Planck’s). The “variable R” is the same in both equations. So it may be dropped, (reduced). And will have:

Qpl^2 / ( 4*pi*ε0 ) = h * C / (2*pi ) and voila Qpl = (h bar*C*4*pi*ε0)^0.5

The same he has made with Newton energy and his own energy, to extract “Planck mass”:
G*M^2 / R = h*C / (2*pi*R) ---- reduced R and voila Mpl = ((h bar *C) / G ) ^0.5 .


The difference in both cases is that factual mass is a variable, instead factual charge is a solid constant.
I see that “Qpl” is an arbitrary fudge and is wrong, caused by necessity to equalize energies. And this doesn’t solve the discrepancy between Planck energy and others.

e^2 / (4*pi*ε0*R) is different with h * C / (2*pi*R) in fact?
Is it right , or wrong ?

It looks here like you are equating the planck charge and the fundamental charge. They are not equal. Nobody has claimed they are equal.
------Right. But, why?


I don't recognize some of these equations and I don't know why you are setting them equal to each other.
------- To compare different kind of energies of electron particles. To find their rate toward each other. And to find if Planck energy correspond to them.

mc^2 is the Einstein rest frame mass-energy equivalence.
-----Yes.
2e^2 / (4*pi*ε0*Re) looks like the electric potential energy for two electrons separated by Re/2, or twice the energy if separated by Re.
------Yes. And they are equal with potential energy of two electric charge in distance Re

Is Re the classical electron radius?
-----Yes

Why should these be equal?
------- The two kinds of energies have the same value.

Where did you get hc/(2*pi*R)? What does that represent?
------ From Planck when he compare potencial electric energy with it’s own.

This is comparison:

me*C^2 =8.18710414*10^-14 J
e^2 / (4*pi*ε0*Re) = 8.187104679*10^-14 J

h*C (2*pi*Re) = h bar*C / Re = 1.121928074*10^-11 J ??


Endy0814
The "E's" don't represent the same thing. All energy, but different forms in different situations.
-------Right. But I wanted to compare the amount of each form of energy.

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-------Right. But I wanted to compare the amount of each form of energy.

 

The question is why? To what end?

 

The energies you are comparing aren't supposed to be equal, as far as I can tell. Which means it will not be at all surprising that they are not equal.

 

The planck charge qp is not the same value as the fundamental charge e. You CANNOT interchange them. The energy using one vs the other will be different.

 

Just as if you use a charge of 1 Coulomb in a problem, you will get a different answer if you solve it with a charge of 2 Coulombs.

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Swansont.

The question is why? To what end?
In Planck area All kind of energies have the same value. Yes! In factual area this is not true.
But in Planck area is changed arbitrarily the value of one of most important protagonist of physics world “electric charge” with “Planck charge”.
Why? Only for the aim to equalize Planck energy ( h * C / 2 * pi / R ) with all kind of energies…. And with that are distorted all kind of Planck constants.
In my thread I am trying to compare Planck constants with, let say Einstein constants, where the only real protagonists are factual electric charge and factual gravity constant.
Why Einstein constants? Because is supposed that in physics formula G*M / R = C2 is the upper mark-area of reality. And the lower mark-- area of reality is h*f = h j.

The energies you are comparing aren't supposed to be equal, as far as I can tell. Which means it will not be at all surprising that they are not equal.
In Planck area they must be. Here is the Controversy in the title of the thread, here is the gist of debate. And if you permit a speculation, even in between areas must be equalization.

The planck charge qp is not the same value as the fundamental charge e. You CANNOT interchange them. The energy using one vs the other will be different.
-----Here is the trick. It will be very appropriate if Planck have righted the equations:

h * C / ( 2 * pi * R ) = (137,036) * e^2 / ( 4*pi*ε * R) or
h * C / ( (2 * pi / α ) * R ) = e^2 / ( 4 pi*ε*R)
(Then leaving physicist to scratch their heads why is this so)
Than using a new fictive physics constant, leaving the others constants unchanged. Using an unwarranted unity when you have a real one, is strange.

 

Just as if you use a charge of 1 Coulomb in a problem, you will get a different answer if you solve it with a charge of 2 Coulombs.
-------I didn’t grasp this example. Then why call 2 coulomb = Qswans. ?

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I refer you to the comments I made before about unit systems. The planck charge is not the same value as the fundamental charge. It's part of a different unit system. There is no reason for any derived quantities like energy to be numerically equal when you change from one system to another

 

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Swansont

I refer you to the comments I made before about unit systems. The planck charge is not the same value as the fundamental charge. It's part of a different unit system. There is no reason for any derived quantities like energy to be numerically equal when you change from one system to another.
----Then why was the need to use a non fundamental charge, for creation a different unit system, instead of the fundamental charge, which I think is the most important unit in physic?
In Einstein system of physic’s laws, are used all fundamental units: e. G. h. C, Bol. and are derived all variables as ultimate constants in Planck area : M, R, T, f, Tem.
This without need to use a “ convenient” charge, as Planck charge, and to distort reality.
It was used Planck charge for creation a system, aiming to “bold” importance oh “h” and to “fade” importance of “factual charge” ? Who know?
Any way! This debate is gone too long without any understanding. Non vale la pena , to continue. You may close this thread.


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----Then why was the need to use a non fundamental charge, for creation a different unit system, instead of the fundamental charge, which I think is the most important unit in physic?

 

 

Because in some calculations it's easier to use different unit systems, and there's no point in making calculations harder by using inconvenient units. You probably wouldn't want to keep track of a plane trip in inches.

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----Then why was the need to use a non fundamental charge, for creation a different unit system, instead of the fundamental charge, which I think is the most important unit in physic?

 

Because others don't think that.

And, if anyone thinks I'm going to buy beer by the half-litre any time soon they are sorely mistaken.

The Pint is the fundamental unit for beer.

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Because others don't think that.

And, if anyone thinks I'm going to buy beer by the half-litre any time soon they are sorely mistaken.

The Pint is the fundamental unit for beer.

It comes in 0.4 litres and 1 litre here in Poland!

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swansont, on 05 Oct 2013 - 5:27 PM, said:

You probably wouldn't want to keep track of a plane trip in inches.
\---- Wrong argument. In 1954 is changed cm.gr.sec.e.system in physics with Kg.m.sec.A., together with a changed dielectric constants. The unit of electric charge is changed, but in the same time is changed too the unity of the energy. From “erg” in “Joule” in conformity of changes of the “bases unit”. So we have the same:

e^2 / Re*ε01 = E erg = (e^2 / ( 4*pi * ε02*Re) = E joule
8.187104762*10^-7 erg = 8.187104762 * 10 ^ -14 Joule

Isn’t this the same amount of energy? Sure is. Because : 1 joule = m^2 *Kg *sec ^-2 *A = 10^2 cm^2 *10^3gr *sec^.2 *e = 10^7 erg.
In Planck charge equations is violated the factual un-equalization of energies, changing arbitrarily the fundamental unity:

Qpl ^2 / ( 4*pi*ε R) = h * C / ( 2*pi*R) compare with

e^2 / (4*pi*ε*R) unequal h*C / (2*pi*R)
So here is not a change of unities. Is a change of concept.

 

ajb
Or at the other end of the scale, light-years.

----- It’s quite different. Using light –years , doesn’t change the fact if we measure it in let say cm.
Jon Kuthber

Because others don't think that.
------- The others have the argument that don’t think that.

And, if anyone thinks I'm going to buy beer by the half-litre any time soon they are sorely mistaken.

The Pint is the fundamental unit for beer.
----- You may buy beer in half-litre :
1 –if the money you pay are comparable with those with pint.
2-- if the beer you by is not donkey pee.
ajb

It comes in 0.4 litres and 1 litre here in Poland!
-----And ?

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swansont, on 05 Oct 2013 - 5:27 PM, said:

You probably wouldn't want to keep track of a plane trip in inches.

\---- Wrong argument. In 1954 is changed cm.gr.sec.e.system in physics with Kg.m.sec.A., together with a changed dielectric constants. The unit of electric charge is changed, but in the same time is changed too the unity of the energy. From “erg” in “Joule” in conformity of changes of the “bases unit”. So we have the same:

e^2 / Re*ε01 = E erg = (e^2 / ( 4*pi * ε02*Re) = E joule

8.187104762*10^-7 erg = 8.187104762 * 10 ^ -14 Joule

Isn’t this the same amount of energy? Sure is. Because : 1 joule = m^2 *Kg *sec ^-2 *A = 10^2 cm^2 *10^3gr *sec^.2 *e = 10^7 erg.

In Planck charge equations is violated the factual un-equalization of energies, changing arbitrarily the fundamental unity:

Qpl ^2 / ( 4*pi*ε R) = h * C / ( 2*pi*R) compare with

e^2 / (4*pi*ε*R) unequal h*C / (2*pi*R)

So here is not a change of unities. Is a change of concept.

 

 

"e^2 / (4*pi*ε*R) unequal h*C / (2*pi*R)" because e and qp are not the same. It's "illegal" to substitute terms when they aren't equal.

 

You are able to convert erg to joules. Why are you having difficulty in converting values of charge? What about converting inches to centimeters? It's conceptually the same problem. 1 inch is 2.54 cm. If the distance is 1 inch, you wouldn't try and use 1 cm in a formula.

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