Post by: PaulTalbot on 16/03/2022 21:20:27
The value of the minimum mass would be Mmin ≈ 1.7206 × 10^^-68 kg, and that of the Hubble constant, H ≈ 72.013 km s-1 Mpc-1. The proton radius can also be calculated using similar assumptions: rp ≈ 0.8264 fm.
Since I am looking for some feedback on this hypothesis, your comments are welcome.
The 4 page preprint is attached (PDF).
Thank you.
Post by: Origin on 16/03/2022 22:57:57
I wonder if the minimum massMinimum mass of what?
the Hubble constant and the proton radius could be calculated precisely using the observed ratio of the electrostatic force to the gravitational force?Electrostatic force of what and the gravitational force of what? You could pick values for the electrostatic and gravitational forces so that they would equal the Hubble constant ant the radius of a proton, but I don't see what use that would be.
I don't see any physical reason that these different phenomena would be related like that.
Post by: Bored chemist on 16/03/2022 23:03:19
So it changes.
The radius of the proton doesn't.
https://en.wikipedia.org/wiki/Hubble%27s_law#Hubble_time
Post by: PaulTalbot on 18/03/2022 18:55:52
Minimum mass of what?The value of Mmin corresponds to the Wesson’s mass (mE ≈ 2 × 10 68 kg), i.e., the quantum of mass calculated by Pr. Wesson [10] in 2003 using the constants c, h and Λ (the cosmological constant).
Post by: PaulTalbot on 18/03/2022 18:58:44
Electrostatic force of what and the gravitational force of what?The ratio, say β, of the electrostatic force to the gravitational force between an electron and a proton is
β = e2 / 4π ε0 G me mp. (4)
ε0 is the vacuum permittivity.
G is the gravitational constant .
me is the mass of the electron.
mp is the mass of the proton.
In the last century, Paul Dirac popularized this β ratio in a cosmological context [8] [9]. It still gets attention because of its high value (β ≈ 2.268 66 × 1039). The gravitational force is tiny compared to other forces.
Post by: PaulTalbot on 18/03/2022 19:32:56
The Hubble constant is pretty close to 1/ the age of the universe.The decreasing value of the Hubble constant is indeed suggested by the conformal model ΛCDM.
So it changes.
The radius of the proton doesn't.
Now, in my opinion, this conformal model is falsified by observation, since the observed value of H is way too high.
Moreover, how do we know if the proton radius is really constant?
Why couldn't it slowly vary with time?
Post by: Halc on 18/03/2022 20:15:54
The value of Mmin corresponds to the Wesson’s mass (mE ≈ 2 × 10 68 kg), i.e., the quantum of mass calculated by Pr. Wesson [10] in 2003 using the constants c, h and Λ (the cosmological constant).Yes, it seems Wesson attempted to determine a quanta of mass.
In the last century, Paul Dirac popularized this β ratio in a cosmological context [8] [9]. It still gets attention because of its high value (β ≈ 2.268 66 × 1039). The gravitational force is tiny compared to other forces.You're giving refereces 8 & 9 which don't correspond to any bibliography.
You are obviously copying directly from the works of others without giving credit. This plagiarism violates site rules.
Please edit your posts with credits properly given.
The decreasing value of the Hubble constant is indeed suggested by the conformal model ΛCDM.That very model suggests that H will actually settle down to a constant value of around 57 km/sec/mpc, which is straight exponential expansion.
Post by: Bored chemist on 19/03/2022 01:39:04
Moreover, how do we know if the proton radius is really constant?Because that would have an effect on the spectra of atoms.
Why couldn't it slowly vary with time?
But we see the emission from atoms from a long time ago, and that (hypothetical) effect is not found
So we know that the radius of the proton has not changed.
So we know , from direct experimental observation, that your idea is wrong.
And now what we do is look at how you react to that fact.
Do you behave like a scientist and thank us for pointing out the glitch which makes your idea impossible, or do you behave like a schoolkid.
Post by: PaulTalbot on 19/03/2022 14:13:11
You're giving refereces 8 & 9 which don't correspond to any bibliography. You are obviously copying directly from the works of others without giving credit. This plagiarism violates site rules.There must be a misunderstanding. The text you refer to is part of the preprint I submitted in my original post (a 4-page attached PDF, relatively quick to read). The bibliography is included in the document.
I am new to this forum, so I thougth I could get some feedback before publishing.
Since the text is already available in the original post, I won't repeat it.
However, for those who just want to have a quick view, please find attached a chart of the assumed relations between
the refered physical constants.
Post by: Origin on 19/03/2022 14:33:02
Since I am looking for some feedback on this hypothesis, your comments are welcome.I looked at your PDF and it appears to me that what you are doing is called numerology. That is you are taking a bunch of unrelated constants and combining them in such a way that you get known quantities. In other words the equations have no physical meaning they are essentially just a bunch of random numbers put together to equal a predetermined quantity.
Post by: Bored chemist on 19/03/2022 14:45:08
You're giving refereces 8 & 9 which don't correspond to any bibliography. You are obviously copying directly from the works of others without giving credit. This plagiarism violates site rules.There must be a misunderstanding. The text you refer to is part of the preprint I submitted in my original post (a 4-page attached PDF, relatively quick to read). The bibliography is included in the document.
I am new to this forum, so I thougth I could get some feedback before publishing.
Since the text is already available in the original post, I won't repeat it.
However, for those who just want to have a quick view, please find attached a chart of the assumed relations between
the refered physical constants.
And now what we do is look at how you react to that fact.OK, so you went with the schoolkid option.
Do you behave like a scientist and thank us for pointing out the glitch which makes your idea impossible, or do you behave like a schoolkid.
Post by: PaulTalbot on 19/03/2022 15:10:29
Because that would have an effect on the spectra of atoms.First, I would like to thank all those who take the time to read this thread. So far, all replies have been and will continue to be thanked, as long as they stay polite.
But we see the emission from atoms from a long time ago, and that (hypothetical) effect is not found
So we know that the radius of the proton has not changed.
So we know , from direct experimental observation, that your idea is wrong.
And now what we do is look at how you react to that fact.
Do you behave like a scientist and thank us for pointing out the glitch which makes your idea impossible, or do you behave like a schoolkid.
I asked a question about the proton radius and I assume you answered it correctly.
Now, if you read the preprint, you will notice that all physical values are referred to as constants, including the proton radius. So, your conclusion about my idea is unjustified. It was just a question.
Please note that I try to answer the replies one by one.
Post by: Origin on 19/03/2022 15:25:20
Now, if you read the preprint, you will notice that all physical values are referred to as constants, including the proton radius. So, your conclusion about my idea is unjustified.When I said you were doing numerology and not physics I was referring to things like this from your paper.
You wrote:
The Hubble constant H (of dimension T-1) would correspond to the minimum frequency:
fmin = H.
Let's think about this for a second, what is the frequency of a constant? There is no frequency of a constant. The Hubble constant is the speed of the expansion of space per Mpc. A constant speed doesn't have a frequency.
You then take the formula for the energy of a photon and substitute in the bogus frequency of the Hubble constant.
None of that makes any physical sense and that is just the first couple of equations.
Post by: Halc on 19/03/2022 15:53:00
Now, if you read the preprint, you will notice that all physical values are referred to as constants, including the proton radius.Despite its name as 'the Hubble constant', it isn't a constant. For a zero energy solution of the universe, H would be exactly 1/t where t is time. The value has the same units as 1/t.
Quote
So, your conclusion about my idea is unjustified.What conclusion exactly was that? I don't remember making one about your idea.
I looked at the paper a bit.
The first 'constant' is this Mmin which lacks a reference. The H value (references given in introduction) is given to far more precision than is justified by the range of values in the references. Looking at really distant galaxies doesn't count since they're way in the past when the value of H was considerably different.
I'm just saying that you're not justifying the sort of precision that the paper claims.
Origin is right, you're substituting variables of different units into equations, rendering them meaningless. You're using the energy formula for a photon and applying it to something that isn't a photon.
Post by: PaulTalbot on 19/03/2022 18:33:18
What conclusion exactly was that? I don't remember making one about your idea.Sorry Halc, I think I mistakenly quoted some text from someone else. The conclusion was that I was a schoolkid because I wondered if the proton radius was really constant. Frankly, I wish I were a schoolkid, so at least, I would be younger. Fortunately, we are on the lighter side of the forum.
Post by: PaulTalbot on 19/03/2022 19:27:14
When I said you were doing numerology and not physics I was referring to things like this from your paper.
You wrote: The Hubble constant H (of dimension T-1) would correspond to the minimum frequency: fmin = H.
...
None of that makes any physical sense and that is just the first couple of equations.
OK, maybe I should have presented myself first and explain why I make such assumptions.
I am a computer scientist, an IT architect or if you prefer, a system designer interested in cosmology and the world we live in. I look at the Universe as a global system, not a simple object. This is why I am so intrigued by the relations between physical quantities.To get the idea, you can look at the attached chart: the MELT-hc network. It is part of a book I wrote on the subject (The Cosmospheric Principle).
This kind of illustration is typically used by system designers, and a network-type structure is quickly detected. The physical quantities (M:Mass, E:Energy, L:Length and T:Time) are represented by circles, and the relations by arrows, each with a boxed formula. Each arrow indicates the direction of the relation expressed by the formula linking two quantities. This direction is arbitrary and can be reversed by changing the formula. For example, the relation E = M c2, Einstein's famous formula can be written M = E/c2.
The MELT-hc network shows known equations that express the wave nature of energy and mass. The value of L then corresponds to a wavelength, and that of T, to the period of this wave. In this model, the relation L = c T links time to space in the same way that E = M c2 links mass to energy. In fact, all the relations of the model have this same universal status.
The relation E = h/T is a reformulation of the Planck-Einstein relation (E = h f), expressed as a function of time (T). The relation L = h/M c is the Compton wavelength associated with a mass M, and so on. These relatively simple relations can be deduced from each other. They involve two extremum values:
• the minimum action (h), and
• the maximum speed (c).
This explains the name of the network: MELT-hc.
Post by: Bored chemist on 19/03/2022 19:33:56
Post by: Bored chemist on 19/03/2022 20:47:41
Post by: PaulTalbot on 19/03/2022 22:04:53
I here try to answer some of them, but one at a time.
Origin is right, you're substituting variables of different units into equations, rendering them meaningless. You're using the energy formula for a photon and applying it to something that isn't a photon.In his 1924 thesis «Recherches sur la théorie des quanta» (Research on the theory of Quanta), Louis de Broglie broadened the interpretation of the Planck-Einstein relation which was then limited to electromagnetic phenomena. It assumes the existence of a periodic phenomenon associated with mass:
«One can thus conceive that because of a great law of Nature, to each piece of energy of proper mass m, is linked a periodic phenomenon of frequency ν such as we get: hν = mc2. This hypothesis is the basis of our system.» (Broglie, 1924)
This hypothesis was quickly adopted by the scientific community and confirmed by the electron diffraction in 1927.
Post by: PaulTalbot on 20/03/2022 14:42:49
OK, so you have the relative strength of these two forces down to 5 significant digits, which is strange since the values are in different units. Gravity between two objects is a function of their mass, and EM force between the same two objects is a function of their charge. So this number you've quoted is fairly random. You do the same calculation between say an electron and a positron and you get a different ratio. I got several different answers for the relative strengths of them, and they can't even agree on the order of magnitude, let along a figure to several digits, which again, being in different units, is largely meaningless.I think you are right about the choice of particles. There seems to be something special about the electron-proton couple, as if they were two sides of the same coin. The same thing applies to the positron-antiproton couple.
In the chapter 2 of his 1974 paper "Cosmological Models and the Large Numbers Hypothesis", Paul Dirac uses this same ratio:
"The electric force between the electron and the proton in a hydrogen atom is e2/r2. The gravitational force between them is Gmp me/r2. Their ratio is the dimensionless number e2/Gmpme. Its value is about 2 x 1039."
In those times, the charge was expressed in esu (electrostatic unit of charge also called statcoulomb or Franklin), so the electric constant was 1 instead of 1/4π ε0. This may lead to a confusion about the dimension of charge, that was then different. The Dirac's paper is attached to this post.
The Dirac's hypothesis about the variation of G over time has not held up well. Nevertheless, some scientists, including myself, follows in the footsteps of Dirac, searching for a meaning of dimensionless numbers.
Post by: PaulTalbot on 20/03/2022 14:45:16
Post by: Origin on 20/03/2022 15:10:44
Nevertheless, some scientists, including myself, follows in the footsteps of Dirac, searching for a meaning of dimensionless numbersI didn't realize you were a scientist. Could you explain how the Hubble constant can have a frequency and what that frequency is? Thanks.
Post by: Bored chemist on 20/03/2022 19:01:16
OK, I am glad that there are so many replies, so thank you again for all those comments.Let me know when you plan to get to this one.
I here try to answer some of them, but one at a time.
The fact that it fails on dimensional analysis is beside the point unless he can explain why something constant is the same as something that's changing.
Post by: PaulTalbot on 20/03/2022 19:35:54
Could you explain how the Hubble constant can have a frequency and what that frequency is? Thanks.The fact is that the dimension of a rate is the same as that of a frequency (T-1). In the case of the Hubble constant, it is expressed as a speed divided by a length so we get [H] = (L T-1) / L = T-1. It is not a frequency as such, but it is assumed to be the smallest possible value of a physical quantity of dimension T-1.
By the way, I consider myself as a low level computer scientist. Since I am not a physicist, I am not qualified to provide pertinent advice on the subject matter. Moreover, my view of the physical world is unconventional.
Post by: PaulTalbot on 20/03/2022 19:56:34
The fact that it fails on dimensional analysis is beside the point unless he can explain why something constant is the same as something that's changing.The fact is that the possible variation of physical constant is a complex and unresolved issue:
https://en.wikipedia.org/wiki/Time-variation_of_fundamental_constants (https://en.wikipedia.org/wiki/Time-variation_of_fundamental_constants)
If some of them are varying over time, it should be very slowly, because there is no obvious observation supporting this hypothesis. I do not pretend to solve this enigma in my preview, so I am just wondering if some of them could vary.
Post by: Bored chemist on 20/03/2022 21:07:30
Yes.The fact that it fails on dimensional analysis is beside the point unless he can explain why something constant is the same as something that's changing.The fact is that the possible variation of physical constant is a complex and unresolved issue:
https://en.wikipedia.org/wiki/Time-variation_of_fundamental_constants (https://en.wikipedia.org/wiki/Time-variation_of_fundamental_constants)
If some of them are varying over time, it should be very slowly, because there is no obvious observation supporting this hypothesis. I do not pretend to solve this enigma in my preview, so I am just wondering if some of them could vary.
But they would all have to vary in carefully specified ways in order to not be obvious.
And your idea doesn't fit with that
Post by: Origin on 20/03/2022 23:39:09
The fact is that the dimension of a rate is the same as that of a frequency (T-1).That is not correct, the dimensions of the rate in this case is L/t.
In the case of the Hubble constant, it is expressed as a speed divided by a length so we get [H] = (L T-1) / L = T-1.You can't just cancel out the Lengths since one of the units is km and the other is Mpc. Using unit analysis you can easily convert Mpc to km and then you will end up with the reciprocal of the age of the universe.
It is not a frequency as such, but it is assumed to be the smallest possible value of a physical quantity of dimension T-1.What do you mean by, "it is assumed to be the smallest possible value of a physical quantity of dimension T-1"?
I agree it is not a frequency and certainly multiplying the Hubble constant by by the Planck constant is not going to give you anything useful.
Post by: PaulTalbot on 21/03/2022 15:51:54
I am as well, although I've never suggested to anybody that I'm a scientist.I have a degree in computer sciences from Université Laval and a specialization in computer security from Université de Sherbrooke. This may not be impressive but enough to confirm my scientific background.
Post by: Origin on 21/03/2022 16:21:15
I have a degree in computer sciences from Université Laval and a specialization in computer security from Université de Sherbrooke. This may not be impressive but enough to confirm my scientific background.It would be much more useful to address the questions about your ideas rather than discussing our educational backgrounds.
Post by: PaulTalbot on 21/03/2022 17:37:26
It would be much more useful to address the questions about your ideas rather than discussing our educational backgrounds.Your are 100% right, this is precisely what I expected.
That is not correct, the dimensions of the rate in this case is L/t.From: https://en.wikipedia.org/wiki/Hubble%27s_law (https://en.wikipedia.org/wiki/Hubble%27s_law)
"Hubble constant is most frequently quoted in (km/s)/Mpc, thus giving the speed in km/s of a galaxy 1 megaparsec (3.09×1019 km) away, and its value is about 70 (km/s)/Mpc. However, the SI unit of H0 is simply s−1, and the SI unit for the reciprocal of H0 is simply the second. The reciprocal of H0 is known as the Hubble time. The Hubble constant can also be interpreted as the relative rate of expansion. In this form H0 = 7%/Gyr, meaning that at the current rate of expansion it takes a billion years for an unbound structure to grow by 7%."
Post by: PaulTalbot on 21/03/2022 17:48:19
What do you mean by, "it is assumed to be the smallest possible value of a physical quantity of dimension T-1"?This is an assumption of the preview. If there is a minimum value for the physical quantities action, charge and temperature, then maybe some other physical quantities may be minimally limited too.
Post by: PaulTalbot on 21/03/2022 18:20:17
My choosing different particles was meant to illustrate the meaninglessness of comparing the one force with another, since they're not functions of the same thing. So there's an insanely large difference between the gravitational and EM attraction between a positron and an electron, yet the gravitational force between a pair of billiard balls is probably more than the EM force between them. So does that mean that gravity is stronger than EM? No, it just means the ratio between them is in different units and is thus a meaningless comparison.In both cases, the attractive force is expressed in newtons (of dimension M L T-2). As for the choice of particles, it is the same one that Dirac made in his paper attached to my previous post. I understand that this may seem arbitrary, but once again:
There seems to be something special about the electron-proton couple, as if they were two sides of the same coin. The same thing applies to the positron-antiproton couple. In the chapter 2 of his 1974 paper "Cosmological Models and the Large Numbers Hypothesis", Paul Dirac uses this same ratio:The following link explains the process in details:
"The electric force between the electron and the proton in a hydrogen atom is e2/r2. The gravitational force between them is Gmp me/r2. Their ratio is the dimensionless number e2/Gmpme. Its value is about 2 x 1039."
https://courses.lumenlearning.com/physics/chapter/18-3-coulombs-law/#:~:text=The%20ratio%20of%20the%20magnitude,G%20%3D%202.27%20%C3%97%2010%2039%20. (https://courses.lumenlearning.com/physics/chapter/18-3-coulombs-law/#:~:text=The%20ratio%20of%20the%20magnitude,G%20%3D%202.27%20%C3%97%2010%2039%20.)
Post by: Origin on 21/03/2022 19:40:26
This is an assumption of the preview. If there is a minimum value for the physical quantities action, charge and temperature, then maybe some other physical quantities may be minimally limited too.That may or may not be true. But there is no reason to think this is true :
The Hubble constant H (of dimension T-1) would correspond to the minimum frequency:
fmin = H.
Post by: Origin on 21/03/2022 19:53:55
However, the SI unit of H0 is simply s−1, and the SI unit for the reciprocal of H0 is simply the second.That's nice, but the bottom line here is treating the Hubble constant like a frequency and multiplying it by Planck's constant is meaningless.
Post by: PaulTalbot on 21/03/2022 20:10:45
That's nice, but the bottom line here is treating the Hubble constant like a frequency and multiplying it by Planck's constant is meaningless.Once again, this is an assumption. If H is the minimum value for any physical quantity of dimension T-1, then this should apply to frequency as well. So there should exist a minimum frequency bounded by the same value.
Post by: Origin on 21/03/2022 21:17:35
If H is the minimum value for any physical quantity of dimension T-1That is a mighty big if. This all still just looks numerology with no basis in anything real.
Post by: PaulTalbot on 22/03/2022 18:14:18
That is a mighty big if. This all still just looks numerology with no basis in anything real.I would like to deal with the numerology question first and then with the "big if" in a next post.
This kind of research looks like solving a puzzle, and yes, some numbers are involved in the process. It may thus be associated to a special kind of numerology, but there are strict rules to comply with:
1 - The assumed relations must comply with dimensional analysis, that is both sides of the equations must have the same dimension. Many questions have been posted on the subject, and it is OK. If you believe that some of the assumed relations do not comply with this rule, please let me know.
2 - The assumed relations should not conflict with already known and validated relations like E=mc2. The use of such equations is highly recommended.
3 - The use of dimensionless numbers like π, α (the fine structure constant) or β (the electrostatic to gravitational ratio) is allowed. The same rule applies to simple coefficients or exponents like 2 or ½. The use of complicated numbers like 127y59α must be justified or ideally avoided. I haven't read about this last rule, but it seems obvious to me.
4 - The assumed relations must comply with observation and ideally could allow some predictions that can be tested. This is part of the scientific process.
Do you agree with those rules?
Post by: Origin on 22/03/2022 19:47:26
Do you agree with those rules?Not really, the important point is to show that the equations actually correspond with a physical phenomena.
Post by: PaulTalbot on 22/03/2022 22:05:23
Not really, the important point is to show that the equations actually correspond with a physical phenomena.And where does that rule come from?
Post by: Origin on 22/03/2022 23:54:46
Common sense? If the equations don't correspond to something physical then it is just math and not physics.Not really, the important point is to show that the equations actually correspond with a physical phenomena.And where does that rule come from?
Post by: PaulTalbot on 23/03/2022 14:24:58
If the equations don't correspond to something physical then it is just math and not physics.I agree with this last formulation "something physical" but not necessarily "a phenomenon".
When we write E=mc2, we assume that mass and energy are related that way, whatever the phenomenon.
The "physical things" are the physical quantities "mass" and "energy". If this relation is right, then all observed physical processes should comply with this relation.
Post by: Bored chemist on 24/03/2022 12:11:42
The assumed relations must comply with dimensional analysis, that is both sides of the equations must have the same dimension. Many questions have been posted on the subject, and it is OK. If you believe that some of the assumed relations do not comply with this rule, please let me know."Could the Hubble constant be related to the proton radius?"
One has units of 1/time and the otehr has units of distance.
Also, one is observably constant and the other changes.
Post by: PaulTalbot on 24/03/2022 21:52:06
Could the Hubble constant be related to the proton radius?"Good question.
One has units of 1/time and the otehr has units of distance.
The assumed relation between H (the Hubble constant) and rp (the proton radius) is
rp = 2 αc / βH, or
H = 2 αc / β rp, which is the same.
α is the fine structure constant (e2 / 2 ε0 h c), and
β, the ratio of the electrostatic to gravitational force previously described and calculated.
Since you are a chemist, you may already know the physical meaning of the value αc. It is the speed of the electron in the ground state of the hydrogen atom. From https://medium.com/starts-with-a-bang/ask-ethan-what-is-the-fine-structure-constant-and-why-does-it-matter-178c1a68471b (https://medium.com/starts-with-a-bang/ask-ethan-what-is-the-fine-structure-constant-and-why-does-it-matter-178c1a68471b)
"If you modeled a hydrogen atom as Bohr did, but took the ratio of a ground-state electron’s velocity and compared it to the speed of light, you’d get a very specific value, which Sommerfeld called α: the fine structure constant."
As for the value of βH, it is the assumed frequency associated with the electron-proton couple (fep) as described in the text of the preview.
So the physical meaning of this assumed relation can be stated:
The proton radius equals twice the speed of the electron in the gound state of the hydrogen atom, divided by the frequncy associated with the electron-proton couple.
As there was a typographic glitch in the previously attached chart of the assumed relations, I attach the corrected version to this post. As you can see "Dimensional Numerology" (this is a term I just invented after the comments in this thread) allows linking physical quantities in new ways.
Only observation can tell us if those relations are right, or if they are the result of coincidences.
Post by: Origin on 25/03/2022 13:52:17
The assumed relation between H (the Hubble constant) and rp (the proton radius) isI tried using your equation above and I did not get the radius of a proton. Could you work out the above equation so I can see where we disagree?
rp = 2 αc / βH
Post by: PaulTalbot on 25/03/2022 14:17:09
I tried using your equation above and I did not get the radius of a proton. Could you work out the above equation so I can see where we disagree?The value of α is 7.29735 × 10-3
That of β is 2.26866 × 1039
and that of H is 2.3338 × 10-18 s-1, which is the SI standard notation for the calculated value.
The result of 2 αc / βH is thus 8.264 × 10-16 m or 8.264 fm, which matches (within 1σ) the proton radius according to recent experiments and calculations [14] [15] [16] [17] (in the preview).
Post by: Origin on 25/03/2022 14:44:15
The result of 2 αc / βH is thus 8.264 × 10-16 m or 8.264 fm, which matchesThere's the issue, the sources I saw all said the radius is about 0.85 fm or an order of magnitude lower than your number. Another question, why are the constants in the numerator multiplied by 2 in the equation?
Post by: PaulTalbot on 25/03/2022 15:45:53
There's the issue, the sources I saw all said the radius is about 0.85 fm or an order of magnitude lower than your number. Another question, why are the constants in the numerator multiplied by 2 in the equation?For the precision of the recent observed values of rp, please refer to the references provided int the text (attached). As for the factor 2 in the equation, this results from the way α is calculated. It is the result of incorporating α in
e2 / ε0 Eep ≈ 0.826 4 fm, (14), which does not include this factor 2.
According to (2) and (5), Eep can be replaced by β h H. Using (9), and the fine-structure constant (α = e2 / 2 ε0 h c), equation (14) of the proton radius then becomes
rp = 2 αc / βH = 2 αc / fep, (15)
Post by: PaulTalbot on 25/03/2022 15:52:04
8.264 × 10-16 m is 0.8264 fm.
Post by: Bored chemist on 25/03/2022 17:23:37
the speed of the electron in the ground state of the hydrogen atomLOL.
Post by: Origin on 25/03/2022 23:02:19
β, the ratio of the electrostatic to gravitational force previously described and calculated.Just to be clear β the ratio of the electrostatic to gravitational force between an electron and a proton. To get a more general relationship you would need to have the masses equal, as has been mentioned before.
I am still not clear on what the relationship is between the fine structure constant, the speed of light, the Hubble constant and β. Why would these constants have anything to do with the diameter of the proton?
As for the factor 2 in the equation, this results from the way α is calculated.Could you explain that a bit more, I still do not understand why the constants in the numerator were multiplied by 2.
Post by: PaulTalbot on 26/03/2022 15:21:16
Could you explain that a bit more, I still do not understand why the constants in the numerator were multiplied by 2.I would be cautious on that question, since I am way out of my comfort zone.
We can get rid of the factor of two by writing:
rp = αc / ½ fep.
This is not unusual as we can see in the calculation of the kinetic energy : Ek = ½ mv2.
As for the physical meaning of that factor ½ preceding the frequency, I think it may be related to the spin of the charged particles (½).
It is as if it took two cycles for the electron-proton couple to fully return to its initial state.
Post by: PaulTalbot on 26/03/2022 18:35:33
Just to be clear β the ratio of the electrostatic to gravitational force between an electron and a proton.Yes, there seems to be something special about the electron-proton couple, as if they were two sides of the same coin.
Post by: PaulTalbot on 26/03/2022 18:37:40
To get a more general relationship you would need to have the masses equal, as has been mentioned before.And why so?
Post by: PaulTalbot on 26/03/2022 19:32:37
I am still not clear on what the relationship is between the fine structure constant, the speed of light, the Hubble constant and β. Why would these constants have anything to do with the diameter of the proton?The proton radius I refer to is the rms charge radius of the proton, where rms stands for "root mean square".
https://en.wikipedia.org/wiki/Charge_radius (https://en.wikipedia.org/wiki/Charge_radius)
"The problem of defining a radius for the atomic nucleus is similar to that of defining a radius for the entire atom; neither atoms nor their nuclei have definite boundaries. However, the nucleus can be modeled as a sphere of positive charge for the interpretation of electron scattering experiments: because there is no definite boundary to the nucleus, the electrons "see" a range of cross-sections, for which a mean can be taken. The qualification of "rms" (for "root mean square") arises because it is the nuclear cross-section, proportional to the square of the radius, which is determining for electron scattering."
The value rp thus refers to the radius of the cross-section of the electron-proton interaction, which has the dimension of an area.
Now, this is precisely what the value fep refers to: the frequency of the electron-proton interaction, that is the exchange of virtual photons between those particles. Of course, this reasoning is somewhat speculative, yet it sounds logical to me.
Post by: Bored chemist on 27/03/2022 13:24:59
Quote from: Bored chemist on 19/03/2022 20:47:41
The fact that it fails on dimensional analysis is beside the point unless he can explain why something constant is the same as something that's changing.
Post by: PaulTalbot on 27/03/2022 14:51:51
The fact that it fails on dimensional analysis is beside the point unless he can explain why something constant is the same as something that's changing.OK, I will try to answer your question in a different way.
First, have a look at the attached modified chart on the assumed relations between physical constants.
There is no more reference to the Hubble constant in this chart. All values are considered constant, so your question becomes irrelevant.
The proton radius is still calculated in the same way and with the same accuracy.
The Wesson's quantum of mass is still calculated in the same way and with the same accuracy.
According to known physics, this should be the result of two unlikely and unrelated coincidences.
Those relations assume the existence of a minimum frequency (fmin).
The big question then arises: How on earth could this value of fmin be the same as that of the Hubble constant?
The answer to that question is simple. The assumption that the Hubble constant is decreasing derives from the ΛCDM conformal model:
https://www.forbes.com/sites/startswithabang/2021/01/05/how-is-the-universe-accelerating-if-the-expansion-rate-is-dropping/?sh=19d6ab984093 (https://www.forbes.com/sites/startswithabang/2021/01/05/how-is-the-universe-accelerating-if-the-expansion-rate-is-dropping/?sh=19d6ab984093)
The fact is that the ΛCDM model is flawed and doomed. It fails to predict a value of H consistent with observation. This could be the subject of another thread.
Post by: Origin on 27/03/2022 18:31:21
How can it be out of your comfort zone if it is something you put in your own equation? It sounds like you are saying you combined a bunch of constants and then put a factor of 2 in the equation to get the answer you want. Assuming that is what you did, that is precisely why I am calling your technique numerology. I can take random constants and combine them in ways to get about anything. The problem is this results of this technique are meaningless.Could you explain that a bit more, I still do not understand why the constants in the numerator were multiplied by 2.I would be cautious on that question, since I am way out of my comfort zone.
This is not unusual as we can see in the calculation of the kinetic energy : Ek = ½ mv2.First of all spin, charge and frequency have nothing to do with Kinetic energy formula. Of course there are numbers in some the equations for physical properties, but they were not arbitrarily thrown in! The '1/2' term is a direct result of the derivation of KE. Energy or work is a force applied over a distance. So integrating F dr will give you the formula for KE.
As for the physical meaning of that factor ½ preceding the frequency, I think it may be related to the spin of the charged particles (½).
What you have done is combined random constants to get a predetermined result. This is going to give a meaningless equation. I did not read much of your paper, but right off the bat you incorrectly identified the Hubble constant as a frequency and the led you down a path that yielded more meaningless results.
If you are interested in physics you need to learn some of the basics before you try to come up with new concepts.
Post by: Bored chemist on 27/03/2022 18:39:49
There is no more reference to the Hubble constant in this chart. All values are considered constant, so your question becomes irrelevant.Then you should ask the Mods to close the thread.
The question has been answered.
"Could the Hubble constant be related to the proton radius?"
No, that was a silly idea.
Post by: Origin on 27/03/2022 18:51:48
The big question then arises: How on earth could this value of fmin be the same as that of the Hubble constant?Are you serious? The reason they are the same is because you arbitrarily defined the minimum frequency as the Hubble constant!
Post by: Origin on 27/03/2022 19:16:29
Obviously if you wanted to compare the forces since the charges are 1 to 1 you would want the masses to be 1 to 1 for a valid comparison. Doesn't that make sense to you?To get a more general relationship you would need to have the masses equal, as has been mentioned before.And why so?
Post by: Bored chemist on 27/03/2022 19:59:17
It's still arbitrary.Obviously if you wanted to compare the forces since the charges are 1 to 1 you would want the masses to be 1 to 1 for a valid comparison. Doesn't that make sense to you?To get a more general relationship you would need to have the masses equal, as has been mentioned before.And why so?
The ratio of the gravitational to electrostatic forces is not the same for a couple of protons as it is for a couple of electrons (By nearly 4 million fold).
So all you can really say is that the electrostatic force is enormously much bigger.
(or, of course, zero if you use uncharged particles).
Since the ratio is a moveable feast, it can't have any numerical significance.
So the OP is still doing numerology.
Post by: Origin on 27/03/2022 23:31:02
The ratio of the gravitational to electrostatic forces is not the same for a couple of protons as it is for a couple of electrons (By nearly 4 million fold).Dooh! You are right of course.
Post by: PaulTalbot on 28/03/2022 15:08:55
How can it be out of your comfort zone if it is something you put in your own equation? It sounds like you are saying you combined a bunch of constants and then put a factor of 2 in the equation to get the answer you want. Assuming that is what you did, that is precisely why I am calling your technique numerology.My comment was about the physical meaning of the factor of 2 (or ½), not where it comes from.
As mentioned in a previous post and in the text of the preview, this factor of 2 results from the inclusion of the dimensionless constant α in the equation:
rp = e2 / ε0 Eep
rp = e2 / ε0 (β H h)
rp = (e2 / ε0 h) / βH
since α = e2 / 2 ε0 hc, we get
rp = 2 αc / βH
This is pure algebra, not numerology.
Post by: Origin on 28/03/2022 18:34:52
Quote from: PaulTalbot link=topic=84397.msg673608#msg673608 date=1648476535
rp = e2 / ε0 Eep
rp = e2 / ε0 (β H h)
This is pure algebra, not numerology.
In the above you are saying:
Where
This is pure numerology. The Hubble constant is not a frequency. This has been pointed out to you several times. Even if it was a frequency it would still be meaningless. If I swing a ball on a string around at a frequency of 5 cycles per second, do you think multiplying that by 'h' will give me anything meaningful?
You have presented an idea and it has been shown to be wrong, not accepting this means you are not doing science you are doing pseudoscience and are entering the land of crankdom. That's too bad, but it happens all the time.
Post by: PaulTalbot on 28/03/2022 21:43:10
As for your last post, sorry Origin, but all you have proven so far is that you are close-minded and that you don't understand the arguments provided by Edington, Dirac, Wesson, Valev and others.
Post by: Bored chemist on 28/03/2022 22:36:26
The Hubble constant is not a frequency.It's not a constant either.
If I swing a ball on a string around at a frequency of 5 cycles per second, do you think multiplying that by 'h' will give me anything meaningful?Well... if it's the resonant frequency of the pendulum then multiplying 5 by h/2 gives you the zero point energy of the system.
But the universe isn't an oscillation; so the point's moot.
The Hubble constant currently has a value near 1/(14 billion years).
But tomorrow it will be slightly different because the universe isn't going to turn round and go back.
If I live for 100 years than the reciprocal of my lifetime will be once per century.
But it's not meaningful to say that I live once per century because that would imply that I live 10 times per millennium and I didn't and won't.
I will live once in the whole of infinite time, so my frequency is 1 in essentially infinity.
If you want to put a number on that it should be zero.
You can do the same thing with the universe.
It's either a one off, or infinitely rare. It has no "frequency" to multiply by h.
And that's the root of the OP's problem.