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Is this (Verlinde-style) lower bound for the fine structure constant correct?

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Two elementary electric charges  \(e\) interact. From the Coulomb force \(F(r)= e^2/ 4 \pi \epsilon_0 r^2\) between them, which decreases as \(1/r^2\), the fine structure constant is defined as:

\(\alpha= \frac{F(r) \ r^2 }{\hbar c}\). This gives \(\alpha = \frac{e^2}{ 4 \pi \epsilon_0 \hbar c}\) , which is Sommerfeld's formula in SI units. The value of \(\alpha\) is measured to be about 1/137.04 and describes the strength of the Coulomb force, or, equivalently, the strength of electromagnetism.

Here, \(c\) is the speed of light, and \(\hbar\) is Planck's constant, and \(r\) is the distance between the charges. Now, a researcher proposes the following argument in two steps:

1. The Coulomb force \(F(r) \) between two elementary charges, i.e. the force due to exchanging *virtual* photons, is surely larger than than the force between two *neutral* (microscopic) *black holes* that appears when they exchange *thermal* photons.

Both forces are electromagnetic in origin, and we neglect gravity effects completely, by assuming extremely small bodies throughout, in the microscopic regime, with negligible masses and/or negligible gravitational interaction. We also assume that the charges are of the same sign, and that the forces are thus repulsive. In particular, the thermal photons from one hot particle push against the other particle and thus induce a force.

2. This last force can be calculated. It is given as

\(F_{BHTP}= \frac{P_{BH}}{c} \frac{A_{BH}/ 2}{ 4 \pi \ r^2 }\)

where \(P_{BH}\) is the power of the black hole Bekenstein-Hawking radiation, \(A_{BH}\) the surface of the black hole, and \(r\) the distance between the two black holes (which is assumed much larger than the black holes, so that space can be approximated as flat).

If we insert the formulae for \(P_{BH}\) and \(A_{BH}\) from the wikipedia (article on Hawking radiation), we get

\(\alpha > \frac{F_{BHTP} \ r^2}{ \hbar c} = \frac{1}{ 7680 \pi} = \frac{1}{ 24127,...}\)

Since alpha needs to be smaller than 1 for many reasons (a simple one: probabilities are always smaller than one), we have deduced:

\(1 > \alpha > \frac{1}{ 24127,...}\)

which is not a great result, but still a result that is better than nothing.

Is this argument correct? Or is there a mistake? In particular, is it correct to assume that

F_(Coulomb) > F _ (neutral black holes)

in the case of negligible gravitation? The algebra of the calculations is correct, but is the argument correct? I have never seen a limit of alpha, ever, and this one, even though it is a rough one, is the first I have ever seen deduced from physical formulae. The issue is really whether the force comparison is valid also for microscopic hot bodies and microscopic black holes. What do you think? The force sequence indirectly assumes that microscopic particles might be "hot"- so it is questionable. But still intriguing.

All this is not my own idea, nor of a friend of mine: I found it via the wikipedia entry on the fine structure constant in this paper by John P. Lestone:  https://arxiv.org/abs/physics/0703151 I simplified the argument in the paper, took away all misleading issues, and corrected the mistakes (there are two factors of 4 pi lost in the paper). The idea is already ten years old but I only found it a few days ago.

---

A few comments:

A. The expression for \(F_{BHTP}\) might not be an equality, because the effective surface  might be different and not exactly given by \(A_{BH}/2\).

B. This reasoning does for electrodynamics the same that Verlinde did for gravity: both assume that a macroscopic force is in fact due to many microscopic degrees of freedom.

C. Also the strength of gravity is conventionally defined using the above expression \(\alpha_G= \frac{F(r) \ r^2 }{\hbar c}\)

Closed as per community consensus as the post is not graduate-level
asked Jul 9 in Closed Questions by Zorro [ revision history ]
recategorized Jul 15 by dimension10
Most voted comments show all comments

$\alpha$ was defined a hundred years ago with a simple formula, which was just an unambiguous notation. Your definitions follow from nowhere and define different things. In particular, the Coulomb force does not contain $\hbar$ nor $c$. What you are doing is not Physics and thus it is worse than nothing.

Vladimir, the usual definition of alpha is the one given in the first formula! Just insert the Coulomb force between two charges, and you get the usual formula for alpha! See the Wikipedia if you do not believe it. Maybe you were thinking about something else?

@anonymous: You are wrong. The usual definition of $\alpha$ was given by Arnold Sommerfeld a hundred years ago. The first formula in OP is wrong since it depends on the force $F$ sign. The worst thing about this formula is that it follows from nowhere. There is no physical calculation where this formula appears naturally. Doing Physics "by analogy" with this formula is misleading. I vote to close this question.

Vladimir,

from the Coulomb law F=e^2 / (4 pi e_0 r^2) and alpha = F r^2 / (hbar c)  you get

alpha = e^2/ (4 pi e_0 hbar c)

which is the usual relation. You can use |F| if you prefer. But that part of the argument is correct.

But maybe there is another error of reasoning, in the sentences after that?

According to Wikipedia, $\alpha$ is "an interaction constant" between a charge and a photon, not between two charges.

Most recent comments show all comments

I have edited the question to take all the comments into account - I hope.

Hi Zorro,

while it is ok to contribute unregistered and anonymously to PhysicsOverflow, the fact that in this thread all of the different anonymous contributions using different names are coming from your IP looks very strange to say the least...

Please do not use different anonymous names in the same thread to falsly mimic a discussion among a group of people.

As some people voiced the opinion that the thread should be closed, I am posting a closevote here just to see what others think.

4 Answers

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Zorro, the inequality between the forces, F_(coulomb) > F _ (neutral black holes via thermal photons), is indeed valid and correct for macroscopic bodies that produce thermal radiation.

Of course, in contrast, a microscopic particle, like an electron, does not produce or emit any thermal radiation at all. But the inequality between the forces does not assume that the elementary charges are microscopic particles; they could be macroscopic bodies loaded with a single elementary charge.

Given this, the limit on alpha deduced in this way seems valid. What a pretty result!

The paper by Lestone that you quote is highly questionable in almost everything it writes about, and you did an admirable job in extracting the only interesting idea inside it. I would have put it aside completely...

answered Jul 14 by Klaus [ revision history ]
edited Jul 14
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I disagree. The calculation does not imply that alpha is larger than 1/7680 pi. There is no argument given for why the Coulomb force should be larger than the radiation pressure from the black hole radiation. You added the greater sign yourself, because you expect it; but you have no reason at all to add it in the first place!

answered Jul 14 by John [ no revision ]

Two charged black hole should interact more strongly than two neutral ones. In the OP's wording: F_Coulomb + F_BHTP > F_BHTP. This says nothing about the relation between F_Coulomb and F_BHTP.

Why should the Coulomb force be larger than the black hole radiation pressure? The effects are so different, and I cannot find a reason why this should be true.

But if an argument can be found, then the limit given by the OP is valid.

John, the charged body need not be a black hole! Your expression F_Coulomb + F_BHTP > F_BHTP does not apply.

Maybe a better argument can be found. Hawking radiation is due to virtual particles near the horizon. Also the Coulomb force is due to virtual particles. Is there a way to show that elementary charges produce more virtual particles than a horizon?

Zorro, virtual particles are always absorbed, the real particles are always scattered.

Compare the Coulomb force with the gravitational force, and you will obtain estimation of which out of them is stronger. $\alpha$ has nothing to do with it.

+ 0 like - 0 dislike

The striking expression alpha > 1 / 7680 pi  indeed follows from the behavior of virtual photons. It should be much more advertized! Probably the expression for F_BHPT should have A/4 (cross section) instead of A/2 (half surface) - but these are minor details.

answered Jul 15 by Lenni [ no revision ]
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What's going on? Where are the moderators? @Dilaton, @dimension10?

answered Jul 15 by Andrey Feldman (600 points) [ no revision ]

What is your opinion on this - where is the mistake?

Zerro, you do not understand the originally established physical meaning of $\alpha=e^2/\hbar c$.

NIST explains the issue nicely on https://physics.nist.gov/cuu/Constants/alpha.html  ; :

The fine-structure constant α is of dimension 1 (i.e., it is simply a number) and very nearly equal to 1/137. It is the "coupling constant" or measure of the strength of the electromagnetic force that governs how electrically charged elementary particles (e.g., electron, muon) and light (photons) interact.





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