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If we calculate the electrostatic potential energy of two protons in a nucleus, it's approximately

$$\frac{e^2}{4 \pi\epsilon_0 (2r)} = 1.15 \times 10^{-13}J$$

where $r$ is the radius of the proton, about $10^{-15}$m, (also the range of the strong nuclear force).

The rest energy of the electron (or positron) is $$m_ec^2 = 8.2 \times 10^{-14}J$$

The same as each other to within a factor $1.4$

Presumably this similarity is known - and explained, perhaps by Quantum Field Theory, but how? Can anyone explain it, mainly in words, without too many complicated calculations?

Is it that the mass of the electrostatic potential energy can be 'stored' as virtual particles (electrons and positrons) when the proximity is small enough and these virtual particles in turn allow the nuclear force to exist - so explaining why it's got the range of about $10^{-15}$m?

Or is there a different reason?

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    $\begingroup$ Coincidences don't have reasons: isn't this how you define them? $\endgroup$
    – Cosmas Zachos
    Commented Oct 12, 2021 at 19:52
  • $\begingroup$ @Cosmas Zachos, it's just a way of saying that it's too suspicious to be a real coincidence, there's likely to be a reason...just wondered if someone could explain it well $\endgroup$
    – John Hunter
    Commented Oct 12, 2021 at 19:54
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    $\begingroup$ No, there is no reason within our present conceptual framework. But it's virtually impossible to prove "there is nothing there". $\endgroup$
    – Cosmas Zachos
    Commented Oct 12, 2021 at 20:04
  • $\begingroup$ @JohnHunter Two things being close doesn't mean there's a relationship. For example, the polar function $r=\frac{e}{2}^\theta$ is very close the Golden Spiral, $r=\phi^\frac {2\theta}{\pi}$, and both use important irrational constants. But this doesn't mean the two constants are closely related in some way. $\endgroup$
    – zucculent
    Commented Oct 12, 2021 at 20:10
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    $\begingroup$ There's no reason for a relation between proton radius and electron mass. $\endgroup$
    – my2cts
    Commented Oct 12, 2021 at 21:40

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Like many of these coincidences, this appears more dramatic than it really is, because it involves aggressively unsuitable units, in this case SI, and thus huge numbers conjuring up specious surprise.

In suitable particle physics natural units, your equation just presents as $$ \alpha~ m_\pi \sim 2 m_e , $$ where α~1/137 , $2m_e\sim 1$MeV, and $m_\pi\sim 140$MeV. The Compton wavelength of the latter is the Yukawa range of the nuclear force, r, or the nucleon confinement radius, etc... all related to the Strong interaction. But the mass of the electron is set by the mysterious Yukawa couplings of the Weak interaction.

No sensible connection between these two has been proffered, in our present conceptual framework.

The point, however, is that the suitable unit-independent dimensionless ratios have little surprise value in them. There is hardly anything here.

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  • $\begingroup$ Thankyou for the answer. The units were not chosen to be 'aggressively unsuitable' - in fact the 'apparent coincidence' seems to remain in different units, i.e $\alpha = \frac{m_e}{m_\pi}$ approximately. If the reason for this isn't known, that's fine, or if you believe it's just a coincidence, that's fine too $\endgroup$
    – John Hunter
    Commented Oct 12, 2021 at 21:42
  • $\begingroup$ Well, there are hundreds of such ratios in HEP. You may find ratios of 140 comparing building heights in a city. $\endgroup$
    – Cosmas Zachos
    Commented Oct 12, 2021 at 21:57
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The electrostatic potential energy of two protons next to each other depends on the elementary charge and on the proton radius, which in turn is determined mostly by the complicated strong interactions between the quarks, whose strength is determined by the corresponding coupling constant.

The mass of the electron comes from the Higgs mechanism; numerically, it's related to the coupling between the electron and Higgs fields, as well as the latter's vacuum expectation value.

We don't yet know of any relation between all these numbers, so it's just a coincidence with no particular reason. Such things are not hard to find if you mess around with numbers.

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  • $\begingroup$ Oh, that's a surprise, thought there would be a known reason for this one - the units match and there's much talk of the Large Number Hypothesis, but not much about this. $\endgroup$
    – John Hunter
    Commented Oct 12, 2021 at 20:18
  • $\begingroup$ Thanks for the answer $\endgroup$
    – John Hunter
    Commented Oct 12, 2021 at 20:29

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