The mass of the proton times its charge radius is very close to $4\hbar/c$. Is this a coincidence?

Question

The claim is $$r_p m_p = 4 L_0 M_0 = 4\hbar/c,$$ where $r_p$ and $m_p$ are the proton's charge radius and mass, and $L_0$ and $M_0$ are the Planck length and mass.

Using the muon measurement $r_p=0.84087 \times 10^{-15}$ m, $m_p=1.6726219 \times 10^{-27}$ kg, $r_p m_p = 1.40646 \times 10^{−42}$ kg m.

$4\hbar/c = 1.40706915 \times 10^{-42}$ kg m, agreement to $0.04\%$. That seems pretty incredible.

Another way to put it is as a "prediction" of $r_p = \dfrac{4\hbar}{m_p c} = .84124$ fm.

With the error bars, the muon measurement is $.84087 \pm .00039$ fm, so call it min $.84048$ fm, max $.84126$ fm. The "prediction" is very close to the max.

The CODATA value for the proton's charge radius (electron measurement) is $.8751 \pm .0061 $ fm, close but not so incredibly close as the muon measurement.

Coincidence?

Answer 1

In one sense, this should not be a coincidence, because the mass and the charge radius are actually determined by more fundamental quantities (quark masses, strong force coupling constant). So given that those quantities have their particular values, in theory it inexorably determines what "mass times charge radius" is going to be.

What should be a coincidence is the appearance of the number 4, when "mass times charge radius" is expressed in natural units. However, since charge radius depends on the distribution of charge inside the proton, it's not inconceivable that, e.g., the distribution of "partons" (quarks and gluons), the distribution of momentum among them, etc., follows some principle which really does imply that the answer is "4 + small corrections". At least, I can't think of a simple reason this should not be so.

For me the deepest fact is that fundamental theory should predict something for this quantity. That prediction might be 4, or it might be e^1.37, but it must be something close to 4.

Update: A comment in chat by @Rococo allows @dandb's observation to be expressed in a very crisp way:

The charge radius of the proton (in muonic hydrogen) is almost exactly four times the reduced Compton wavelength of the proton.

Update 2017: Via P.R. Silva (eqn 6), I have run across a heuristic model of the nucleon in which M = 4/R (in natural units). Here R is the radius of the bag in the "bag model". See Xiangdong Ji, "Mass of the hadron", slide 20. I have not found where this argument originates, but a remark in a 1994 paper by Ji (see paragraph beginning "In the chiral limit...", on the final page) hints at it.

Answer 2

Another attractive feature of this conjecture is that it is similar to another conjecture related to hadrons that is known to be true: that the spin of a hadron is equal to the sum of the spins of the quarks in the hadron (which come in discrete half integer increments), even though non-quark partons in the hadron have non-zero spins that "magically" cancel out in the total for reasons that are not well understood (i.e. the "proton spin crisis")). Until we understand why this is the case for spin in hadrons, we can't rule out that this conjecture is exactly true for related reasons.

I'd also note that the in lots of other areas of the Standard Model (e.g. some of the more obscure relationships between electroweak constants in the Standard Model), there are lots of known exact relationships between Standard Model, so it wouldn't be a priori unreasonable to wonder if there was such a relationship here, particularly given that charge radius is an electroweak phenomenon. I don't think that there is any known Standard Model constant relationship that can explain why this conjecture should be true, but the precision of the the conjectured relationship is sufficiently great that it isn't unreasonable to entertain the possibility that it is exactly or exactly subject to small corrections, true.

Answer 3

No, that’s probably not a coincidence. The theoretical model that may explain your discovery that $M_{p} R_{p} = \frac{4 h}{2 \pi c}$, where $M_{p}$ and $R_{p}$ are the proton’s mass and charge radius, is that the proton may be composed of four photons each of mass equal to one quarter of the proton’s mass. That’s where the $4$ in your equation may come from. The wavelength of a photon of mass equal to one quarter of the proton’s mass can be calculated from $L = \frac{h}{m c}$, and is $5.28563 ~ fm$. The circumference of the proton, using its 2018 CODATA radius value, is: $2 \pi \left( 0.8414 ~ fm \right) = 5.28667 ~ fm$. So, the wavelength of a quarter proton mass photon, and the proton’s circumference calculated from its 2018 CODATA radius value are almost identical. They differ by only $0.02 \%$, or $1$ part in $5000$. Could the proton be composed of four quarter proton mass photons? To decide, one needs to calculate the volume of a photon of mass equal to a quarter of the proton’s mass and compare that volume to the proton’s volume. Will four such photons fit inside the proton?

Using Maxwell’s model of the shape of a light wave - a sine curve from $0$ to $2 \pi$, with amplitude $r$ - the volume of its enclosing cylinder is its wavelength times the area of its base circle of radius $r$: $V = \left( 2 \pi r \right) \left( \pi r^{2} \right) = 2 \pi^{2} r^{3}$. That volume is almost $5$ times larger than the 3D volume of the proton, so the four photons cannot fit in the proton’s 3D interior, but it is exactly equal to the surface volume of a 4-sphere of radius equal to the proton’s radius. (The surface volume of a 4-sphere is: $S = 2 \pi^{2} r^{3}$, which is exactly the same formula as the formula for the volume of the photon’s enclosing cylinder.) And, since photons are bosons, all four can occupy the same space at the same time. So, it’s possible the proton is composed of four quarter proton mass photons, all occupying the surface volume of a 4-sphere of radius equal to the proton’s radius, in the 4D space surrounding the 3D location of the proton. If such is the case, if $100 \%$ of the mass of the proton is found in the 4D space surrounding the proton, that means the interior 3D volume of the proton is devoid of matter. The proton may be hollow with respect to our 3D space, according to this model. It could be that the surface of the proton is not the surface of a 3D object, as is currently thought, but the location of the intersection of the proton’s mass (which may be in the surface of a 4-sphere of radius equal to the proton’s radius) with our 3D space (aka, the Higgs field). If true, this invalidates the quark theory model of hadron structure, which posits that the 3D interior of the proton is where point particle quarks and the strong force are located. So, no. I don’t believe the relationship you found is a coincidence at all. It may be a consequence of the proton having a structure different from what it is currently believed to have.