The Q = 4 Proton: Proton Radius, Proton–Electron Mass Ratio, and the Phonon Speed Limit - Version 2

One of the most compact algebraic structures in the Super Golden / TOTU framework begins with a quantized circulation condition and ends by connecting the proton length scale, the proton-to-electron mass ratio, the Rydberg constant, the fine-structure constant, the Bohr radius, and the proposed phonon speed limit.

The central relations are

$$\boxed{r_p=\frac{4\hbar}{m_pc}}$$ $$\boxed{\frac{m_p}{m_e} =\frac{\alpha^2}{\pi r_pR_\infty}}$$

and

$$\boxed{ v_u = c\sqrt{\frac{\pi}{2}r_pR_\infty} }.$$

What is especially useful is that these equations can be reduced into a closed family of equivalent algebraic relations. This post develops that chain explicitly and also separates what is assumed, what is derived, and what still requires physical justification.

---

1. Start with quantized circulation

For a superfluid-like phase field,

$$\Psi=\sqrt{\rho}\,e^{i\theta},$$

the velocity is

$$\mathbf v=\frac{\hbar}{m}\nabla\theta.$$

Single-valued phase closure requires

$$\oint\nabla\theta\cdot d\mathbf l=2\pi Q,$$

where $Q$ is an integer winding number.

Therefore

$$\oint\mathbf v\cdot d\mathbf l = Q\frac{h}{m}.$$

For a circular vortex of radius $r$,

$$2\pi rv=Q\frac{h}{m}.$$

Since $h=2\pi\hbar$,

$$\boxed{mvr=Q\hbar}.$$

For the proton, take

$$m=m_p, \qquad v=c.$$

Then

$$\boxed{ r(Q)=Q\frac{\hbar}{m_pc}. }$$

If the proton occupies the $Q=4$ circulation sector,

$$\boxed{ r_p=\frac{4\hbar}{m_pc}. }$$

Using the CODATA proton reduced Compton wavelength,

$$\bar\lambda_p = \frac{\hbar}{m_pc} = 2.10308910051\times10^{-16}\ {\rm m},$$

gives

$$\boxed{ r_p = 4\bar\lambda_p = 8.41235640204\times10^{-16}\ {\rm m} }$$

or

$$\boxed{ r_p=0.8412356402\ {\rm fm}. }$$

The numerical constants used here are consistent with the current published 2022 CODATA set maintained by NIST.

---

2. Equivalent proton-radius relations

Because

$$\bar\lambda_p=\frac{\hbar}{m_pc},$$

the $Q=4$ proton immediately satisfies

$$\boxed{r_p=4\bar\lambda_p}.$$

Using the ordinary proton Compton wavelength

$$\lambda_{C,p} = \frac{h}{m_pc} = 2\pi\bar\lambda_p,$$

we also obtain

$$\boxed{ r_p=\frac{2}{\pi}\lambda_{C,p}. }$$

The proton circumference therefore has the particularly simple form

$$2\pi r_p = 8\pi\bar\lambda_p = 4\lambda_{C,p},$$

so

$$\boxed{ 2\pi r_p=4\lambda_{C,p}. }$$

The circulation relation itself becomes

$$\boxed{ m_pc\,r_p=4\hbar. }$$

Multiplying by $c$,

$$\boxed{ m_pc^2r_p=4\hbar c. }$$

This suggests a natural proton energy scale

$$E_0\equiv\frac{\hbar c}{r_p}.$$

Using $r_p=4\hbar/(m_pc)$,

$$\boxed{ E_0=\frac{m_pc^2}{4}. }$$

Numerically,

$$E_0\approx234.568\ {\rm MeV}.$$

Thus another equivalent form is

$$\boxed{ r_pE_0=\hbar c. }$$

The associated angular-frequency scale is

$$\omega_0=\frac{c}{r_p} =\frac{m_pc^2}{4\hbar}.$$

---

3. Introduce the corresponding electron $Q=4$ length

Apply the same $Q=4$ Compton scaling to the electron:

$$r_e^* = \frac{4\hbar}{m_ec}.$$

I use $r_e^*$ deliberately. This is a derived $Q=4$ electron length scale, not a claim that the electron has a measured classical geometric radius.

Thus

$$\boxed{ r_e^*=4\bar\lambda_e. }$$

Because

$$m_pr_p=4\frac{\hbar}{c}$$

and

$$m_er_e^*=4\frac{\hbar}{c},$$

we have

$$\boxed{ m_pr_p=m_er_e^*. }$$

Therefore

$$\frac{m_p}{m_e} = \frac{r_e^*}{r_p}.$$

Define

$$\mu\equiv\frac{m_p}{m_e}.$$

Then

$$\boxed{ \mu=\frac{r_e^*}{r_p}. }$$

This is the mass-radius inverse scaling of the $Q=4$ construction.

---

4. Bring in the Bohr radius

The Bohr radius is

$$a_0 = \frac{\hbar}{m_ec\alpha}.$$

Therefore

$$\frac{\hbar}{m_ec} = \alpha a_0.$$

So the electron $Q=4$ length becomes

$$r_e^* = 4\frac{\hbar}{m_ec} = 4\alpha a_0.$$

Hence

$$\boxed{ r_e^*=4\alpha a_0. }$$

Using the CODATA Bohr radius,

$$a_0 = 5.29177210544\times10^{-11}\ {\rm m},$$

gives approximately

$$\boxed{ r_e^* = 1.5446371\times10^{-12}\ {\rm m}. }$$

The proton-to-electron mass ratio can therefore already be written as

$$\boxed{ \mu = \frac{4\alpha a_0}{r_p}. }$$

---

5. Connect the Bohr radius to the Rydberg constant

For the infinite-mass Rydberg constant,

$$R_\infty = \frac{\alpha^2m_ec}{4\pi\hbar}.$$

Rearranging,

$$\frac{\hbar}{m_ec} = \frac{\alpha^2}{4\pi R_\infty}.$$

But

$$\frac{\hbar}{m_ec} = \alpha a_0,$$

so

$$\boxed{ a_0=\frac{\alpha}{4\pi R_\infty}. }$$

Now substitute this into

$$r_e^*=4\alpha a_0.$$

Then

$$r_e^* = 4\alpha \left( \frac{\alpha}{4\pi R_\infty} \right),$$

which reduces to

$$\boxed{ r_e^* = \frac{\alpha^2}{\pi R_\infty}. }$$

This is the key bridge between the electron scale and the Rydberg constant.

---

6. Derive the proton-to-electron mass ratio

Since

$$\mu=\frac{r_e^*}{r_p},$$

and

$$r_e^* = \frac{\alpha^2}{\pi R_\infty},$$

we obtain

$$\boxed{ \mu = \frac{\alpha^2} {\pi r_pR_\infty}. }$$

Therefore

$$\boxed{ \frac{m_p}{m_e} = \frac{\alpha^2} {\pi r_pR_\infty}. }$$

Using

$$\alpha = 7.2973525643\times10^{-3},$$ $$R_\infty = 10\,973\,731.568157\ {\rm m^{-1}},$$

and

$$r_p=0.8412356402\ {\rm fm},$$

gives

$$\boxed{ \frac{m_p}{m_e} \approx1836.1526734, }$$

matching the CODATA proton-electron mass ratio

$$1836.152673426(32).$$

The numerical values of $\alpha$, $R_\infty$, $a_0$, and $m_p/m_e$ are from CODATA 2022.

---

7. The closed algebraic identity

The mass-ratio equation can be written especially compactly as

$$\boxed{ \alpha^2 = \pi r_pR_\infty\mu. }$$

This single dimensionless identity can be inverted in several useful ways:

Quantity	Equivalent expression
Proton-electron mass ratio	$\displaystyle \mu=\frac{\alpha^2}{\pi r_pR_\infty}$
Proton radius	$\displaystyle r_p=\frac{\alpha^2}{\pi R_\infty\mu}$
Rydberg constant	$\displaystyle R_\infty=\frac{\alpha^2}{\pi r_p\mu}$
Fine-structure constant	$\displaystyle \alpha=\sqrt{\pi r_pR_\infty\mu}$

This makes the internal structure of the relation transparent.

---

8. Closing the loop back to $Q=4$

Now substitute the standard Rydberg relation

$$R_\infty = \frac{\alpha^2m_ec}{4\pi\hbar}$$

into

$$r_p = \frac{\alpha^2} {\pi R_\infty\mu}.$$

Since

$$\mu=\frac{m_p}{m_e},$$

we obtain

$$r_p = \frac{\alpha^2} { \pi \left( \frac{\alpha^2m_ec}{4\pi\hbar} \right) \left( \frac{m_p}{m_e} \right) }.$$

Canceling $\alpha^2$, $\pi$, and $m_e$,

$$\boxed{ r_p=\frac{4\hbar}{m_pc}. }$$

Thus the mass-ratio equation and the $Q=4$ radius equation close exactly.

Equivalently,

$$\boxed{ \frac{m_pc\,r_p}{\hbar}=4. }$$

Define the observational or geometric circulation index

$$Q_{\rm eff} \equiv \frac{m_pc\,r}{\hbar}.$$

Then the predicted proton radius gives exactly

$$\boxed{Q_{\rm eff}=4}.$$

---

9. The phonon speed-limit equation

The related phonon/sound-speed relation is

$$\boxed{ v_u = c \sqrt{ \frac{\pi}{2}r_pR_\infty }. }$$

The combination under the square root is dimensionless because

$$[r_p]={\rm length}, \qquad [R_\infty]={\rm length}^{-1}.$$

Now use

$$\pi r_pR_\infty = \frac{\alpha^2}{\mu}.$$

Then

$$\frac{\pi}{2}r_pR_\infty = \frac{\alpha^2}{2\mu}.$$

Therefore

$$\boxed{ \frac{v_u}{c} = \frac{\alpha}{\sqrt{2\mu}}. }$$

Since

$$\mu=\frac{m_p}{m_e},$$

this becomes

$$\boxed{ \frac{v_u}{c} = \alpha \sqrt{ \frac{m_e}{2m_p} }. }$$

Thus

$$\boxed{ v_u = \alpha c \sqrt{ \frac{m_e}{2m_p} }. }$$

Numerically,

$$\boxed{ \frac{v_u}{c} \approx1.20419\times10^{-4} }$$

and

$$\boxed{ v_u\approx3.6101\times10^4\ {\rm m/s} }$$

or

$$\boxed{ v_u\approx36.10\ {\rm km/s}. }$$

This is the same fundamental-constant form reported in the published condensed-matter analysis Speed of sound from fundamental physical constants, where it appears as an approximate upper scale for sound speed in condensed phases.

---

10. An unexpected Rydberg-energy form

Define the Rydberg energy

$$E_R=hcR_\infty.$$

Using

$$R_\infty = \frac{\alpha^2m_ec}{4\pi\hbar}$$

and $h=2\pi\hbar$,

$$\boxed{ E_R = \frac{\alpha^2m_ec^2}{2}. }$$

Now start from the phonon relation,

$$\left(\frac{v_u}{c}\right)^2 = \frac{\pi}{2}r_pR_\infty.$$

Insert

$$r_p=\frac{4\hbar}{m_pc}.$$

Then

$$\left(\frac{v_u}{c}\right)^2 = \frac{\pi}{2} \frac{4\hbar}{m_pc} R_\infty = \frac{2\pi\hbar cR_\infty} {m_pc^2}.$$

Since

$$2\pi\hbar=h,$$

we obtain

$$\boxed{ \left(\frac{v_u}{c}\right)^2 = \frac{hcR_\infty}{m_pc^2}. }$$

Therefore

$$\boxed{ \left(\frac{v_u}{c}\right)^2 = \frac{E_R}{m_pc^2}. }$$

Or, equivalently,

$$\boxed{ v_u^2=\frac{E_R}{m_p}. }$$

Hence

$$\boxed{ E_R=m_pv_u^2. }$$

This is an algebraic identity within the combined relations; it should not be confused with the ordinary Newtonian kinetic-energy expression $\tfrac12mv^2$.

Using

$$E_0=\frac{m_pc^2}{4},$$

we can also write

$$\boxed{ \left(\frac{v_u}{c}\right)^2 = \frac{E_R}{4E_0}. }$$

This links the proton $Q=4$ energy scale directly to the atomic Rydberg scale.

---

11. More useful inverse relations

The phonon equation also allows the proton radius to be recovered from the sound-speed ratio:

$$\boxed{ r_p = \frac{2}{\pi R_\infty} \left(\frac{v_u}{c}\right)^2. }$$

The proton-electron mass ratio can be written

$$\boxed{ \mu = \frac{\alpha^2} {2(v_u/c)^2}. }$$

The fine-structure constant can be written

$$\boxed{ \alpha = \sqrt{2\mu}\, \frac{v_u}{c}. }$$

And the Rydberg constant becomes

$$\boxed{ R_\infty = \frac{2}{\pi r_p} \left(\frac{v_u}{c}\right)^2. }$$

So the same algebra can be entered from several different directions.

---

12. A compact relation map

The entire chain can be summarized as

$$\boxed{ Q=4 }$$ $$\Downarrow$$ $$\boxed{ r_p = 4\frac{\hbar}{m_pc} = 4\bar\lambda_p = \frac{2}{\pi}\lambda_{C,p} }$$ $$\Downarrow$$ $$\boxed{ m_pc\,r_p=4\hbar }$$

while for the electron

$$\boxed{ r_e^* = 4\frac{\hbar}{m_ec} = 4\alpha a_0 = \frac{\alpha^2}{\pi R_\infty}. }$$

Therefore

$$\boxed{ \mu = \frac{m_p}{m_e} = \frac{r_e^*}{r_p} = \frac{\alpha^2}{\pi r_pR_\infty}. }$$

And consequently

$$\boxed{ \frac{v_u}{c} = \sqrt{\frac{\pi}{2}r_pR_\infty} = \frac{\alpha}{\sqrt{2\mu}} = \alpha\sqrt{\frac{m_e}{2m_p}} = \sqrt{\frac{E_R}{m_pc^2}}. }$$

That is the algebraic core of the construction.

---

13. Comparison with the measured proton charge radius

The $Q=4$ circulation radius is

$$r_{Q=4}=0.84123564\ {\rm fm}.$$

A 2026 precision atomic-hydrogen determination reported the proton rms electric charge radius as

$$r_{E}=0.8406(15)\ {\rm fm},$$

while the cited muonic-hydrogen value is

$$0.84060(39)\ {\rm fm}.$$

The new atomic-hydrogen result is therefore in the same narrow $0.84\ {\rm fm}$ region as the $Q=4$ length.

If we simply form

$$Q_{\rm obs} = \frac{m_pc\,r_E}{\hbar},$$

using $r_E=0.8406(15)$ fm gives approximately

$$\boxed{ Q_{\rm obs}=3.9970\pm0.0071. }$$

This is numerically consistent with $Q=4$.

But one distinction is essential.

The experimental proton radius is the electromagnetic rms charge radius defined by the slope of the electric form factor,

$$r_E^2 = -6 \left. \frac{dG_E(Q^2)}{dQ^2} \right|_{Q^2=0}.$$

The circulation calculation derives a geometric or dynamical vortex radius.

Therefore

$$\boxed{ r_{\rm vortex}=r_E }$$

is not yet a theorem.

A complete theory would need to derive the proton electromagnetic current

$$J^\mu_{\rm EM}$$

and show that its form factor produces the same rms radius.

That remains an important falsifiable step.

---

14. What is independent and what is algebraically equivalent?

This distinction matters.

Once

$$Q=4$$

and

$$v=c$$

are supplied, the circulation equation gives

$$r_p=\frac{4\hbar}{m_pc}.$$

Once the standard definitions of $R_\infty$, $a_0$, $\alpha$, and $m_p/m_e$ are then used, the relation

$$\frac{m_p}{m_e} = \frac{\alpha^2}{\pi r_pR_\infty}$$

follows algebraically.

Conversely, if that mass-ratio equation is taken as the starting relation and the standard Rydberg formula is inserted, it returns

$$r_p=\frac{4\hbar}{m_pc}.$$

So these two expressions form an exact consistency loop:

$$\boxed{ r_p=\frac{4\hbar}{m_pc} \quad\Longleftrightarrow\quad \frac{m_p}{m_e} = \frac{\alpha^2}{\pi r_pR_\infty} }$$

given the standard Rydberg identity.

That is mathematically powerful, but it also means they should not be counted as two statistically independent predictions.

Likewise, the phonon relation

$$v_u = c\sqrt{\frac{\pi}{2}r_pR_\infty}$$

reduces to

$$v_u = \alpha c \sqrt{\frac{m_e}{2m_p}},$$

the known fundamental-constant sound-speed expression.

The value of the construction is therefore in the unification of the algebraic structure, while the deeper dynamical challenge remains explaining why the proton occupies the $Q=4$ sector and why the geometric circulation radius should equal its electromagnetic charge radius.

---

Conclusion

The $Q=4$ proton construction produces the compact radius

$$\boxed{ r_p=\frac{4\hbar}{m_pc} =0.8412356402\ {\rm fm}. }$$

The corresponding electron $Q=4$ length is

$$\boxed{ r_e^* = \frac{4\hbar}{m_ec} = 4\alpha a_0 = \frac{\alpha^2}{\pi R_\infty}. }$$

Their ratio gives

$$\boxed{ \frac{m_p}{m_e} = \frac{r_e^*}{r_p} = \frac{\alpha^2}{\pi r_pR_\infty}. }$$

The same dimensionless combination then generates the phonon speed limit,

$$\boxed{ v_u = c\sqrt{\frac{\pi}{2}r_pR_\infty} = \frac{\alpha c}{\sqrt{2(m_p/m_e)}} }$$

with

$$\boxed{ v_u\approx36.10\ {\rm km/s}. }$$

Finally,

$$\boxed{ \left(\frac{v_u}{c}\right)^2 = \frac{E_R}{m_pc^2} }$$

connects the atomic Rydberg energy directly to the proton mass-energy scale.

The resulting algebraic network is remarkably compact:

$$\boxed{ Q=4 \;\longrightarrow\; r_p \;\longleftrightarrow\; \frac{m_p}{m_e} \;\longleftrightarrow\; \alpha,R_\infty,a_0 \;\longrightarrow\; v_u. }$$

The outstanding physics question is no longer whether these equations are mutually consistent—they are. The sharper question is whether a microscopic proton theory can derive the $Q=4$ sector, the electromagnetic form factor, and the phonon dynamics independently rather than assuming them. That is the point at which this algebraic closure becomes a genuine physical derivation rather than a highly constrained consistency relation.