Is there a classical radius for the proton?

Question

High energy photons scatter off electrons via Compton scattering. The differential cross section of this scattering is given by the Klein-Nishina formula, which contains within it the classical radius of the electron, $r_{e}$,

$$\frac{d\sigma}{d\Omega} = \frac{1}{2}r_{e}^{2}\left(\frac{\lambda}{\lambda’}\right)^{2} \left[\frac{\lambda}{\lambda’} + \frac{\lambda’}{\lambda}-\sin^{2} \theta\right]$$

Have the equivalent experiments been done with the proton? Is there a "classical proton radius"? The closest equivalent I can find is the proton charge radius, which stems from electron-proton scattering.

Edit: The differential cross section of high energy photons hitting protons found here (Figs. 10 and 11) looks quite similar to data from Compton scattering off an electron (pdf, Fig 2). This seems to suggest both are governed by similar equations, meaning there is a classical radius for the proton.

you can define it; the electron is presumably a point charge yet the classical radius is forced upon it, so you can do it for proton too. It will just be smaller than the proton's actual radius — naturallyInconsistent, May 05 '23 at 19:18
@hft The classical radius of the proton should be the same as the classical radius of the electron. The classical electron radius depends on the mass of the electron. The equivalent concept for the proton would use the proton mass. Both would be the radius at which the particle’s electrostatic field energy equals its mass-energy. — Ghoster, May 05 '23 at 19:21
Related: Is the zero distance limit of Coulomb's law something to worry about? — mmesser314, May 05 '23 at 22:00

hft · Answer 1 · 2023-05-07T16:44:08.947

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Is there a classical radius for the proton?

You can determine the classical proton radius in the same way as you can determine the classical electron radius. Just substitute the mass of the proton for the mass of the electron.

Ignoring geometric factors, the classical proton radius can be estimated in the same was as the classical electron radius as: $$ R_{\text{CL}} = \frac{e^2}{4\pi \epsilon_0M_pc^2}\approx 10^{-18}\text{meters}\;, $$ where $e$ is the charge of the proton, $M_p$ is the mass of the proton, $c$ is the speed of light, and $\epsilon_0$ is that dumb constant that shows up in SI electrostatic units.

Update:

In natural units, where $c=\hbar=1$, we can also write the classical proton radius as: $$ R_{\text{CL}} = \frac{\alpha \hbar c}{M_pc^2} = \frac{\alpha}{M_p}\;, $$ where $\alpha \approx 1/137$.

And we can write the classical electron radius as: $$ r_{e} = \frac{\alpha \hbar c}{M_ec^2} = \frac{\alpha}{m_e}\;. $$

So, in these units, a factor of the classical radius is equivalent to $\alpha$ divided by a factor of the respective mass.

Looking at section 8.4 of these notes, the elastic Mott Scattering formula for electron-proton scattering at high energy is: $$ \frac{d\sigma}{d\Omega} = \frac{\alpha^2}{4p_e\sin^4(\theta/2)}\left(\cos^2(\theta/2) - \frac{(\Delta p)^2}{2 M_p^2}\sin^2(\theta/2)\right) $$ $$ = \frac{1}{4p_e\sin^4(\theta/2)}\left(\alpha^2\cos^2(\theta/2) - \frac{R^2_{\text{CL}}(\Delta p)^2}{2}\sin^2(\theta/2)\right)\;, $$ which is Eq. (8.5) of the linked reference, and where the classical proton radius can be seen on the far RHS above.

For comparison, in the low energy limit, this formula reduces to Eq. (8.6) in the linked reference: $$ \frac{d\sigma}{d\Omega} = \frac{\alpha^2}{4m_e^2v_e^4\sin^4(\theta/2)} $$ $$ \frac{r_e^2}{4v_e^4\sin^4(\theta/2)}\;. $$

edited May 07 '23 at 16:44

answered May 07 '23 at 00:12

hft

19,536

Yes, mathematically a similar value can be calculated. I'm more interested in whether it has any physical meaning, as the classical electron radius appears to. – WaveInPlace May 07 '23 at 00:29
You asked: "Is there a classical radius for the proton?" – hft May 07 '23 at 02:14
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"Physical meaning" questions are often answered with opinions and baloney. But, in short, no, there is no "physical meaning" other than the radius of a sphere whose electrostatic self energy is equal to the rest mass of the proton. – hft May 07 '23 at 02:15
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The classical electron radius is already close to a femtometer, which would be the usual scale we associated with a proton. The classical proton radius is about 2000 times smaller. So, seemingly quite unrelated to "physically meaningful" length scales. – hft May 07 '23 at 02:18
"Physical meaning" questions are often answered with opinions and baloney. That was why I included all the details of the Klein-Nishina equation. The classical electron radius appears to be quite important in photon-electron scattering. I'm wondering if there's a similar value important in photon-proton scattering. – WaveInPlace May 07 '23 at 12:07
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@WaveInPlace I have updated my answer to provide one example of electron-proton scattering and where the classical proton radius appears, and how it goes away in the low energy limit. – hft May 07 '23 at 16:45
It might be worthwhile emphasizing to the OP that this formal classical radius is a hundredth of a millifermi, so five orders of magnitude smaller than the actual, measured size of the proton.... – Cosmas Zachos May 08 '23 at 00:28
@CosmasZachos Five orders of magnitude smaller? Or three orders of magnitude? I indicated in my answer that this formal classical proton radius is about 2000 times smaller than the classical electron radius (because of the $M_p/m_e$ factor). And the classical electron radius is about 2 fm. – hft May 08 '23 at 01:05
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Right, but the electron has no compositeness size, zero, while the proton is a hadron of size about one fermi… trying to give meaning to the classical proton radius is thus a bit absurd, as you suggest… – Cosmas Zachos May 08 '23 at 02:13

score 0 · Answer 2 · answered May 06 '23 at 02:50

Here is how I would go about this.

If I know (from scattering experiments) the diameter of a certain nucleus, and I know the number of nucleons contained within that nucleus, and I assume the nucleus is approximately spherical, then geometry and algebra will yield an estimate of the "classical" proton diameter, which the margin of this page is too small to contain.

score 0 · Answer 3 · edited Jan 30 '24 at 18:31

$$r_{p} = \frac{{m_{e} \alpha^2}}{{m_{p} \pi R_{\infty}}}=8.41235641(34 \pm26)\cdot10^{-16} \: \text m$$

Based on torque balance, this was known to be the most accurate classical formula for the proton radius prior to the empirical correction by better than 4% that occurred recently. It is still within the best current empirical correction.

The derivation is: $$m_{e}=\frac{2 h R_{\infty}}{c \alpha}$$ $$m_{e} r_{e}=\frac{2 h}{c \pi}$$ $$m_{e} r_{e} = m_{p} r_{p}$$ Eliminating $c$, $h$ and $r_{e}$ we have: $$m_{p} \pi R_{\infty} r_{p} = m_{e} \alpha^2$$ Solve for $r_{p}$: $$r_{p} = \frac{{m_{e} \alpha^2}}{{m_{p} \pi R_{\infty}}}=8.41235641(34 \pm26)\cdot10^{-16} \:\text m$$

Is there a classical radius for the proton?

3 Answers3