San José State University
Thayer Watkins
Silicon Valley
& Tornado Alley

A Comparison of the Properties
of a Muonic Hydrogen Atom
and an Electronic One

Experimental physicists have found an interesting puzzle. Protons can form hydrogen atoms with muons as well as electrons. The puzzle is that the measured dimensions of the proton in a muonic hydrogen atom are smaller than than those of a proton in an electronic hydrogen atom.

First note that a hydrogen atom revolves at such a rapid rate, many millions of times per second, that all that can be observed about the proton is its dynamic appearance. In the case of the proton in a hydrogen atom that may be an oblate spheroid or a torus dpending upon the orbit radius relative to the radius of the particle. .

An electron has a mass only 1/1836 of the mass of a proton so the center of mass of a proton-electron atom is close to the center of the proton and hence the proton has relatively little motion compared to that of the electron. The muon, on the other hand, has a mass 208 times that of an electron so its mass is approximately 1/9 of the mass of a proton. The center of mass of the muon-proton system is distinctly outside of the proton and the proton circles that center of mass just as the muon does.

The subatomic particles execute periodic motion so rapidly that all that can be observed of them is their dynamic appearance. What follows here is an analysis which is essentially an extension of Niels Bohr's analysis of a hydrogen atom. That analysis is marvelously successful in explaing the spectrum of hydrogen. However the analysis is put in most general terms so it applies to positronium as well as electronic and muonic hydrogen atoms.

The Dynamics of a
Two Charged Particle System

Let the characteristics of the two particles be labeled 0 and 1. The masses, orbit radii and magnitudes of the charges are given as m, r and q, respectively. It is assumed that the charges of the particles are opposite. The force of attraction between them is given by

F = kq0q1/(r0+r1

The balances of that attractive force with the so-called centrifugal forces are given by

kq0q1/(r0+r1)² = m0ω²r0 = m1ω²r1

where ω is the rate of rotation of the system.

The second equality above implies

m0r0 =m1r1
and hence
r1/r0 = m0/m1

This is just the condition for determining the center of mass of the system.

Angular Momentum

The angular momentum of the two particle system is quantized as

L = m0ωr0² + m1ωr1² = nh

where h is Planck's constant divided by 2π and n is a positive integer.

This condition may be expressed more succinctly as

L = Jω = nh

where J is the moment of inertia of the system and J=m0r0² + m1r1².

Note that, since m1r1=m0r0

J = m0r0²[1 + r1/r0] = m0r0²[1 + m0/m1]


The kinetic energies of the particles are

K0 = ½m0ω²r0²
K1 = ½m1ω²r1²

Note that

K1/K0 = m1r1²/(m0r0²) = r1/r0 = m0/m1

The total kinetic energy of the system K is given by

K = ½Jω²

The potential energy of the system is kq1q0/(r0+r1). This can be reduced to

V = kq1q0/[r0(1+r1/r0) = kq1q0/[r0(1+m0/m1)]

Since ω is equal to L/J

K = L²/(2J) = n²h²/(2J)

The total energy of the system can then be expressed as

E = n²h²/(2J) + kq1q0/[r0(1+m0/m1)]
which reduces to
E = n²h²/[2m0r0²(1 + m0/m1)] + kq1q0/[r0(1+m0/m1)]
and further to
E(1+m0/m1) = n²h²/[2m0r0²] + kq1q0/r0

If this equation is divided by m0 the result is

E(1/m0+1/m1) = n²h²/[2(m0r0)²] + kq1q0/[m0r0]

This is a quadratic equation in 1/(m0r0). The expression (1/m0 + 1/m1) can be represented as 1/μ, where μ is called the reduced mass of the system. The equation determining (1/m0r0) is then

h²/[2(m0r0)²] + kq1q0/[m0r0] − E/μ = 0.

The equation determing (1/m1r1) is the same, as it should be since m1r1 = m0r0.

The solutions to this equation are

(1/m0r0) = (1/m1r1) = [−kq1q0 ± [(kq1q0)² + 2n²h²E/μ]½/(n²h²)

The analysis can be limited to the minimum case of n=1 and the + of the ±. For a hydrogen atom q1q0=e² where e is the electrostatic charge. Furthermore ke² can be represented as αhc where α is the fine structure constant, h is Planck's constant divided by 2π and c is the speed of light. These specializations reduce the equation to

(1/m0r0) = (1/m1) = [−αhc ± [(αhc)² + 2h²E/μ]½/h²

This may be rearranged to

(h/m0r0) = [−αc/2π ± [(αc/2π)² + 2E/(α²cμ]½
or, equivalently
(2πh/(αm0r0)) = (2πh/(αm1r1)) = [1 + 8π²E/(α²c²μ)½
and further to
r0 ≅ αc²μ/(2πm0h)
r1 ≅ αc²μ/(2πm1h)

The Effect of the Substitution
of a Muon for the Electron
in a Hydrogen Atom

Now let r1e and r1m be the values of r1 for the radii of the proton's orbit for the electron and the muon in a hydrogen atom. From the above formula

r1m/r1e ≅ (μme)

Let particle 1 be the proton. When particle 0 is an electron the reciprocal of the reduced mass of the atom is proportional to 1+1/1836. When it is a muon the recirocal of reduced mass is proportional to 1/9+1/1836. This is a reduction to roughly 1/9. Thus

r1m/r1e ≅ = (9/1)

As with intuition the proton has a much larger orbit radius with a muon than does the proton with an electron. So the puzzle is even more extreme.

(To be continued.)

Dedicated to my long term friend Lydia
for alerting me to this puzzle
but with more graditude for her perfecting
the department she was chairman of.

HOME PAGE OF applet-magic
HOME PAGE OF Thayer Watkins,