# Bohr's model and positronium

I was trying to get an order of magnitude estimate for the radii and energies of positronium using the Bohr's model. I did find a few places where they have used the reduced mass to replace the electron mass in the equations. See this for example. However, I don't see why using the reduced mass works. What is the reasoning behind this?

I tried a different approach. If the positron and electron are in a orbit of radius $r$ about their common centre of mass then the force of attraction felt by each is

$$F=\frac{e^2}{4r^2}=\frac{e(e/4)}{r^2}$$

where working in cgs units makes coulomb's constant 1. $e$ is the magnitude of electronic charge.

So, the electron for all intents and purposes "feels" as if there is a positive charge of magnitude $\frac{e}{4}$ around which it is revolving. And therefore, we must replace $Z$ by $e/4$ in all equations for hydrogen-like species.

This gives results that are scaled by a factor of $4$ from the hydrogen atom. Whereas using the reduced mass gives results scaled by a factor of $2$. Why is this approach incorrect?

• I think you forgot that the mass of your combined particle is twice that of the electron. The reduced mass approach does this correctly. Jan 26, 2016 at 16:52
• @CuriousOne: I don't see what you mean by the "combined particle". As soon as we replace the positron with an equivalent (in terms of the force on the electron) positively charged entity of charge $e/4$ at the centre of mass, the positron is excluded from the analysis. Jan 26, 2016 at 17:29
• The effective "particle" that orbits around the center of mass has the mass of electron and positron combined. That's what reduced mass does. I don't know why you don't want to use it? It is the correct approach to all dynamics problems of this kind, not just to the positronium. Jan 26, 2016 at 17:31
• @CuriousOne: I'm not that familiar with applying reduced mass. Besides, I'm much more interested in finding out the mistake in my approach. Jan 26, 2016 at 17:33
• The mistake in your approach is that you are not using the reduced mass. Jan 26, 2016 at 17:38

Your proposal will work as long as mass and charge enter into the expression in a ratio 1:-2. Find an expression with a different ratio (like electric potential, or radiated power ...) and you're sunk.

The fine structure constant $$\alpha = \frac{e^2}{\hbar c} \tag{Gaussian units}$$ is an example of such an expression and controls the fine structure of the spectrum, so there is solid experimental evidence that can tell the two approaches apart.

As for why the reduced mass, it comes from dividing the motion of the system (of two particles) into the motion of the center-of-mass and motion of the constituent particles relative the center of mass. This division is a standard part of the theory of central force motion and shows up in computing planetary orbits as well as in computations of the structure of simple atoms.

In the case of the hydrogen atom (and treating the problem classically, which is wrong but will do for the purposes of this question), the center of mass is essential at the position of the proton, and electron moves in a path with radius approximately the separation of the two particles (while the proton barely moves at all). In the case of positronium, the center of mass is halfway between the two particles and they both move through circles with radius of half the separation.

The particles in a positroinum atom more along smaller circles than the electron in the hydrogen atom when subject to the same force (because the separation is still the same), which means they move as if they had a smaller mass and were moving around an immobile force center. That smaller mass is the "reduced mass" of the system. (And it really is a property of the system not of either particle.)

• As I understand you have proved that my approach is incorrect. Could you kindly elaborate a bit on the flaw in my reasoning? I still don't understand where I went wrong. Thanks. Jan 26, 2016 at 17:28

Your mistake is that the other particle has energy too. In the hydrogen atom, the nucleus is just sitting there, but in positronium, the positron has an equal amount of kinetic energy. (You are also cutting the potential energy in half.) Doubling all energies will give the right result.

As other people have said, it's much much easier to do this with reduced mass. Any correct derivation that doesn't use the idea of reduced mass will, at some point, contain all of the complexity required to derive reduced mass in the first place!

• What about the radius? I would not halve the radius I got from my derivation just because there are two particles. Jan 27, 2016 at 2:56