# Why are the classical electron radius, the Bohr radius and the Compton wavelength of an electron related to each other?

Using the definition of the fine-structure constant $\alpha = \frac{4 \pi \epsilon_0 \hbar c}{e^2}$ and the Compton wavelength of an electron $\lambda_c = \frac{h}{m_e c}$ the classical electron radius $r_e$ and the Bohr radius $a_0$ can be expressed like $$r_e = \alpha \frac{\lambda_c}{2\pi}$$ $$a_0 = \frac{1}{\alpha} \frac{\lambda_c}{2\pi}$$

This means e.g. that the classical electron radius can be expressed in terms of the Bohr radius as $r_e = \alpha^2 a_0$.

Isn't that peculiar? Why should the classical radius of the electron and the distance of an electron to the nucleus in an atom be related to each other? And why are both multiples of the Compton wavelength?

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It is not surprising that both $r_e$ and $a_0$ are multiples of the Compton wavelength: any two positive lengths are multiples of each other. While it is true that there is more to this than that simple statement, the essential fact is that since those three lengths are composed simply and out of the same basic ingredients, there is very little leeway for how they can be different.

Let's have a look at these quantities: $$\lambda_C=\frac{2\pi\hbar}{mc},\,\, r_e=\frac{e^2}{mc^2}\text{ and }a_0=\frac{\hbar^2}{m e^2},$$ in Gaussian units, where $m$ is the electron mass. $%These are, respectively, the characteristic length scales of the photon momentum that matches an electron, the electron's rest mass as electrostatic energy, and the basic quantum mechanical electrostatic problem for the electron.$

Notice that they are all inversely proportional to the electron rest mass, though for different reasons: heavier electrons would require beefier photons to deflect them; they have higher rest mass and would need a more compact spherical charge to match; and a higher $m$ effectively reduces $\hbar$ in the hydrogenic Schrödinger equation, making it harder to get to the quantum regime.

Given that, you have three lengths that are determined by the three constants $\hbar$, $c$ and $e$. That's enough constants to make three different lengths, but they are few enough that any quotient must be a function of the unique dimensionless combination of these constants - the fine structure constant, $$\alpha=\frac {e^2} {\hbar c}.$$ Thus, it is necessary that any two of these three lengths must be multiples of the third and of $\alpha^{\pm1}$ (modulo $2\pi$).

This constant, however, is particularly important. It is the natural measure of the strength of electromagnetic interactions: it gives, as a pure number, the electromagnetic coupling $e^2$ between two unit charges, in natural relativistic units where $\hbar=c=1$. Thus, while the relations you remark on are algebraically necessary, it doesn't mean they are devoid of physical content:

• $r_e=\alpha \lambda_C/2\pi$ says that for a more strongly interacting QED a more loosely bound spherical electron would suffice to match the rest mass energy.
• $a_0=\frac 1\alpha \lambda_C/2\pi$ says that for a more strongly interacting QED the proton would hold its hydrogenic electron in tighter orbits.

Both of these are indeed most naturally phrased in terms of the Compton wavelength, as it is the characteristic quantum-relativistic length scale of the electron, and does not depend on any particular physical interaction, whereas the other two do - and are therefore obtained from the first via the strength of that interaction.

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The three lengths you are considering are all built using only the $e$, $m_e$ and the fundamental constants. If you look at the definitions, you can notice that they all have the form $\frac{something}{m_e}$.

It's clear then that you can get one length from the others just multiplying by some factor of $e$ and fundamental constants; since all the quantities are lengths, the factor must be dimensionless: it must be a power of $\alpha$, times some number.

The $2\pi$ factors come from the fact that $\lambda_C$ involves $h$ and the others $\hbar$.

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Concerning the compton wavelenght $\lambda_c$ and the classical electron radius $r_e$, you can synthetize this way: $$h\nu=m_ec^2=\frac{e^2}{4\pi \varepsilon_0\, a_0}$$ The first equality represents the interaction between a photon an electron which transfer their energy from one to another (compton scattering). This gives the compton wavelength, typical range of the electron-photon interaction.

The second equality represents the energy of the electron matching the potential energy it would experience in the classical electric potential of a point charge (another electron for instance). This gives the classical electron radius, that I would interpret as the typical range for the electron-electron interaction.

Thus, the relationship between $\lambda_c$ and $r_e$ is $\alpha$ because it is a measure of the strength of electrostatic interaction.

Now, I am not sure about the Bohr radius part. This is not as direct because it involves the quantization of angular momentum.

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