# Stability of Matter

One of the most important problems to inspire the creation of quantum mechanics was the model of the Hydrogen atom. After Thompson discovered the electron, and Rutherford the nucleus (or Kern, as he called it), the model of the Hydrogen atom was refined to one of the lighter electron of unit negative elementary charge orbiting the larger proton, of unit positive elementary charge. However, it was well known that classical electrodynamics required that accelerated charges must radiate electromagnetic waves, and therefore lose energy. One question of interest is what determines the rate $\rho\!$ of this radiation and how high is this rate?

The parameters that would describe this radiation for an electron in Bohr's model, in which the electron is assumed to travel in quantized circular orbits around the nucleus, are the radius of the orbit $r_0\!$, the angular velocity ω, the charge of the particle $e\!$, and the speed of light, $c\!$. Therefore the rate of energy emission can be written by the function of those factors

$\rho=\rho(r_0,\omega,e,c)\!$.

However, far away from the atom, $\rho\!$ can only depend on the the dipole moment, so we can express the rate as

$\rho(er_0, \omega, c) \!$

Since light is energy, we are looking for how much energy is emitted per unit time: $[\rho]=\frac{\text{energy}}{\text{time}}.$ Knowing this much already imposes constraints on the possible dependence of $\rho\!$ on $er_0\!$, $\omega\!$, and $c\!$. We now use dimensional analysis to construct a quantity with units of energy.

From potential energy for coulombic electrostatic attractions: $\text{energy}=\frac{e^2}{\mbox{length}}\!$

Since we are considering $er_0\!$ as one parameter, let's multiply by $r_0^2\!$ and divide by $\text{length}^2\!$. Also, the angular velocity has units of frequency, so to obtain a quantity with units of energy per unit time, we simply multiply our result by the angular velocity, obtaining

$\frac{\text{energy}}{\text{time}}=\frac{e^2 }{\text{length}}\frac{r_0^2}{\text{length}^2}*\omega.$

In addition, $c/\omega \!$ has units of length, so we can write

$\frac{\text{energy}}{\text{time}} \sim \frac{e^2r_0^2 }{(c/\omega)^3} \omega = \frac{e^2r_0^2 }{c^3}\omega^4\sim\frac{1}{r_0^4 }.$

Therefore, as the dipole loses energy by radiating, the radius of the electron's orbit decrease. That is, the rate of emission increases as the radius decreases. As a result, classical physics predicts that atoms should collapse because the electron will radiate away all of its energy.