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. For an electron that moves in circular orbit about the more massive nucleus under the influence of the Coulomb attractive force, here is a simple non-relativistic model of this classical system:

Where $r\!$ is the orbital radius, and we neglect the motion of the proton by assuming it is much much more massive than the electron.

The question is: What determines the rate $\rho \,\!$ of this radiation and how fast is this rate?

The electron in the Bohr's model involves factors of radius $r_{0}\!$, angular velocity $\omega \!$, charge of the particle $e\!$, and the speed of light, $c\!$. Therefore the rate of the energy dissipation 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 follows

$\rho (er_{0},\omega ,c)\!$

What is the dimension of $\rho \,\!$?

Essentially, since light is energy, we are looking for how much energy is passed in a given time: $[\rho ]={\frac {\text{energy}}{\text{time}}}\!$

Knowing this much already imposes certain constraints on the possible dimensions. By using dimensional analysis, let's construct something 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 ${\text{length}}^{2}\!$. Also, the angular velocity is in frequency, so to get the above equations in energy/time, just multiply it with the angular velocity, then we have

${\frac {\text{energy}}{\text{time}}}={\frac {e^{2}}{\text{length}}}{\frac {r_{0}^{2}}{{\text{length}}^{2}}}*\omega$

In addition, since $c/\omega \!$ gives the dimension of length, we can write

${\frac {\text{energy}}{\text{time}}}\sim {\frac {e^{2}r_{0}^{2}}{(c/\omega )^{3}}}\omega ={\frac {e^{2}r_{0}^{2}}{c^{3}}}\omega ^{4}\sim {\frac {1}{r_{0}^{4}}}\!$

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 decrease. As a result, classically the matter should collapse.