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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \rho\,\!} of this radiation and how high is this rate?
The electron in the Bohr's model involves factors of radius Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r_0\!} , angular velocity Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \omega\!} , charge of the particle Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e\!} , and the speed of light, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c\!} . Therefore the rate of the energy dissipation can be written by the function of those factors
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \rho=\rho(r_0,\omega,e,c)\!} .
However, far away from the atom, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \rho \!} can only depend on the the dipole moment, so we can express the rate as follows
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \rho(er_0, \omega, c) \!}
What is the dimension of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \rho\,\!}
?
Essentially, since light is energy, we are looking for how much energy is passed in a given time: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [\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: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \text{energy}=\frac{e^2 }{\mbox{length}} \!}
Since we are considering Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle er_0\!} as one parameter, let's multiply by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r_0^2\!} and divide Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\text{energy}}{\text{time}}=\frac{e^2 }{\text{length}}\frac{r_0^2}{\text{length}^2}*\omega }
In addition, since Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c/\omega \!} gives the dimension of length, we can write
- Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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 decrease. As a result, classically the matter should collapse.