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In a recent discussion in this thread I wrote:
I assumed that this non-QM explanation was wrong and that I was missing something obvious somewhere. I have tried to figure out why this is not at least a plausible explanation. I can't.
We know from celestial mechanics that when orbiting masses trade potential for kinetic energy, they adopt a smaller radius of orbit and speed up. Since the object that they are orbiting is large compared to the radius of orbit, a sufficiently reduced orbit radius means that they crash.
But in the case of gravity (black holes excepted) the speed of the orbiting object is not enough to change its mass appreciably (ie v<<c). So there is no minimum limit to the radius of orbit.
It is a very different matter with an electron orbiting a proton. The orbital speed at an atomic radius [itex] \approx 10^{-12} m.[/itex] is relativistic, as I have shown. If it trades potential energy for kinetic, it speeds up but its mass increases as [itex]\gamma = (1 - v^2/c^2)^{-1}[/itex] increases so the radius of orbit approaches a limit that is greater than the radius of the proton.
What is wrong with this argument?
AM
Using classical mechanics and electomagnetism, work out the speed that an electron would have to have in order to orbit a hydrogen nucleus at a distance of 10^-12 m:
[tex]F_c = \frac{m_ev^2}{r} = \frac{kq_e^2}{r^2}[/tex]
[tex]v = \sqrt{\frac{kq_e^2}{mr}}[/tex]
where:
[itex]r = radius of orbit = 1e-12 m[/itex]
[itex]k = 9e9 Nm^2/C^2[/itex]
[itex]q_e = 1.602e-19 C.[/itex]
[itex]m_e = 9.1e-31 kg[/itex]
v works out to 1.6e7 m/sec or about 5% of the speed of light.
Now work out what the radius of orbit could be if the electron traveled at the speed of light. This would obviously be the minimum orbital radius permitted by relativity. It would take an infinite amount of energy for an electron to get arbitrarily close to the speed of light.
I get r = 2.5e-15 m. or 2.5 Fermi units
The radius of a proton is about .5 Fermi. To reach a 2.5 Fermi radius of orbit, the electron would need an infinite amount of energy. So an orbiting electron simply can't get enough energy to crash into the nucleus!
[tex]F_c = \frac{m_ev^2}{r} = \frac{kq_e^2}{r^2}[/tex]
[tex]v = \sqrt{\frac{kq_e^2}{mr}}[/tex]
where:
[itex]r = radius of orbit = 1e-12 m[/itex]
[itex]k = 9e9 Nm^2/C^2[/itex]
[itex]q_e = 1.602e-19 C.[/itex]
[itex]m_e = 9.1e-31 kg[/itex]
v works out to 1.6e7 m/sec or about 5% of the speed of light.
Now work out what the radius of orbit could be if the electron traveled at the speed of light. This would obviously be the minimum orbital radius permitted by relativity. It would take an infinite amount of energy for an electron to get arbitrarily close to the speed of light.
I get r = 2.5e-15 m. or 2.5 Fermi units
The radius of a proton is about .5 Fermi. To reach a 2.5 Fermi radius of orbit, the electron would need an infinite amount of energy. So an orbiting electron simply can't get enough energy to crash into the nucleus!
I assumed that this non-QM explanation was wrong and that I was missing something obvious somewhere. I have tried to figure out why this is not at least a plausible explanation. I can't.
We know from celestial mechanics that when orbiting masses trade potential for kinetic energy, they adopt a smaller radius of orbit and speed up. Since the object that they are orbiting is large compared to the radius of orbit, a sufficiently reduced orbit radius means that they crash.
But in the case of gravity (black holes excepted) the speed of the orbiting object is not enough to change its mass appreciably (ie v<<c). So there is no minimum limit to the radius of orbit.
It is a very different matter with an electron orbiting a proton. The orbital speed at an atomic radius [itex] \approx 10^{-12} m.[/itex] is relativistic, as I have shown. If it trades potential energy for kinetic, it speeds up but its mass increases as [itex]\gamma = (1 - v^2/c^2)^{-1}[/itex] increases so the radius of orbit approaches a limit that is greater than the radius of the proton.
What is wrong with this argument?
AM