Radiative collapse of an electron in a classical atom

[LauritsT]

Hello!

How do I find how much energy does electron radiate during one revolution if the energy radiation rate is given by Larmor Formula: dE /dt = [(− 2/3)* e²* a²]/ c³. Should I use the chain rule? At the moment I only want a hint how to solve this.

With best regards

 

[mfb]

You can find the time of a classical revolution and multiply the energy loss per time by that time. Then compare the result to the overall energy. If it is small the approach works, if it is large then dE/dt will change during the orbit and you cannot use this simple approach. Then you have to solve a differential equation as a depends on t now.

 

[LauritsT]

You can find the time of a classical revolution and multiply the energy loss per time by that time. Then compare the result to the overall energy. If it is small the approach works, if it is large then dE/dt will change during the orbit and you cannot use this simple approach. Then you have to solve a differential equation as a depends on t now.

Thanks! So very roughly speaking, the overallenergy that the electron should have is K~ e² and the energy lost per revolution is E~ e⁵ / (m^3/2 * r * c³). So since the energy lost per revolution is much smaller (~10³⁰ times smaller) then the approach works. In the case when dE/dt is large: the differential equation already has t as a variable as I understand, since dE=...dt, yes?

 

[mfb]

Yes, you'll get $dE=c (a(t))^2 dt$ with some constant c - but you don't know a(t), so you probably want to rewrite that as $dE=c (a(E))^2 dt$, plug in a(E) and then solve for E(t). But with a small energy loss, that is not necessary.

 

[Vanadium 50]

So since the energy lost per revolution is much smaller (~10³⁰ times smaller)

That doesn't sound right to me. This leads to atoms stable for thousands of years, when I thought it was well under a microsecond.

 

[LauritsT]

That doesn't sound right to me. This leads to atoms stable for thousands of years, when I thought it was well under a microsecond.

Thanks for the tip! I've read also that electron should fall onto the nucleus in 10^-11 seconds. But since it the energy of only one full rotation with respect to the whole energy (kinetic) then I didn't think of it as a fault right away. I will check it once more.

//edit: just to be clear my result was before dE/K = 10^-30 where dE is energy lost in one revolution and K is kinetic energy. But once again, I will check my calculations

 

[mfb]

I get nW, or 10 GeV/s. At a few PHz classical revolution frequency, this is a few µeV per revolution. Not 10³⁰ times smaller, but still small enough to make the initial approximation reasonable.
Calculated for a distance of 1 Ångström.