# Radiative collapse of a classical atom

I'm trying to determine the time it takes for a classical electron to spiral towards a proton, but I can't seem to get an answer on the right timescale. Using the energy of the electron $$E = -\dfrac{ke^2}{r}+\dfrac{1}{2}mv^2$$ and assuming the force on the electron is $$\dfrac{mv^2}{r} = \dfrac{ke^2}{r^2}$$ gives $$E = -\dfrac{1}{2}\dfrac{ke^2}{r},$$ then differentiating $$\dfrac{dE}{dt} = \dfrac{dE}{dr} \dfrac{dr}{dt} = \dfrac{1}{2}\dfrac{ke^2}{r^2}\dot{r}$$ and setting equal to the Larmor formula gives $$\dfrac{1}{2}\dfrac{ke^2}{r^2}\dot{r} = -\dfrac{2}{3}\dfrac{e^2a^2}{c^3} = -\dfrac{2}{3}\dfrac{e^2k^2e^4}{c^3m^2r^4}$$ or $$r^2dr = -\dfrac{4}{3}\dfrac{ke^4}{c^3m^2}dt$$ Integrating from some initial radius $$r_i$$ to some final radius $$r_f$$ gives $$t=-\dfrac{1}{4}\dfrac{c^3m^2}{ke^4}\big(r_f^3-r_i^3\big)$$ Plugging in $$r_i = 1\rm{e}-10\rm{m}$$ and $$r_f = 1\rm{e}-15\rm{m}$$, gives a spiral time of $$0.944$$ seconds. However this doesn't seem to agree with timescales I've seen online, giving something around $$1.1\rm{e}-10$$ seconds. It seems like I might be off by a factor of $$1/k$$ but I can't see where I might have messed up the algebra. This has been bugging me for a few days, so any help is greatly appreciated!

• What is $k$ in whatever unit system your using for Larmor's formula...? Commented Apr 12, 2020 at 23:16

If $$k = 1/4\pi \epsilon_0$$, then I think Larmor's formula is $$\frac{dE}{dt} = \frac{2e^2a^2}{3c^3}k$$