Electron sprials from r_i to r_f

[doggydan42]

Homework Statement

Find an expression for the time it takes for an electron to spiral in from an initial radius $r_i$ to a final radius $r_f$. Write your answer in terms of $r_i$, $r_f$, $m_e$, e, and c.

Homework Equations

Larmor Formula:
$$\frac{dE}{dt} = -\frac{2}{3}\frac{e^2a^2}{c^3}$$

Potential Energy:
$$V = -\frac{e^2}{r}$$

Magnitude of the force:
$$F = \frac{e^2}{r^2}$$

In another part of the problem, the velocity for radius r is calculated:
$$v(r) = \sqrt(\frac{e^2}{m_er})$$

The energy is also calculated to be:
$$E = -\frac{1}{2}\frac{e^2}{r}$$

The Attempt at a Solution

I tried solving it as a separable differential equation.
Most of the terms on the right were constants, so I focused on $e^2a^2$.
Since it was circular motion, $F = e^2/r^2 = m\frac{v^2}{r} = ma$, so $a = \frac{v^2}{r}$.
So, $a^2 = \frac{v^4}{r^2}$, using the formula for v(r),
$$a^2 = \frac{(\sqrt(\frac{e^2}{m_er}))^4}{r^2} = \frac{(\frac{e^2}{m_er})^2}{r^2} = \frac{e^4}{r^4m_e^2}$$
Since $E =E = -\frac{1}{2}\frac{e^2}{r}$, $\frac{e^2}{r} = -2E$, so
$$a^2 = \frac{(-2E)^2}{m_er^2}$$
So,
$$e^2a^2 = \frac{(-2E)^2e^2}{m_er^2} = \frac{(-2E)^3}{m_er} = \frac{-8E^3}{m_er}$$
Also, $\frac{1}{r} = \frac{e^2}{re^2} = \frac{-2E}{e^2}$
So $$e^2a^2 = \frac{-8E^3}{m_e}\frac{-2E}{e^2} = \frac{16E^4}{m_ee^2}$$
Plugging this into the initial equation:
$$\frac{dE}{dt} = -\frac{2}{3}\frac{16E^4}{c^3m_ee^2} = -\frac{32}{3}\frac{E^4}{c^3m_ee^2}$$
Rearranging the terms:
$$\frac{3c^3m_ee^2}{32E^4}dE = -dt$$
Integrating both sides:
$$\int_{E_i}^{E_f} \frac{3c^3m_ee^2}{32E^4} dE = \int_{t_i}^{t_f} -dt$$
Solving the integral:
$$-\Delta t = \left[\frac{-c^3m_ee^2}{32E^3}\right]_{E_i}^{E_f}$$
or
$$\Delta t = \left[\frac{c^3m_ee^2}{32E^3}\right]_{E_i}^{E_f}$$

From there I could plug in $E_i = E(r_i)$, and $E_f = E(r_f)$

My concern is that the problem stated we would need differential equations, so I was unsure if this was a good approach.

 

[TSny]

Your work looks good to me, except I think you dropped the square on the electron mass when going from the first to the second equation in your solution.