Electron sprials from r_i to r_f

  • #1
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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.
 
Last edited:

Answers and Replies

  • #2
TSny
Homework Helper
Gold Member
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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.
 
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