# Minimum energy of an electron with the Larmor formula

## Homework Statement

Is there a minimum value for the total energy of the electron (in this analysis)?

The previous parts:
Use Larmor formula to find ##\frac{|\Delta E|}{K}##, where ##|\Delta E|## is the energy lost per revolution.
the result is ##\frac{8\pi v^3}{3c^3}##.

##\frac{v(r)}{c}## was caluclated for r = 50 pm, 1pm, and 1 fm.

Lastly, ##t(50 pm \rightarrow 1 pm)## was calculated.

## 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 time it takes for the electron to spiral from ##r_i## to ##r_f## is:
$$t(r_i\rightarrow r_f) = \frac{m_e^2c^3}{4e^4}((r_i)^3-(r_f)^3)$$

## The Attempt at a Solution

I have been trying to look at the formulas and see how if one variable increase what happens to dE/dt. If there was a minimum energy, there would be some r such that dE/dr = 0. This would be at negative and positive infinity radii. Pluggin the infinity into ##E = \frac{-e^2}{2r}## gives E = 0. So as the radius approaches infinity, the energy reaches a maximum, 0.

As the radius approaches 0, however, E approaches -infinity from the right, and infinity from the left. So would the would we be able to say there is a minimum energy at the limit of E as r approaches 0+?

Is there a better way to go about solving this problem?

Would the fact that E goes to -infinity mean that that there is no minimum energy? The electron would keep getting closer to the center of the orbit, never reaching, but decreasing in energy all the while?

nrqed