- #1

doggydan42

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## 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?