# Electron under gravity

• Reshma

## Homework Statement

An electron at rest is released from rest and falls under the influence of gravity. In the first centimeter, what fraction of potential energy lost is radiated away?

## Homework Equations

Lienard-Wichert potential for the electron of charge e is given by:
$$\phi = \frac{e}{R(1 - \beta \cdot \hat R)}$$

In this case the charge is accelerated by a gravity (a = g).

## The Attempt at a Solution

Reference: Electrodynamic Radiation by Marion and Heald

The problem hasn't mentioned whether the speed of the electron is relativistic.

If the speed of the electron is less than c ($\beta << 1$) then $R(1 - \beta \cdot \hat R) \rightarrow 0$ and the potential can be written as:
$$\phi = \frac{e}{R}$$
The accelerated field can be given as:
$$\vec E = \frac{e}{c^2 R^3}\left((\vec R \cdot \vec g)\vec R - R^2\vec g\right)$$

I don't know how the potential energy can be calculated here. If the direction of g and R are the same, shouldn't E = 0?

I assumed a non-relativistic case here. Am I going wrong here?

The problem hasn't mentioned whether the speed of the electron is relativistic.
Do you really think that something can reach relativistic speeds after falling for 1 centimeter?

Hint: Look up the Larmor formula for energy radiated by an accelerating charge.

Do you really think that something can reach relativistic speeds after falling for 1 centimeter?

Hint: Look up the Larmor formula for energy radiated by an accelerating charge.

Thanks, I got it! Here goes:
Considering the motion of the electron along the y-axis.
The dipole moment is $\vec p = -ey\hat y$ and $y^2 = {1\over 2}gt^2$
Hence,
$$\vec p = -{1\over p}get^2\hat y$$

$$\ddot{\vec p} = -{1\over p}ge\hat y$$

By Larmor's formula, the radiated power is given by,

$$P = {2\over 3c^3}(ge)^2$$

Now, the time t it takes to fall a distance 'h' is given by: $h = {1\over 2}gt^2$

$t = \sqrt{{2h\over g}}$

So the energy released in falling a distance 'h' is: E = Power x time

$$E_{rad.} = Pt = {2g^2e^2\over 3c^3}(\sqrt{{2h\over g}})$$

Meanwhile the potential energy lost is PE = mgh.

$$f = {E_{rad}\over E_{pot}} = {2g^2e^2\over 3c^3}(\sqrt{{2h\over g}} \times {1\over mgh} = {2e^2\over 3mc^3}(\sqrt{{2g\over h}})$$