- #1

Reshma

- 749

- 6

## 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:

[tex]\phi = \frac{e}{R(1 - \beta \cdot \hat R)}[/tex]

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 ([itex]\beta << 1[/itex]) then [itex] R(1 - \beta \cdot \hat R) \rightarrow 0[/itex] and the potential can be written as:

[tex]\phi = \frac{e}{R}[/tex]

The accelerated field can be given as:

[tex]\vec E = \frac{e}{c^2 R^3}\left((\vec R \cdot \vec g)\vec R - R^2\vec g\right)[/tex]

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?