# Cherenkov radiation in Jackson

1. Feb 9, 2010

1. The problem statement, all variables and given/known data
Determine how many quanta with wavelengths between 4000 and 6000 angstroms are emitted per centimeter of path in Lucite by a 1-Mev electron, where the index of refraction for Lucite is 1.50 in this range.

This is question 14.16 in Jackson's Classical Electrodynamics

2. Relevant equations

This is equation 14.133 from Jackson's
dI(omega)/dx = [(e^2)*(omega)/(c^2)]*[1-(1/(beta^2)*(dielectric constant of Lucite))]

Where I(omega) is the total energy radiated per frequency, and x is the path length

3. The attempt at a solution

Well, this seemed easy enough. Take the whole thing, divide by planck's constant times omega, and we should have something like ergs/cm. But checking the units, this isn't true. In fact, the equation Jackson gives for energy per path length seems to be in units of work, since

(e^2)*(omega)/(c^2) gives (ergs)*(cm)*(s-1)/(cm^2/s^2) = (ergs)*(seconds)/(cm) and so I am very confused how this formula seems to be working out.