Cherenkov radiation in Jackson

[Zadig]

Homework Statement

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

Homework 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

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.

 

[Jhero]

Thank you for your question. I will try my best to help you with this problem.

First, let's break down the equation given in Jackson's Classical Electrodynamics:

dI(omega)/dx = [(e^2)*(omega)/(c^2)]*[1-(1/(beta^2)*(dielectric constant of Lucite))]

dI(omega)/dx represents the change in energy radiated per frequency (I(omega)) over a certain path length (x). This can also be written as dE(omega)/dx, where E(omega) is the total energy radiated per frequency.

(e^2)*(omega)/(c^2) represents the energy radiated per frequency (I(omega)) by a single electron. This is found by dividing the square of the electron's charge (e^2) by the square of the speed of light (c^2), and multiplying by the frequency (omega).

[1-(1/(beta^2)*(dielectric constant of Lucite))] represents the correction factor for the dielectric medium (Lucite) through which the electron is passing. This takes into account the effect of the medium on the energy radiated by the electron.

Now, to find the total number of quanta emitted per centimeter of path in Lucite by a 1-Mev electron, we need to integrate the equation above over the wavelength range of 4000 to 6000 angstroms. This can be done by using the relationship between wavelength and frequency (omega = 2*pi*c/lambda) and converting the units to ergs/cm:

dE(lambda)/dx = [2*pi*(e^2)*c/lambda^2]*[1-(1/(beta^2)*(dielectric constant of Lucite))]

To find the total energy radiated per wavelength (E(lambda)) in ergs/cm, we need to divide the above equation by the wavelength (lambda):

dE(lambda)/dx = [2*pi*(e^2)*c/lambda]*[1-(1/(beta^2)*(dielectric constant of Lucite))]

Finally, to find the total number of quanta emitted per centimeter of path in Lucite by a 1-Mev electron, we need to integrate the above equation over the wavelength range of 4000 to 6000 angstroms:

Number of quanta emitted per cm = ∫dE(lambda)/dx*d(lambda