Quality of an oscillating electron

[PragmaticYak]

Homework Statement

This problem is 3-16b) from French, Vibrations and Waves.

According to classical electromagnetic theory an accelerated electron radiates energy at the rate

K = (Ke^2 a^2)/(c^3)

where K = 6 × 10^9 N m^2/C^2, e = electronic charge (C), a = instantaneous acceleration (m/s^2), and c = speed of light (m/s).

b) What is the Q of the oscillator?

Relevant Equations

It was found in part a) that in one cycle, the oscillator radiates

ΔE = (8Ke^2 π^4 ν^3 A^2)/(c^3)

away, where A is the amplitude of the oscillator and ν is its frequency, in Hz. Also,

Q = (ω_o)/(γ)
E(t) = E_o e^(-γt)
ω_o = 2πν

The oscillator's initial energy can be found by considering when all of its energy is potential energy.

E_o = (1/2)kA² = (1/2)mω²A² = (1/2)m_e(2πν)²A² = 2m_eπ²ν²A²

where m_e is the mass of an electron. With this in mind, the energy dissipated after one cycle is given by

ΔE = E(0) - E(1/ν) = E_o - E_oe^-γ/ν = E_o(1 - e^-γ/ν)

Solving for γ, we see that

ΔE/E_o = 1 - e^-γ/ν

1 - ΔE/E_o = e^-γ/ν

-γ/ν = ln(1 - ΔE/E_o)

Assuming that ΔE/E_o is very small, we can make the very good approximation

-γ/ν = ln(1 - ΔE/E_o) ≈ -ΔE/E_o

Thus

γ = νΔE/E_o = ν * (8K²e²π⁴ν³A²)/(c³) * (1)/(2m_eπ²ν²A²) = (4Ke²π²ν²)/(m_ec³)

Calculating Q, finally,

Q = ω_o/γ = 2πν/γ = 2πν * (m_ec³)/(4Ke²π²ν²) = (m_ec³)/(2πνKe²)

However, the answer in the back of the book says

Q = (m_ec³)/(4πνKe²)

I cannot figure out where the error in my work is.

 

[TSny]

I don't see any mistakes in your work. It looks to me like your result for $Q$ agrees with Feynman's result $Q = \large \frac{3 \lambda mc^2}{4 \pi e^2}$ given in formula (32.10) of Vol. I of his lectures.

Here it is important to note that Feynman uses $e^2$ to denote $q^2/(4\pi \epsilon_0)$, where $q$ is the value of the charge in Coulombs. $\lambda$ is the wavelength of the radiation. French's constant $K$ is equal to $1/(6 \pi \epsilon_0)$ as you can see if you compare French's expression for the radiated power with the first formula given here.

Thus, starting with Feynman's expression for Q and replacing $e^2$ by $e^2/(4 \pi \epsilon_0)$, replacing $\lambda$ by $c/\nu$, and replacing $\epsilon_0$ by $1/(6 \pi K)$ you get your result. So, it appears that French's result is half of the correct result.

 

[PragmaticYak]

Thank you!