Radiation from an ideal LC tank circuit

confuted

Consider an ideal LC tank circuit with some initial conditions such that oscillations take place. I am trying to find the amount of power radiated per cycle due to the accelerating charges (I realize that this should come out to be a very small value).

Setup and solve the relevant differential equations, and you'll get
[tex]i(t)=Ae^{j\omega t}+Be^{-j\omega t}[/tex]
with [tex]\omega=\frac{1}{\sqrt{LC}}[/tex]. Here I am using i(t) as current, and [tex]j=\sqrt{-1}[/tex]. For appropriate initial conditions, we can take
[tex]i(t)=A\sin{\omega t}[/tex].

Now the instantaneous current is equal to some charge density multiplied by the instantaneous velocity of the charge carriers:
[tex]i(t)=\rho v[/tex]. (Is this a valid assumption?)

So the instantaneous acceleration of the charge carriers is
[tex]a(t)=\frac{dv}{dt}=\frac{d}{dt}\frac{i(t)}{\rho}=A\omega\cos{\omega t}[/tex]
The time average of the square of this acceleration is
[tex]a^2=<a^2(t)>=\frac{A^2\omega^2}{2}[/tex]

Now (in Gaussian units), the Larmor radiation formula is
[tex]P=\left(\frac{2}{3}\right)\frac{e^2<a>^2}{c^3}=\left(\frac{1}{3}\right) \frac{e^2A^2\omega^2}{c^3}[/tex] (*)

We could model this (as far as the circuit is concerned) as an effective resistance. The power dissipated in a resistor is
[tex]P=i^2*R[/tex]
Taking time averages and substituting in (*),
[tex]R=<P>/<i^2>=\left(\frac{2}{3}\right)\frac{e^2\omega^2}{c^3}[/tex]

All seems well and good, except that Feynman (Feynman Lectures, Volume 1, Section 32.2) derives a similar formula for charge oscillating under SHO. His result seems to be
[tex]P=\frac{1}{3}\frac{e^2\omega^4}{c^3}[/tex] (I've dropped his x₀² factor.)

Where's the discrepancy?

confuted

Sorry to bump my thread, but I feel like it didn't get much attention because of LaTeX being down after the move. Does anybody have some insight here?

atyy

I can imagine a fudge. Take Q(t)=Acos(wt). Then you can differentiate twice to get another power of w.