- #1
confuted
- 28
- 0
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
i(t)=Ae^{j\omega t}+Be^{-j\omega t}
with \omega=\frac{1}{\sqrt{LC}}. Here I am using i(t) as current, and j=\sqrt{-1}. For appropriate initial conditions, we can take
i(t)=A\sin{\omega t}.
Now the instantaneous current is equal to some charge density multiplied by the instantaneous velocity of the charge carriers:
i(t)=\rho v. (Is this a valid assumption?)
So the instantaneous acceleration of the charge carriers is
a(t)=\frac{dv}{dt}=\frac{d}{dt}\frac{i(t)}{\rho}=A\omega\cos{\omega t}
The time average of the square of this acceleration is
a^2=<a^2(t)>=\frac{A^2\omega^2}{2}
Now (in Gaussian units), the Larmor radiation formula is
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} (*)
We could model this (as far as the circuit is concerned) as an effective resistance. The power dissipated in a resistor is
P=i^2*R
Taking time averages and substituting in (*),
R=<P>/<i^2>=\left(\frac{2}{3}\right)\frac{e^2\omega^2}{c^3}
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
P=\frac{1}{3}\frac{e^2\omega^4}{c^3} (I've dropped his x02 factor.)
Where's the discrepancy?
Setup and solve the relevant differential equations, and you'll get
i(t)=Ae^{j\omega t}+Be^{-j\omega t}
with \omega=\frac{1}{\sqrt{LC}}. Here I am using i(t) as current, and j=\sqrt{-1}. For appropriate initial conditions, we can take
i(t)=A\sin{\omega t}.
Now the instantaneous current is equal to some charge density multiplied by the instantaneous velocity of the charge carriers:
i(t)=\rho v. (Is this a valid assumption?)
So the instantaneous acceleration of the charge carriers is
a(t)=\frac{dv}{dt}=\frac{d}{dt}\frac{i(t)}{\rho}=A\omega\cos{\omega t}
The time average of the square of this acceleration is
a^2=<a^2(t)>=\frac{A^2\omega^2}{2}
Now (in Gaussian units), the Larmor radiation formula is
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} (*)
We could model this (as far as the circuit is concerned) as an effective resistance. The power dissipated in a resistor is
P=i^2*R
Taking time averages and substituting in (*),
R=<P>/<i^2>=\left(\frac{2}{3}\right)\frac{e^2\omega^2}{c^3}
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
P=\frac{1}{3}\frac{e^2\omega^4}{c^3} (I've dropped his x02 factor.)
Where's the discrepancy?
Last edited by a moderator:
Take Q(t)=Acos(wt). Then you can differentiate twice to get another power of w.