# Radiation by an accelerated charge

An elastically bound electron vibrates in simple harmonic motion at frequency $$\omega$$ with amplitude $$A\;.$$

Find the average rate of loss of energy by radiation.

So I think I can use Larmors formula for this but what would I put for the acceleration because I don't think I can just put acceleration $$-A\sin(\omega t)$$ here (for position $$x=A\sin(\omega t)$$) since it loses energy and so the amplitude should decrease, and if I put $$A$$ in as a function of time $$A(t)$$ how then would I calculate the average?

For damped motion I get this:

$$x= Ae^{-\alpha t}\cos\omega t\\ \ddot{x}=Ae^{-\alpha t}(\alpha ^2 \cos(\omega t)- \alpha \omega \cos(\omega t)+\omega^2 \sin(\omega t)) \\ \langle P_\text{rad}\rangle= \left\langle \frac{e^2\ddot{x}^2}{6\pi\epsilon\cdot c^3}\right\rangle= \frac{e^2}{6\pi\epsilon\cdot c^3}\;\frac{\omega}{2\pi}\int_{t}^{t+\frac{2\pi}{\omega}} \ddot{x}^2 dt'$$

where $$P_\text{rad}$$ is the power radiated using Larmors formula

This problem is from Morin and Purcell; the next part of the problem is

If no energy is supplied to make up the loss, how long will it take for the oscillator's energy to fall to $$1/e$$ of its initial value?

which seems to imply that the amplitude decreases exponentially but how would you derive that?

The equation for the displacement you have written appears to be the transient response of a damped harmonic oscillator. Usually what you would be dealing with here is the steady state response of a weakly damped, forced harmonic oscillator. But you are probably over-thinking the problem - if it says it is SHM, then there is no damping.

$$x = A \sin \omega t$$ $$\ddot{x} = -\omega^2 A \sin \omega t$$ $$< \ddot{x}^2 > = \frac{1}{2} \omega^4 A^2$$

Plug this into Larmor's formula.

If you assume it is a simple elastic oscillation, then the total energy is known in terms of the oscillation amplitude and frequency, and then I suppose you could say $$\frac{1}{2} \frac{d}{dt} m \omega^2 A(t)^2 \simeq - \frac{e^2}{6 \pi \epsilon_0 c^3} \frac{1}{2} \omega^4 A(t)^2$$ This is an approximation because you are assuming that the amplitude doesn't change very much over one cycle, so you can use the average $\ddot{x}^2$ to calculate the radiative losses (i.e. assumes weak damping). $$\frac{dA}{dt} = -\frac{e^2 \omega^2}{12 \pi \epsilon_0 c^3 m} A$$

Hence $$A = A(0) \exp(-\beta t),$$ where the time constant $$\beta = \frac{e^2 \omega^2}{12 \pi \epsilon_0 c^3 m}\ \ s^{-1}$$

For the approximation to be valid we can check retrospectively that $\beta \ll \omega$.

• so for the next part if I use that x to calulate the energy the energy is constant, which contradicts the question Dec 13, 2015 at 20:08
• @physicsnoob1000 see edit. Dec 13, 2015 at 20:15

When I understand you correctly, you want to know what the acceleration of the charge looks like?

Well I think that your explenation is a bit contradictory. If the motion was simple harmonic then you should actually be able to say that $$a = -A\cdot sin(\omega t)$$ But since the electron is elastically bound I would guess that this implies that you can add a damping-term. So you should solve the equation of motion for a damped oscillator. From that you can get the acceleration I guess.

Maybe have a look here: http://www.eecs.berkeley.edu/~attwood/srms/2007/04_new_2007.pdf

• hey Ive just edited my post, I get that thing in the picture for the average and Dec 13, 2015 at 19:38