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?


2 Answers 2


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 $\beta \ll \omega$.

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

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

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