Calculate the radiation damping of a charged particle attached to a spring of natural frequency ω0, driven at frequency ω.

Question

Calculate the radiation damping of a charged particle attached to a spring of natural frequency [latex]ω_0[/latex], driven at frequency ω.

Accepted Answer

The equation of motion is

[latex]m\ddot{x}=F_{spring}+F_{rad}+F{driving}=-m\omega ^{2}_{0}x+m au \overset{...}{x} +F_{driving}. [/latex]

With the system oscillating at frequency ω,

[latex]x(t)=x_{0}\cos(\omega t +\delta ),[/latex]

so

[latex] \overset{...}{x} =-\omega ^{2}\dot{x}. [/latex]

Therefore

[latex]m\ddot{x} + mγ\dot{x} + m\omega ^2_0x =F_{driving},[/latex] (11.83)

and the damping factor γ is given by

[latex]γ = ω^2τ.[/latex] (11.84)

[When I wrote [latex]F_{damping} = −γmv,[/latex] back in Chap. 9 (Eq. 9.152), I assumed for simplicity that the damping was proportional to the velocity. We now know that radiation damping, at least, is proportional to [latex]\ddot{v}[/latex]. But it hardly matters: for sinusoidal oscillations any even number of derivatives of v would do, since they’re all proportional to v.]

[latex]F_{damping} = −mγ \frac{dx}{dt} .[/latex] (9.152)

Question 11.4: Calculate the radiation damping of a charged particle attach...

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