Past exam of the theoretical physics I course of the University of Cambridge / 2017 / 4 / b / Solution

At $\epsilon =0$ the algebraic poles are $\boxed{{\omega }_{\pm }^{(0)}=\pm 1}$; a retarded limiting prescription displaces them infinitesimally below the real axis. Set ${\omega }_s=s+{\delta }_s$, with $s=\pm 1$ and ${\delta }_s=O(\epsilon )$. The denominator becomesConsequentlyFor positive $\epsilon$ these two poles move below the real axis, giving slowly damped oscillations. For negative $\epsilon$ they move above it and the slow oscillations grow. This is only the slow part of the radiation-reaction oscillator pole structure; the third pole must still be examined.

The sign can be checked independently by multiplying the equation by $\dot{x}$. With $E=({\dot{x}}^2+x^2)/2$,For periodic motion the total-derivative near-field term averages to zero, so positive $\epsilon$ gives a radiative energy loss, consistent with the slow poles' damping. Negative $\epsilon$ acts as energy gain in this model.

For the comparison equation as printed, $\ddot{x}+x-\epsilon \dot{x}=f$, the response denominator instead is $1-{\omega }^2+i\epsilon \omega$, with exact polesThus ordinary positive damping in that sign convention requires $\boxed{\epsilon <0}$, the opposite sign from slow radiation damping in the third-derivative equation. The word “damped” does not override the explicitly printed minus sign. With the usual equation $\ddot{x}+x+\gamma \dot{x}=f$, positive $\gamma$ gives the familiar lower-half-plane poles.