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

Use exactly the supplied inverse Fourier transform convention $e^{-i\omega \tau }$, with $\tau =t-t^{\prime }$. Then ${\partial }_t\mapsto -i\omega$ and ${\partial }_t^3\mapsto i{\omega }^3$. The transformed equation isThis algebraic inverse still needs a pole prescription to select the time-domain solution, particularly when a pole is on the real integration contour.

A retarded Green function obeys $G(\tau )=0$ for $\tau <0$ andFor a bounded or suitably tempered response represented by the real-axis Fourier integral, analyticity in the upper frequency half-plane is required, with suitable behaviour at infinity. For negative $\tau$, $e^{-i\omega \tau }$ decays on a large upper semicircle. If that half-plane has no singularities, contour closure gives zero response before the source. Lower-half-plane poles give factors decaying for positive time. Real poles are approached from below for a retarded prescription; at $\epsilon =0$, for example, $G_R(\tau )=\Theta (\tau )\sin \tau$.

The causality versus stability of a Green function distinction matters: an unstable initial-value problem can still have a strictly causal response that grows exponentially. Such a response has no ordinary real-axis Fourier transform. Its inverse Laplace transform must run on a frequency contour above its growth poles. Thus “upper poles violate causality” is correct only with the stable/tempered real-axis Fourier assumptions, not as an unconditional theorem about differential equations.

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.

Substitute $\omega ={\alpha }_3/\epsilon +\cdots$ into the cubic denominator. Its leading terms areThe nonzero large root obeys $1+i{\alpha }_3=0$, soThe sum of the three roots is exactly $i/\epsilon$. Combining this with the two slow-root expansions gives the refinement ${\omega }_3=i/\epsilon +i\epsilon +O({\epsilon }^3)$.

For physical radiative sign $\epsilon >0$, the fast pole lies high in the upper half-plane. Its homogeneous mode is $e^{-i{\omega }_{3}t}\simeq e^{t/\epsilon }$, a runaway. It prevents a bounded retarded response with the real-axis Fourier prescription even though the two slow poles are damped. If $\epsilon <0$, the fast pole decays but the slow pair grow. No nonzero real sign puts every pole in the lower half-plane.

There are two distinct boundary choices. A strictly causal initial-value Green function exists and includes the growing mode for $\epsilon >0$; it must be represented by a Laplace/Bromwich contour above that pole rather than the stated ordinary Fourier integral. Alternatively the real-axis inverse gives a bounded stationary no-runaway response and preacceleration: for $\tau <0$, upper closure encloses the fast pole and yields a nonzero response before the force. Its leading contribution isThis follows from the upper-pole residue $\mathrm{Res}G\simeq -i\epsilon$ and the factor $i$ from upper contour closure. Eliminating runaway growth by a future boundary condition therefore sacrifices strict retardation on the short scale $\epsilon$.

The singular root also shows why a perturbative expansion around only $\omega =\pm 1$ misses the problem. For a low-frequency effective description, order reduction of a radiation-reaction oscillator uses $\stackrel{...}{x}\simeq \dot{f}-\dot{x}$ in the small correction and gives $\ddot{x}+\epsilon \dot{x}+x=f+\epsilon \dot{f}$. For positive $\epsilon$ this approximate second-order equation has a causal damped response and discards the fast mode outside its intended regime; it does not make the original third-order equation simultaneously bounded and retarded.