Solution

ID: past-exam-of-the-theoretical-physics-i-course-of-the-university-of-cambridge/2017/4/c/solution

Substitute into the cubic denominator. Its leading terms are
The nonzero large root obeys , so
The sum of the three roots is exactly . Combining this with the two slow-root expansions gives the refinement .
For physical radiative sign , the fast pole lies high in the upper half-plane. Its homogeneous mode is , a runaway. It prevents a bounded retarded response with the real-axis Fourier prescription even though the two slow poles are damped. If , 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 ; 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 , upper closure encloses the fast pole and yields a nonzero response before the force. Its leading contribution is
This follows from the upper-pole residue and the factor from upper contour closure. Eliminating runaway growth by a future boundary condition therefore sacrifices strict retardation on the short scale .
The singular root also shows why a perturbative expansion around only misses the problem. For a low-frequency effective description, order reduction of a radiation-reaction oscillator uses in the small correction and gives . For positive 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.

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