 Solution

Metric-sign qualification. With this paper's $(-,+,+,+)$ metric signature, b givesThe minus sign in the printed Larmor formula would therefore make a positive radiative loss negative. A concrete counterexample is an instant of rest with nonzero acceleration: $F^{\alpha }=(0,m\mathbf{a})$, so the printed expression is $-q^2{a}^2/(6\pi {ϵ}_0{c}^3)$. That sign is appropriate for the opposite metric signature. With the convention already fixed in question 1, positive emitted power instead isAlternatively the printed negative quantity can be interpreted as the particle's signed radiative energy change, rather than the positive loss magnitude.

For fixed rest mass, resolve the acceleration parallel and perpendicular to the velocity, as in b. This givesWriting $C_q=q^2/(6\pi {ϵ}_0{c}^3)$, the Liénard radiated-power formula becomes $P_{\mathrm{rad}}=C_q({\gamma }^6{a}_{\parallel }^2+{\gamma }^4{a}_{\perp }^2)$. For uniform circular motion, the acceleration is perpendicular to the velocity; for straight-line motion it is parallel. Comparing the same particle at equal relativistic energy and equal laboratory acceleration magnitude $a$ therefore givesThis particular comparison makes the linear loss larger. The familiar larger loss from bending applies to a different comparison: at equal applied laboratory force magnitude $F$, $a_{\perp }=F/(m\gamma )$ and $a_{\parallel }=F/(m{\gamma }^3)$, giving $P_{\mathrm{circular}}=C_q{\gamma }^2{F}^2/m^2$ and $P_{\mathrm{linear}}=C_q{F}^2/m^2$. Stating which quantity is held fixed is essential.