The magnetobremsstrahlung force $f^i$ acting on a particle is proportional to $$f^i\propto\left[\frac{d^2 u^i}{ds^2}-u^iu^k\left(\frac{d^2u_k}{ds^2}\right)\right],$$ where the second term could be rewritten in the form $(F_{kl}u^l)(F^{km}u_m)u^i$ that is a standard form for the radiated $4$-momentum or radiated energy for $i=0$ (that comes from the expression $d{\cal E}=(2e^2/3c^3)w^2dt$, where $w$ is a particle acceleration and ${\cal E}$ is a particle energy.
Energy loss could be found also by integrating $f^i\,ds$ over the world line. After integration the first term of $f^i$ goes to zero because $du^i/ds=0$ at infinity and the second term gives obviously the standard radiation form $(F_{kl}u^l)(F^{km}u_m)u^i$.
I am wondering why do we have to integrate over the whole particle world line (or may be average the energy losses over time) to find the correct expression for the radiated energy? Where in determining energy losses by the general expression $d{\cal E}=(2e^2/3c^3)w^2dt$ we neglected the first term $(d^2u^i/ds^2)$ of the radiation force $f^i$? Is it coming from the point that $d{\cal E}=(2e^2/3c^3)w^2dt$ is derived for the wave zone, i.e. far from the charge? If yes, how to express this explicitly?
What about the energy loss rate at a given point of the world line (it should take into account the full expression for $f^i$)?
