Complete and Partial Transfer of Energy in Bremsstrahlung Should Include Rotational and Vibrational Kinetic Energies

In complete braking achievement the rotational and vibrational as well as the linear kinetic energies of the charged particle results in a photon: $h\nu = 1/2mv^2 + 1/2I\omega^2 + 1/2kx^2.$ In partial transfer of kinetic energies of the deccelerating particle the resulting photon is $h\nu = [(1/2mv^2)_1 + (1/2I\omega^2)_1 + (1/2kx^2)_1] - [(1/2mv^2)_2 + (1/2I\omega^2)_2 + (1/2kx^2)_2]$. The linear kinetic energy of the charged particle is$1/2mv^2$, the rotational kinetic energy is $1/2I\omega^2$ and the vibrational kinetic energy is given by $1/2kx^2$.