My question is motivated by this paper. They are measuring the mass of a molecular ion in a Penning trap, and they are able to see a difference due to the fact that the molecule gets polarized (the motion is classical and non-relativistic). I was able to derive their result, for an induced electric dipole, using the Lagrangian:
$$L = mv^2/2 + \alpha E^2/2$$
where $\alpha$ is the polarization and $E$ is the electric field. If we use the fact that $E = vB$ we can see from the form of the Lagrangian, if we take the derivative with respect to $v$:
$$\frac{\partial L}{\partial v} = mv + \alpha v B^2 = (m+\alpha B)v$$
From this we get an effective mass of:
$$m+\alpha B^2$$
which is consistent with their result. However, I was wondering, if we assume that the molecule is highly (or fully) polarized and not just weakly, instead of $\alpha E^2/2$ we have simply $d E$, where $d$ is the intrinsic dipole moment of the molecule. However, assuming $d$ is constant, which is (very close to being) true for a fully polarized molecule, we have $d E = dvB$ which gives:
$$\frac{\partial L}{\partial v} = mv + dB$$
now we can't factor out $v$ anymore and thus it's not clear anymore how to count $dB$ towards the mass of the molecule. However, intuitively, I would expect that the higher the polarization, the higher the shift in the measured mass. What am I doing wrong, or how should I interpret the $dB$ term in this case? Thank you!
