- #1
kelly0303
- 573
- 33
Hello! My question is motivated by this paper (also attached below). 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!
$$L = mv^2/2 + \alpha E^2/2$$
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!
