A Is linear polarization not expected from an astrophysical maser

1. Jan 18, 2017

I was reading this text where they state: "Many cosmic masers are observed to have both circular and linear polarization although the reasons for linear polarization are not well understood."

Just curious, aren't there any molecules such as OH, CH4, or water masers that have transitions that emit linearly polarized waves? Why is the mechanism behind linear polarization in masers an outstanding question?

Perhaps on a related note, have astrophysical masers ever had measurements of rotation measure (RM)? Although these masers can arise in a variety of conditions (e.g. near stars, pulsar), they are typically in environments with strong EM fields, correct? So would we be able to observe an RM associated with a maser?

2. Jan 21, 2017

jambaugh

The issue is *selective* emission limited (or at least preferential) to one direction of linear polarization. Since the emission polarization due to atomic transitions is the difference in spin states the question reduces to the question of what limits these transitions. To answer your question about whether they can emit linearly polarized waves, of course they can, in any polarization and likewise any circularly polarized mode as well. You would in a general case get a completely randomized emissions, 50/50 left and right circular and 50/50 vertical and horizontal polarized and 50/50 rotated vertical vs rotated horizontal, and so on with elliptical cases too.

I am not in any way up to speed on the astrophysics and what follows is rank speculation but... for example consider atoms with an odd atomic number in a strong magnetic field and for transitions from excited to lower orbital modes such that there is already an electron occupying the lower orbit. The lower occupied electron modes being themselves polarized by the magnetic field, you would see the only de-excitation transitions possible are into the opposite spin state. Only excited states with spin in the same direction as the occupied lower state would be able to de-excite (so it can flip as it drops to the lower energy state in order to emit a spin-1 photon) and the polarization vector would be in the direction of this difference in spins.

As to whether the emitted photon's polarization is linear or circular, that depends on whether the photon is emitted parallel to or perpendicular to this background magnetic field and resultant polarization vector. I would guess all directions are equally likely for spontaneous emission but we're talking masers here so the amplification would likely be in the direction of the strong magnetic field if it were say the polar field of a pulsar or something like that. You'd have magnetic confinement of the plasma in that direction and thus a preference for circular polarization in a given direction.

This is probably totally wrong but it is, I imagine, the kind of thing they're talking about w.r.t. polarization mechanisms.

3. Jan 22, 2017

Thank you for the response. And so are forbidden transition simply ones corresponding to the transitions where the excited and ground state electrons may have differences beside simply a spin difference of 1?

And just to clarify a point, even for laboratory lasers, if you're considering a particular transition such as the magnetic dipolar 21 cm transition in Hydrogen, you would see an equal amount of signal at every possible elliptical polarization (including the special cases of circular and linear)?

To clarify one more point, the following paper says:

"HI 21 cm (magnetic dipole) transition with a similar study for the OH 18 cm (electric dipole) line."

But equation (2) in the paper shows that the 18 cm line is only given by the electric field (E) and polarization (P), while for the 21 cm line, they use essentially an analogous set of equations but now with E and P being replaced by the magnetic field (B) and magnetization (M), respectively. But shouldn't an electric field be able to cause (or observed from) the 21 cm line? And shouldn't a magnetic field be able to cause (or be observed from) the 18 cm line as well simply based on consideration of Maxwell's equations?

4. Jan 23, 2017

Ken G

Remember that the 21 cm line would make a terrible maser, because it is a forbidden transition. That is typical of spin flips. A maser needs a reasonable optical depth in order to stimulate emission, so I should think would be restricted to allowed transitions, i.e., transitions where the angular momentum in the photon comes from rotational angular momentum in the molecule, rather than from electron spin. But as was said above, if there is nothing to break the symmetry, you expect 50/50 in both circular polarizations, without any phase correlations, so that's unpolarized light. To get polarization, you need to break the symmetry, typically with a magnetic field. As soon as you have a magnetic field, then looking along that field, opposite polarizations have a different handedness relative to that field, which could cause one or the other circular polarization to receive preferential treatment at any given observed frequency. Or, when looking perpendicular to the field, the field direction can pick out one of the linear polarizations, in contrast to 90 degrees from that. This generally causes shifts in the resonant frequencies, so you can see correlations between polarization and observed frequency. I don't know why they say it is hard to understand the linear polarization, perhaps the problem is that the simple expectations are not borne out in the actual data.

5. Jan 23, 2017

Are not the modes from stimulated emission correlated so that if the initial seed photon had a linear polarization along a certain axis that the next photon would be more likely to have the same polarization state?

6. Jan 23, 2017

Ken G

Yes, stimulated emission always creates a second photon identical to the first.

7. Jan 23, 2017

With regards to the 21cm transition (Figure 2), is the change in angular momentum here between the initial and final states (regardless of $m_F$) always 1? It is due to the hyperfine structure resulting from the spin of the proton and the electron, and there are obvious energy differences when going from $F = 1$ and either $m_F = 1, 0$ or $-1$ to $F=0$; but is the angular momentum always considered to be $\Delta F$ here which is simply 1? Wouldn't the different $m_F$ lead to different angular momenta for emitted photons when going back to $F=0$?