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Mossbauer effect: neglectable recoil or absolute no recoil?

I stumbled on the Mossbauer effect. From the wikipedia article, I cannot tell for sure if the momentum of recoil is really zero, or only neglectable.

From what I do (think I) understand, the theoretical energy of the recoil is very, very small, and is quantified into phonons that have a minimal energy. Hence, when the recoil energy would be less than the minimal energy of a phonon, there would be absolutely no recoil.

If my understanding is correct, wouldn't that violate the conservation of momentum (with all implications like being able to build a space drive that doesn't lose mass)?

Answer

The energy given to the lattice is much reduced, but not zero.

In a two-body interaction, if you push the bodies apart the ratio of the masses corresponds with the change in KE of the two bodies.  The more mass one of the objects has, the less energy the interaction gives to it (even though the momentum change is identical).

In the nuclear decay, one of the "objects" is a gamma ray.  If the other is an atomic nucleus, then it can get a significant fraction of the energy.  If however you are able to treat the crystal lattice as the other object, the much greater mass means that almost all the energy goes into the gamma ray.

The lattice still receives the entire impulse or change in momentum due to the interaction.  Just that the energy associated with it is tiny.

I didn't recognize that the big portion of your question was about the phonons.

> the theoretical energy of the recoil is very, very small, and is quantified into phonons that have a minimal energy.

Almost.  The energy of the recoil can be partitioned into vibrational energy within the lattice and bulk motion of the entire lattice.  Normally the energy term is dominated by these vibrations, which can be quantized into phonons.  In certain conditions, the vibrational energy is too low and is excluded, so the vibrational energy/phonons *is* exactly zero.  But the entire lattice still receives momentum from the interaction.  Because the mass of the lattice can be quite large, this will drive the change in velocity (and therefore the actual energy received) down to very low levels, but not theoretically zero.

> ...phonons take vibrational energy that is inherently different to energy associated with momentum change, and all the wikipedia fuzz about phonons is not to lose /another/ part of the energy to vibrations!

Rather than say that the vibrational energy is inherently "different", I would tend to think of it more like another "container" for the energy.

If that container is excluded at the time of the interaction, the the only locations for it are the gamma ray and the KE of the entire lattice.  If that container is available, then it can hold some of the energy and the the range for the gamma is much less constricted.

Edit Summary

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$\begingroup$ I understand that part, but what then is the fuzz about phonons? $\endgroup$
– Gyro Gearloose
Commented Apr 11, 2019 at 16:35
$\begingroup$ Added a bit that might help. $\endgroup$
– BowlOfRed
Commented Apr 11, 2019 at 16:51
$\begingroup$ Well, that's closer. But how can there be a change in momentum (as tiny as it may be) without a change in energy? $\endgroup$
– Gyro Gearloose
Commented Apr 11, 2019 at 16:56
$\begingroup$ There can't. I hope I don't suggest that it would be possible in my answer. The last sentence says it is not zero. $\endgroup$
– BowlOfRed
Commented Apr 11, 2019 at 17:00
$\begingroup$ Well, that's exactly the point I don't understand: (no need to say everything is tiny): low energy is quantified into /no/ phonons, so no energy is exchanged but still momentum and in accordance, energy is exchanged. This is my point of confusion. $\endgroup$
– Gyro Gearloose
Commented Apr 11, 2019 at 17:04

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