4
$\begingroup$

Can solids be cooled to a sufficiently low temperature such that sound can travel through them without acoustic impedance?

It seems that if I had a solid in the shape of a torus, sitting in ambient vacuum, cooled to absolute zero, and then decided to introduce a compression along a circular slice (rest of the solid still at 0), then this wave should propagate indefintely looping around the torus.

Can this behavior occur at $T_c > 0$ if one takes into account Bose-Einstein condensation of phonons?

$\endgroup$
5
$\begingroup$

To some extent phonons are already condensed. Usually they are thought of as the Goldstone modes of broken translational symmetry (ignore optical modes for simplicity). This means the superflow of phonons is constant movement of the crystal itself. In other words, the crystal can move at a constant velocity indefinitely (assuming no outside forces at work). In the torus geometry, the entire crystal will rotate indefinitely without dissipation.

However, I believe what you are really interested in is Bose condensation of the nuclei of the lattice themselves. This is actually what Ultracold atom people achieve regularly. However in their case they are dealing with gases, not solids!

But can you actually see this in a normal solid (say Lithium?)

The answer is probably yes in theory, but in practice it is no. As a rule of thumb for bose condensation, you need to have the thermal de Broglie wavelength of the atoms be at least on the order of the atomic spacing itself. For electrons, this is achieved trivially even at extremely high temperatures because of their tiny mass. However for nuclei, which have masses that are thousands of times heavier, this temperature would need to be in the nano kelvins at the highest (pico kelvin in reality). At that point, you could possibly get in the regime of atomic condensation, but usually you need to get even colder.

Currently, the coldest you can get a solid is in the milli to (high) micro kelvin regime. Getting into nano or pico kelvin is completely out of reach for an ordinary solid. Perhaps it will be possible in the future someday.

$\endgroup$
3
$\begingroup$

I think there is an important difference between electrons and phonons that has to be considered here. The superconducting transition entails the formation of a many-particle state, where electrons attract each other and bind together through phonon-mediated interactions. In effect what is happening is that the motion of the electrons and lattice vibrations interfere coherently to form a quasiparticle that can move freely through the many-body vacuum. At finite temperature the main source of resistance in a metal is electron-phonon scattering, but this source is eliminated when you form a coherent electron-phonon state, so resistance drops basically to zero (although scattering of Cooper pairs by impurities and lattice defects is still possible, and will give a small residual resistivity).

However, there is no equivalent process for phonons. There is no mechanism through which they will interfere coherently with their scatterers (mainly, lattice defects, anharmonic effects, and electrons) to create quasiparticles that propagate freely. As mentioned by user157979, what phonons can do at low temperatures is form a Bose-Einstein condensate, which is a completely different type of object, with different properties.

$\endgroup$
  • $\begingroup$ There is an equivalent mechanism for phonons, that mechanism is Bose condensation of the ions themselves. However at that level the concept of a phonon is ill-defined. Numerically you're right that this probably is infeasible in want realistic material, but it is a possibility. $\endgroup$ – KF Gauss Jul 26 '17 at 14:59
  • $\begingroup$ So, what you are describing is propagation of sound in a Bose-Einstein condensate? Would you get lossless propagation in that case? I don't actually know much about the topic, could you recommend some references? $\endgroup$ – David Ruiz-Tijerina Jul 26 '17 at 23:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.