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Do phonons conserve momentum or not? |
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| Jun3-12, 02:12 PM | #1 |
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Do phonons conserve momentum or not?
I'm reading Kittel and on p101 he says
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| Jun3-12, 02:15 PM | #2 |
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In inelastic neutron or x-ray scattering, the momentum is transferred onto the entire lattice. Since that has a huge mass, the effect is zip. However, the energy/momentum spectrum measured by the neutron is the phonon dispersion curve, as if the neutron was scattered of a massive particle.
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| Jun3-12, 02:32 PM | #3 |
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Hm, but suppose we have an incoming electron with wavevector k and that is reflected with wavevector k', in the process creating a phonon with wavevector K, then Kittel writes down the momentum conservation law as [itex]\mathbf k = \mathbf k' + \mathbf K + \mathbf G[/itex] where G is a reciprocial lattice vector and which, according to Kittel, corresponds to the [itex]\hbar \mathbf G[/itex] recoil momentum of the whole lattice as a result of the collision.
My problem is then: if that K wavevector doesn't correspond to a real momentum [itex]\hbar \mathbf K[/itex], then why do we say that "the true momentum of the whole system is rigorously conserved"? |
| Jun4-12, 03:33 AM | #4 |
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Do phonons conserve momentum or not?
I think the recoil momentum of the whole lattice has to be [itex]\hbar(K+G)[/itex].
The phonon is a displacement wave delocalized on the whole lattice, therefore [itex]\hbar K[/itex] should not be considered a "real" momentum. Saying that only [itex]\hbar G [/itex] is the recoil momentum of the lattice does not make any sense at all. In case of doubt, check another textbook, e.g. Ashcroft and Mermin. |
| Jun4-12, 07:49 AM | #5 |
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That is an interesting question. The clue is the distinction between "true" momentum and "pseudo-" momentum or crystal momentum (although I think the first term is more appropriate).
While a phonon carries pseudo momentum it does not carry true momentum (at least as long as k is not 0). Pseudomomentum also is well defined and conserved in fluids, i.e. in materials where the corresponding translational symmetries aren't broken to a discrete subgroup as in crystals. Hence in fluids G won't appear. In a medium you can distinguish between translations as generators of symmetry of both the excitations and the medium (which gives rise to momentum as a conserved quantity) and of translations of only the excitations relative to the medium (which gives rise to quasi-momentum). See e.g. http://arxiv.org/pdf/cond-mat/0012316.pdf or this article by Peierls: www.jstor.org/stable/10.2307/79058 or http://arxiv.org/pdf/0710.0461.pdf |
| Jun9-12, 04:28 PM | #6 |
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The law of conservation of momentum is only valid in a system which is invariant to translation. If the Hamiltonian varies with spatial position, then the conservation of momentum can’t be true. A phonon is a vibrational or rotational excitation. A phonon can also be visualized as a quasiparticle that is traveling in a medium. The phonon to some extent satisfies the de Broglie relations, although it is not a true particle. Visualizing the phonon as a particle is a way to take into account the harmonic part of the elastic potential of the medium. However, the phonon undergoes “collisions” due to the anharmonic part of the potential. If the medium was really homogeneous, then conservation of momentum would always be satisfied in collisions. However, the medium is seldom homogeneous. The inhomogeneities cause variation from the law of conservation of momentum. In an ideal crystal, the inhomogeneities are periodic in space. That means the Hamiltonian of the crystal is invariant to displacements of specific size, which are called reciprocal vectors. Thus, the conservation of momentum is only conserved modulo the reciprocal vectors. Thus, collisions between phonons in a crystal sometimes have a momentum added or subtracted by a reciprocal vector. If the reciprocal vector is not zero, the collision places the phonon in a different branch. Such a collision is referred to as an “umklapp collision”. Here are some links to help you. http://en.wikipedia.org/wiki/Umklapp_scattering “Umklapp scattering (also U-process or Umklapp process) is the transformation, like a reflection or a translation, of a wave vector to another Brillouin zone as a result of a scattering process, for example an electron-lattice potential scattering or an anharmonic phonon-phonon (or electron-phonon) scattering process, reflecting an electronic state or creating a phonon with a momentum k-vector outside the first Brillouin zone. Umklapp scattering is one process limiting the thermal conductivity in crystalline materials, the others being phonon scattering on crystal defects and at the surface of the sample.” http://en.wikipedia.org/wiki/Brillouin_zone “In mathematics and solid state physics, the first Brillouin zone is a uniquely defined primitive cell in reciprocal space. The boundaries of this cell are given by planes related to points on the reciprocal lattice. It is found by the same method as for the Wigner–Seitz cell in the Bravais lattice. The importance of the Brillouin zone stems from the Bloch wave description of waves in a periodic medium, in which it is found that the solutions can be completely characterized by their behavior in a single Brillouin zone.” http://www.oeaw.ac.at/smi/uploads/me...9-Lesson12.pdf “Invariance Principles and conservation laws Example 8 Invariance under translation (moving origin) ~x ! ~x+ ~ x0, leads to conservation of momentum.” |
| Jun9-12, 04:55 PM | #7 |
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In Umklapp scattering, the crystal lattice gains the negative of the reciprocal lattice added during the scattering event. So the "true momentum" is conserved. Almost always, the translation of the entire crystal is ignored. The displacement of the entire crystal due to phonon scattering is generally too small to measure, anyway. However, the displacement of the crystal is theoretically real. The translation of the crystal is unimportant for calculating thermal resistivity. When calculating resistivity, one only has to acknowledge that Umklapp scattering takes place. |
| Jun9-12, 10:57 PM | #8 |
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If there is a phonon in the crystal, then the atoms are vibrating. However, the lattice vectors are not portrayed as vibrating. Each lattice point is the time averaged position of an atom or molecule, not an instantaneous position. The position has to be averaged over an integration time much larger than the period of the vibration. Similarly, the momentum of each atom in a vibrating crystal changes rapidly. However. the momentum of a phonon is constant between collisions. This only makes sense if the momentum of a phonon is time averaged over a very long time. Similarly, the center of mass has to be averaged over a long time compared to the periods of vibration. Therefore, the pseudomomentum of a phonon is the time averaged momentum of the vibrational mode modulo a reciprocal vector. In Umklapp scattering, the time averaged momentum of the phonons is not conserved. However, the time averaged center of mass changes momentum enough to compensate for the momentum of the phonons. Making the vibrational mode into a quasiparticle is like time averaging. The energy of a phonon is precisely determinable. So the uncertainty in time of the phonon has to be very long. Therefore, just in order to use the concept of quasiparticle one has to assume a very long integration time. So consider the lattice points as time averages of the position of averages. If the lattice moves at a constant velocity, it means that the time averaged position of the center of mass has that velocity. That is where the "real momentum" is going. |
| Jun12-12, 02:46 AM | #9 |
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| Jun12-12, 05:52 PM | #10 |
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Consider an finite crystal at a finite temperature with a two vibrational modes excited above and beyond the thermal phonons. Assume that the two vibrational mode are traveling waves moving in opposite directions with the same frequency. The crystal is very big compared to the lattice constant, but the lattice does not extend into infinity. Assume that the entire crystal is on an horizontal plane. It appears to me that the momentum of each separate atom is changing with time. The momentum is going to vary rapidly because of the phonons, both the nonthermal traveling wave and the thermal phonons. Yet, the lattice is usually visualized as being stationary. The phonons travel through the stationary lattice. I propose that this is a time averaged lattice. The position of each lattice point is actually a time averaged position. The integration time has to be finite, but it can be very long. Claiming that a phonon has a specific momentum does not make sense without some time averaging. The motion of the atoms associated with the vibrational mode are changing at every instant of time. Therefore, the total momentum of the vibrational mode has to be changing all the time. The momentum of the phonon can't be well defined unless the momentum is being time averaged over a finite time longer than the period of the vibration. |
| Jun12-12, 06:39 PM | #11 |
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If there was no time averaging, then a phonon couldn't have a slowly varying momentum associated with it. However, this doesn't specify the momentum of what. Every unit cell in a crystal has to have a center of mass. The momentum carried by any mode of vibration can be divided into two parts. One component of momentum is the periodic vibration of the center of mass of the unit cell. The other component of momentum is associated with the motions of each atom in the unit cell relative to the center of mass of the unit cell. Therefore, I suggest that what is referred to as the pseudomomentum is the time averaged component of momentum associated with the vibrations of the center of mass. There is another component of momentum associated with the internal structure of the unit cell. An elastic collision between two phonons results in the conservation of pseudomomentum. This is merely the time averaged momentum carried by the wave of oscillating center of masses. An Umklapp collision between two phonons results in the total momentum of the colliding phonons being changed by one reciprocal lattice. Thus, some of the momentum associated with the center of mass of the unit cell gets becomes associated with the relative motions of the atoms in the unit cell. Thus, the real momentum can be conserved while the pseudomomentum is not conserved. |
| Jun13-12, 03:01 AM | #12 |
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| Jun13-12, 04:53 AM | #13 |
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For a transverse phonon the motion of the individual atoms is perpendicular to the pseudo-momentum. Trying to derive the pseudo-momentum or the real momentum transfer from the time-averaged motion of the individual atoms does not make sense.
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| Jun13-12, 11:10 AM | #14 |
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I don't have a reference nor have I worked out the mathematical details for a phonon in a crystal. However, I can visualize how a traveling wave in a crystal can move the center of mass of the crystal. I know of two good analogs for "the real momentum" of a phonon. Ocean waves have a pressure associated with them. It is sometimes said that ocean waves don't move water. Most of the motion of water in an ocean wave is in a circle. Because of the circular motion, the time averaged current caused by an ocean wave is unexpectedly small. However, it is not zero. The Stokes current is a term describing the time averaged flow of water in an ocean wave. Electromagnetic waves are comprised of oscillating electromagnetic fields. At any one instant, the force on an electrons caused by an electromagnetic wave can be large. However, the time averaged force of a traveling electromagnetic wave is not zero. The combination of electric and magnetic fields results in a component of force on the electron which is not zero. This results in a traveling electromagnetic waving having a radiation pressure. A traveling wave mode in a crystal would have to be similar "Stokes current". If two mirror image traveling waves where excited in a crystal, the superposition would be a standing wave. There is no "Stokes current" for a standing wave. However, a single traveling wave mode would have to have a nonzero "Stokes current" associated with it. There would have to be a time averaged translation of the crystal atoms associated with the traveling wave. Imagine a traveling wave in a finite crystal. The crystal has boundaries. The traveling wave has to be emitted at the beginning of the crystal and absorbed on the other end of the crystal. Spatial averaging the momentum of the entire crystal, end to end, won't be zero. The momentum at the boundaries are not balanced. I just realized that this may be the real answer to the OP's question. I admit that I hadn't thought it out completely. However, let me try again. I am trying to learn this, too. The OP's question got me interested. I discovered that I don't know my field as thoroughly as I though. So I am trying out different definitions of "momentum". Feel free to correct me if you find a flaw. The momentum of a phonon represents the momentum of the atoms spatially averaged over a unit cell when a traveling wave passes through it. The traveling wave causes pressure on the boundaries of the unit cell. If there is nothing holding it in place, the unit cell has to accelerate due to the "radiation pressure" of of the traveling vibrational mode. An elastic collision between phonons does not change the acceleration of the unit cell. An Umklapp reaction takes place when another unit cell exerts a force that stops the first unit cell from accelerating. Therefore, I think the "pseudo momentum" of a phonon is really associated with the "radiation pressure" of a traveling vibrational mode on a unit cell. If the boundaries of the crystal are not fixed by external forces, a crystal can move because of the traveling waves. "True momentum" is associated with the radiation pressure of traveling waves on the entire crystal. |
| Jun26-12, 03:03 AM | #15 |
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I was out for vacation but thought about the question for some time. Now I think that the conservation of crystal momentum is not a consequence of translational invariance (at least not directly) but of invariance under the permutation of identical nuclei. It is well known that some subgroup of the "complete group of nuclear permutations & inversion" (CNPI) underlies the point group symmetry classification of rotations, vibrations and electronic states in molecular physics, see e.g.
http://books.google.de/books/about/M...d=FZEI7VmjNyMC and it is clear that the vibrational labels of a crystal can be obtained in the limit of sufficiently large molecules. However, I don't know any condensed matter book which treats vibrations on that basis. The relevant permutations would be permutations where neigbouring atoms along a lattice vector are interchanged (cyclically if periodic boundary conditions are assumed). In the limit of an infinitely large crystal, the corresponding group is isomorphical to the translation group by lattice vectors. The operations act like passive translations on the electronic and vibrational part, however, in contrast to real translations, they won't affect the center of mass of the crystal. |
| Jul4-12, 03:40 AM | #16 |
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Yesterday I found the time to go through the Appendix on crystal momentum in Ashcroft and Mermin. Indeed they derive crystal momentum from the permutation of the indices of the ion cores.
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| Jul5-12, 01:32 PM | #17 |
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The clue is in the determination whether phonon is a particle (or quantized-wave) or merely a quantum by which the vibrating lattice interacts with other objects (like electron, neutron, etc.)
It is only a quantum so it does not "carry" physical momentum, as it is a quantum of physical mentum/energy by itself. What carry physical mentum/energy are the lattice and the other objects like electrons, neutrons, which interact by exchanging (or not exchanging!) physical mentum/energy in units of phonon. While this may eliminate the original contradiction (?), what about photon? Is photon merely a quantum? So the question is really a difficult one. |
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