# Umklapp Scattering & Conservation of Energy

1. Oct 10, 2013

After looking at Umklapp scattering, I believe I have finally gotten most of it down, but a few things are still not clear.

1) Momentum is not conserved for certain phonon collisions, and a certain number of reciprocal lattice vectors are transferred to the crystal lattice:

h-bar(k1+k2)=h-bar(k3 + G)

But if momentum is transferred to the lattice, which is called the thermalization of the lattice, isn't energy transferred to the lattice as well?

2) And if energy is transferred to the lattice. Why would the energy of the phonons still be conserved? It seems to me I am missing something.

3) If the lattice is "thermalized," what exactly is the consequence of this? Does it heat up/dissipate heat, similar to what happens with electrical resistance.

2. Oct 10, 2013

### UltrafastPED

Yes, the missing energy is thermalized in the lattice. But the energy that was carried away was carried by a phonon ... so phonon energy was conserved at the point of the u-event.

3. Oct 10, 2013

### DrDu

In deed. You have to be carefull to distinguish real momentum from crystal momentum. Phonons don't carry real momentum at all.
On the other hand, e.g. absorption of a photon will change the momentum of a crystal by a very small amount p. If the total lattice was at rest (i.e. its energy E=0), the energy carried by the lattice afterwards will be $E=p^2/M$. As M is practically infinitely large, E=0 even afterwards.
So conservation of real momentum is never a problem and does not interfere with energy conservation.
In the process of thermalization, the total energy remains conserved. However the distribution of the single phonons becomes Boltzmann.