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Long wavelength dispersion relation in a solid
In solid state physics, perhaps the most visited two elementary excitations are phonons (vibration) and magnons (spin wave). It is interesting to notice that the long wavelength dispersion relation are different for them, i.e.
phonon: 
magnon: 
,
where is the frequency and 
is the wave vector.
One can go back to the derivation of those two excitations and find the mathematical reason. On the other hand, the origin can be resorted to the fundamental difference between the two equation of motions.
phonon: 
magnon: 
,
where x is the displacement,
is the angular momentum, 
is the spring constant and 
is the torque.
In solid, the restoring force for an oscillator comes from the imbalance of the interaction between neighbors. For long sinusoidal wave, it is always propotional to 
.
Hence, the final dispersion relation will be decided by the equation of motion. Since for phonon, it involves second derivative, one gets ; for magnon, the first derivative gives .
Change of supersaturation
In gas-solid phase transition, the phase boundary as a function of 
divides the two phases. In equilibrium, if a 
combination falls on the solid side, the matter is in its solid phase. On the other hand, at inequilibrium, for a finite amount of time, a system can be with a combination on the solid side but still remain in gas phase. This is called supersaturation. As shown in the Fig below for the points 
and if they are in gas phase.
The quantitative definition of supersaturation is
,
where 
is the pressure corresponding to the that on the phase boundary line for the temperature 
. This is actually the chemical potential difference between state and state 
in gas phase.
It is useful to know the dependence of on and 
. In other words, 
. Here we will start with the differential form of the relation, i.e. 
and 
. The second partial differential can be found from the definition:
,
while /Tneeds some effort to find.
To find , we look at
![\mu (P,T2)- \mu (P,T1)=[\mu(P,T2)-\mu(P'',T2)]-[\mu(P,T1)-\mu(P',T1)]\\
=[\mu(P,T2)-\mu(P,T1)]-[\mu(P'',T2)-\mu(P',T1)]](https://www.google.com/chart?cht=tx&chf=bg,s,FFFFFF00&chco=000000&chl=%5Cmu%20%28P%2CT2%29-%20%5Cmu%20%28P%2CT1%29%3D%5B%5Cmu%28P%2CT2%29-%5Cmu%28P%27%27%2CT2%29%5D-%5B%5Cmu%28P%2CT1%29-%5Cmu%28P%27%2CT1%29%5D%5C%5C%0A%3D%5B%5Cmu%28P%2CT2%29-%5Cmu%28P%2CT1%29%5D-%5B%5Cmu%28P%27%27%2CT2%29-%5Cmu%28P%27%2CT1%29%5D)
From Gibbs-Duham equation:
The first term
.
Or
The second term is the chemical potential change along the phase boundary.
Since on the phase boundary, one has
where 
is the enthalpy change between solid and gas phase per mole, which does not vary too much with the temperature.
Again using Gibbs-Duham equation,
.
Therefore,
Or
.
Finally:
.
Or
d\Delta\mu=RTdlnP-hdlnT
Differential pumping
Problem:
The governing equation of the problem is the following:
P2=G/S (1)
C(P1-P2)=P2S (2)
Note that P1>>P2, so the equation (2) can be written as
CP1=P2S.
In fact, if we believe these equation are true, the quantity P1 is not affected by the pumping speed Sas long as G is kept constant and P1>>P2. So changing the speed of the pump (e.g. by closing the gate valve upstream of pump to some extend) does not affect the pressure P1, at lest to the first order.
On the other hand, if all the quantities except P2are unknown, there is not way to know the pressure P1.
