n=1 n=2 n=3 n=n= E dN/dE Does the density of states vary through the continuum?

vxvx vyvy vzvz Classically, for free particles E = ½ mv 2 = ½ m(v x 2 + v y 2 + v z 2 ) Notice for any fixed E, m this defines a sphere of velocity points all which give the same kinetic energy. The number of “states” accessible by that energy are within the infinitesimal volume (a shell a thickness dv on that sphere). dV = 4  v 2 dv

Classically, for free particles E = ½ mv 2 = ½ m(v x 2 + v y 2 + v z 2 ) We just argued the number of accessible states (the “density of states”) is proportional to 4  v 2 dv dN dE  E 1/2 some constant

T  absolute zero The Maxwell-Boltzmann distribution describes not only how rms velocity increases with T but the spread about  in the distribution as well

With increasing energy (temperature) a wider range of velocities become probable. There are MORE combinations of molecules that can display this total E. That’s exactly what “density of states” describes. The absolute number is immaterial. It’s the slope with changing E, dN/dE that’s needed.

For a particle confined to a cubic box of volume L 3 the wave number vector is quantum mechanically constrained The continuum of a free particle’s momentum is realized by the limit L 

For a LARGE VOLUME these energy levels may be spaced very closely with many states corresponding to a small range in wave number The total number of states within a momentum range 

The total number of final states per unit volume or From E=p 2 /2m  dE=p dp/m  dE = v dp

number of final states per unit volume whose energies lie within the range from E to E+dE Giving (FINALLY) the transition rate of