How are the energy and speed distributions of classical particles related?

[sachi]

We are given that
n(E) dE = A(E^0.5) dE
gives us the no. of particles between E and E+dE, where E is the energy of a particle and A is a constant. We are told to model the particles as classical particles i.e E=0.5m(v^2)
We need to find the speed distribution. This is of the form N(v)dV, where N is a function of v. We were given a solution that says substitute dE = mv dv into our first eq'n and we get A (E^0.5) mv dv = A ((m^3)/2)^0.5) (v^2)dv. Therefore we conclude that N(v)dv = A ((m^3)/2)^0.5) (v^2)dv. This means that we initially have to assume that n(E)dE = N(v)dv , and I am not sure about the logic of this. There must be some physical principle about why the distribution of particles with respect to one variable can be transformed into another.

Thanks v. much.

 

[Physics Monkey]

You know how many particles have energy between E and E + dE, and you want to find how many particles have speed between v and v + dv. But you also know that the range (v,v+dv) of speeds corresponds to a certain range (E,E+dE) of energies, right? In this case you would have $$E = mv^2/2$$ and $$dE= m v dv$$. What does this say? It says that all the particles with speed (v,v+dv) also have energy (E,E+dE) provided we identify E and dE as I did above. Now write this statement in mathematical language: "number of particles with energy in (E,E+dE)" = $$n(E)dE$$, and "number of particles with speed in (v,v+dv)" = $$N(v)dv$$. Both of these are definitions, and the connection between them comes when you realize that these two numbers should be equal if E and dE are connected to v and dv as above.