23 Nov 2019

Relevant information:

For a normal distribution function, F = Aexp(-βE), β =1/(kT), E =1/2mv^2, A is normalization factor, T is temperature, kis theboltzmann's constant, v is velocity, and m is the mass.

For a fermi distribution function, F =1/(B1exp(-βE)+1).<- This is what I need tosolve

For a boson distribution function, F =1/(B2exp(-βE)-1).<- This is what I need tosolve

For a normal distribution, the relationship between themostprobable velocity and temperature, v = (2kT/m)^(1/2), To getthisequation, the book used 3D space. F d3v=Aexp(-1/2βmvx2-1/2βmvy2-1/2βmvz2)d3v.A new function for one axis is defined G dv=A'exp(-1/2βmx2) and to normalizethisfunction, the integral from -infinity to +infinity of functionG isdefined as 1. So, A'(2π/(βm))^(1/2) = 1 and A'is(βm/(2π))^(1/2). Since function F is in 3D space, A=(βm/(2π))^(3/2). The book defined the volume of aspherical shellwith velocity v to be 4πv2dv, so Fdv =4πAexp(-1/2βmv2)v2 dv. To findthe mostprobable speed of the distribution function, thederivative is takenand it is set to 0. So it looks likethis:exp(-1/2βmv2)(2v)-1/2βm(2v)exp(-1/2βmv2)v2=0. The book skipped a lot of steps but the tricky matheventuallyreduced to v = (2kT/m)^(1/2). <- This is the typeofrelationship I am looking for


I do not know how the book did this math but I have to dosomethingsimilar with the Fermion and boson functions to get thevelocity totemperature relationships for each of thefunctions.

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