Maxwell Distribution: Deriving Mean and Max Speed

AI Thread Summary
The discussion revolves around deriving the mean speed (v-bar) and maximum speed (v-max) of particles in a Maxwell distribution. The mean speed is expressed as v-bar = ((8kT)/(pi*m))^(1/2), while the maximum speed is given by v-max = ((2kT)/m)^(1/2). Participants discuss the need for standard integrals and differentiation techniques to solve the problem, particularly focusing on the distribution function p(v) proportional to v^2 exp(-mv^2 / 2kT). There is an emphasis on using the product rule for differentiation to find the maximum speed and setting up integrals for calculating the mean speed. The conversation highlights the importance of understanding probability functions in these calculations.
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Homework Statement




show that the mean speed of the particles (v-bar) and the speed of the maximum of the distribution are given by

v-bar=((8kT)/(pi*m))^(1/2)
and
v-max=((2kT)/m)^(1/2)

Homework Equations





The Attempt at a Solution


tried looking thru noted and textbook. i sussed that one is supposed to use standard integrals, but no notes or textbook is telling me how or where
please help, thanks in advance
 
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Do you have an expression for the distribution function? How would normally find the maximum of a function?
 
thanks for the suggestion
ok, so the distribution function is
p(v) is proportional to v^2 exp(-mv^2 /2kT)
so i differentiate it and get
dp/dv is proportional to (-m/2kT)*2v^3 exp(-mv^2 /2kT)=0
hmm
 
Not a bad start, that's the right idea, but you didn't differentiate correctly.

p(v) is proportional to v2·e-mv2/2kT,
or in other words (a function of v)·(another function of v).

Use the product rule for differentiation to find the derivative.
 
i got it
thanks a lot
still clueless on the mean speed tho
 
The mean of a quantity is often calculated by using it's probability function and setting up an integral ... are you familiar with that method?
 

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