Most likely speed in Maxwell-Boltzmann distribution

  • Thread starter Vrbic
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  • #1
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Homework Statement


What is the most likely speed in Maxwell-Boltzamann distribution?

Homework Equations


[itex] f(v)dv=4\pi(\frac{m}{2 \pi kT})^{3/2}v^2Exp(-\frac{mv^2}{2kT})dv[/itex]

The Attempt at a Solution


I know I need maximum of f(v) -> [itex]\frac{df}{dv}=0[/itex]. But it is not trivial to do. I found some solution where they said: [itex]\frac{d}{dv^2}(\ln[v^2Exp(-\frac{mv^2}{2kT})])=0[/itex]. But I don't know how they arrive to it. Could somebody advise?
 

Answers and Replies

  • #2
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Oooou sry, it is easy. I made a mistake I derivate f(v) - function without v2 term. Otherwise, why they can rewrite this problem to logarithm and derivative according to v2.
 
  • #3
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They just felt like its easier to differentiate the natural log of f, rather than f itself. Both f and its natural log have a maximum at the same value of v.
 
  • #4
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They just felt like its easier to differentiate the natural log of f, rather than f itself. Both f and its natural log have a maximum at the same value of v.
Thank you for your response. So can I say generally that if I apply some monotonic function to other, that extremes stay at same point?
 
  • #5
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Thank you for your response. So can I say generally that if I apply some monotonic function to other, that extremes stay at same point?
What are your thoughts on this?

Chet
 
  • #6
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What are your thoughts on this?

Chet
What are you asking me now? How or my oppinion if it is true?
 
  • #9
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Suppose g(y) is a monotoically increasing function of y, and y(x) is a function of x with a maximum. Then, by the chain rule,
$$\frac{dg}{dx}=\frac{dg}{dy}\frac{dy}{dx}$$
dg/dy is always positive, so dg/dx has a zero derivative at the same location where dy/dx has a zero derivative.
 
  • #10
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Suppose g(y) is a monotoically increasing function of y, and y(x) is a function of x with a maximum. Then, by the chain rule,
$$\frac{dg}{dx}=\frac{dg}{dy}\frac{dy}{dx}$$
dg/dy is always positive, so dg/dx has a zero derivative at the same location where dy/dx has a zero derivative.
Very nice ;) It is true what I said.
 

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