1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Most likely speed in Maxwell-Boltzmann distribution

  1. Dec 24, 2015 #1
    1. The problem statement, all variables and given/known data
    What is the most likely speed in Maxwell-Boltzamann distribution?

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

    3. 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?
     
  2. jcsd
  3. Dec 24, 2015 #2
    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.
     
  4. Dec 24, 2015 #3
    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.
     
  5. Jan 6, 2016 #4
    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?
     
  6. Jan 6, 2016 #5
    What are your thoughts on this?

    Chet
     
  7. Jan 6, 2016 #6
    What are you asking me now? How or my oppinion if it is true?
     
  8. Jan 6, 2016 #7
    I'm asking to see if you can reason it out mathematically.
     
  9. Jan 6, 2016 #8
    No. I just ask :)
     
  10. Jan 6, 2016 #9
    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.
     
  11. Jan 6, 2016 #10
    Very nice ;) It is true what I said.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Most likely speed in Maxwell-Boltzmann distribution
Loading...