Most likely speed in Maxwell-Boltzmann distribution

[Vrbic]

Homework Statement

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

Homework Equations

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

The Attempt at a Solution

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

 

[Vrbic]

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.

 

[Chestermiller]

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.

 

[Vrbic]

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?

 

[Chestermiller]

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

 

[Vrbic]

What are your thoughts on this?

Chet

What are you asking me now? How or my oppinion if it is true?

 

[Chestermiller]

What are you asking me now? How or my oppinion if it is true?

I'm asking to see if you can reason it out mathematically.

 

[Vrbic]

I'm asking to see if you can reason it out mathematically.

No. I just ask :)

 

[Chestermiller]

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.

 

[Vrbic]

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.