# Most likely speed in Maxwell-Boltzmann distribution

## 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?

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
Mentor
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.

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
Mentor
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

What are your thoughts on this?

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

Chestermiller
Mentor
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.

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

Chestermiller
Mentor
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.

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.