# Most likely speed in Maxwell-Boltzmann distribution

1. Dec 24, 2015

### Vrbic

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

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

3. 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?

2. Dec 24, 2015

### 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.

3. Dec 24, 2015

### Staff: 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.

4. Jan 6, 2016

### Vrbic

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. Jan 6, 2016

### Staff: Mentor

What are your thoughts on this?

Chet

6. Jan 6, 2016

### Vrbic

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

7. Jan 6, 2016

### Staff: Mentor

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

8. Jan 6, 2016

9. Jan 6, 2016

### Staff: 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.

10. Jan 6, 2016

### Vrbic

Very nice ;) It is true what I said.