Most probable velocity from Maxwell-Boltzmann distribution

[TensorCalculus]

Homework Statement

Derive the formula $v_mp = \sqrt{\frac{2k_bT}{m}}$

Relevant Equations

$f(v)=4\pi(\frac{m}{2k_BT})^{\frac 3 2} v^2 e^{\frac{-E}{k_BT}}$
product rule: $\frac{d}{dx} (uv) = u'v + v'u$

This wasn't really a homework problem: I just randomly realised that I had been using the formula $v_{mp} = \sqrt{\frac{2k_bT}{m}}$ without actually knowing where it came from, so I decided to try and derive it. I got pretty close but I think I made some sort of silly mistake because the answer I got was the negative of what I should have gotten. I spent quite a while staring at it yesterday trying to figure out what went wrong, to no avail, and tried the same thing today... I fear I have tunnel vision. The mistake is probably a really small and dumb one, but I'm having quite a bit of trouble finding it

Using the idea that $v_mp$ would be when the plot of the Maxwell-Boltzmann distribution is at a maximum for that given temperature and mass, and the derivative is 0 at maxima:
$$ \frac{d[f(v)]}{dv} = 0$$
$$\frac d {dv} (4\pi(\frac{m}{2k_BT})^{\frac 3 2} v^2 e^{\frac{-mv^2}{2k_BT}}) = 0$$
$$\frac d{dv} (v^2 e^{\frac{-mv^2}{2k_BT}}) = 0$$
$$v^2(\frac{-mv}{k_BT})(e^{\frac{-mv^2}{2k_BT}}) + 2v(e^{\frac{-mv^2}{2k_BT}}) = 0$$
$$v(-\frac{mv}{k_BT}) + 2 = 0$$
$$v^2 = \frac{2}{-\frac{m}{k_BT}} = -\frac{2k_BT}{m}$$
$$v=\sqrt{-\frac {2k_BT}{m}}$$
Somehow I got a negative inside the square root: where did I go wrong?

 

[kuruman]

where did I go wrong?

See figure on the right. Where did the $v^2$ in the last equation come from and why is there no sign change when you move the $2$ to the right hand side?

It looks like you did too much algebra in your head.

 

[anuttarasammyak]

From

$$\frac{d}{dv} F = 2v \frac{d}{dv^2} F = 0 $$
It is easier to calculate.

 

[TensorCalculus]

View attachment 364375See figure on the right. Where did the $v^2$ in the last equation come from and why is there no sign change when you move the $2$ to the right hand side?

It looks like you did too much algebra in your head.

The $v^2$ came from pulling the v out from the numerator of the fraction in the first line.
I see my mistake now how did I not spot it . All I had to do was realise that subtracting the two meant that it would be negative on the other side... whoops...

View attachment 364376


From
View attachment 364377

$$\frac{d}{dv} F = 2v \frac{d}{dv^2} F = 0 $$
It is easier to calculate.

yeah... I don't know how I managed that.
What do you mean by the last bit where you talk about that being easier to calculate?

 

[anuttarasammyak]

Formula in ( ) , I referred as F, has v^2 only, no single v. Try replacing v^2 with x in my proposal. The equation would become easier to handle.

 

[TensorCalculus]

Ah right: I'll give it a shot tomorrow morning, thank you for the suggestion!