Difference between Maxwell-Boltzmann distribution and Maxwell-Boltzmann velocity distribution I read on Wikipedia that the Maxwell-Boltzmann distribution is given by:
$$f(v) = \sqrt{\left(\frac{m}{2 \pi kT}\right)^3}\, 4\pi v^2 e^{- \frac{mv^2}{2kT}}....(1)$$
In Wikipedia's words: "This probability density function gives the probability, per unit speed, of finding the particle with a speed near $v$". Now later on, they define the "distribution for velocity vector" as:
$$f_\mathbf{v} (v_x, v_y, v_z) =
\left(\frac{m}{2 \pi kT} \right)^{3/2}\exp \left[-\frac{m(v_x^2 + v_y^2 +v_z^2)}{2kT}\right]....(2)$$
and this gives "The probability of finding a particle with velocity in the infinitesimal element $[dv_x, dv_y, dv_z]$ about velocity $v = [v_x, v_y, v_z]$." which, for one dimension reduces to:
$$f_v (v) =\sqrt{\frac{m}{2 \pi kT}}\exp \left[\frac{-mv^2}{2kT}\right].$$
but since the speed just equal to the absolute value of the velocity in one dimension, after making the appropriate change of variables $u=|v|$ in $(1)$, we get:
$$f_u (u) =\sqrt{\frac{2m}{ \pi kT}}\exp \left[\frac{-mu^2}{2kT}\right]....(3) $$
where $u$ ranges from $0$ to $\infty$.


*

*But shouldn't $(3)$ correspond to $(1)$ since the interpretaions, as mentioned (by Wikipedia, at least) are the same? why are the equations different? 

*Also how can the probability of finding a particle in the velocity neighborhood about $v=0$ be the highest but probability of finding a particle in the speed neighborhood of $u=0$ be $0$ ? i.e , physically, why does $(2)$ peak at $0$ when $(1)$ is $0$ at $0$?

 A: You have to go back to see where your equation $(1)$ came from.
For one dimension the probability of particles having a velocity between $\vec v_{\rm x}$ and $\vec v_{\rm x}+ d\vec v_{\rm x}$ is given by 
$$f(\vec v_{\rm x})\;d\vec v_{\rm x} =\sqrt{\frac{m}{ 2\pi kT}}\exp \left[\frac{-mv_{\rm x}^2}{2kT}\right]\,d\vec v_{\rm x}$$
The speed distribution is given by 
$$f(v_{\rm x})\;dv_{\rm x} =2\,\sqrt{\frac{m}{ 2\pi kT}}\exp \left[\frac{-mv_{\rm x}^2}{2kT}\right]\,dv_{\rm x}=\sqrt{\frac{2m}{ \pi kT}}\exp \left[\frac{-mv_{\rm x}^2}{2kT}\right]\,dv_{\rm x}$$
the factor $2$ being there because the speed (magnitude) of $\vec v_{\rm x}$ is the same as that of  $-\vec v_{\rm x}$.
This is in agreement with your equation $(3)$.
Equation $(1)$ came from the idea that in three dimensions there is no preferred direction so
$f(\vec v_{\rm x},\vec v_{\rm xy},\vec v_{\rm z})\,d\vec v_{\rm x}d\vec v_{\rm y}d\vec v_{\rm z} = f(\vec v_{\rm x})d\vec v_{\rm x}\,f(\vec v_{\rm y})d\vec v_{\rm y}\,f(\vec v_{\rm z})d\vec v_{\rm z}$
You now have to count all the speeds which are the same ie the magnitude of the velocity $v$ is the same where $v^2 = v^2_{\rm x}+v^2_{\rm y}+v^2_{\rm z}$.
In this three dimensional case the volume of a shell of radius $v$ and thickness $dv$ is $d\vec v_{\rm x}d\vec v_{\rm y}d\vec v_{\rm z}= 4 \pi v^2 dv$ is being considered which results in your equation $(1)$.  
$$f(v) \,dv = \sqrt{\left(\frac{m}{2 \pi kT}\right)^3}\, 4\pi v^2 \exp \left[\frac{-mv^2}{2kT}\right] \,dv$$
The equivalent distribution for two dimensions with an area of a ring of radius $v$ and thickness $dv$ is $d\vec v_{\rm x}d\vec v_{\rm y}= 2 \pi v\, dv$ and $v^2 = v^2_{\rm x}+v^2_{\rm y}$ is 
$$f(v) \,dv = \left(\frac{m}{2 \pi kT}\right)\, 2\pi v\, \exp \left[\frac{-mv^2}{2kT}\right] \,dv$$
