# Maxwell velocity distrubution: why is it effectively dependant on magnitude?

I'm trying to understand how Maxwell velocity distribution is derived.

I'm using Chapter 2 of this PDF as a base.

I'm trying to understand this part:

With the definition of the probability density distribution above , we can determine the probability of a particle having the components of velocity $$v_x + dv_x,v_y+dv_y,$$ and $$v_z+dv_z$$ as : $$\Omega(v_x,v_y,v_z)\ dv_x\ dv_y\ dv_z\ =\ \left[f(v_x)\ dv_x\right]\left[f(v_y)\ dv_y\right]\left [f(v_z)\ dv_z\right]$$ $$\Omega(v_x,v_y,v_z) dv_x\ dv_y \ dv_z\ = \ \left[f(v_x) \ f(v_y)\ f(v_z)\right]\ dv_x \ dv_y \ dv_z$$ An assumption regarding the nature of this distribution :the gas is isotropic such that the direction of the particle movement does not affect the properties of the fluid.In this sense,the velocity distribution is effectively dependent on the magnitude of the velocity.

The magniude of the velocity is: $$|v|\ =\ \sqrt {v_x^2\ +\ v_y^2 +\ v_z^2}$$ $$\qquad$$ We now consider the natural logarithm of the distribution function: $$ln \ \Omega \left(v_x,\ v_y,\ v_z\right )\ =\ ln\ f(v_x) \ + \ ln \ f(v_y)\ +\ ln\ f(v_z)$$ $$ln\ \Omega(v) = \ ln\ f(v_x)\ +\ ln\ f(v_y)\ +\ ln\ f(v_z)$$

Why can we replace vector components with magnitude in $$\Omega$$ and get the same result? What is the actual assumption from the nature of this distribution which allows us to do this?

P.S: This question is derived from this question on MathStackExchange.

What you are trying to find is a distribution $$\Omega (\vec v)$$. In the source given by you the author choses to give a heuristical derivation of the maxwell distribution. The first assumption made is that the distribution $$\Omega(\vec v)$$ is seperable in the components so $$\Omega(\vec v) = f(v_x)f(v_y)f(v_z)$$ which for an non-interacting ideal gas is a kind of intuitive assumption since we see no reason why the distibution of the $$x$$-component should be dependent on any of the other components.
The next assumption made is that the propability distribution is not dependent on the direction of the velocity only on its magnitude $$|\vec v|=\nu$$. This can be pictured perfectly fine, since if one direction would be more propable then another the liquid as a whole would move in this direction.