Interpretation of the Boltzmann distribution and link with Maxwell-Boltzmann distribution [duplicate]

Question

I'm having trouble interpreting the boltzmann distribution and making the link with the Maxwell-Botlzmann distribution. I have found several discussions dealing with this problem :

The shape of the Maxwell-Boltzmann distribution

How can the Maxwell-Boltzmann distribution be reconciled with the Boltzmann distribution?

Why do velocities obey the Boltzmann distribution?

but none of them gave me a satisfying answer.

First, let's talk about the Boltzmann distribution : $$\begin{align} P(E) \propto e^{-E/k_B T} \tag{1} \end{align}$$. For me, this expression is a probability density function that allows us to find the probability of a particle having an energy between two bounds by integrating the expression between the two desired bounds (provided that the function is normalised, i.e. that the integral between the two extreme energy levels is equal to 1). I don't see the point of evaluating this expression for any value of E, all we're interested in is calculating the integral to get a probability, but I've read several people saying that the expression corresponds to a probability, which confused me, for me the expression is a probability density and not a probability. My first question is: is my interpretation of the Boltzmann distribution correct?

My second question concerns the Maxwell-Boltzmann distribution $$\begin{align} P(v) \propto v^2 e^{-mv^2/2k_B T} \tag{2} \end{align}$$. I don't understand why we need a new expression for speed. Speed is associated with kinetic energy so I thought we could simply replace the E in expression (1) with $mv^2/2$ and thus get an expression for the probability density of finding a particle at some speed but it doesn't work. it does work for the velocity vector which I find even more confusing. My second question is: why the speed distribution depends on the Maxwell-Boltzmann distribution while the velocity vector distribution depends on the Boltzmann distribution.

Sorry if my interpretations are off the mark. I'm trying to understand the basics of thermodynamics by myself thanks to the book "Concepts in Thermal Physics" but it is not easy.

Answer 1

Yes, it is a probability density. It should really be written as $$\tag1P(v_x,v_y,v_z)\mathrm dv_x\mathrm dv_y\mathrm dv_z\propto e^{-\frac{m\vec v^2}{2k_BT}}\mathrm dv_x\mathrm dv_y\mathrm dv_z$$ because then it becomes very clear that this simplifies into $$\tag2P(v)\mathrm dv\propto4\pi v^2e^{-\frac{mv^2}{2k_BT}}\mathrm dv$$ You can think of it as Equation (1) being the probability density to be in a particular microstate, whereas Equation (2) is a slight macrostate grouping of the same thing. They are extremely related and obviously treated as different aspects of the same one thing most of the time.