Change of variables in an expression involving differentials I'm trying to derive Maxwell-Boltzmann's distribution using the statistics of the classical canonical ensemble. Doing some operations, I have found out that the probability that a molecule of a gas has a position between $\boldsymbol{r}$ and $\boldsymbol{r}+d^3\boldsymbol{r}$ and momentum between $\boldsymbol{p}$ and $\boldsymbol{p}+d^3\boldsymbol{p}$ is given by $(1)$:
$$P(\boldsymbol{r}, \boldsymbol{p}) d^{3} \boldsymbol{r} \space d^{3} \boldsymbol{p}=\frac{1}{V}\left(\frac{\beta}{2 \pi m}\right)^{3 / 2} e^{-\beta \varepsilon} d^{3} \boldsymbol{r} \space d^{3} \boldsymbol{p}  \tag{1}$$
where $\boldsymbol{r}=(x,y,z)$, $d^{3} \boldsymbol{r}=d x \space d y \space d z$, $\boldsymbol{p}=(p_{x}, p_{y}, p_{z})$, and $d^{3} \boldsymbol{p}=d p_{x} \space d p_{y} \space d p_{z}$.
I would like to derive the probability in terms of the velocity, $P(\boldsymbol{r}, \boldsymbol{v})$, from the expression $(1)$. However, I'm getting a wrong result just by replacing $\boldsymbol{p}$ by $\boldsymbol{v}$ and $d^3 \boldsymbol{p}$ by $d^3\boldsymbol{v}$... Could please anyone give me a hint about how should this kind of change of variables be done?
 A: If you define $P(r,p)$ as the probability that you'll find a molecule with position between $r$ and $r + d^3r$ and momentum within $p$ and $p + d^3p$, what's true is that
$$P(r,p) = \frac{1}{V}\left(\frac{\beta}{2\pi m}\right)^{3/2}d^3r d^3 p$$
without the differentials on the LHS. Think of it this way: if you were to just consider the position of the molecule in the container of volume $V$, the probability you find it in any box of side length $dr$ is just
$$P(r) = \frac{dr^3}{V}$$
since that's the fraction of the volume you're considering. The probability is infinitesimal, and it should be, since you're only considering an infinitesimal piece of the phase space. Now you should find that the change of variables $p \to mv$ produces the desired result.
A: Define a new probability function $\tilde{P}$ that gives the probability to find the particle within $d^3 x \,d^3v$ as
$$
(\text{probability within } d^3x\, d^3v)=\tilde{P}(\mathbf{r}, \mathbf{v}) \,d^3 x \,d^3v.
$$
Since this is just a change of variables, the probabilities must be the same:
$$
\tilde{P}(\mathbf{r}, \mathbf{v}) \,d^3 x \,d^3v = P(\mathbf{r}, \mathbf{p}) \,d^3 x \,d^3p.
$$
Now substitute in the RHS $\mathbf{p}\to m\mathbf{v}$ (both in the argument of $P$ as in the differential) to find
$$
\tilde{P}(\mathbf{r}, \mathbf{v}) \,d^3 x \,d^3v = P(\mathbf{r}, m\mathbf{v})m^3 \,d^3 x \,d^3v,
$$
and since the differentials are now the same,
$$
\tilde{P}(\mathbf{r}, \mathbf{v}) = m^3 P(\mathbf{r}, m\mathbf{v}).
$$
