Derivation of the Maxwell-Boltzmann speed distribution

Question

We're currently deriving the Maxwell-Boltzmann speed distribution, but I'm struggling to squeeze out the right answer.

For the first exercise we have to derive the fraction of molecules travelling between speed $v$ and $v+\mathrm{d}v$.

Using the following equation: \begin{equation} q(v_x)\propto \mathrm{e}^{-m v_x^2/2kT}, \end{equation} (with $q$ the fraction of molecules travelling between $v_x$ and $v_x+\mathrm{d}v_x$) and \begin{equation} q(v_x,v_y,v_z)=q(v_x)\cdot q(v_y)\cdot q(v_z), \end{equation} I get $q(v)\mathrm{d}v\propto\mathrm{e}^{-mv^2/2kT}\mathrm{d}v$.

For the second part, we have to derive the expression for the region in velocity space comprised between $v$ and $v+\mathrm{d}v$. Because we're going from vector space to scalars: $v=|\vec{v}|$, I can represent all speed scalars in the plane, to justify $A=4\pi r^2$. Evaluating this last expression, I find $A=4\pi\left[(v+\mathrm{d}v)^2-v^2\right]$ for the desired region enclosed between limits $v$ and $v+\mathrm{d}v$.

Ideally both of these answers would now multiply to the Maxwell-Boltzmann speed distribution (minus normalization factor), but this doesn't seem to be the case: \begin{equation} p(v)\mathrm{d}v\propto\mathrm{e}^{-mv^2/2kT}\mathrm{d}v\cdot 4\pi\left(2v\mathrm{d}v +\mathrm{d}v^2\right). \end{equation} (Where $p$ should be the Maxwell-Boltzmann distribution without normalization.) Contrary to most derivations, the way the question is asked very much disincentivizes using the expression $A=4\pi v^2$. How can I justify using this anyway ? It's clearly not the region enclosed by the limits given in the exercise. Secondly, the wording in the exercise explicitly mentions to multiply these two results, resulting in another factor $\mathrm{d}v$ that I can't explain.

I'd very much appreciate some guidance on what I'm missing.

Answer 1

To pass from the 1st part of your exercise to the 2nd part should not be difficult. Imagine a 3-dimentional space in which on the $x$-axis represents the $x$ projection of the velocity vector, the $y$ axis represents the y projection, and the $z$ axis the $z$ projection. You already calculated the probability of having velocities in a small cube of dimensions $dv_x, dv_y, dv_z$, with the cube positioned at the end of the vector $(v_x, v_y, v_z)$. You got

$$q(v_x, v_y, v_z)dv_x dv_y dv_z = C exp[(−mv^2)/2kT]dv_x dv_y dv_z.$$

where $C$ is the normalization constant. Now, you have to distinguish between probability and DENSITY of probability, i.e. probability per UNIT VOLUME of your space. The density of probability is equal to

$$C \exp[(−mv^2)/2kT].$$

Notice that this density of probability is independent of the direction of the velocity.

Now, you are asked to calculate the probability of having the velocity between $|v|$ and $|v| + d|v|$, where $|v|$ is the length of the velocity vector. Consider therefore a sphere of radius $|v|$ in the velocity space, and around it another sphere of radius $|v| + d|v|$. All the velocity vectors, no matter in which direction they are directed, end-up between these two spheres if their length is between $|v|$ and $|v| + d|v|$. So, what is the volume confined between these two spheres? It is $4\pi|v|^2 d|v|$. The density of probability (per unit of volume in the velocity space) you already calculated, and found it independent of the direction of the velocity. Then, to find the probability $dP$ in all the volume between the two spheres you multiply the probability per unit volume, with the volume. You get

$$dP = C \exp[(−mv^2)/2kT]4\pi|v|^2 d|v|.$$

What you are requested is to calculate $dP/d|v|$. Obviously, it is equal to

$$C \exp[(−mv^2)/2kT]4\pi|v|^2.$$

Good luck,

Sofia

Answer 2

We know that $q(v_x)\propto \mathrm{e}^{-m v_x^2/2kT}$. Thus, the probability $q(\vec{v})$ that a particle has velocity $\vec{v}=(v_x,v_y,v_z)$ is: $$q(\vec{v}) \propto \mathrm{e}^{\frac{-m (v_x^2+v_y^2+v_z^2)}{2kT}} dv_xdv_ydv_z$$

Now, You want to know the probability $q(v)$ that a particle has speed between $v$ and $v+dv$. For that, you have to integrate the Boltzmann distribution over all $v_x$,$v_y$,$v_z$ such that $v_x^2+v_y^2+v_z^2=v^2$ is a constant. Thus,

$$q(v) \propto \int_{|\vec{v}|=v}\mathrm{e}^{\frac{-m (v_x^2+v_y^2+v_z^2)}{2kT}} dv_xdv_ydv_z$$ $$= \mathrm{e}^{\frac{-m v^2}{2kT}} \int_{|\vec{v}|=v} dv_xdv_ydv_z$$ Inthe $2^{nd}$ step, I could take out the exponential term since it is a constant, as I am integrating over the condition |$\vec{v}|=v$ is a constant.

Now, all I have to do is to evaluate $ \int_{|\vec{v}|=v} dv_xdv_ydv_z$. To do this, I'll go into a 3D coordinate system, where the $x-$axis represents $v_x$, $y-$axis represents $v_y$ and $z-$axis represents $v_z$. In this system, $dv_xdv_ydv_z$ represents the volume element $d\tau$. Notice that, the set of all points such that $v_x^2+v_y^2+v_z^2=v^2$ is a sphere of radius $v$. Thus, the required integral is just $4\pi v^2dv$. As a result, we now get our final expresion,

$$q(v) \propto 4\pi v^2 \mathrm{e}^{\frac{-m v^2}{2kT}}$$

I hope this clears your doubt. If not, please comment or send me a message, and I'll be happy to explain.