# What is the difference between Maxwell distribution and Boltzmann distribution?

I am learning Physical Chemistry now. And on the print professor uploaded was a sentence "Compare Maxwell distribution with the Boltzmann distribution"

What I know is only -mv^2/kT is written in Maxwell distribution and -E/kT is written in Boltzmann distribution. What does this mean?

First of all, review what you learnt, if the professor asked you for this he probably explained it somewhere! When you are done, here's the rest of the answer:

Briefly, Maxwell's distribution is a "sub-part" of Boltzmann's for which we focus only on the velocity distribution of the particles.

As you probably know, the state of the sistem is defined by its position $(x, y, z)$ and its velocity $(v_x, v_y, v_z)$ with magnitude (speed) $v=\sqrt{v_x^2+v_y^2+v_z^2}$.

Boltzmann's distribution is the probability distribution of the state of the system, i.e. $P_B(x, y, z, v_x, v_y, v_z)dx dy dz dv_x dv_y dv_z$ is the probability of finding a particle in a given position with a given speed. Thus it depends on both the position and the velocity of the system.

If we are only interested in the velocities of the system, we can obtain Maxwell's distribution $P_M$ from Boltzmann's as $$P_M(v_x, v_y, v_z) =\int dx dy dz\; P_B(x, y, z, v_x, v_y, v_z)$$ i.e. by summing (integrating) over all possible positions.

Since in general (for systems which are normal enough) we can write the energy $E$ as a sum of the kinetic energy $K(v_x, v_y, v_z)={1\over 2}mv^2$ (related to the velocity/speed) and of the interaction potential $U(x, y, z)$, so that $E=K+U$, we get that (neglecting pre-factors): $$P_B(x, y, z, v_x, v_y, v_z)\propto e^{-E/kT} = e^{-K(v_x, v_y, v_z)/kT}e^{-U(x, y, z)/kT}$$ and from the definition above, integrating over all possible positions

$$P_M(v_x, v_y, v_z) \propto \int dx dy dz\; e^{-{1\over 2}mv^2/kT}e^{-U(x, y, z)/kT} = e^{-{1\over 2}mv^2/kT} \int dx dy dz\; e^{-U(x, y, z)/kT} = Ae^{-{1\over 2}mv^2/kT}$$ where $A$ is a constant coming from the integration over all positions. And this is what we call a Maxwell distribution (which can also be derived in other ways, of course, but this one is the one that links it with Boltzmann's).

EDIT: often, when one talks about Boltzmann's distribution, he or she is referring to the probability distribution $\rho(x, y, z)$ of the positions only (e.g. in an harmonic or gravitational potential). Separating the position and velocity parts is (formally) easy when they are not coupled as in the example above, i.e. $U$ does not depend on the velocity and $K$ does not depend on the position, so that, doing the opposite of what we did above i.e. integrating over the veocities:

$$\rho(x, y, z)\propto \int dv_x dv_y dv_z\; e^{-{1\over 2}mv^2/kT}e^{-U(x, y, z)/kT} = Be^{-U(x, y, z)/kT}$$

where $B$ contains the result of the integration and all the constant factors. Often one refers to $\rho$ as Boltzmann's distribution.

• this is very clear and very useful, thanks for taking the time to spell it all out. – niels nielsen Mar 25 '18 at 0:05
• no problem, I also edited a little bit to improve the answer. – JalfredP Mar 25 '18 at 12:33