# Positioning of the root-mean-squared speed, average particle speed and most probable speed on the Maxwell-Boltzman distribution curve

I'm inquiring if there is any intuitive reasoning behind the positioning of $v_{rms}$, $v_{mp}$, and $v_{avg}$ in their positionings on a Maxwell-Boltzman distribution such as the one below: I was thinking since $v_{rms}$ has to do with squared speeds, it would maybe make sense to be largest? Although that can be flat wrong. But other than that, I can't seem to work out the intuitive sense behind this.

• Warning: your intuition is not the same as mine. Really, one should not ask for intuition from other people. So, the answer is as follows: the most-probable is, of course, the peak of the distribution. Since the distribution is not uniform, the average will not be the peak, and ends up being on the higher velocity side given the non-uniformity. Then, as you say, the rms value is the square root of the average of the squared velocity, that ends up being even further to the right. Not intuition, just math. – Jon Custer Oct 24 '17 at 19:05

That $v_\text{rms}\geq \bar{v}$ can be seen from the Steiner-König-Huygens theorem:

$$v_\text{rms}^2 = \bar{v}^2 +\sigma^2 \Longrightarrow v_\text{rms} = \sqrt{\bar{v}^2 +\sigma^2},$$

where $\sigma$ is the standard deviation (always a positive quantity). A proof of the theorem can be found in Wikipedia and, yes, as intuitions go, it's valid to say it is so because it has to do with squared values.

Note it's not always true, though, that $v_\text{rms}$ is larger than $v_\text{mp}$. That's a characteristic of the Maxwell-Boltzman distribution.

As for the meanings of these averages:

• $v_\text{mp}$: by definition of probability distribution, the most probable value of $v$ is the position of the peak of the distribution - that's the most common speed in your sample of particles.

• $\bar{v}$: simple arithmetic average, there's an equal number of particles faster and slower than $\bar{v}$; and it might not coincide with $v_\text{mp}$ whenever the distribution is not symmetric.

• $v_\text{rms}$: that's the mean that appears in the equations for the pressure and kinetic energy, the same way $I_\text{rms}$ shows up in the average electrical power for AC current - the RMS shows up in essentially whatever quantity that depends on the square of some any other quantity.