In chemistry, we define the root-mean-square speed as

$\sqrt{\bar{u^2}}$ = $\sqrt{\frac{3\text{RT}}{\text{M}}}$

A student asked me why we can't just remove the square root symbol. And aside from "because this is how we define it", I didn't actually have a reason.

So, I'm hoping someone can shed some light on why the above equation is used and not:

$\bar{u^2} = \frac{3\text{RT}}{\text{M}}$

In case it is important, we use this equation to determine the rms speed of a gas. It depends on the temperature (T) and the molecular mass of the gas (M). R is a constant value. I understand we don't just use the average because in a set of gases, they move in a random direction so the average is 0. But, by squaring isn't that issue resolved, without the square root?

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18  
$\bar{u^2}$ does not have units of velocity. – Michael McGovern yesterday
    
@MichaelMcGovern $\bar{u^2}$ has units of energy over mass. That's equal to $\frac{m^2}{s^2}$ and gives the right units. – Omry yesterday
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The cheeky answer is "because then it isn't RMS, it's just MS". – Tim Pederick yesterday
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@Omry: How is $m^2/s^2$ the right units for velocity? It's the right units for velocity squared, which I'm sure is Michael's point... – Tim Pederick yesterday
    
see also: Root mean square vs average absolute deviation?, stats.stackexchange.com/questions/7447/… – Florian Castellane yesterday
up vote 23 down vote accepted

$T$ has units of Kelvin (K). The gas constant $R$ has units of Joule/mole/K. The molecular mass $M$ has units of kg/mole. Also remember that a Joule is $\mathrm{kg.m^2/s^2}$. So the units of $3RT/M$ are $$\mathrm{\frac{kg\,m^2\,K}{s^2\,mole\,K\,kg\,mole^{-1}}=\frac{m^2}{s^2}}$$ which is a velocity squared. So taking the square root gives the correct units for a velocity.

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This is true, but I think it's worth noting that you can't take the square root, but you can take it squared. i.e., since $\bar{u}$ is an average, $\bar{u}^2 \neq \bar{(u^2)}$, but you can get $\bar{u^2}$ without the root. – Omry yesterday
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I always tell students to write out units, I suppose I should have done the same :) – J M yesterday

$$\overline{u^2}=\frac{3RT}{M}$$ is perfectly valid.

It is just that it gives you an "average squared speed", which is an unusual physical quantity. You will prefer an average speed (more precisely rms speed) by taking the square root.

$$\langle u\rangle:=\sqrt{\overline{u^2}}.$$


As you noted, the ordinary average is of no use, as $\overline u=0$. Then you see that

$$\overline{u^2}=\overline{(u-\overline u)^2}=\sigma_u^2$$ is the variance of the speed, while

$$\langle u\rangle =\sigma_u$$ is the standard deviation, which has the same dimension as $u$.

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Ordinary average of the speed is not 0, the speed is non negative. If I recall correctly, the average speed in this case (Boltzmann distribution) is $\sqrt{8RT/(\pi m)}$. – Juris 16 hours ago
    
@Juris: you must be speaking of the absolute speed, which indeed must be positive. – Yves Daoust 15 hours ago
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in kinematics velocity is the vector and speed is the length (absolute value/magnitude) of that vector. In the context of this question speed (not velocity) is what is described by these formulae and RMS speed is the only case where the value coincides with the respective value of velocity (RMS velocity). – Juris 4 hours ago

While working out the units points it out nicely, one can also consider that in:

$\sqrt{\bar{u^2}}$ = $\sqrt{\frac{3\text{RT}}{\text{M}}}$

The value $\sqrt{\bar{u^2}}$ in the whole is the actual result, and the square root is just part of the way "RMS" is written out. The value $\bar{u^2}$ is uninteresting for practical purposes.

It may be more clear if the definition is spelled out explicitly, for example like this:

  1. The speed $u$ of individual gas molecules is essentially random. However, there are useful statistical properties for the root mean square speed of a large set of molecules. We can call this root mean square speed $r$, and define it as $r = \sqrt{\bar{u^2}}$

  2. If we know the temperature and properties of the gas, we can calculate $r$ as $r = \sqrt{\frac{3\text{RT}}{\text{M}}}$.

Which removes the temptation to "simplify" the definition of RMS.

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The simple answer is because RMS is "Root of Mean value of Squares", it's the definition.

The reason you can't skip the square root is because it will not be a good representative of the concept of mean value, and that's because one have squared the quantities in the first place.

More deeper explanation why one does this is because the velocity squared is proportional to the kinetic energy of the particle. If they have the same mass the RMS of the velocities correspond to the average energy of the particles. So the RMS of the velocities is the velocity a particle with average energy has.

Similar reason is behind RMS value of voltage. Because the power produced is $P=UI$ and $I$ is proportional to $U$ according to Ohm's law the power is proportional to $U^2$ and therefore the RMS value corresponds to the (constant) voltage that results in the same power as the average power.

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I think this question is more like a physics question. If you wanted the mathematics, it's already answered - you can't always drop the SQRTs, but you can in this case and the formula would still be correct. But why don't we do that?

If you have a bunch of gas particles you might want to somehow describe their average speed with a single number. Well, maybe not really the average but a speed that is characteristic to that particular gas in the state considered. How can we do that?

  • Average velocity. As you correctly mention, averaging over all velocities you get 0 in many cases. This is useful in some cases, for example you can use $\bar{\boldsymbol{v}}=(0,0,0)$ as a boundary condition when deriving velocity distribution.

Let's look at the distribution of how many molecules move with certain speed. You can see that actually almost none of the particles have zero speed or velocity. You might conclude that the average velocity is not that great metric to characterize a state of a gas. If the gas fills a container, it's average velocity doesn't care about the state, only about the movement of the container. So let's just average over the graph linked above, shall we?

  • Aveage speed. The speed is magnitude (absolute value) of velocity. You could take the average (mean) speed of gas molecules.
  • Most probable speed. When you looked at the graph... you might actually want to use the speed where the peak (maximum) is. That would describe the curve well, wouldn't it?
  • Root mean square speed. Let's take the square of speed. Find the mean value. Take the root. Sounds like a nightmare, but this is actually the most useful metric. It is the speed that characterizes the energy of the gas.

The molecules of gas moves with different speeds. Each of the molecules have some kinetic energy. You could calculate the total or the average energy if you could measure each speed and do quite a lot of maths (not doable in a lifetime for a reasonable container of gas). However, if all of the gas molecules were moving with a certain speed that we call root mean square speed their total and average kinetic energy would be the same as it really is. As energy is usually what we care about the most in physics/chemistry, this is the speed that describes the speeds of a gas in a way that is useful for us.

If you care about the others:

  • Average speed describes the momentum in the same way (total/average momentum would stay the same if the molecules would all move with the mean speed).
  • Average velocity describes the motion as a whole (motion of the center of mass if all the particles are of equal masses).
  • Most probable speed says that more molecules have about that speed than any other speed. To be precise you should choose interval, let's say consider number of molecules having a speed +/- 10m/s. Than the number pf molecules having speed in that interval will be the highest if the speed is the most probable speed. That is the best usefulness for this number that I can come up with.

Some stuff for further (yet introductory) reading on wikipedia.

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Your student is quite right in seeing that the formula $\sqrt{x}=\sqrt{y}$ is equivalent to the formula $x=y$.

The question is, which of these two equivalent formulas is easiest to use?

In this case we are interested in calculating an average speed. And the Root-Mean-Square is the type of average we are going to use.

We are not interested in the Mean-Square on its own and there is no point in calculating it.

Therefore the formula gives the RMS speed directly, even if it looks a bit redundant.

A different question is which formula is easiest to remember. If the student thinks that it more intuitive and easier without the square root signs, then they can go ahead and learn it that way. They just have to be aware that the exam questions will ask for the ROOT-Mean-Square. If they answer with just the Mean-Square that will be a wrong answer.

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The student is NOT correct, and nor are you or the OP, because the formula is actually "sqrt(x^2) = sqrt(y^2)". For a single value, this gives x=abs(y); and for the mean you need to integrate positive and negative portions separately. The key to RMS is that it copes with the signal going negative, which a simple "average" fundamentally will not do. – Graham yesterday
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@Graham Neither $R$ nor $T$ nor $M$ are ever going negative. And as $u$ is considered positive and directionless, neither does it. The main point to keep in mind is that the formula is about (the square root of) $\overline{ u^2}$ and not $\overline u$ or $\overline{\vec u}$ (which would certainly be $0$) – Hagen von Eitzen 7 hours ago
    
@HagenvonEitzen Thanks for that context on the original equation. I'm coming from a background in electronics and engineering, where negative values are not just possible but expected. – Graham 3 hours ago

It depends on what you are trying to do.

If you are, for example, using the RMSE as a cost function that you are trying to minimise, then it is indeed a waste of computational resources and human brain power to take the square root. You can minimise the mean square error instead. The position at which the mean square error is minimal is identical to the position at which the root mean square error is minimal.

However, if you are trying to communicate an error or uncertainty, chances are they are more familiar with 10 m/s than with 100 m²/s².

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The square root of a square always results in a positive number.
Usually this is desired.

For example: When dealing with sine waves, RMS is used for an average because the normal average of a sine wave is zero (the negative peaks cancel out the positive peaks), but the RMS results is an actual number because the negative values are 'rectified' positive.

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Zero isn't "an actual number"? – David Richerby 19 hours ago

Because if you remove 'Square Root', its unit will no longer be speed ( in this case ) i.e ${m^2}$/$sec^2$

This term $\bar{u^2}$ represents 'Mean of Squared Component', aka 'ms' from 'rms'. Hence to get the 'root mean square' value, you need to take square of 'ms'.

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