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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    $\begingroup$ $\bar{u^2}$ does not have units of velocity. $\endgroup$ – Michael McGovern Jan 16 '17 at 4:11
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    $\begingroup$ The cheeky answer is "because then it isn't RMS, it's just MS". $\endgroup$ – Tim Pederick Jan 16 '17 at 8:04
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    $\begingroup$ @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... $\endgroup$ – Tim Pederick Jan 16 '17 at 8:12
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    $\begingroup$ see also: Root mean square vs average absolute deviation?, stats.stackexchange.com/questions/7447/… $\endgroup$ – Florian Castellane Jan 16 '17 at 10:47
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    $\begingroup$ The answer you accepted explains nothing. We want to know the RMS not because we want units of velocity! The RMS is not a kind of velocity whatsoever. The next-top-voted answer is wrong; see my comment. Juris' answer is the most accurate explanation (agreeing with mine). If you don't believe me, ask a professional mathematician in real life and not over the internet. $\endgroup$ – user21820 Jan 18 '17 at 12:50

$T$ 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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    $\begingroup$ 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. $\endgroup$ – Omry Jan 16 '17 at 4:31
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    $\begingroup$ I always tell students to write out units, I suppose I should have done the same :) $\endgroup$ – J M Jan 16 '17 at 4:31
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    $\begingroup$ This answer is not wrong, but it misses the broader context of the issue, which is that RMS is a concept defined and used way more broadly than this one formula. See also my answer for that part. $\endgroup$ – Brick Jan 17 '17 at 17:28
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    $\begingroup$ Simple is good if it's correct. In this case it seems like you're feeding them a line that's convenient for you rather than helpful to them. Certainly dimensional analysis is helpful and important, which is why I said this answer isn't quite wrong. (I mentioned the consistency of units as well.) The answer is still misleading. I've taught freshman physics, so I both understand and abhor the point you're making here since you seem to be shirking your responsibility to educate rather than appease. @JM $\endgroup$ – Brick Jan 20 '17 at 20:56
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    $\begingroup$ @David My objection is to everyone taking a narrow view to their own domain of something used ubiquitously across many, many fields. RMS is defined in ALL fields by the left-side of that equation. The fact there is a square root on the right side is an accident of this application. The students should at least be made aware of connections across fields. This is exacerbated by the phrasing of the question, which makes it appear that the equation stated "defines" RMS velocity. That's not true. RMS velocity is defined by the left side of the equation only - The right side is derived from physics. $\endgroup$ – Brick Jan 23 '17 at 0:49

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.


I think this question is more like a physics question. If you wanted the mathematics, it's already answered - you can square both sides 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 similar way (total/average magnitude of 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, also the momentum of the system).
  • 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.

  • $\begingroup$ "you can't always drop the SQRTs" What does this mean? Could you give an example? $\endgroup$ – quid Jan 17 '17 at 22:47
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    $\begingroup$ That $A=B$ is not equivalent to $A^2 = B^2$ is clear to me. But drop the roots is not the same as square both sides. Except if you meant "omit/drop squares" rather than square roots. I am curious about $\sqrt{x}= \sqrt{-5}$ what $x$ other than $-5$ does make it true? $\endgroup$ – quid Jan 17 '17 at 23:08
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    $\begingroup$ "Dropping the roots is a jargon for squaring both sides" So if you know that $2a= x+y$ and you conclude from this that $4a^2= x^2 +2xy +y^2$ then you'd refer to this as drop the square roots? Really. Seems rather backwards to me then. Why not just say "square both sides"? If the roots made sense to begin with you can always drop them. (The sole issue might be if the root is there to indirectly restrict the domain.) By the way can I now conclude $x=-5$ if I know $\sqrt{x}=\sqrt{-5}$. It seems to me I can. $\endgroup$ – quid Jan 18 '17 at 0:16
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    $\begingroup$ What is your definition of $\sqrt{x}$? And what is the definition of $\sqrt{-5}$? As I commented elsewhere "I think if you have any sane notion of √ such that √x and √y are well-defined to begin with, then √x=√y will be equivalent to x=y. (I'd be curious to hear a counter-example.) " $\endgroup$ – quid Jan 18 '17 at 0:43
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    $\begingroup$ Maybe to be more clear, if you define $\sqrt$ as some principal branch, say in polar coordinates with angle in $[0,2\pi)$, you just half the angle, which is compatible with you saying $\sqrt{-5}= i \sqrt{5}$ , then there is just no complex number whatsoever whose square root is $-i\sqrt{5}$, as they all have argument less than $\pi$. Yet, if you define $\sqrt{x}$ as any number whose square is $x$, then it is not clear why $\sqrt{-5}$ cannot be $-i \sqrt{5}$ too. At which point $\sqrt{-5}$ might not equal $\sqrt{-5}$! $\endgroup$ – quid Jan 18 '17 at 0:58

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.


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.


The RMS of $u$ is defined by the left side of the equation, for whatever physical quantity $u$ you're considering. It's just coincidence in this case that the right hand side of the equation, which expresses this in terms of physical parameters of the system, also has a square root. You could find many examples where that's not the case and this particular "simplification" wouldn't make sense. As other have noted, the "root" and the "square" part of the definition are chosen to ensure that you get an answer in the same units as $u$ itself.


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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    $\begingroup$ 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. $\endgroup$ – Graham Jan 16 '17 at 11:30
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    $\begingroup$ @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 $0$) $\endgroup$ – Hagen von Eitzen Jan 17 '17 at 7:15
  • $\begingroup$ @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. $\endgroup$ – Graham Jan 17 '17 at 10:42
  • $\begingroup$ @Graham the negative numbers are tangential. I think if you have any sane notion of $\sqrt{}$ such that $\sqrt{x}$ and $\sqrt{y}$ are well-defined to begin with, then $\sqrt{x}=\sqrt{y}$ will be equivalent to $x= y$. (I'd be curious to hear a counter-example.) $\endgroup$ – quid Jan 17 '17 at 23:32
  • $\begingroup$ @quid: Logically, if you have the assumption that there exists $r,s$ such that $r^2 = x$ and $s^2 = y$, and $\sqrt{x},\sqrt{y}$ is one such $r,s$ respectively, then indeed the equivalence is correct. But without specifying the assumptions one would be leading students into a pit. This incorrect attitude in teaching is far more pervasive and damaging than you might think. Just look at all the posts on MathEd SE and you will be forced to conclude that nearly all of students' illogical thinking arises from such inadequate instruction. $\endgroup$ – user21820 Jan 18 '17 at 3:26

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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