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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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    $\bar{u^2}$ does not have units of velocity. – 2017-01-16
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    @MichaelMcGovern $\bar{u^2}$ has units of energy over mass. That's equal to $\frac{m^2}{s^2}$ and gives the right units. – 2017-01-16
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    The cheeky answer is "because then it isn't RMS, it's just MS". – 2017-01-16
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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... – 2017-01-16
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    see also: Root mean square vs average absolute deviation?, http://stats.stackexchange.com/questions/7447/root-mean-square-vs-average-absolute-deviation – 2017-01-16
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    Because then it's not the root mean square any more, it's just the mean square. – 2017-01-16
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    @TimPederick That's not cheeky. That's a succinct, correct, and helpful explanation of what's going on. – 2017-01-17
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    Analogous to the situation in signal processing where we use the square root when we think of voltage (or current) and leave it off when we think of power (for example for Power Spectral Density). – 2017-01-17
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    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. – 2017-01-18

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$T$ 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. – 2017-01-16
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    I always tell students to write out units, I suppose I should have done the same :) – 2017-01-16
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    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. – 2017-01-17
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    @Brick While I can understand that there is a broad context of applicability to the concept of RMS, I did frame it in a very specific setting, namely, studying chemistry as it related to the speed of molecules with changing temperature and mass. In the broad context, maybe the accepted answer doesn't do that justice. However, for the purposes of expressing this concept to students in a 1st semester general chemistry course, an answer that ties to units (something they are very clear on) is very helpful for them. – 2017-01-20
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    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 – 2017-01-20
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    @Brick I don't agree. From a purely numerical point of view, $x$ and $x^2$ can be thought of as measuring the same scalar quantity. This is, among other things, the reason for the usual simplification in Lagrange multiplier problems. Furthermore, dimensional analysis, while it can never be used to prove that an equation is correct, most certainly can conclusively demonstrate that an equation is incorrect - as is exemplified in the present case, where the formula without the square root cannot possibly be an average *velocity*. – 2017-01-22
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    @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. – 2017-01-23
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$$\overline{u^2}=\frac{3RT}{M}$$ 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.

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

You have noted that the ordinary average of speed is $0$. It is not correct, the ordinary average of speed is $\sqrt{\frac{8RT}{\pi M}}$. Possibly you mixed it up with the average value of velocity $\boldsymbol{u}$ or one of velocities components (for example $u_x$) for which it is true.

You can also see that the RMS of a component

$$\overline{u_x^2}=\overline{(u_x-\overline u_x)^2}=\sigma_{u_x}^2 = \frac{RT}{M}$$ is the variance of the component, while

$$\langle u_x\rangle =\sigma_{u_x} = \sqrt{\frac{RT}{M}}$$ is the standard deviation, which has the same dimension as $u_x$.

And you can use this to find the standard deviation of velocity:

$$\langle \boldsymbol{u} \rangle =\sigma_{\boldsymbol{u}} = \sqrt{\frac{3RT}{M}}$$

which is the statistical interpretation of what that RMS speed actually is.

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    Comments are not for extended discussion; this conversation has been [moved to chat](http://chat.stackexchange.com/rooms/51985/discussion-on-answer-by-yves-daoust-why-cant-we-remove-the-sqrt-from-rms). – 2017-01-18
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    @user21820 I have suggested an edit. If Yves refuses to fix or delete the answer, at least community should fix it... But I don't think this answer is worth keeping because all of the ideas below the line are not relevant. We don't use RMS speed because it's the standard deviation of the velocity. We use it because it is the speed that represents the energy of the gas. – 2017-01-18
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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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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.

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    "you can't always drop the SQRTs" What does this mean? Could you give an example? – 2017-01-17
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    @quid there are some comments by Graham and Omry discussing that. You can only drop the roots (square both sides) if there are no minus signs disturbing you. both $x=-5$ and $\sqrt{x}=\sqrt{-5}$ would change meaning if you squared both sides. – 2017-01-17
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    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? – 2017-01-17
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    @quid "Dropping" is not an operation in mathematics. Dropping the roots is a jargon for squaring both sides (which is also a jargon...). If you solve a problem in the real numbers and get to $\sqrt{-5}$, there is no solution at all, the math is broken. You can't just square it and proceed. $\sqrt{x}=uncomputable$ should not imply $x=-5$. – 2017-01-18
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    "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. – 2017-01-18
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    @quid skipping the details the main point is that $x=-5$ is not equivalent to $\sqrt{x}=\sqrt{-5}$. The second is equivalent to $\sqrt{x}=i\sqrt{5}$ while from the first you could get both $\sqrt{x}=\pm i \sqrt{5}$. – 2017-01-18
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    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.) " – 2017-01-18
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    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 $\sqrt{-5}$! – 2017-01-18
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/52026/discussion-between-quid-and-juris). – 2017-01-18
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    @quid is right. Suppose that in some context $x = y$. Then indeed it may not be valid to deduce $\sqrt{x} = \sqrt{y}$. After all, $x$ could be a tree, and trees don't have square-roots. However, in any ordinary mathematical context where $\sqrt{x} = \sqrt{y}$, the very meaning of that statement **includes** that $\sqrt{x},\sqrt{y}$ both have a value and are equal, and (ordinarily) $\sqrt{x}^2 = x$ and $\sqrt{y}^2 = y$, thus clearly $x = y$ by **basic equality rules**. Apart from that, your answer is still probably the best one here. Please clarify if you don't understand our point. – 2017-01-19
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    Following your precedent, I've edited your answer to fix it. If you disagree, you're welcome to roll-back and explain further to @quid and myself. Thanks! – 2017-01-19
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    @user21820 I don't object and even like your edit. It now simply states that you can drop the roots here and does not even mention the controversial other (unrelated) cases that I and quid (and you) currently disagree at. – 2017-01-19
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    Thanks! It's not mathematically controversial though, so feel free to ping us in the chat-room to fully resolve it if you wish. – 2017-01-19
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    Thank you for the very nice abstraction to the concepts embedded within and related to RMS. I think some of the details in this answer would be lost on this particular level and group of students in the course, but it certainly helps me appreciate the relevance of the RMS. – 2017-01-20
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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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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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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. – 2017-01-16
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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 $0$) – 2017-01-17
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    @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. – 2017-01-17
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    @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.) – 2017-01-17
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    @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. – 2017-01-18
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    @user21820 if you do not specify such assumptions you rather not use the square-root symbol in an equation of the form under discussion. There is also a danger is saying some things do not work that actually do work. (Some here, maybe not you, seem not to appreciate that it is in fact possible and common to have a square-root function on the complex numbers for which this works perfectly fine; the principle branch, and that's also what most would in fact use intuitevly. Those that see issue are somehow "too clever by half") – 2017-01-18
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    @quid: My approach ensures that students can know their mathematics and not only convince themselves but also remain convinced even if I say that they are wrong. This is ultimately necessary for any student to truly grasp rigorous mathematical proof. In this case, without the assumptions stated somewhere it is simply wrong pedagogy. If *Stig* had instead repeated the equivalence stated in the question I would not disagree, since in the context of the question it is indeed well-defined and true. But Stig stated in the manner of generalization, which makes it wrong. I can say "Pigs can fly!". – 2017-01-18
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    @quid: Pigs can fly on an airplane... Anyway discussion of complex exponentiation is vastly beyond the level of the asker, who needs to first get a proper grip on basic logic. It is never the right thing to tell students things without giving full justification or at the very least say when justification is not given. And theorems stated as fact better be precise and correct. If not one should say "informally" or "roughly speaking". Anyway we should continue somewhere else if you like to discuss pedagogy further! =) – 2017-01-18
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    @user21820 the problem arises for those that have an incomplete not to say false unedrstanding of the square-root for (complex) numbers. Either $\sqrt{x}= \sqrt{y}$ is meaningless to being with (in the given context) or it is equivalent to $x=y$. The problem is some kind of chimera of those that know there are "monsters" lurking somewhere around there without knowing their stuff well enough to realize there are none at this precise point. It might still be unwise to take somebody ill-prepared down that road as they may get lost, but the specific thing is just not false, but true. – 2017-01-18
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    To stress the point that might get lost in the indirection: there is no problem for those that have *no* understanding of complex roots. Abstractly there is a fundamental difference between taking sqrt and squaring. *If* the former makes sense as a function it must be injective. If it does not make sense as a function, it's not clear how it could be ever part of an equation in that form to begin with. – 2017-01-18
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/51988/discussion-between-user21820-and-quid). – 2017-01-18
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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.

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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'.