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In dimensional analysis, it does not make sense to, for instance, add together two numbers with different units together. Nor does it make sense to exponentiate two numbers with different units (or for that matter, with units at all) together; these expressions make no sense:

$$(5 \:\mathrm{m})^{7 \:\mathrm{s}}$$

$$(14 \:\mathrm{A})^{3 \:\mathrm{A}}$$

Now my question is plainly this: why do they not make sense? Why does only multiplying together numbers with units make sense, and not, for instance, exponentiating them together? I understand that raising a number with a unit to the power of another number with a unit is quite unintuitive - however, that's not really a good reason, is it?

Kevin Reid
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Jubilee
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  • Could you confirm that all the values being considered are read (rather than complex)? There are some subtle discussions to be had if 'complex' values are allowed. – Philip Oakley Aug 02 '16 at 21:06

5 Answers5

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A standard argument to deny possibility of inserting dimensionful quantities into transcendental functions is the following expression for Taylor expansion of e.g. $\exp(\cdot)$:

$$ e^x = \sum_n \frac{x^{n}}{n!} = 1 + x +\frac{x^2}{2} + \dots\,.\tag1$$

Here we'd add quantities with different dimensions, which you have already accepted makes no sense.

OTOH, there's an argument (paywalled paper), that in the Taylor expansion where the derivatives are taken "correctly", you'd get something like the following for a function $f$:

\begin{multline} f(x+\delta x)=f(x)+\delta x\frac{df(x)}{dx}+\frac{\delta x^2}2\frac{d^2f(x)}{dx^2}+\frac{\delta x^3}{3!}\frac{d^3f(x)}{dx^3}+\dots=\\ =f(x)+\sum_{n=1}^\infty\frac{\delta x^n}{n!}\frac{d^nf(x)}{dx^n},\tag2 \end{multline}

and the dimensions of derivatives are those of $1/dx^n$, which cancel those of $\delta x^n$ terms, making the argument above specious.

dmckee --- ex-moderator kitten
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    Just a trivial addition: a general power $x^y$ may be written as $\exp(y \ln x)$ so it has the same problem if $y$ fails to be dimensionless. ... In a similar way, the exponents should always be Grassmann-even (not "fermionic"), and so on. – Luboš Motl Mar 28 '11 at 06:44
  • Got a downvote today, which I suppose is reasonable given the miserable state of this answer. @Jubilee, could you at least un-accept it? So that something actually defensible could appear on the top. – dmckee --- ex-moderator kitten Apr 05 '15 at 17:41
  • @dmckee I'm confused. Is this answer wrong? If so, why don't you delete it? Hope this isn't rude. – pentane Jul 14 '15 at 14:45
  • @dmckee Why not present the main part of the paper in the answer? The Taylor expansion fallacy is quite an interesting fact (although it has some problems too). I was looking if someone had already written such an explanation (and was going to do it myself if not), but since you've found it, it'd be very useful if you put it into your answer. – Ruslan Jul 14 '15 at 15:45
  • @Ruslan I had been hoping that alancalvitti would write a answer based on the paper as he is the one who brought it to my attention. Then time passed and I forgot about this business. I should bring in the findings of the paper, but I don't know when I might get around to it. So I'm making the answer Community Wiki and anyone who cares enough can get it done. – dmckee --- ex-moderator kitten Jul 14 '15 at 23:39
  • @dmckee I've changed your answer almost completely (using your Community Wiki license), please have a look at it. – Ruslan Jul 15 '15 at 04:28
  • @Ruslan and dmckee, isn't it a bad idea to change an answer "almost completely" after 32 people have voted on it and it has been accepted? – pentane Jul 15 '15 at 12:56
  • @pentane Generally yes, but as Jubilee has not returned it seems that this answer is going to remain accepted unless I wield moderator powers to delete it completely. I'd have done that if alancalcitti had written an answer based on the paper. – dmckee --- ex-moderator kitten Jul 15 '15 at 14:04
  • I think it is dishonest to change an answer completely and keep the 30 upvotes that the community ascribed to the previous answer attached to the new answer. It does not represent the community's views. Why don't we delete this answer and have Ruslan post it as a new answer for the community to vote on from scratch? Clearly you can delete it, as you already have once. – pentane Jul 15 '15 at 14:32
  • @pentane If you feel strongly about it, why not take it to meta so that we can see how the user base feels about it? As aside the delete/un-delete was a test when I noticed that I had been assume that I couldn't delete the answer because of the accept (which is the usual case) but had been ignoring my moderator status (which ought to empower me to delete anything). – dmckee --- ex-moderator kitten Jul 15 '15 at 14:34
  • Don't take any of this personally; I think you're one of the coolest people on this site. But as a mod I guess I expect you to be a role model. If someone posts a highly upvoted question that they realize is wrong, they should delete it and post the right answer. They shouldn't edit the already upvoted answer to be "right." That's not fair to the people who didn't get a chance to read whatever they were putting their stamp of approval on. Or am I wrong in your view? – pentane Jul 15 '15 at 14:42
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    @dmckee: it's not an invalid argument and the paper is available on an author's Dropbox. In fact, the argument is a standard one also used to deal with, e.g. $\exp(\hat A)$ where $\hat A$ is a matrix: since we do not have a literal meaning to this expression, we define it by the Taylor expansion of the pure function. That Taylor expansion, it turns out, violates dimensional analysis and there is no better way to do it. The author's counterfactual "if the argument were dimensioned then this expansion would be OK" is a red herring. – CR Drost Jul 17 '15 at 13:30
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    @ChrisDrost actually yes, I was thinking that counterargument is flawed. It says that it's possible to define the function consistently with dimensional analysis. But by inconsistency of $(1)$ definition, the consistent one wouldn't be analytical continuation in any sense. It may not even be continuous at $0$ (which is dimensional and dimensionless at the same time). But in that case we could just call the function anything, not e.g. $\exp$ or $\sin$ as it was originally. If you feel like improving the answer, feel free to do it — it's Community Wiki :) – Ruslan Jul 17 '15 at 14:16
  • @ChrisDrost Please see thew edit history. Concerning who wrote which parts of the answer. – dmckee --- ex-moderator kitten Jul 17 '15 at 14:16
  • The comment on the Matta paper is incorrect. The authors do not conclude that dimensioned valued can be used with transcendental functions. They are merely criticizing the argument from Taylor expansion from Wikipedia, not denying the conclusion. Quoting: "If we write the formal Taylor expansion, [the equation] shows that, should $f (x)$ be dimensioned, then every term in the expansion has the same dimensions as $f(x)$, because the dimensions of $(\partial x^n/1) \times (1/\partial x^n)$ cancel." (Emphasis mine). – Mark H Aug 04 '16 at 23:38
  • Continuing in the same paragraph: "The reason for the necessity of including only dimensionless real numbers in the arguments of transcendental function is not due to the dimensional nonhomogeneity of the Taylor expansion, but rather to the lack of physical meaning of including dimensions and units in the arguments of these function." – Mark H Aug 04 '16 at 23:41
  • @MarkH Quite so. I hadn't noticed that text had been added. I've put it back to something like the last version I wrote before making the whole Community Wiki. – dmckee --- ex-moderator kitten Aug 04 '16 at 23:44
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    @dmckee Personally, I find the Taylor expansion argument convincing for why certain functions cannot take dimensioned values as inputs. The paper's argument only shows that Taylor expansions are valid for functions whose output is dimensionless. The function still needs to be able to take dimensioned input in the first place, which is what the Taylor expansion argument purports to show is impossible for transcendental and other functions. – Mark H Aug 04 '16 at 23:52
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(I know I am answering an old question, but I think the following is a nice way to explain to young students.)

You don't need to know Taylor expansions. Simply remember the definition of the exponential. It satisfies the differential equation

$$ \frac{\text d y}{\text d x} = y(x) $$

According to this, the derivative of $\text e^x$ has the same dimension as $\text e^x$. Therefore, $x$ should be dimensionless, since the derivative of $\text e^x$ has the dimension of $\text e^x$ divided by $x$. (This assertion comes from the definition of the derivative as a limit and it is also suggested by the $\text d / \text d x$ notation.)

Roan
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    Unlike the taylor expansion argument which is not correct, I think yours is the correct one, one could generalize it to other functions like $ \sin(x), \cos(x), \log(x)$ satisfying $\dfrac{d^2y(x)}{dx^2}=-y(x), \dfrac{dy(x)}{dx}=\dfrac{1}{x}$ respectively. – Omar Nagib Jul 24 '17 at 06:59
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One further point to note, is that strictly one is just saying that the exponent is dimensionless, not that it does not contain expressions with dimension. So for example we could have some expression like $X=a^{(E/E_0)}$ where the exponent for a is a ratio of energies.

There are several restrictions on the space (sometimes viewed as a vector space) of dimensional quantities: for example units are raised to rational, but not irrational values. This allows a theorem: The Buckingham $\Pi$ Theorem to form.

Roy Simpson
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Because of the way an exponential is defined. By an expression like $a^b$ we mean to say that the quantity $a$ is multiplied $b$ times with itself. Thus an expression like $(5m)^{7s}$ would mean $5m$ multiplied "7 seconds" times with itself, which is meaningless.

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    A mathematically naive view of exponentiation, but okay. – Mark Eichenlaub Mar 28 '11 at 01:53
  • @MarkEichenlaub Could you briefly explain exponentiation in the way that you mean? – Mark C Sep 07 '11 at 05:42
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    @Marc C Exponentiation is a limiting process. It's fundamentally in idea from analysis. Unless your exponent is an integer, it doesn't make sense to say that you're just multiplying a thing by itself a certain number of times. I'm not sure why I made the comment six months ago, though. It wasn't very important to this question. – Mark Eichenlaub Sep 07 '11 at 10:03
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    As @Mark said, it's a very naive way of looking at exponentation. The same (flawed) logic could be used to say that only natural numbers (0,1,2,...) can be exponents. Even MATRICES can be exponents, and so can clifs. – Abhimanyu Pallavi Sudhir Jul 28 '13 at 07:20
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The community wiki answer seems to have become an inconclusive mixture of opinions, followed by a long comment thread that is difficult to interpret. The paper being referred to there is Matta et al., http://pubs.acs.org/doi/pdf/10.1021/ed1000476 . The Matta paper claims to correct "common misconceptions and errors," but in fact much of their own reasoning is specious.

As Matta points out, there is no reason that a transcendental function must take a unitless input and give a unitless output. For example, let f(t)=(1 meter)exp[t/(1 second)]. This is a perfectly sensible transcendental function, and it takes a unitless input and gives a unitless output. If you take its Taylor series, you will find that the coefficients of the series have the right units so that f can be defined, if you wish, in terms of its Taylor series.

All you can say along these lines is that a lot of the standard functions require unitless inputs and give unitless outputs if you define them by their Taylor series. This is by no means a conclusive argument in all cases, both because we can have functions other than the standard ones (such as the f defined above) and because not all functions need to be or even can be defined in terms of Taylor series.

A good example is the square root function. We wouldn't want to define it in terms of its Taylor series about x=0, because it doesn't have such a Taylor series. If we wanted to be perverse, then we could define it in terms of its Taylor series about some point b>0. Then all that would happen would be that if b had units, so would the coefficients in the Taylor series.

When dealing with logs and exponents, it is not obvious nonsense to do things like taking logs of unitful quantities. For example, you can say that ln(5 meters)=ln(5)+ln(meters).

Matta complains that log(meters) doesn't make sense, because what power y would you raise e to in order to get meters? All they have really proved here is that y is not a quantity that fits into the algebra of unitful quantities. This is a weak argument, since by introducing unitful quantities, we have already extended the algebra of the reals. For example, if we have three base units (m, kg, s), then the algebra of unitful quantities is isomorphic to the direct product RxQxQxQ. For example, 7 newtons would be represented by the 4-tuple (7,1,1,-2), where the second through fourth entries are the exponents of the base units, and the group operation for multiplication is defined in terms of multiplying the first entry and adding the others. So it's perfectly reasonable to imagine extending this algebra to include things like ln(meters). A more cogent objection would be that this algebra doesn't have nice properties, e.g., it isn't a field.

Matta points out correctly that there are perfectly good alternatives to writing things like ln(5 meters). For example, one can write ln[(5 meters)/(1 meter)], and this is the style preferred by the journal in which the paper was published. But this is merely a matter of style, not logic.