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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?

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up vote 33 down vote accepted

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, and it appears to be OK to insert dimensionful quantities into any analytic function.

From this we may conclude that it does mathematically make sense to insert dimensionful quantities into transcendental functions. But usually it doesn't make physical sense.

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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 6:44
Thank you for a very intuitive explanation! – Jubilee Mar 28 '11 at 13:27
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 Apr 5 '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

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.

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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.)

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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 1:53
@MarkEichenlaub Could you briefly explain exponentiation in the way that you mean? – Mark C Sep 7 '11 at 5:42
@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 7 '11 at 10:03
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. – centralcharge Jul 28 '13 at 7:20
@MarkEichenlaub, you're right that exponentiation is a limit and it started in analysis, but categorically, an exponential $A^B$ is the set of functions from $B$ to $A$. – alancalvitti Jan 29 '15 at 17:17

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