Fundamental question about dimensional analysis

Question

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?

Answer 1

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.

Answer 2

(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.)

Answer 3

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.

Answer 4

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.