# Dimensional analysis with derivatives, logs, exponents and trigonometric functions

How should we do dimensional analysis when we have derivatives, logs, exponents and trigonometric functions in an equation. Should we assume that the operands are pure dimensionless numbers? Coming from a chemist's perspective, the simplest example is that of pH is defined as the negative logarithm of hydrogen ion concentration. Lets us take the unit of concentration as molarity = moles/Liter. One way to circumvent this is that some people say one can normalize the concentration by dividing by a unit concentration. As a result, the log operates on a dimensionless number and pH is then dimensionless.

For example Hecht*, in his celebrated Optics writes "It's necessary to introduce the constant k simply because we cannot take the sine of a quantity that has physical units. The sine is the ratio of two lengths and is therefore unitless. Accordingly, kx is properly in radians, which is not a real physical unit"

He is talking about a sinusoid = sin k (x-vt).

What is the opinion of physicists on dimensional analysis when a equation involves transcendental functions and derivatives?

Thanks.

*E. Hecht, Optics, 2002

• Mathematics isn't right or wrong depending on people's personal opinions, and you don't prove things in math by appealing to authorities and describing textbooks as "celebrated". – alephzero Jun 14 at 21:44
• Are you implying that there are no opinions in the field of mathematics and nothing can be praised? – M. Farooq Jun 14 at 22:20
• I don’t understand why derivatives are in your list. If $x$ is in meters, $d/dx$ is in inverse meters $(m^{-1})$. – G. Smith Jun 14 at 23:22
• @G. Smith, It is a long story. I was working on signal resolution enhancing approach and the equation looked like: Resolution enhanced signal = Original signal - K (second derivative), where K is a small positive number. One of the chemistry theoreticians argued in a conference that this is like subtracting acceleration from distance. I disagreed because my K (a real number) can be adjusted to cancel the units of the second derivative. One can expand the function as Taylor series, what about the units then? – M. Farooq Jun 15 at 0:20
• @M.Farooq, you are correct regarding your comment on resolution enhanced signal. "K" is a number, but it must also have units associated with it to ensure that your equation is dimensionally consistent. And note - if you apply a Taylor series expansion to something in the real world, even that equation must be dimensionally consistent. – David White Jun 15 at 19:59

The main reason why the argument of transcendental functions has to be dimensionless is that these functions are actually the convergence function of an infinite power series. Because the terms in the series have different powers, it is impossible to add the terms if there is a dimension attached.

As an example, consider the exponential function $$f(x)= e^x$$. Its power series is: $$f(x)=1+x+\frac{x^2}{2}+\frac{x^3}{3!}+...$$

If we have an argument $$x=1 \text{m}$$, for instance, where $$\text{m}$$ is the dimension, then we have:

$$f(1 \text{m}) = 1 + \text{m} + \frac{\text{m}^2}{2} + \frac{\text{m}^3}{3!}+...$$

which we cannot add. From here the requirement of having a dimensionless argument.

• Yes, this is where the fallacy of units starts. How to get rid of the dimensions in a Taylor series? – M. Farooq Jun 15 at 0:23
• Well, the point is that the argument $x$ is dimensionless to begin with. You cannot get rid of the dimension from the argument when you expand the power series. – Alonso Perez Lona Jun 15 at 1:36

It is not as simple as it seems. We tell the children they should never take a transcendental function of a dimensional quantity and that keeps them safe. (Derivatives, as @G.Smith points out, are not a problem and do not belong in this discussion.)

But if it is true that, say $$p=mv$$ it is manifestly true that $$\sin p =\sin (mv)$$. That is an acceptable equation. We could even (if we wanted to, I can't think why) say $$\sin p = \sin m \cos v + \cos m \sin v$$. It looks horrible, but if you punch the numbers into a calculator or computer program they will work.

So if you have such a function (sin, cos, log, exp and similar) the argument is probably dimensionless, but there can be exceptions in some weird cases.

• I agree, undergraduate texts rarely talk about this issue. In case you missed the derivative story, I was working on signal resolution enhancing approach and the equation looked like: Resolution enhanced signal = Original signal - K (second derivative), where K is a small positive number. One of the chemistry theoreticians argued in a conference (not in public though) that this is like subtracting acceleration from distance. I disagreed because my K (a real number) can be adjusted to cancel the units of the second derivative. I published this work, but still that thought lingered in my mind. – M. Farooq Jun 15 at 20:02

As a follow up to Alonso Perez response, I found an article in the Journal of Chemical Education, "Can One Take the Logarithm or the Sine of a Dimensioned Quantity or a Unit? Dimensional Analysis Involving Transcendental Functions" 2011, 88, 65.

The authors provide an example of the Taylor series and highlight the point that it is possible to look at the Taylor series, while including the dimensions, using the equation (19).

Quote from the above mentioned article:

I would instead say that the way dimensional analysis works in the first place is based on the idea that you cannot take an arbitrary function of a dimensionful quantity, only a dimensionless quantity.

A dimensioned quantity can be regarded in a couple of ways. In the SI units, a dimensioned quantity lives in $$\mathbb R \times \mathbb Q^7$$, where the real number is a sort of magnitude and the $$\mathbb Q^7$$ are rational-number exponents applied to the seven SI base units. The algebra defines an operation $$(r_1, q_1)+(r_2, q_2) = (r_1 + r_2, q_1)$$ which is a partial function, it only is meaningful when $$q_1=q_2,$$ and an operation $$(r_1, q_1)\times (r_2, q_2) = (r_1 \times r_2, q_1 + q_2)$$ which is a full function meaningful even if $$q_1 \ne q_2.$$

Taking an arbitrary function of such a dimensioned quantity is technically definable even when $$q\ne [0,0,0,0,0,0,0].$$ One could simply define that $$\operatorname{lift}[f](r, q) = (f(r), q).$$ However this does lead to some very strange things like $$x \times x \ne x^2.$$ This approach does not really capture the “why” we are doing what we are doing, just the “what” we are doing.

A stronger statement which encapsulates the “why” would state that a dimensioned quantity is an equivalence class of unit-value pairs, so for example one kilogram is secretly a set that looks like, $$\{(\text{mg}, 10^6), (\text{g}, 10^3), (\text{kg}, 1), (\text{lb}, 2.20462),\dots\}.$$ Indeed these could also be stated as functions from units to reals, if that helps.

When you define a thing as an equivalence class, you can “lift” functions from the source domains to the equivalence classes only when they preserve the equivalence relation, in which case we say that the lifting is “well-defined.”

This captures the “why” much better as it gives a reason, for example, why addition can be well-defined only when the dimensions are the same.

Arbitrary functions do not have a clear well-defined way to be applied to these equivalence classes directly, of course, unless the unit is the empty unit and hence the equivalence class is just the number itself, $$\{(\bullet, \pi)\}$$ for example.

The result is that if you take all of your dimensionful input parameters $$p_{1,2,3\dots n}$$ you will find that you can form only a few dimensionless parameters $$D_{1,2,3\dots \ell}$$, and then for any physical quantity $$P$$ you will find that the most general expression can only be $$P = p_1^{q_1}~p_2^{q_2}~\dots p_n^{q_n}~f\big(D_1, D_2, \dots D_\ell\big),$$ where the choice $$\{q_1\dots q_n\}$$ are just any arbitrary choice of rational numbers which gets you the same dimensions as $$P$$, and $$f$$ is some arbitrary function of the dimensionless parameters.

So for example if something is falling from space to Earth, you might be interested to figure out the time it takes to do so; this depends on the starting distance $$r_0$$ from Earth, and the mass of earth $$M$$ and the gravitational constant $$G$$ and its initial speed $$v_0$$. There is only one dimensionless constant and so the fall time must have some representation like $$t = \sqrt{\frac{r_0^3}{GM}} ~f\left(\frac{r_0 v_0^2}{G M}\right),$$ for some function $$f$$. The reason I know this is that no other arbitrary functions are well-defined.