Understanding units and the units of the derivative operator

Question

Suppose that $f$ is a function from unit $A$ to $B$, then what is the unit of $f'(x)$?. We can do $f'(x)\Delta x$ to get an estimate of $f(x + \Delta x)$. Since the latter has unit $B$, so has the former. $\Delta x$ has unit $A$, so $f'(x)$ has unit $B/A$. So far so good, but what if $A$ is the unit of temperature? Intuitively it does not make sense to give $\Delta x$ the same unit as temperature, because a temperature difference cannot be used interchangeably with an absolute temperature. In particular it does not make sense to do $f'(x)\cdot y$ where $y$ is an absolute temperature. Are there extensions to the system of units that let you recognize that as an error just by looking at the units, i.e. a system of units where $\Delta x$ has a different unit than $x$? Similarly, adding two absolute temperatures is invalid. Is there a system of units that lets you only add two relative temperatures to get a new relative temperature or a relative plus an absolute tempurature to get a new absolute temperature?

Now for a more general and more vague question:

I was reading the page on extensions to dimensional analysis on Wikipedia. It describes a system of units that seems to capture the rotational invariance of physical laws. All physical laws should be rotationally invariant, and Siano's system of units makes sure that if a law is not rotationally invariant then it has a unit error (i.e. it does not matter what we define as our zero angle). Is this a correct intuition? Are there similar extensions to the unit system that capture other physical invariances? It seems like the standard units capture scale invariance (i.e. it does not matter what mass we call 1 kg as long as we do it consistently), and the system that I'm after in the first paragraph should capture translation invariance (it does not matter which point we define as zero). Of course there are other symmetries. Is there a system of units that lets you easily check from the units whether a law satisfies these symmetries?

Another limitation of our current unit system is that some mathematical laws are not expressible. For example $e^{ab}=(e^a)^b$ is not expressible if $a$ and $b$ have units. Is there a way to solve this problem?

Answer 1

As you correctly stated at the beginning, the units of $f'(x)$ are easily seen from writing the derivative as $$ f'(x) = \frac{df}{dx} $$ so the units are the same as units of $f/x$. However, your concerns about the temperature can't be justified. The units of absolute temperature and the temperature difference are the same. In particular, the international system of units, SI, uses 1 kelvin for both.

One may also work with temperature scales that are not absolute, e.g. the Celsius degrees. The absolute zero doesn't correspond to 0 °C; in this sense, these scales are "nonlinear". However, a Celsius degree is still a unit of temperature as well as temperature difference. As a unit of temperature difference, 1 °C and 1 K are the very same thing. You can surely never forget or omit units such as degrees (of temperature) from physical quantities, whether they are computed as derivatives or not.

Similarly, adding two absolute temperatures is invalid.

It may be unnatural or useless in most physical situations (see Feynman's "Judging Books By Their Cover") but it is a valid procedure when it comes to the units. After all, absolute temperatures are just energies per degree of freedom so adding absolute temperatures isn't much different from adding energies which is clearly OK.

In particular it does not make sense to do $f'(x)\cdot y$ where $y$ is an absolute temperature.

It makes perfect sense. Thermodynamics is full of such expressions. For example consider $f(x)=S(t)$, the entropy as a function of time. Then $S'(t)\cdot T$ is a term that appears in the rate of change of some energy according to the first law of thermodynamics.

Quite generally, it is not sensible to single out temperature in these discussions. The same comments hold for distances, times, or pretty much any other physical quantities. Take time. One may consider the "current year". It's some quantity whose unit is 1 year. (Similarly, the position of something in meters.) And one may consider durations of some events whose units may also be years. It is the same unit. It is obvious that the difference $A-B$ i.e. any difference has the same units as $A$ as well as $B$. In my "current year" analogy, $Y=0$ corresponds to the birth of Jesus Christ, a random moment in the history of the Universe. That's totally analogous to $t=0$ °C, the melting point of ice. But in both cases, the time differences or temperature differences have the same units as the quantities from which the differences were calculated – such as temperature (whether they're absolute or not) or dates.

It wouldn't make any sense to have different units for quantities and their differences because dimensional analysis would cease to hold: one could no longer say, among other things, that the units of $A-B$ are the same as units of $A$ or $B$ separately.

It is very correct that one cannot calculate a sensible value of $\exp(a)$ if $a$ is dimensionful i.e. if it has some nontrivial units. Such an exponential would be adding apples and oranges, literally. Express $\exp(a)$ as the Taylor expansion, $1+a+a^2/2+a^3/6+\dots$. If $a$ fails to be dimensionless, each term has different units so it's not dimensionally correct to add them. For this reason, all exponentials (and, with a somewhat greater tolerance, logarithms) in physics are exponentials of dimensionless quantities (which have no units). The desire to avoid physically (and mathematically) meaningless quantities such as exponentials of dimensionful quantities is one of the very reasons why we use units and dimensional analysis at all. It is not a "problem"; it is a virtue and the very point of these methods.