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We all know that $x \cdot \mathrm{cm}$ is "$x$ times $\mathrm{cm}$" but I was wondering about what is $x^{\mathrm{cm}}$ ($x$ to the power $\mathrm{cm}$) can be define this $x^{\mathrm{cm}}$ thing? For example what is $2^{\mathrm{cm}}$? Do physics have an answer?

I just have a little thought about it. We can write this as $e^{\mathrm{cm} \ln x}$ and we have a Maclaurin series for $f(x)=e^x$ function so we can write like that. But if we write like that we will have $\mathrm{cm}^4$, $\mathrm{cm}^5$, $\mathrm{cm}^6$, $\ldots$? and I don't know what that units mean.

Sangchul Lee
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Atlas Say Cesur
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2 Answers2

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An expression like $e^{1 \text{ cm}}$ simply does not work: as you're noticing, there is no way to make sense of the units. What happens in the Taylor series expansion is an example of that: you'd get terms with $\text{cm}, \text{cm}^2, \text{cm}^3$ in them all added together, but it's invalid to add things with different units.

In expressions like $a^x$, or $\ln x$, the value of $x$ simply has to be unitless.

You occasionally see exponentials and logarithms in physics equations, but the result will always either be unitless or can be converted into a unitless quantity. For example, in the Wikipedia article on thermodynamic equations you see a lot of $\ln \frac{V_1}{V_2}$ or $\ln \frac{T_1}{T_2}$ where the units inside the logarithm (volume or temperature) cancel. Sometimes people carelessly write these as $\ln V_1 - \ln V_2$ or $\ln T_1 - \ln T_2$ instead, which is technically a difference of two meaningless expressions, but we can make sense of this if we combine the two logarithms.

For another example, in Newton's law of cooling we see a factor of $e^{-t/\tau}$ where $t$ and $\tau$ have units ($t$ is time and $\tau$ is a time constant) but $-t/\tau$ is unitless.

Misha Lavrov
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  • Alternatively, define the above symbols such that they are unitless; for example, define the duration as $t$ seconds instead of as $t;$ in this way, $(\ln V_1 - \ln V_2)$ becomes meaningful. – ryang May 05 '22 at 17:42
  • I think that's not best done here, because expressions like $e^{-t/\tau}$ and $\ln \frac{V_1}{V_2}$ are always dealing with ratios. If the duration is $t$ seconds and $t$ is unitless, then you might be tempted to write $e^{-t}$, but this is equivalent to $e^{-t/1 \text{ s}}$ in the old system; you've implicitly snuck in a baseline of $1$ second and you might not realize it. (It works if you know what you're doing, though.) – Misha Lavrov May 05 '22 at 17:46
  • pH in chemistry is odd though. https://en.wikipedia.org/wiki/PH – badjohn May 05 '22 at 17:59
  • @badjohn The Wikipedia article actually comments a bit on this issue: "More correctly, the thermodynamic activity of $H^+$ in dilute solution should be replaced by $[H^+]/c_0$, where the standard state concentration $c_0 = 1 \text{ mol}/\text L$. This ratio is a pure number whose logarithm can be defined." – Misha Lavrov May 05 '22 at 18:24
  • @MishaLavrov Yes but it feels like a cheat similar to your comment on sneaking in 1s. People sometimes try to argue away units e. g. talking of Avogadro's constant as a number since it is just a count of some entities. – badjohn May 05 '22 at 18:31
  • I am not a physicist, so arguing about what is good or bad physics feels a bit over my head. But I think sneaking in $c_0$ or sneaking in $1\text{ s}$ is bad if you forget that you're doing it and accidentally say something meaningless; it is fine if - when it matters - you remember that the reference constant is there. – Misha Lavrov May 05 '22 at 18:34
  • @MishaLavrov I am not a physicist or chemist either. Clearly it works, pH is a useful measurement, it just feels a bit untidy to me. – badjohn May 05 '22 at 18:36
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In the comments Sangchul Lee provides a link to the "Dimensional homogeneity" section of the wikipedia page for "Dimensional analysis" which precicely hits the nail on the head.

Only commensurable quantities (physical quantities having the same dimension) may be compared, equated, added, or subtracted...

Thus, dimensional analysis may be used as a sanity check of physical equations: the two sides of any equation must be commensurable or have the same dimensions...

This has the implication that most mathematical functions, particularly the transcendental functions, must have a dimensionless quantity, a pure number, as the argument and must return a dimensionless number as a result.

Basically, this says that we can look at an expression like "$L=5\mathrm{in}+10\mathrm{cm}$" and make sense of it. Even though the mix of units is awkward we know that both terms on the right are units of length and thus commensurable; that isn't to say that either "$L=15\mathrm{in}$" or "$15\mathrm{cm}$" are correct interpretations of that expression... just that everyone should get concordant results if asked to draw a line of length "$L$" on their paper. Conversely, an expression like "$X=5\mathrm{in}+3\mathrm{kg}$" is not commensurable because "Draw a line of length X." is nonsensical due to the impossibility of drawing a line "3kg long"; whilst likewise "Measure out a sample of mass X." is nonsensical due to the impossibility of trying to "weigh out 5in".

Now, to the specific problem at hand: $x^\mathrm{cm}$

You are correct to notice that this can be expanded as a series, and also that the series expansion you get feels nonsensical. (Because it is!)

The reason is because, like in the "$X=5\mathrm{in}+3\mathrm{kg}$" example, that $$ x^\mathrm{cm} = \sum e^{ln(x)\mathrm{cm}} = 1 + ln(x)\cdot\mathrm{cm}+ \frac{ln(x)^2}{2}\cdot\mathrm{cm}^2 + \frac{ln(x)^3}{6}\cdot\mathrm{cm}^3\dots $$ has units that are non-commensurable on the right; it is trying to reconcile a length, an area, a volume, and so on... which just cannot be done any more-so than "Draw a line of length $2\mathrm{cm}^3.$" can be done.

This is exactly what that wikipedia sentence "most mathematical functions... must have a dimensionless quantity... as the argument" is about. The only way that a series expansion will be commensurate is if all it's expanded terms are commensurate with one another... which for almost all functions will mean that we need dimensionless ratios in the argument rather than dimensioned units.

DotCounter
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  • Possible idea: If there is no unit, then try $$1=\text{cm}^0$$ to experimentally get a $0$d unit – Тyma Gaidash May 05 '22 at 20:24
  • @TymaGaidash While the first term in this particular expansion could technically be written out as $\frac{log(x)^0}{0!}\cdot \mathrm{cm}^0$... I'd worry that things like $0!$ and $\mathrm{cm}^0$ are too likely to derail the conversation in directions unhelpful to the OP's original question. – DotCounter May 05 '22 at 22:13