Exponential or logarithm of a dimensionful quantity?

Question

I have a unit measure, say, seconds, $s$. Furthermore let's say I have a dimensionful quantity $r$ that is measure in seconds, $s$. What is the unit measure of $e^r$? ($1/r$ is in $Hz$.)

My question is general, how to find the unit measure of a transformation function $y=f(x)$ where $x$ takes some known unit measure. I give above two functions $f(\cdot)=e^\cdot$ and $f(.)=1/\cdot$.

Unless $r(s)$ is unitless, $e^r$ doesn't make much sense (see, for example, its definition in terms of the power series) — Kyle Kanos, Apr 25 '14 at 15:42
@Jika: As Kyle mentioned, in physics it is impossible to have something of the form $e^r$, unless $r$ is unitless. If obtain something of the form $e^r$ where $r$ is not unitless, that means you made a mistake somewhere. Try to check your math for errors. — DumpsterDoofus, Apr 25 '14 at 15:46
ln(42m)= ln(42)+ln(m) so when dividing ln(10km/1km) = ln(10)+ln(km)-(ln(1)+ln(km)) = ln(10)-ln(1) = ln(10) It will still be dimmisionLess Hence, kevin's argument is invalid . — Dhruv Chadha, Oct 10 '21 at 18:55

Nephente · Accepted Answer · 2019-02-26T07:35:17.790

31

The only sensible rule when working with units is, that you can only add together terms which carry the same unit. Say $ [x]=[y] $, then $x+y$ is unit-wise a valid statement. You may also multiply arbitrary units together. Whether that is physically sensible is another question. Obviously you cannot add, e.g meters and seconds, but multiplying to form $m/s$ as a unit for velocity is a valid operation.

From that follows, that the argument of the exponential must not carry a unit, because the exponential is defined as a power series. $$ e^x =\sum_{n=0}^\infty \frac{x^n}{n!}$$ If $x$ were to carry a unit, say meters, one would add (schematically) $m+m^2+m^3+\cdots$, which is nonsenical.

If you encounter an exponential, a sine/cosine, logarithm,... in physics you will find almost always that its argument, which must be dimensionless, is a product of often two conjugate variables. Examples are time and frequency, or distance and momentum.

edited Feb 26 '19 at 07:35

answered Apr 25 '14 at 15:58

Nephente

3,278

23

People do sometimes write things like $\log E$ and so on, with $E$ being a quantity with units, but that's understood to be a shorthand for $\log\frac{E}{E_0}$ for some reference value $E_0$ whose value is irrelevant. (And as far as I'm concerned, it's generally better to write out $\log\frac{E}{E_0}$ explicitly.) – David Z Apr 25 '14 at 16:42
1

I would say always, not almost always, although it might not be readily apparent. You might have $(e^r)^k$, where $k$ has units of inverse meters. With logs you might have $(\ln{r} - \ln{k})$. An author might choose to drop the second term if it is of no consequence, but that only serves to add to the confusion. – garyp Apr 25 '14 at 16:44
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Another common notation (especially in axis labels) is something like $\log(d/\mathrm{m})$ (in this case the $E_0$ David Z mentions is just "one metre"), or $\log(v/c)$ (the "$E_0$" is a dimensional constant with the same dimensions as the quantity of interest). – Kyle Oman Apr 25 '14 at 20:13
Where can I learn more about conjugate variables? – Neil G Dec 20 '19 at 21:00

score 16 · Answer 2 · edited Mar 07 '19 at 02:20

See "what's the logarithm of a kilometer" for a discussion about that. As David Z also said in the comment here, using the logarithm of a dimensionful quantity is actually quite reasonable.

This is not true for the exponential. The power series definition "proves" that, however the same argument would also work for the logarithm. Personally I don't like treating the Taylor series as anything more than a useful calculation tool. The "more fundamental" (of course there's no such metric) definition is as a solution to the differential equation $\tfrac{\mathrm{d}\exp}{\mathrm{d}x} = \exp(x)$. Which tells you right away $$ \tfrac{[\exp]}{[x]} = [\exp] \qquad \Rightarrow\quad [x] = 1. $$ Note that this does not come out when using the analogous definition of the logarithm: $$ \frac{\mathrm{d} \ln}{\mathrm{d}x} = \frac{1}{x} \qquad \Rightarrow \quad \tfrac{[\ln]}{[x]} = \tfrac{1}{[x]} \qquad \Rightarrow \quad [x] =\:? $$ Of course, both equations only define the functions up to gauge of an initial value. For $\ln(1) = 0$ to make sense, you certainly need the argument to be dimensionless. But as long as you only consider differences between logarithms, the gauge cancels anyway!

score 1 · Answer 3 · answered Feb 18 '17 at 23:28

Consider:

Conc = 100 mg/mL

Log10( Conc ) = 2

What if I express Conc as mg/dL? Then Conc2 = 1 mg/dL (note that this is the same, just different units with a different quantity):

Log10( Conc2 ) = 0.

Log10( Conc ) ne Log10( Conc2 ) and we have an issue, unless we retain units. Why wouldn't log10 mg/mL be any less reasonable than mg/mL?

Second, consider this argument from the internet: to take the logarithm, the quantity must be dimensionless, thus divide by the unit (under what justification?):

Log10( 10 km / 1 km )

The issue is:

Log10( 10 km / 1 km ) = Log10( 10 km ) - Log10( 1 km )

-Kevin

"consider this argument from the internet"? Source would be helpful. — tusky_mcmammoth, Jul 02 '17 at 15:41

Exponential or logarithm of a dimensionful quantity?

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