1
$\begingroup$

Through my years in nuclear engineering, it has always been the case that in physical relations, the arguments of transcendental functions, e.g., the exponential in the law of radioactive decay, $N=N_0 e^{-\lambda t}$, should be dimensionless so that the outcome is correspondingly dimensionless.

In purely empirical relations, however, I have observed that it's okay for transcendental functions to have dimensional arguments as long as they fit the results of experiments. For example, the energy spectrum of prompt fission neutrons is fitted by: $$\chi(E)=0.453 e^{-1.036E}\text{sinh}\sqrt{2.29E}.$$ Or does the 1.036 in the exponential have units of ${\text{eV}}^{-1}$?

Is my observation correct? And if so, what is the basis of all that? I tried to dig into dimensional analysis, but I couldn't understand much.

Update on Jan 14, 2020:

I've found an insight on Wikipedia that I think will, hopefully, enrich the discussion.

A problem with transcendental functions is that: applying a non-algebraic operation to a dimensional quantity creates paradoxical results. For instance, $\text{log}_a(5L)=\text{log}_a(3)+\text{log}_a(L)$, where $L$ is the dimension of length, and $a$ is an arbitrary base.

This also raises the question: is $\text{log}_a(L)$ dimensional? Does a transcendental function become dimensional when fed with a dimensional argument?

$\endgroup$
2
$\begingroup$

You are correct, in the case you cite, 1.036 has dimension eV${}^{-1}$ (assuming $E$ is measured in eV).

All functions can be written in terms of dimensionless arguments, not just the transcendental ones. It's common to write equations in dimensionless form, where all variables are dimensionless, that is, the ratio of some measured quantity to a "characteristic value" of the same dimension that sets the scale.

In your case the characteristic value is $E_0 = (1/1.036) eV$, and it sets the scale. Values of $E$ less than $E_0$ are "small", those greater are "large". When $E=E_0$ the value of the function is of the order of unity (one).

Every function can be written with dimensionless arguments. For "normal" rational, power law, algebraic, etc functions, the characteristic value can be factored out if desired, and the dimensionless quality is less evident.

| cite | improve this answer | |
$\endgroup$
7
$\begingroup$

For a theory to make sense, it has to make useful predictions regardless of the units you use.

For algebraic formulae, that usually ends giving you constants in the formula with different values. Newton’s gravity looks exactly the same in English and metric units, except the constant G has a different numeric value.

But that doesn’t work for transcendental functions. $e^{24}$ when working in inches and $e^{2}$ when working in feet can’t be made equal with a constant in front of the expression; that number would be different for different input values, hence not constant.

Instead, we specify what units the argument should be. “Must be in feet”, if that makes the numbers work, provides a convention.

Sometimes we’ll actually write that as $$e^{x/(\rm{1 ft})}$$ Then if x is in other units, you’re reminded to convert.

Sometimes, this happens automatically. For example a lifetime: $$e^{-t/\tau}$$ Both of those quantities have units, and we’re used to making them consistent before dividing. Once we do that, it becomes “3 lifetimes” and the units we measure in no longer matter: the transcendental argument is again unitless.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ So for empirical relations we have to specify the units along with the relation. I see that a lot but I couldn't make the connection! Thanks so much! $\endgroup$ – Mohamed Anwar Dec 28 '19 at 3:20
2
$\begingroup$

$$e^x = 1 + x + x^2 + ...$$

Therefore if $x$ has units, $e^x$ doesn't even make sense by dimensional analysis. The argument of log, and any trigonometric function is also unitless for the same reason.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Note that by your argument this is even true for polynomials. $\endgroup$ – Anders Sandberg Dec 28 '19 at 8:05
  • $\begingroup$ Yes. Should that not be the case? Whoever downvoted, if my answer is wrong for some reason I'd like to understand. $\endgroup$ – doublefelix Dec 28 '19 at 12:31
  • $\begingroup$ Also note that it is only true for polynomials if the coefficients are unitless, as they automatically are in e^x and such. $\endgroup$ – doublefelix Dec 28 '19 at 12:52
  • $\begingroup$ @doublefelix If the coefficients of the polynomial have dimensions, then their sum must have dimensions. One can then divide both sides by a quantity of that dimension and then the entire equation is dimensionless. And if you divide by some quantity characteristic of the problem (a length, an energy, ...) then the resulting equation is universal in the sense that it will work for situations of any "size", and any system of units. On top of all that, a universal dimensionless equation makes much nicer graphs! No units on the axes, and all values are near unity. $\endgroup$ – garyp Dec 28 '19 at 20:47
  • $\begingroup$ I don't see the relevance to the question at hand @garyp $\endgroup$ – doublefelix Dec 28 '19 at 23:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.