4
$\begingroup$

I have a distribution of microparticles that follows a lognormal distribution. The cumulative distribution function thus is given by:

$$ F_X(x;\mu,\sigma) = \frac12 \operatorname{erfc}\!\left(-\frac{\ln x - \mu}{\sigma\sqrt{2}}\right) $$ $$ \mu = ln(M) + \sigma^2 $$

Now, the plot of the distribution function should be exactly the same no matter if the particle diameter $x$ is given in micrometers or meters (as long as I adapt the x-axis accordingly of course). However, this only works if I only convert $x$ and $M$, while not touching the numerical value of $\sigma$, and I don't understand why. $F_X$ has to be unitless, so $x$, $\mu$ and $\sigma$ should all have the same unit, right?

$\endgroup$
  • 1
    $\begingroup$ Actually, $\ln x,\mu$ and $\sigma$ have the same unit. $\endgroup$ – Glen_b Mar 9 '14 at 14:26
  • $\begingroup$ Right. But what unit would $\sigma$ have when e.g. $M$ and $x$ are given in meters? $\endgroup$ – akid Mar 9 '14 at 14:31
5
$\begingroup$

It's perhaps a somewhat subtle and interesting question.

That it may be subtle can be seen from the different positions here (though most of the conclusions are identical).

The answer is that $\ln(x)$, and hence, $μ$ and $σ$, are unit-free.

This paper might be of some help:

Matta, Massa, Gubskaya & Knoll, (2011),
Can One Take the Logarithm or the Sine of a Dimensioned Quantity or a Unit?
Dimensional Analysis Involving Transcendental Functions,
Journal of Chemical Education, Vol 88, No. 1

which explains that properly considered, the logs are taken of ratios relative to unit rate constants, which necessarily remove the units. It also has useful discussion of some other functions.

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

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.