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?

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    $\begingroup$ 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

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.

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