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I just read Accurate predictions of coexistence in natural systems require the inclusion of facilitative interactions and environmental dependency in which they defined the following property:$^1$

$$\rho = \frac{e^{\alpha_{ij}}e^{\alpha_{ji}}}{e^{\alpha_{ii}}e^{\alpha_{jj}}}.\tag{11}$$

The problem is the $\alpha$ are (I think) not dimensionless. They do not explicitly talk about units in the article however the $\alpha$ come from a difference equation that is essentially equivalent to the time discrete Lotka Volterra model. I would say the unit of $\alpha$ is $\frac{m^2}{\text{individual}}$ (for this question we can just assume it is).

Is the $\rho$ than even defined? Can we argue that the unit of the denominator and the numerator will cancel themselves out and $\rho$ is dimensionless? Are there any attempts to define $e^{\mathrm{m}}$?

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$^1$ $\alpha$ is strictly speaking called $\alpha'$ in the article.

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    $\begingroup$ I'm not sure what Farcher had in mind, but I do think it would be useful to mention the title of the article and make that a link to the page, instead of just having the link text be the word "article". That being said, there's nothing wrong with what you did; a link presumably allows people to find the original article and that should be fine. It just would be better to have more information. $\endgroup$
    – David Z
    Sep 5, 2018 at 7:25
  • $\begingroup$ @DavidZ Sorry, in the original post I did not see that the word article was highlighted to show that it was a link to the document in question. The other problem for me is that I cannot access the article in question. $\endgroup$
    – Farcher
    Sep 5, 2018 at 8:09
  • $\begingroup$ In principle, this expression would not make sense if the $\alpha$ had units. You can rewrite it as $\rho = e^{\alpha_{ij}+\alpha_{ji}-\alpha_{ii}-\alpha_{jj}}$ so you would be stuck with the same problem. Another way to see this, is that the coefficient $\rho$ is not invariant under rescaling the units. $\endgroup$
    – Tob Ernack
    Jul 11, 2019 at 15:20
  • $\begingroup$ But how do you know the units of $\alpha$? I had a look at the paper, and they don't seem to be very explicit about the units for the $\alpha'$. They call them "interaction coefficients" or something like that. They use a model from another paper, and I looked this other paper up too, and they also don't explicitly mention the units of $\alpha$. They say that they are not the same as the "relative interaction coefficients in the Lotka-Volterra equation", but I don't really know what this means. $\endgroup$
    – Tob Ernack
    Jul 11, 2019 at 15:23
  • $\begingroup$ @TobErnack The Lotka-Volterra equations (similar as the annual plant model) are in pronciple applicable to any biological system. Hence $\alpha$ does not have a unit per se, rather the unit of $\alpha$ depends on the system measured. For their system the $\alpha$ is unitless, however for my system the unit of $\alpha$ would be $cm^{-3}$, so I wondered whether I could use their method. Turns out not. $\endgroup$
    – Jürg Merlin Spaak
    Jul 15, 2019 at 5:09

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One can write out the exponential as \begin{equation} e^m = \sum_k \frac{1}{k!} m^k, \end{equation} so this expression would yield a series of terms which all have different units (namely, a different power of $m$). Such a series would not make physical sense.

Quantities are often made dimensionless before being plugged into an exponential by dividing out some arbitrary reference value. In your case, the choice of reference value would only affect $\rho$ by raising it to some power. Such an effect can be irrelevant, depending on the context; in this case, I suppose it would be.

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