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One of the prerequisites of the Buckingham π theorem is that the physical law in question should be unit-free. I couldn't find an example of a physical law that is not unit-free. Is there such thing?

Added later: Here is a definition from Applied Mathematics by Logan:

The physical law $$f(q_1, \ldots, q_m) = 0$$ is unit-free if for all choices of real numbers $\lambda_1, ... , \lambda_n,$ with $\lambda_i > 0$, we have $f(\bar{q}_1, \ldots , \bar{q}_m) = 0$, if, and only if $f(q_1, \ldots, q_m) = 0$.

(Note that the definition is not quite self-contained. You might want to click on the above link to look up how $\lambda$'s relate to $\bar{q}$'s, it's spread over a page of the book.)

So I'm asking because the wording in textbooks somehow implies (the way I read it) that being unit-free is a property of a physical law, i.e. like differentiability is a property of a function. So I naturally wanted to see a counterexample.

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    $\begingroup$ At current completely unit free laws are not possible. We have, for instance, absolutely no connection between (rest) mass and electric charge. One can not replace one by the other in calculations, hence there have to be different units. Maybe one day there will be a formula that will link the charge and the mass spectrum, but as of now they are completely different physical phenomena. $\endgroup$
    – CuriousOne
    Commented May 20, 2016 at 2:26

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I'm not sure if this gives you the answer you want in terms of a rigorous physics explanation, but when I studied dimensional analysis, the main point was to ensure all sides were equal when it came to units. Essentially, the equations, and laws, needed to be unit-free in order to make sense. You can't be left with something where a unit is equal to a non-unit.

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    $\begingroup$ Nitpick: There's a difference between units and dimensions, and I think you mean dimensions, not units. $\endgroup$
    – DanielSank
    Commented May 20, 2016 at 3:52
  • $\begingroup$ No, he does mean units. I read the Buckingham pi theorem, it seems to say that if a law is say f=Ma and another is f = Mm/r^2, then you can write each as unitleess $\endgroup$
    – Bob Bee
    Commented May 20, 2016 at 4:04
  • $\begingroup$ By saying f/ma = 1 and the same for the other equation. You can combine them as ma/[Mm/r^2] = 1. I don't see the value, just a way to balance units. $\endgroup$
    – Bob Bee
    Commented May 20, 2016 at 4:14
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    $\begingroup$ There is a difference between dimensions and units, but it's fairly minor, and generally they can be used interchangeably for this explanation - but yes, technically I mean dimensions. For the value in the exercise, it's more just a useful tool to make sure your equation is balanced and you haven't forgotten something (like using r instead of r^2 in that example - you'd be left with something at the end so you know there's a problem) $\endgroup$
    – SeanD
    Commented May 20, 2016 at 6:55

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