Do all physical equations have a polynomial and a non-polynomial part?

Asked Oct 14 '19 at 18:42

Active Oct 14 '19 at 18:42

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Let $Q,Q_1,...,Q_n$ be $n+1$ physical quantities associated with some physical dimensions. Assume, the physical equation

$$Q = f(Q_1,...,Q_n) $$

is true. While discussing methods of dimensional analysis, my professor noted, that all physical equations of the above form can be written as

$$Q = \tilde{f}(Q_1,...,Q_n) Q_1^{p_1}...Q_n^{p_n}$$

for some real exponents $p_1,...,p_n$.

Is this true? Why should this be the case? So far, I have a very hard time, wrapping my mind around dimensional analysis.

asked Oct 14 '19 at 18:42

Joker123

What is $\tilde{f}(Q_1,...,Q_n)$? – Dietrich Burde Oct 14 '19 at 18:45
This is not a mathematical statement. You need some kind of physical postulates. – ajotatxe Oct 14 '19 at 18:45
1

The professor was probably referring to Buckingham π theorem. – Jean-Claude Arbaut Oct 14 '19 at 18:46
The only physical postulate seems to be that the dimension on the left and right side of the equation is the same. – Joker123 Oct 14 '19 at 18:47
This obviously does not work as written if $p_1\neq 0$ and there is any combination of values of $Q,Q_2,\ldots,Q_n$ for which $Q \neq 0$ and $Q=f(0,Q_2,\ldots,Q_n).$ I think there are some implicit additional assumptions. – David K Oct 14 '19 at 19:00

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