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Suppose we want to calculate a velocity. We identify all the relevant dimensions on which the solution could depend and write out and solve the equation $$l/t= a^x \, b^y \, c^z$$ where $a$, $b$, and $c$ are those relevant dimensions, and $l$ stands for length and $t$ for time.

I am having trouble understanding why this works. My thought process has been as follows; We seem to be looking for some function from $\mathbb{R}^n \rightarrow \mathbb{R}$ where $n$ is the number of relevant units. I guess we can assume this function exists since for any set of values the relevant parameters take we should be able to predict the velocity. If the function has the form written above I would say that it has to be dimensionally consistent since if not the solution would be dependent on the definition of one or more units which is obviously not the case (does someone have a better explanation for this?). I don't understand why we can write this function in the form $a^x \, b^y \, c^z$. How do we know the function doesn't take a different form?

Jemlin95
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You can only write the function in that form if there is only one relevant parameter of each dimension. For example, say your problem has two length scales, say sides of a rectangle, x,y. Then there is a dimensionless ratio x/y and then you couldn't rule out, a priori, any functional prefactor f(x/y).

In short, dimensional analysis works in the simple cases where there is no dimensionless constant relevant to the problem.

Requested elaboration: In your original statement, consider that there is a 4th parameter d, which has the same dimensions as a. If $a^x b^y c^z$ is a dimensionally correct expression, then so is $d^x b^y c^z$. and then so is their sum $a^x b^y c^z + d^x b^y c^z =a^x b^y c^z \left(1+\left(\frac{b}{a}\right)^x \right)$

There is no longer a unique expression which is dimensionally correct. In fact there is quite a lot of freedom. The following is also dimensionally correct:

$f(a/b)a^x b^y c^z $

for any dimensionless function $f$.