# Why does dimensional analysis work? [duplicate]

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?

• Comments are not for extended discussion; this conversation has been moved to chat. – ACuriousMind Sep 26 '17 at 0:10

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 $f$.

• Could you ellaborate a bit on why this is? – Jemlin95 Sep 25 '17 at 20:23
• edited the answer to do so – Mr.Weathers Sep 25 '17 at 20:45
• thanks for elaborating. What I still don't quite understand is why in the case where each dimension has only one parameter the function must be of the form c(a^x * b^y....) where c is dimensionless. – Jemlin95 Sep 25 '17 at 20:51
• well you can only change the dimensions of a quantity by raising it to a power or multiplying by other dimensionful quantities. You cannot add two quantities unless they have the same dimension. So this tells us that we must consider objects of the form $a^x b^y c^z$ which have the correct dimensi0ns, and then build the most general function out of these dimensionally correct building blocks. In the case where there are no dimensionless parameters, there is only one such block. – Mr.Weathers Sep 25 '17 at 20:55
• @Jemlin95 The function doesn't have to look like that. Blind dimensional analysis to find functions works in simple cases, but not in complicated ones. – J. Murray Sep 25 '17 at 20:59