I have a rather fundamental question about the Buckingham $\pi$ theorem. They introduce it in my book about fluid mechanics as follows (I state the description of the theorem here, because I noticed in my search on the Internet that there are many different forms of this theorem) :
if we have a physical relation between some physical quantities $X_1$, $X_2$, …, $X_n$:
$$\phi(X_1,X_2,…,X_n) = 0$$
Then we can reduce this to a relation between $m$ dimensionless quantities, where $m = n-r$, with $r$ the rank of the dimensional matrix (the matrix formed by tabulating the exponents of the fundamental dimensions of $X_1$, $X_2$,…). So we get $m$ dimensionless quantities: $\pi_1, \pi_2, …, \pi_m$. And we have: $$\psi(\pi_1, …, \pi_m) = 0$$
These dimensionless quantities are formed by taking a core group of $r$ physical quantities from the $n$ quantities $X_1, …, X_n$ (for instance $X_1, …, X_r$). Now we can form:
$$\pi_1 = X_1^{a_1}*X_2^{b_1}*…*X_r^{f_1}*X_{r+1}$$
$$\pi_2 = X_1^{a_2}*X_2^{b_2}*…*X_r^{f_2}*X_{r+2}$$
and so on. These exponents are now determined so that the $\pi_1, \pi_2, ..$ are dimensionless.
So this was the description of the Buckingham $\pi$ theorem they mention in my book. First of all I wonder if this description is correct and complete, because I saw many different forms on the Internet that make the understanding of the theorem rather confusing.
Secondly, I tried to apply this theorem (in a way of getting familiar to it) to Newton's universal law of gravity (in a way of trying to derive the law, by the Buckingham $\pi$ theorem): $F = \frac{Gm_1m_2}{r^2}$. So while doing this, I see that there is a relation between $F, m_1, m_2$ and $r$ : $\phi(F,m_1,m_2,r) = 0$. The rank of the dimensional matrix $r$ is 3. So we have only one dimensionless number $\pi$. But if I take now as core group $m_1, m_2, r$, there is no way of making this number dimensionless (in the description of the theorem they don't mention that this can be the case). So I have to include $F$ in the core group and if I take $F,m_1, m_2$ as core group then $\pi_1 = m_1/m_2$ and I get $\phi(m_1/m_2) = 0$. This makes no sense to me, so I guess something is wrong in the way I understand this theorem, and I hope someone can explain to me what I'm missing.
Best Answer
Not sure if this will answer your queries, but I did get intrigued by the implied question:
Could Newton have derived his theory of gravitation from dimensional analysis?
Newton had a hunch that planetary motion is nothing more than a direct result of the same effect that makes objects drop to earth: the presumed universal attraction between all masses. He wanted to derive the consequences of this hypothesis. Available to him were his three laws of motion, as well as two important observations:
1) the empirical fact that different masses undergo the same acceleration (observed by Galileo)
2) Kepler's third law of planetary motion: the square of the orbital period of a planet is proportional to the cube of the radius of its orbit around the sun.
The first observation tells Newton he has to consider the acceleration $g = F_g/m$ ($m$ being the planetary mass) due to gravity rather than the force $F_g$ itself. The second observation tells him that a constant (which we will refer to as Kepler's constant) $K$ will probably play a role in the law of gravitation that he is seeking. This constant is the product of the square of the angular frequency $\omega$ with the cube of the radius $r$ of the planetary orbital motion: $K = \omega^2 r^3$. And finally, apart from $g$ and $K$, the distance $r$ is expected to feature in the equation sought.
So, Newton was seeking a relation $f(K, r, g) = 0$. Kepler's constant $K = (2 \pi)^2 AU^3 yr^{-2}$ has dimensions $[L^3 T^{-2}]$, the distance $r$ has dimension $[L]$, and the acceleration $g$ has dimension $[L T^{-2}]$. So one dimensionless parameter can be defined: $r^2 g/K$. This directs Newton towards an equation of the form $g \propto K/r^2$. In terms of the force $F_g = m \ g$, this reads:
$$F_g \propto K \frac{m}{r^2}$$
However, according to Newton's third law, this force needs to be mutual between the two bodies. If $M$ denotes the sun's mass, it follows that $K = G M$ with $G$ a universal constant equal to $(2 \pi)^2 M_{sol}^{-1} AU^3 yr^{-2}$. The result is an equation symmetric in both masses:
$$F_g \propto G \frac{M \ m}{r^2}$$
Obviously, we can replace the proportionality sign with an equal sign provided we absorb any mathematical factors hidden from the dimensional analysis into the constant $G$.