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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.

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    $\begingroup$ More on Buckingham $\pi$ theorem: physics.stackexchange.com/search?q=buckingham $\endgroup$ – Qmechanic Aug 14 '13 at 14:21
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    $\begingroup$ "there is a relation between $F, m_1, m_2$ and $r$ : $\phi(F,m_1,m_2,r) = 0$" -- you should include $G$. $\endgroup$ – Johannes Aug 14 '13 at 14:37
  • $\begingroup$ I don't completely understand though why I should include $G$, because there exists also a relation $\phi$ without $G$. Also, it seems a bit strange to me if I have to include $G$, because I thought that Buckingham-$\pi$ theorem is used to develop an equation, when it is only known that there is a relation between some physical quantities. I don't see how the theorem would be useful if we have to include the constants, because it seems to me that there is no way in knowing these, before the formula is already derived $\endgroup$ – Rayman Aug 15 '13 at 17:38
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    $\begingroup$ @Rayman - including the dimensionful constant $G$ is no different from including the dimensionful constant $g$ in the derivation of the period of a simple pendulum (the standard example for how to apply the $\pi$-theorem). $\endgroup$ – Johannes Aug 15 '13 at 18:32
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    $\begingroup$ @Rayman you are right about the constant. There is no way to know about it from start, since it's basically fixing the problem of the dimensions in equation. I have a feeling it is not possible to derive newton's law with dimensional analysis same way as it is not possible to define F=ma, simply because there are situations where F is not equal M*A. Problem is that initial transformations between dimensions are based on THESE law's. Otherwise there is no way to define force through acceleration, it is simply nonsense. For example forces exist there where are no accelerations. $\endgroup$ – Asphir Dom Aug 16 '13 at 13:28
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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 $G$.

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  • $\begingroup$ I think using Kepler's law here is not justified to be called dimensional analysis. You need to use parameters of the system (size, from, mass etc.) not its dynamical (already observed) relations. These relations already uncover you part of the answer. Thus in this case its rather a tool to rewrite the known rather than a method to see the new. IMO. $\endgroup$ – Asphir Dom Aug 16 '13 at 13:57
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    $\begingroup$ It is possible to find out Newton's law just starting from Galileo's observation and ellipse as an universal orbit form for all planets? (not using third law of Kepler) $\endgroup$ – Asphir Dom Aug 16 '13 at 14:04
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    $\begingroup$ @Asphir - that seems rather dogmatic. Can't think of a single reason why one would allow static properties and forbid dynamic properties in dimensional analysis. Furthermore, I don't agree that the above derivation 'rewrites the known'. The key inventive step $K=GM$ renders the outcome far more general than Kepler's law. $\endgroup$ – Johannes Aug 16 '13 at 14:58
  • $\begingroup$ What i mean is that first step where you were trying to solve f(K,r,g)=0 is to my opinion all extent of dimensional analysis. With it you could reach only point of $g\approx\frac{K}{r^{2}}$. Further arguments already involve usual mechanical line of thought, same as solving it. I don't see how you made leap $K=GM$, just to make it symmetric? $\endgroup$ – Asphir Dom Aug 16 '13 at 19:11
  • $\begingroup$ @AsphirDom You arrive at $K=GM$ by stating that, since $F_g = mg$ by actio=reactio one must also have $F_g = Mg'$, where $g'$ denotes the gravity caused by mass $m$ acting on $M$. Therefore, $K\propto M$ and the proportionality factor is labelled $G$. Of course $G$ might still be a function of $K,r,g$ without further investigation... $\endgroup$ – Tobias Kienzler Jan 8 '15 at 12:48

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