Buckingham-$\Pi$ theorem application: the case of only 0 or 1 dimensionless groups?

Question

In dimensional analysis, we might consider a problem like:

$$ f(q_1, q_2, ..., q_{n-1}) = q_{n} $$

where $q_i$ are physical quantities of interest. In order to determine how many $\Pi$ groups can be formed, we consider how many "fundamental" quantities are present in the problem. Let us say there are $k$ such fundamental quantities, then we will have $j = n - k$ dimensionless groups, and we can write the above relationship as:

$$ f(\Pi_1, \Pi_2, ..., \Pi_{n-k-1}) = \Pi_{n-k}$$

So consider this in the case of an example, the equation of motion in a 1-D overdamped system, with $\eta$ being the viscosity (dimensions $MT^{-1}$), velocity $\rm{d}x/\rm{d}t$ with units $LT^{-1}$, and force $F$ with units $MLT^{-2}$:

$$\eta \frac{\rm{d}x}{\rm{d}t} = F$$

Or, I can rewrite the equation in a form more familiar in the context of dimensional analysis, where don't care about the actual relationship between quantities: $$ f(\eta, \frac{\rm{d}x}{\rm{d}t}, F) = 0$$

Here, we have $n = 3$ variables, and one may say that we also have $k = 3$ fundamental dimensions $M$, $L$ and $T$. Then, we have $j = 0$! What now?

$$ f(1, 1, 1) = 0$$

What is dimensional analysis telling me here? Is dimensional analysis telling me that I can write the relationship between variables in a totally dimensionless form?

Okay, maybe I can dance around this by saying that I really only have $k = 2$ fundamental dimensions, if I let them be $MT^{-1} = [\eta]$, and $LT^{-1} = [\dot{x}]$. Okay, so now I have $j = 1$, so I can form one $\Pi$ group, let me call it $\Pi_0$. Skipping the work, $\Pi_0$ will correspond to $F = \eta\dot{x}$.

This is a major hint that dimensional analysis is telling me something really interesting. In cases where we have $j \leq 0$, we can redefine "fundamental dimensions" to have $j > 0$. Then, dimensional analysis gives us a...major hint about how the variables are related? I am sure there is a better (more insightful) way to put what I have I just learned -- but what is it?

Answer 1

You've made a mistake in saying

Here, we have $n = 3$ variables, and one may say that we also have $k = 3$ fundamental dimensions $M$, $L$ and $T$.

but it's a subtle one.

To see why the implication does not hold, consider three different variables, with dimensions $[A]=ML/T$, $[B]=ML/T$ and $[C]=ML/T$. Here you have three variables, and three independent dimensions $M$, $L$ and $T$... but of course you cannot infer from their dimensions alone that $A$, $B$ and $C$ are themselves independent. In my example, $n=3$, $k=1$, I can form $n-k=2$ independent pi groups, $B/A$ and $C/A$, and any physically meaningful equation $f(A,B,C)=0$ can be rephrased as $F(B/A,C/A)=0$. You can see how it would go if you add more variables $D,E,\ldots$ with the same dimensionality.

(Note that this is quite a common situation. For example, $B$ and $C$ might be the momentum of two particles which are joined in a collision to give a total momentum $A$. The correct relationship between them is then $F(b,c)=b+c-1=0$.)

The way to test for this is via the dimensionality matrix. In your case, with $[F]=ML/T^2$, $[\eta]=M/T$ and $[v]=L/T$, the dimensionality matrix is $$ D=\begin{array}{cl} \begin{array}{ccc}F&\eta&v\end{array} & \\ \left(\begin{array}{ccc}1 & 1 & 0 \\ 1 & 0 & 1 \\ -2 & -1 & -1 \end{array}\right). &\begin{array}{c} M\\ L\\ T \end{array} \end{array} $$ Its rank is $k=2$, which means that you really only have two independent dimensionalities. Any two will do, but the clearest choice is probably $[\eta]$ and $[v]$, in which case $[F]=[\eta][v]$ is obviously dependent. Since there's $k=2$ independent dimensions and $n=3$ variables, you can form $n-k=1$, i.e. a single pi group, which is $F/\eta v$ or its inverse. Your initial physically-meaningful equation $g(F,\eta,v)=F-\eta v=0$ can thus be rephrased as $$G(F/\eta v)=F/\eta v-1=0.$$

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Note in particular that the pi groups need to be dimensionless. This is the whole point of the pi theorem and the game doesn't really make sense unless you do that. Thus

Skipping the work, $\Pi_0$ will correspond to $F = \eta\dot{x}$.

is not correct.

Similarly,

cases where $j=n-k<0$

don't make sense. You cannot encompass $k>n$ independent dimensions using only $n$ variables. This is a fundamental fact of linear algebra: a set of $n$ vectors (in dimensionality space) can only span a subspace of dimension $k\leq n$.

It may be worthwhile exploring the other end of the dimensionality scale to clarify what's going on - the case where $n=k$. Here all your dimensions are physically independent - you might have a length, a time and a mass, say. There is then nothing you can say about them that will be physically meaningful, and you can't form a single pi group. The only allowed statement is $F()\equiv0=0$, which is an empty statement. You can define new variables (like, say, momentum) but until you do you're stuck with nothing.