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Apparently, you can't use dimensional analysis to derive the formulae of quantities which depend on more than three other, different, composite quantities. Why is this so? The argument I've seen for it is that four different quantities would bring four variables into play (the powers of each of the quantities) but with only just the three equations that would result once we tidy it up into the terms of fundamental quantities, the system would not be solvable.

But it aren't just the most common M, L, and T that are fundamental (if they were, it makes sense why it'd work out that way), so isn't it possible that there could be a formula in terms of four composite quantities, which can be represented in terms of four different fundamental quantities, which would then give a solvable system? Then again, isn't it also possible that there are quantities depending only on three other quantities, but which when represented in fundamental quantities, only use two fundamental quantities? In such a case, the resulting system wouldn't be solvable.

In short, it doesn't seem like the number of initial composite quantities (the number of unknowns) is the only factor that matters; the number of fundamental quantities (the number of equations) that the expression is finally reduced to, also plays a role. In general, whether the resulting system is consistent with a single solution is the condition we need to be true to use dimensional analysis.

Where did I go wrong here for the rule to remain correct?

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Actually, you're right: according to the Buckingham $\pi$ theorem, you can avoid this problem as long as the number of quantities of which you take powers doesn't exceed the number of independent physical dimensions, which can be more than $3$. The dimensions usually considered are length $\mathsf{L}$, mass $\mathsf{M}$ and time $\mathsf{T}$, of respective Planck units$$\sqrt{\frac{G\hbar}{c^3}},\,\sqrt{\frac{c\hbar}{G}},\,\sqrt{\frac{G\hbar}{c^5}}$$viz.$$[c]=\mathsf{LT^{-1}},\,[G]=\mathsf{L^3M^{-1}T^{-2}},\,[\hbar]=\mathsf{L^2MT^{-1}}.$$We therefore can't get away with e.g. a second mass scale without the results of dimensional analysis becoming non-unique. But adding charge $\mathsf{Q}$, we can get away with a Planck charge $\sqrt{\frac{c\hbar}{k_C}}$ in terms of the Coulomb constant $k_C:=\frac{1}{4\pi\varepsilon_0}$, the electromagnetic analogue of $G$, viz. $[k_C]=\mathsf{L^3MQ^{-2}T^{-2}}$. Similarly, adding temperature $\mathsf{\Theta}$ gives a Planck temperature $\sqrt{\frac{c^5\hbar}{Gk_B^2}}$; note $[k_B]=\mathsf{L^2MT^{-2}\Theta^{-1}}$.

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  • $\begingroup$ I think Buckingham pi is a couple levels above me, so is it just that yes, dimensional analysis may fail to give correct formula in many cases due to specific conditions, and that it's best used to check whether the equation is correct at at least the dimensional level? $\endgroup$
    – harry
    Commented Oct 20, 2021 at 20:54
  • $\begingroup$ @harry Dimensional analysis can always check whether two given quantities have the same dimension. The main point of the theorem is how many "independent" dimensionless quantities you'll get when there are more quantities than dimensions. But the flip side of that is when this doesn't happen, as discussed here. As for whether dimensional analysis gives the "correct" result, that's never guaranteed. Try the Newtonian KE as a function of $m,\,v$: up to scaling the result has to be $mv^2$, but the $\frac12$ factor can't be guessed this way. $\endgroup$
    – J.G.
    Commented Oct 20, 2021 at 21:18
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I think it's useful here to step back from the specific numbers (three, four) in your answer, and think about this more abstractly.

Let's say you want to relate a quantity $X$, to a given mass scale $M$. If you know $X$ has units of mass, then dimensional analysis will tell you $X=M$ up to some dimensionless factor.

But now let's say $X$ is actually some combination of two different mass scales, $m$ and $M$. Now dimensional analysis doesn't tell you if $X=m$, or $X=M$, or $X=m^{3/5} M^{2/5}$, or some other combination. I think the numbers three and four from your answer, probably refer to the case where $X$ depends on a length, mass, and time; but if you have (say) two time scales, one mass, and one length, you will find yourself in an analogous situation.

The deep issue here is that there is a dimensionless ratio $m/M$, which need not be order 1. Therefore, while we can say something like $X=f(m/M) M$ for some dimensionless function $f(m/M)$, we have no reason to think that $f$ will be some constant, order 1 function.

One example of this appearing is in gravitational waves. The frequency of two coalescing binary black holes evolves in time, and the evolution of the frequency should depend on a dimensionful combination of the two masses somehow. It turns out that the right dimensionful combination that enters into the frequency evolution is the chirp mass \begin{equation} M_c = \frac{(m_1 m_2)^{3/5}}{(m_1 + m_2)^{1/5}} \end{equation} which you would not have guessed from dimensional analysis.

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  • $\begingroup$ By different mass or time scales, do you mean different units, like metres and kilometres or seconds an hours? Thanks for the answer! $\endgroup$
    – harry
    Commented Oct 20, 2021 at 20:49
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    $\begingroup$ @harry Say you want to know the energy of a rocket ship that's moving linearly and also spinning. The mass is $M$, it's physical size is $R$, the distance it travels is $L$, the time it takes to get from beginning to end is $T$, and the time it takes to perform one rotation is $t$. There are several different quantities like $M (R/T)^2$ and $M (L/t)^2$ that have units of energy. You need more than dimensional analysis input to realize the two meaningful combinations are $M (L/T)^2$ and $M (R / t)^2$. Here there's one mass scale $M$, two length scales $L$, $R$, and two time scales $T$, $t$. $\endgroup$
    – Andrew
    Commented Oct 20, 2021 at 20:59

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