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It seems to me that although dimensional analysis is useful for demonstrating when a physics equation is wrong (when the dimensions are inconsistent), there is little justification for how it is often used to show that a given equation is correct. I'll motivate this claim with an example I went over in a class I'm TAing. Generally speaking, my question is: How (and under what conditions/assumptions) can we know that a functional relationship predicted by dimensional analysis, specifically by Rayleigh's method (https://en.wikipedia.org/wiki/Rayleigh%27s_method_of_dimensional_analysis), is the unique functional relationship that can hold between a given set of quantities?

Consider a column of stationary fluid in Earth's gravitational field. We'd like to find a formula for $\Delta P$, the increase in pressure that is measured when moving from the surface of the fluid to a point at depth $h$. Assume that we know that $\Delta P$ depends on only three variables: The depth $h$, the mass density $\rho$ of the fluid, and the acceleration of gravity $g$. These quantities have dimensions of:

$$\Delta P : [M][L]^{-1}[T]^{-2}$$ $$\rho : [M] [L]^{-3}$$ $$g : [L][T]^{-2}$$ $$h : [L]$$

Here, $[M]$ is mass, $[L]$ is length, and $[T]$ is time. Now, the standard argument in dimensional analysis goes as follows (this is basically Rayleigh's method of dimensional analysis). We suppose the equation for $\Delta P$ has the form

$$\Delta P = k \rho^a g^b h^c,$$

where $k$, $a$, $b$, and $c$ are all dimensionless constants. Now, for this equation to be meaningful, the dimensions on the left and right sides must be the same. To ensure this, we can replace each quantity in $\Delta P = k \rho^a g^b h^c$ with its corresponding dimensions, yielding:

$$[M][L]^{-1}[T]^{-2} = ([M] [L]^{-3})^a ([L][T]^{-2})^b ([L])^c = [M]^a [L]^{-3a+1+c} [T]^{-2b}$$

For the dimensions to be the same on both sides, we have to have $a=b=c=1$, and thus $\Delta P = k \rho g h$ for some dimensionless $k$.

At this point, we've certainly shown that $\Delta P = k \rho g h$ is a possible, dimensionally consistent formula for the pressure change. But I don't see how this formula is at all unique. For me, the biggest unjustified assumption enters when we assume that $k$ must be dimensionless. Even if we know that the only variable dimensionful quantities that $\Delta P$ depends on are $\rho$, $g$, and $h$, I don't understand how it follows that $\Delta P$ can't depend on one or more dimensionful constant. But the values we get for $a$, $b$, and $c$ will depend on what dimension we suppose that $k$ has. So if we can't pin down the dimension (or lack thereof) of $k$, then we can't fix $a$, $b$, and $c$, and so we've made no progress beyond the initial assumption $\Delta P = k \rho^a g^b h^c$.

When considering Newton's law of gravitation from the perspective of dimensional analysis, this ambiguity manifests itself in the opposite way. In order to derive the proportionality $F = G m_1 m_2 / r^2$ in the same way as we did $\Delta P = k \rho g h$ (by assuming that $F = G m_1^a m_2^b r^c$), we must assume that our constant $G$ has dimensions $[L]^3 [M]^{-1} [T]^{-2}$. Why is a dimensionful constant reasonable in this case, but not when deriving $\Delta P = k \rho g h$? I think this is the sticky point that this one asker was getting at in their question: Dimensional analysis - When can you introduce constants that make dimensions compatible?

In that question, the answerer provides a method to determine the dimensions of $G$ to make Newton's law dimensionally consistent, but this is only after having shown that $F \propto m_1 m_2 / r^2$, without dimensional arguments. But this is a far more modest result than what is often claimed to be a consequence of dimensional considerations - namely results like $\Delta P = k \rho g h$, or $T \propto \sqrt{l/g}$ for the period of a pendulum as a function of $g$ and its length $l$.

So is there an insight missing in my analysis, which would clear up these ambiguities and ensure that dimensional analysis produces a unique functional form for the relationship between a set of quantities? Or is dimensional analysis just often used in ways that aren't always justified?

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  • $\begingroup$ Have a look at the Buckingham's $\pi$ theorem. $\endgroup$ – Massimo Ortolano Sep 7 '16 at 19:19
  • $\begingroup$ @MassimoOrtolano my understanding is that invoking the pi theorem doesn't alleviate the concerns i've discussed. i think it again comes down to the assumption that you only have a certain number of dimensionful quantities in the problem. while this assumption gives me the right answer in the fluid pressure example, it leads me astray when considering newton's law - unless i arbitrarily assume that the relation between $F$, $m_1$, $m_2$, and $r$ also has some constant $G$ with units $M^{-1} L^3 T^{-2}$. but why is it right to have this extra dimensionful constant in one case but not the other? $\endgroup$ – Wade Hodson Sep 7 '16 at 20:01
  • $\begingroup$ Sorry, maybe I've commented too hastily: I'll reread your question more carefully tomorrow with fresh mind... bedtime here now. $\endgroup$ – Massimo Ortolano Sep 7 '16 at 20:04
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Your last suggestion is correct : the use which you are making of dimensional analysis is not justified.

Dimensional Analysis has two purposes : (1) to check that equations or terms in equations are commensurate; and (2) to find combinations of quantities with particular dimensions or no dimensions at all. It is not able to derive physically meaningful formulas, and certainly not a unique formula, which is what you seem to be expecting it to do.

The 2nd purpose can be achieved using Rayleigh's Method, or Buckingham's Pi Theorem which is a more formal version of it. For this purpose the method works, and it is easy to see why it works. That to me is sufficient justification.

In your 1st example, Rayleigh's Method tells you that the formula $\Delta P=k\rho gh$ (where $k$ is a dimensionless constant) is dimensionally consistent, but it cannot tell you that this formula is physically meaningful or correct for the application you have in mind. It cannot tell you what value of $k$ you should use. It can only tell you what dimensions $k$ must have for the formula $\Delta P=k\rho gh$ to be dimensionally consistent.

In the 2nd example, the method cannot tell you that the force of gravitational attraction between two masses should be $F=Gm_1m_2/r^2$. The method does not generate any dimensionless combinations of the quantities $F, m_1, m_2, r$. This is not a failure of Rayleigh's Method because no dimensionless combinations of these variables can be derived by any method. As with your 1st example, dimensional analysis can only tell you what dimensions the constant $G$ must have for the formula $F=Gm_1m_2/r^2$ to be dimensionally correct.

The strengths and limitations of the method are stated in the wikipedia article :

The Buckingham π theorem provides a method for computing sets of dimensionless parameters from given variables, even if the form of the equation remains unknown. However, the choice of dimensionless parameters is not unique; Buckingham's theorem only provides a way of generating sets of dimensionless parameters and does not indicate the most "physically meaningful".

Two systems for which the [dimensionless] parameter [are equal] are called similar. As with similar triangles, they differ only in scale. They are equivalent for the purposes of the [unknown] equation. The experimentalist who wants to determine the form of the equation can choose the most convenient [system to investigate].

Most importantly, Buckingham's theorem describes the relation between the number of variables and [the number of] fundamental dimensions.

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Dimensions are a human construct, any argument based on dimensional analysis can always be rewritten as a scaling argument in a different system of units where we would not have introduced these dimensions (e.g. natural units). A good example is the derivation of the classical limit of special relativity when we work in natural units and therefore we don't have a dimensional c available that would make the derivation trivial as I've explained here.

As you can read there, this derivation is more complicated because you now need to think about how to approach the desired scaling limit. The result is then more illuminating, you can now derive $E = m c^2$ as a consequence of the scaling argument where c is now a dimensionless scaling parameter, while some other relations don't contain c. So, we don't need to introduce dimensions and a dimensional c a priori and then derive such results on the basis of dimensional analysis. Doing that would beg the question of where these dimensions and the dimensional c come from in the first place.

Now the reverse is not true, not all scaling arguments will be consistent with dimensional analysis performed in some fixed set of units (e.g. SI units), but they will still be correct arguments (they can be re-interpreted as a dimensional analysis argument in another set of units were dimensions are introduced in a way that's incompatible compared to SI units). A simple example is a derivation of the ground state energy of the hydrogen atom. In SI units we can write down the dimensionally correct equation:

$$E = m c^2\tag{1}$$

but this is not what we want, we want to be able to take the $c$ to infinity limit (which we can re-interpret in natural units as taking the scaling limit to corresponding to the classical limit). Now we know that in SI units the fine structure constant is dimensionless, so we're free to multiply the r.h.s of (1) by any power of $\alpha$, within SI units any power is as good as any other power, dimensionally there is no distinction:

$$E = \alpha^n m c^2\tag{2}$$

But I'm free to re-interpret this equation in natural units where I can introduce a dimensionless c which plays the role of a scaling parameter which also appears in $\alpha$. I then demand that in the $c\to\infty$ limit Eq. (2) remains regular. This means that $c$ has to drop out, so $n = 2$.

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  • $\begingroup$ i'm not sure i understand how this answer is relevant to the specific problem i posed. would you mind elaborating on that, maybe by explaining how your analysis relates to either of the scenarios i mentioned (the fluid pressure equation or newton's law of gravity)? $\endgroup$ – Wade Hodson Sep 8 '16 at 21:20

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