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Usually in all discussions and arguments of scaling or solving problems using dimensional analysis, the dimensionless constant is indeterminate but it is usually assumed that it is of order 1.

  1. What does "of order 1" mean? 0.1-10?
  2. Is there any way, qualitative or quantitative, to see why the dimensionless constant is of order 1?
  3. Are there exceptions to that? I mean cases where the dimensionless constant is very far from 1? Could you give some examples? Can such exceptions be figured out from dimensional analysis alone?
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Here are two examples of where dimensional estimates fail:

Divergent expressions

I have a laser pointer pointed at a wall directly facing me, 1 meter away. I turn the laser pointer by 90 degrees over the course of 1 second. What is the average speed of the laser spot during the turn? The dimensional analysis argument is 1meter/1second, which would be roughly right if I didn't have a singular function. Similar examples are resonant phenomena, where you have a lot more response than you would predict from dimensional analysis.

Exponential growth:

Suppose I have 20 particles in a box of mass M, in volume V, with total energy E. When will they all return to within 10% of their starting positions/velocities? If you ignore the dimensionless 19, then you will get the recurrence time for one particle, which is shorter by an enormous amount.

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One notable class of exceptions are what are called "hierarchy problems" in particle physics. For example, if you identify the Planck mass as a fundamental energy scale, you end up with huge dimensionless numbers which don't have an obvious explanation (i.e. ratio of Planck to electroweak scale, etc.). Explaining these large (or small depending on how you look at it) dimensionless numbers is a major focus of particle theory.

Another fun counterexample: see 't Hooft's original computation of the instanton effective interaction in 1976. After a lengthy calculation, the answer turns out to have a dimensionless factor $2^{10}\pi^{6}$ in front, resulting from a slew of Gaussian normalizations and such.

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Well, of course you have to pick the quantities in your dimensional analysis right.

Example: Use dimensional analysis to estimate the potential energy of a star, hold together only by gravitation.

Solution: Newtons gravitational constant $G$ better show up somewhere. This requires us to include something with units $kg^2 / m$. We can get this by inserting $M^2$, with $M$ the mass of the star, and by inserting $1/R$, with $R$ the radius of the star. Thus, the potential energy is estimated to be $$E_\text{pot} \approx -G \frac{M^2}{R}$$ which is off by a factor $3/5$.

I could, of course, have inserted the mass of a hydrogen atom and then everything would be off by many orders of magnitude...

There is no general guarantee that the constant is of order one. But:

Why it often comes out as "order of 1" is, I believe, the following:

In many cases, the true solution involves an integral over a variable $x$ where the function to be integrated is of the form $f(x) \sim x^n$. In physics, $n$ is small in most cases, so the integration gives a factor $\frac{1}{n}$. Other corrective factors that get ignored in dimensional analysis are $\pi$, $2\pi$ or $4\pi$, i.e., some small integer multiples of $\pi$.

A counterexample... the fine-structure constant $\alpha = \frac{1}{137}$, maybe? But this constant itself can be obtained through dimensional analysis, so I am not sure.

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  • $\begingroup$ So it seems that as long as you are identifying the right degrees of freedom on a specific scale (which is the mass of earth but not of individual atoms in your example) you will get roughly the dimensionless constant of order one. I still do not know why this is so tho, because still, in the example of integration you mentioned I can argue that the over all dimensionless constant before performing the integration is of order 1, why is that? Another thing, had we used mass of an average atom instead of M, what else should have been modified to get roughly the same order of magnitude PE? $\endgroup$
    – Revo
    Commented Aug 12, 2011 at 16:19
  • $\begingroup$ Well, you also need the right physical model. I recall that the specific heat capacity of a metal is off by a factor of $100$ if you don't take into account that only electrons near the Fermi energy can contribute to it. $\endgroup$
    – Lagerbaer
    Commented Aug 12, 2011 at 17:58
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    $\begingroup$ And if you used the mass of an average atom, then it would make sense to also include the number of atoms in the star. Which brings you back to $M$ :) $\endgroup$
    – Lagerbaer
    Commented Aug 12, 2011 at 17:58
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Notable exception: Reynolds number.

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To add to the other answers, it should be noted that when formulating laws and relationships it is custom to use only unitary values of the relevant physical quantities.

$F=ma$, $V=IR$ etc could just as well have had a constant inserted, and if you shift from SI units you will need to include some. Therefore lack of numerical constants at the beginning leads to a lack of constants at the end of the analysis...

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  • $\begingroup$ No that is different. $\endgroup$
    – Revo
    Commented Aug 16, 2011 at 15:16
  • $\begingroup$ How so @Revo ? Could you expand? $\endgroup$
    – Nic
    Commented Aug 16, 2011 at 19:15
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The answer follows from considering why we use dimensional quantities in the first place. So, given that the laws of physics can be formulated in a dimensionless way, how come we've ended up with using dimensional quantities. The answer is that physical quantities that we can readily measure at the macroscopic scale have certain scaling properties, and these scaling properties are not going to be the same for quantities. This means that if a simple relation exists between two quantities at the fundamental level, the corresponding relation between the rescaled quantities expressed in the units that macroscopic observers would express their experimentally accessible results in, may involve extremely small or large constants, and would therefore not be discovered early on by the first few generations of physicists. The missing relations make it necessary to assign incompatible dimensions to the physical quantities

When the missing relations are later found, they appear with dimensional constants that when expressed in the conventional units, will be extremely large or small. Compared to these typically extremely large numbers, the values of dimensionless constants will typically be much smaller. If extremely large or small dimensionless constants still appear, you can often frame that in terms of some other scaling effect. So, just like the macroscopic World we live in is approximately an infinite scaling limit that you could just as well describe using dimensionless scaling parameters (just start by using natural units $c=\hbar = G = 1$ and then replace these constants by dimensionless scaling parameters), some other scaling limit will involve some other dimensionless constant becoming very large or small.

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One more exception: weak localization.

Second class of exceptions (Lagerbaer named first where interaction constant is involved) originates from two characteristic lengths/times/... in the system. [Function of] their ratio may appear in the answer.

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A notable exception would be Debye's model for low temperatures. The molar heat capacity is then approximated as

$ C = 234 \ k \ \left( \frac{T}{T_D} \right)^3 $

with k the Boltzmann constant and $T_D$ the Debye temperature.

I think the biggest reason why constant are always so small is that if they're not small, they're absorbed in other quantities.

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Another exception: the factors can depend on the number of dimensions. Esp in models with extra dimensions you gather easily some (2 pi)^d that can be a factor 1000 or larger.

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  • $\begingroup$ Most probably this is the correct reason. Because now I am thinking that the technique of dimensional analysis would not work in worlds with different dimensions, since the volume and surface area of a sphere in N dimensions depend on N, so the dimensionless constant would be "naturally" very different from 1. $\endgroup$
    – Revo
    Commented Aug 16, 2011 at 15:09

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