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After some reading on dimensional analysis, it seems to me that only rational exponents are considered. To be more precise, it seems that dimensional values form a vector space over the rationals. My question is, why do we restrict ourselves to rational exponents? While it's true that I have never seen a physical equation with irrational exponents, is there any real guarantee that it will never occur? What are the benefits of such a restriction?

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5 Answers 5

up vote 5 down vote accepted

There are many natural examples of quantities with fractional non-rational dimensions. These are ubiquitous in macroscopic physics, because the scaling laws of nature are emergent properties that don't usually care about differentiability. It is only in cases where you demand smoothness that you restrict to integer dimensions.

The phenomenon discussed below are discussed at much greater length in Mandelbrot's "The Fractal Geometry of Nature", and in his publications.

Levy Diffusion

suppose a particle is undergoing Levy diffusion, a process by which it hops from place to place with a jump whose distribution has a tail, meaning that the probability density of a hop of length l is:

$$\rho(l) \propto {1\over l^{1+\beta}}$$

where $0<\beta<2$. Then, if you take the limit of many steps, the analog of the central limit theorem guarantees that the probability of finding the particle at position x obeys a Levy diffusion equation, which is easiest to write in a spatial Fourier transform:

$${\partial \rho(k) \over \partial t} = - D k^\beta \rho(k)$$

The constant D is the Levy analog of the diffusion constant, and it has dimensions {L^\beta\over T}. The interpretation of D is that, when multiplied by t and extracting the $\beta$ root, it gives the typical displacement scale after time t. It is a natural quantity with fractional dimension.

Processes which undergo Levy diffusion are relatively common. For example, advection of a dust particle by a turbulent flow (see this experimental article:http://citeseerx.ist.psu.edu/viewdoc/summary?doi=

Hausdorff measures

When you try to define the "length" of a fractal shape, you run into the problem that the length is infinite. The way around this is to define a notion of length per length-scale, and to ask exactly how the length blows up.

In mathematics, the most general way to do this, which works for essentially any set, is called the Hausdorff measure. In practice, you can define a simpler version in terms of box-counting scaling which is sufficient for ordinary fractals.

Consider the coastline of England, and ask: given a boxes of length \epsilon, how many do I need to cover it up? The minimum number N diverges as a power of \epsilon:

$$ N(\epsilon) \approx {A\over \epsilon^\alpha}$$

The coefficient of the divergence, A, is the Hausdorff measure, and A has units of $L^{\alpha}$, and $\alpha$ can be anything between 1 and 2 for a coastline, and between 0 and 3 for a physical shape, like a cloud boundary, or a diffusion-limited-aggregation cluster.

Three dimensional quantum fields

Consider a scalar field in 2+1 dimensions, with a quartic self-interaction. At short distances, the correlation function of the field blows up. The scale dimension of the field is defined by this power:

$$\langle \phi(x)\phi(y)\rangle = {1\over |x-y|^{1+2\nu}}$$

This might seem surprising if you are used to fields having canonical dimensions. The scalar in 2d has dimension 1/2 right? Wrong. The canonical dimension 1/2 only describe a free scalar theory or a cut-off theory with no short-distance self-interactions. After a successful continuum-limit by renormalization, the fields get new dimensions, which do not involve the dimensional quantity $\epsilon$, which has gone to exactly zero. Then the fields must be defined so that the left side and the right side of the equation above are dimensionally consistent.

The exponent $\beta$ is very close to 0, the correction is quadratic in "d-4" (I hate this traditional name for the expansion parameter of the Wilson-Fisher point), but it is nonzero, and if you had to take bets, it will certainly be irrational.

In four dimensions, ordinary local quantum fields only get logarithmic corrections to their classical scaling values, so the scaling of the field at short distances is either canonical (like in QCD), or nonsensical (like in QED). But in 3 dimensions, you get good anomalous dimensions, and fields have a true scale-invariant continuum limit.

Other fields

If you are willing to leave physics, the quantities of finance often have Levy behavior. If you consider the dollar a unit of nature, you can find fractional exponents there too.

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I've chosen to accept this question instead as I feel the answer is more concrete than the previously accepted one. I appreciate the examples in particular. Still, I wish that the other aspect of my question: "Why is it beneficial to restrict dimensional analysis to rational exponents?" was discussed more. –  EuYu Sep 5 '11 at 9:06
It is not beneficial. –  Ron Maimon Sep 5 '11 at 13:47
The phrase "fractional non-rational dimensions" is bothering me. –  David Zhang Feb 9 at 4:11

A counterexample: In second order phase transitions of 3-dimensional systems, the critical exponents can be irrational numbers. If $|T-T_c|$ is taken to be of dimension 1, then the correlation length, heat capacity, susceptibility etc. can have irrational dimensions.

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I'm not aware of any mathematical proof that they are irrational, but I'm also not aware of any reason that they should be rational. –  felix Aug 7 '11 at 23:28
Interesting! E.g., this paper prd.aps.org/abstract/PRD/v60/i8/e085001 gives a critical exponent of -0.01294±0.00060 (theory), α=-0.01285±0.00038 (exp) for a certain system. But clearly an experiment can never determine whether a number is rational or irrational. I wonder if the one in the model can be proved to be rational or irrational -- too bad the article is paywalled. –  Ben Crowell Aug 8 '11 at 2:31
Hi Ben, thanks for the link. So they've done it to 7 loops! Is it theoretically (not computationally) possible to approach the true answer by going to arbitrarily many loops? Or will the answer eventually diverge because QFT involves asymptotic rather than convergent expansion? –  felix Aug 8 '11 at 7:44
@felix: I believe that such critical quantities are a priori dimensionless: critical physical quantities are always of the form $t := (T - T_c)/T_c$ etc. –  Gerben Aug 8 '11 at 9:47

There's no reason I can't get irrational exponents of a unit by algebraic manipulation. For example, starting from $F=ma$, I could derive $F^\pi=m^\pi a^\pi$.

One concrete reason why I would not expect to see such things arise in any well-motivated physical context is that a system of units typically includes units for time and distance, and each of these can naturally occur as a negative number. Since the laws of physics are supposed to make well-defined predictions, we expect the equations representing them to be single-valued. But if you take non-integer powers of quantities that may be negative, you run into problems with branch cuts. I would therefore expect all the laws of physics to involve only integral exponents (as Misha has observed), and there is therefore no obvious reason why I would obtain irrational exponents through algebraic manipulation, except if I was just trying to be perverse.

Many examples involving irrational exponents can be massaged into a form where they don't involve irrational exponents. For instance, if I was presented with $F^\pi=m^\pi a^\pi$, I could simplify it to $F=ma$. To get an equation involving only multiplication and exponentiation that can't be reduced to its simplest form, I need a system of units with more than one base unit, and an equation with different irrational exponents. E.g., I could define "funkocity," a measure of the speed of motion, as $u=x^\pi/t^e$. The exponents in this equation can't be rationalized.

This leads to a stronger justification for the fact that we don't encounter irrational exponents. The SI has as many as three base units (or 4 if you count the Coulomb) only because of historical circumstances. If I'm doing calculations in special relativity, I'm going to use a system of units with $c=1$, and in GR I'd also set $G=1$, which gives geometrized units. In geometrized units, there is only one base unit, and therefore every expression involving only multiplication and exponentiation can reworked into a form where the exponents are rational.

So I would say that the fundamental reason we don't need irrational exponents in physics is that we have two fundamental theories of physics right now, GR and the standard model. In GR, the natural system of units is one with $c=1$ and $G=1$, so there is only one base unit. In the standard model, the natural system has $c=1$ and $\hbar=1$, so again we only have one base unit.

What this argument fails to rule out is examples of equations with exponents that can't be rationalized because they have other operations such as addition and subtraction. For example, if we had an equation $a^\pi=b^\pi+c^\pi$, with $a$, $b$, and $c$ all having units of the base unit, then there is no way to rationalize the exponents. The only reason I can offer for the fact that such examples don't come up is to go back to the argument in the second paragraph.

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I am not going to downvote, but only because this is such engrained stuff for people that there are no natural fractional dimensions. This is completely false. The coefficient of Levy diffusion has fractional exponent. The coastline constant: the quantity that determines the "length" of a coastline, is also an irrational exponent, and there are many other examples. Benoit Mandelbrot discovered these fractional nonrational scaling laws. –  Ron Maimon Aug 28 '11 at 7:05

Ok, first answer was incorrect.

However, the idea may be reformulated in a way it is correct. The way you get powers in a dimensional analysis is restricted with physics laws. Usually, in laws only integer powers of physical values appear. Thus you get rationals out of integers. If you find a law with more sophisticated dependency (actually, I can hardly imagine such a law. Probably, in spaces with non-integer dimensionality) you might get any powers in principle.

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Roy Sympson PSE-here pointed to The Buckingham Π Theorem.
You should consider measurable "fundamental units" like "time, distance, velocity, mass, momentum, energy, and weight", and clearly $c,G,\hbar$ do not belong to these fundamental base units. The dimension equation of $\left[G\right]={M^{-1}}{L^{3}}{T^{-2}}$, $\left[c\right] = LT^{-1},\left[h\right] = ML^{2}T^{-1}$.
In math you have the liberty to construct the axioms and consistently derive from those. In Physics we are constrained by how the universe manifests itself or in ,other words, "we model upon data".
To find an irrational number in the exponent is not the point, e.g. $e^{t/k}$, where $t$ is valid for any $t$ including irrationals, but $k$ is also a time constant and the exponent is dimensionless, and this is the key point.

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to the downvoter: can you explain? please. –  Helder Velez Aug 8 '11 at 15:19
-1--- the answer to the question is that there are natural quantities with irrational dimensions. This was discovered by Benoit Mandelbrot, and exploited by him to find many novel relations. –  Ron Maimon Aug 28 '11 at 7:08

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