Scale invariance and the power law

Posted by David Zaslavsky on February 7, 2012 4:15 AM — Edited May 8, 2012 12:32 PM

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With all the fuss about SOPA and PIPA, plus having actual work to do, I haven’t been able to write anything in a while. So I figured it’s time for some good old fashioned physics. Today I would like to introduce the concept of a scale invariant (or dilatation invariant) function. This special class of functions is defined by the property that when you scale (or dilate) the argument of the function by some factor, it’s equivalent to scaling the value of the function by some related factor.

$$f(\lambda x) = C(\lambda) f(x)$$

$C(\lambda)$ is something that depends on $\lambda$, but not on $x$.

To understand why physicists find scale invariant functions so fascinating, we have to go way back to the definition of an analytic function, the power series expansion. Pretty much every mathematical function used in physics can be expressed as some power series, like this:

$$f(x) = c_0 + c_1 x + c_2 x^2 + c_3 x^3 + \cdots$$

Even if the function isn’t normally written like this, it’s enough that it can be. (The sine and cosine, for example.)

Now, what happens when you plug some value with units into this expansion? You get an ever-increasing series of different powers of the unit, for example

$$f(\SI{3}{m}) = c_0 + 3 c_1 \si{m} + 9 c_2 \si{m^2} + 27 c_3 \si{m^3} + \cdots$$

But you can’t add all these terms with different units together. It doesn’t make sense to add a length to an area or a volume! This means that, in order to get something sensible out of the function, you can only plug in pure numbers. If you start with a quantity with units, you have to divide it by some reference value first. So $x$ can’t be a length, but it could be a ratio of lengths, like $\frac{l}{L}$.

$$f\biggl(\frac{l}{L}\biggr) = c_0 + c_1\frac{l}{L} + c_2\frac{l^2}{L^2} + c_3\frac{l^3}{L^3} + \cdots$$

You might have noticed, though, that I kind of pulled that length $L$ out of thin air. What’s so special about $L$, anyway? If I choose $L = \SI{1}{m}$, what’s to say I couldn’t have chosen $L = \SI{2}{m}$, or $L = \SI{0.004}{m}$?

Does it really make a difference, though? Here’s how we can find out. Suppose I started with one particular choice of $L$, and I want to switch to another one, $\lambda L$. Look at how that changes the dimensionless ratio $x$:

$$x = \frac{l}{L} \to \frac{l}{\lambda L} = \lambda^{-1}x$$

This transformation is nothing other than a dilatation — a scale transformation. Dilatations are important because they correspond to unit changes.

Now, what happens to the power series for $f(x)$ under this scale transformation?

$$f\biggl(\frac{l}{\lambda L}\biggr) = c_0 + \lambda^{-1}c_1\frac{l}{L} + \lambda^{-2}c_2\frac{l^2}{L^2} + \lambda^{-3}c_3\frac{l^3}{L^3} + \cdots$$

Every term of the series is changed separately, by a different amount. That makes it impossible to get away with choosing a different unit, at least not without changing the form of the function. For this reason, we say that an arbitrary power series is not scale invariant. In other words, it does matter what reference value I choose, because if I change the reference value, I’m going to change the form of the function.

Since essentially all functions in physics can be expressed as power series, what we’ve discovered is that most functions used in physics are not scale invariant. In other words, if you want to plug in a quantity with units, you have to choose a reference value, and furthermore, you have to choose the right reference value to make the math work.

Fortunately, the reference value you need to use will usually pop up in your equations somewhere. For example, in my analysis of the terminal velocity of a person in atmospheric free fall, the terminal speed $v_T$ was a convenient scale for velocity and the “vacuum drop height” $h_0$ was a convenient scale for length. So I wrote the equation in terms of the dimensionless ratios $\frac{v_y}{v_T}$ and $\frac{y}{h_0}$, and it wound up looking a lot simpler. Or, suppose you’re working on a crystal lattice. The lattice spacing gives you a convenient length scale which you can use as your reference value. And if your equations don’t suggest a particular reference value, we have standardized unit systems that you can take your reference values from: meters for length, kilograms for mass, Newtons for force.

OK, so far, so good. Where do scale invariant functions fit into all this? Well, remember that the defining property of a scale invariant function is that when you rescale the parameter, you simply rescale the function. And as we saw, a change of units is nothing but a dilatation, a rescaling. So with a scale invariant function, when you switch from one reference value ($L$) to another one ($\lambda L$), you wind up simply multiplying the function by a constant,

$$f\biggl(\frac{l}{\lambda L}\biggr) = C\bigl(\lambda^{-1}\bigr)f\biggl(\frac{l}{L}\biggr)$$

That sort of change can usually be ignored, because you’re probably setting it equal to some other ratio of a quantity with units to its own reference value. The point is that with a scale invariant function, it doesn’t matter which reference value you choose. This is what makes these functions so interesting, that they simply refuse to pin you down to a specific reference value.

Now that we know why scale invariant functions are special, let’s find some examples. The canonical one is the power law, a function of the form

$$f(x) = ax^k$$

for constants $a$ and $k$. Strictly speaking, this is an example of a power series, but it avoids the problem with adding different units because there is only one term in the series. The scale invariance of this function is easy to verify:

$$f(\lambda x) = a(\lambda x)^k = \lambda^k f(x)$$

What about other scale invariant functions? Are there any others? Here’s a cute way to find out. I’ll start with the scale invariance condition,

$$f(\lambda x) = C(\lambda) f(x)$$

and take the logarithm of both sides.

$$\ln f(\lambda x) = \ln C(\lambda) + \ln f(x)$$

I’ll also introduce a new function $F(x)$ defined as $F(\ln x) = f(x)$. This gives

$$\ln F(\ln\lambda + \ln x) = \ln C(\lambda) + \ln F(\ln x)$$

which can be rearranged to

$$\frac{\ln F(\ln\lambda + \ln x) - \ln F(\ln x)}{\ln\lambda} = \frac{\ln C(\lambda)}{\ln\lambda}$$

In the limit as $\ln\lambda\to 0$, the left side becomes a derivative,

$$\ud{\ln F(\ln x)}{\ln x} = \lim_{\ln\lambda\to 0}\frac{\ln C(\lambda)}{\ln\lambda}$$

If you look at the right side, on the other hand, it’s independent of $x$, so it’s not a function. That leaves basically two possibilities: either the limit converges to some constant, or it doesn’t. In the latter case, $F$ is not differentiable, which means that as a physicist, I don’t particularly care about it. That leaves the first option as the only way to generate a scale invariant function. In other words, all scale invariant functions have to satisfy this differential equation:

$$\ud{\ln F(\ln x)}{\ln x} = k$$

where $k$ is some constant. Integrating the equation gives

$$\ln F(\ln x) = k\ln x + a$$

$$F(\ln x) = f(x) = ax^k$$

Presto, all scale invariant functions are power functions. (Mathematicians, feel free to leave exotic counterexamples in the comments)

Power functions of this form pop up all over physics, especially in the study of phase transitions and renormalization, but they’re also very common in probability theory and statistics. In every case, they’re usually associated with some sort of surprising behavior. The Wikipedia article is a great place to start reading about some of the many interesting applications of these scale-invariant functions.