# Why is it natural to look for solutions involving dimensionless quantities?

While studying the Heat Equation, I got stuck in a statement in my book. It says:

We have seen that the combination of variables $\displaystyle \frac{x}{\sqrt{Dt}}$ is not only invariant with respect to parabolic dilations but also dimensionless. It is then natural to check if there are solutions of the Heat Equation involving such dimensionless group. (...) it makes sense to look for solutions of the form:

$\displaystyle u^{*}(x,t)=\frac{q}{\sqrt{Dt}}U\left( \frac{x}{\sqrt{Dt}} \right)$

where $U$ is a (dimensionless) function of a single variable."

Why is the bold text so? I don't see why it is intuitive to search for such solutions.

• – rob Aug 20 '14 at 2:05
• For completeness, it would be good if you could specify which book you are using. – Emilio Pisanty Aug 20 '14 at 9:09

Let's say your goal is to describe the shape of some object, such as a box.

You could create a completely arbitrary ruler and measure the three axes of the box, coming out for example with lengths of 11.72, 23.44, and 35.16 of your arbitrary ruler units.

Or you might look at your results more closely and think hmm, something is going on here, since the second two lengths are exact multiples of the first one.

So instead of using arbitrary units, you make the box into its own ruler by choosing pairs of sides and looking at the ratios of their lengths. By the rules of division, such ratios are of course dimensionless, since for example your original arbitrary units would cancel out in the ratio.

But look at the result! Instead of (11.72, 23.44, 35.16), you have stripped out the arbitrary noise created when you defined and arbitrary ruler. As a direct result, your "theory of box" instead becomes a much more interesting sequence such as (1, 2, 3) or (1/3, 2/3, 1), depending on which side you pick as your divisor (or "1") unit of natural length. Such simpler ratios immediately suggest that the box is composed of smaller, more fundamental units.

Physics, and in particular the Standard Model, is of course far more complicated that a simple box. But it too is rich in same-to-same ratios (about two dozen total) that measure different magnitudes of identical quantities. For those pairs, you can again use the ratio trick to strip out any "arbitrary ruler" noise and come up with more fundamental, and hopefully insightful, values for fundamental "size and shape" constants that define the exact "shape" of the Standard Model.

For example, the absolute invariance that special relativity imparts to the speed of light makes c into a brain-dead obvious choice for the natural unit of anything involving velocity. Thus c=1, and any other velocity simply uses that unique velocity as its divisor. The charge of an electron is another good divisor choice, though there you can see that there is some wiggle room in the choices, since you could also argue that the -1/3 charge of a down quark might in some way be more "fundamental."

• nice, but what if one dimension is time and another is space how would they cancel out in a ratio? – Nikos M. Aug 20 '14 at 2:07
• In that somewhat special case you use c, which is itself a ratio of space over time, as the bridge for defining a self-consistent shared unit, giving one nanosecond of time equaling about 0.3 meters of space. – Terry Bollinger Aug 20 '14 at 2:16
• of course, the comment was just a means for clarification in the answer – Nikos M. Aug 20 '14 at 2:17
• It's a good question, since it points out that there is a certain degree of art or even convenience in some of the choices. – Terry Bollinger Aug 20 '14 at 2:44

Dimensionless equations have the advantage that they work for any value of the parameters. They are scale invariant. So the solution in terms of a single dimensionless variable applies to all values of $D$ and $t$.

It also allows the definition of characteristic values for the dynamic variables. In your example, one could say $u_0$ = $\frac{q}{\sqrt{Dt}}$, and $x_0 = \sqrt{Dt}$. These set the scale of both the independent and dependent variable. It allows, for example, one to say when a value is "big" (greater than the characteristic value), or "small" (less than the characteristic value). It's a little strange in this case because these characteristic values change with time. Things are a bit clearer when they are constants. Then you would have, for example, a characteristic length, period.

Now if we define dimensionless variable, say $\upsilon = u(x,t)/u_0$ and $\chi = x/x_0$, we can write the much simpler, and more universal equation $$\upsilon(\chi, t) = U(\chi)$$ and "large" and "small" are set by whether or not $\chi$ and $\upsilon$ are greater than or less than unity. A further feature is that a plot on dimensionless axes will have values on the $x$ and $y$ axis that center around $1$. The plots are cleaner, and also universal.

These kinds of dimensionless solutions are a convenience, but they don't add anything new to the theory. Whether or not it is natural to look for them ... well, that is in the eye of the beholder. Personally, I do express equations in dimensionless form whenever possible.

In the case of the heat equation, we see that as time marches on the characteristic length gets larger, that is, the "hot region" grows, as the square root of the time variable, and the size of that region is roughly $x_0$. Similarly, the characteristic heat variable (perhaps energy density) decreases ... "gets cooler" ... as the square root of the time variable, and its value is roughly $u_0$.

These solutions are preferred because they directly embody the scale invariance of the equation. In general, when a physical problem has some sort of symmetry - like the parabolic dilation invariance of the heat equation - then this establishes a corresponding action of the symmetry group on the solutions. The canonical forms based on dimensionless parameters are often eigenfunctions of this symmetry.

With the heat equation, for example, $$\frac{\partial u}{\partial t}=D\frac{\partial^2 u}{\partial x^2},$$ the nontrivial symmetry is a combined dilation between space and time: on a problem that's twice the size, the solution will be the same except the heat diffusion will take four times as long. To be more precise, if you change coordinates to $x'=a x$ and $t'=a^2 t$, for some scale parameter $0<a<\infty$, then the equation is exactly the same: $$\frac{\partial u}{\partial t'}=D\frac{\partial^2 u}{\partial x'^2}.$$ This means that for any given solution $u_1(x,t)$ of the heat equation, you can always construct an infinite family of solutions, $$u_a(x,t)=a\, u_1(a x,a^2t).$$ (I have added a scale factor in $u$ as well, to keep the conserved total heat $q=\int_{-\infty}^\infty u_a(x,t)\,\text dx$ constant, though this is not strictly that necessary.)

Now, these solutions are probably different to the original one unless that one is completely flat, but as physicists we're not very interested in that and we want to remove this superfluous degree of freedom. (I should note that mathematicians are even more trigger-happy on that, too.) The problem, though, is that since they're all equivalent it's hard to pick one representative of the class. Moreover, in general, all the $u_a$ will be linearly independent, so they are in some sense different.

This is where canonical forms of the type $$u^*(x,t)=\frac{q}{\sqrt{Dt}}U(x/\sqrt{Dt})$$ come in. It should be clear that these solutions are eigenfunctions of the dilation symmetry operator $L_a$, which means that all the corresponding $u_a$ are linearly dependent, which greatly simplifies the corresponding equivalence class.

On a more formal setting, we have a bunch of symmetry operators which commute with each other and with the differential operator $L=D\tfrac{\partial^2 }{\partial x^2}$, which means that we can expand any solution of $L=0$ as a linear combination of the eigenfunctions of the $L_a$ and - crucially - the time evolution under $\tfrac\partial{\partial t}=L$ will preserve the coefficients of this expansion.

On a more pragmatic footing, it is nice that we're guaranteed the existence of such solutions because of the symmetry of the equation, but looking for them is useful because it reduces the dimensionality of the problem: we now have to solve a one-dimensional ODE rather than a two-dimensional PDE, and this is a vastly easier problem. The price we need to pay is that there is no guarantee that the special solutions we find will match the initial conditions we may need to impose, but this is generally something we're OK with, because we can use linear combinations of these special solutions to solve for arbitrary initial conditions.

Because it is easier for dimensionless quantities to be combined in arbitrary polynomial terms (or other terms e.g exponential) with no loss (or extra) factors. Think like "characteristic times" used in exponential factors.

Especially quantities appearing in solutions of the form $e^{a} = \sum_0^\infty \frac{a^n}{n!}$, one can see why a dimensionless quantity $a$ is necessary here. Else there would be problem matching the proper dimensions (in an infinite amount of powers).

In a sense they are invariant quantities suitable for expressing solutions in terms of them.

In an analogy with tensor analysis, if we assume for a minute that each quantity that has a dimension (e.g mass, time, etc.) can be a vector or tensor, then a dimensionless quantity can be scalar invariant.

The key concept is that the units of measurements you chose to use make no difference to the physical behaviour. I.e you expect the solution behaviour to be independent of the units of measurement used. This means you should get the same solution if we chose to measure $x$ is in meters, and $t$ is in seconds and $D$ is in $m^2/s$ as if we chose to measure $x$ is in furlongs and $t$ is in fortnights and D is in $furlongs^2/fortnight$.

By looking for a solution in terms of the dimensionless parameters we ensure this is occurs. In fact this is the only way it can be ensured.

Intuitively a lot of equations we use to describe physical phenomena such as $exp(x)$ take in only dimensionless variables. Moreover in physics we're always taking the derivative or the log of things, and it works out to be much simpler if you're not always having to deal with a dimensioned constant factor that drops out every time you take the derivative.