Problem Solving with Dimensional Analysis

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Problem Solving with Dimensional Analysis

Dimensional analysis is the technique of analyzing relationships through their base quantities. I demonstrate the power of this approach by approximating a Gaussian integral without calculus.

Published

11 February 2023

A dimension is a physical quantity<br>that can be measured, while a unit is a specific amount of a physical<br>quantity. For example, time is a dimension, which can be<br>measured in temporal units such as hours or seconds.

Dimensional analysis is the technique of analyzing and simplifying equations<br>by expressing terms using their dimensions rather than their numerical<br>values. Because equations should be dimensionally consistent, meaning that the<br>dimensions on both sides of an equation are equivalent, dimensional analysis is<br>useful for making inferences about functional forms or the dimensions of<br>variables. Note that this is distinct from analyzing units, although that<br>is also a useful problem-solving technique.

In my experience, dimensional analysis is particularly useful when coming at a problem where even the structure or basic terms in the problem are<br>unknown. For example, I often find myself using dimensional analysis when<br>reading unfamiliar code. It does not clarify everything, but it often<br>provides an “initial hook” by allowing me to quickly understand<br>variables’ dimensions or units.

The goal of this post is to demonstrate the power of dimensional analysis by<br>working through the example that first convinced me while reading (Mahajan, 2010): simplifying a Gaussian integral.

Gaussian integral

Consider the integral

I=∫−∞∞e−ax2dx.(1)<br>I = \int_{-\infty}^{\infty} e^{-a x^2} dx. \tag{1}<br>I=∫−∞∞​e−ax2dx.(1)

Imagine that we didn’t know calculus. We just think of integrals as sums and<br>dxdxdx as “a little bit of xxx”. And imagine that we had never seen this equation<br>before. Let’s use dimensional analysis to make sense of it.

Perhaps the simplest thing we can say is that both sides of the equation must<br>have the same dimensions. It must be dimensionally valid. Let’s use the notation<br>[x][x][x] to denote the dimensions of xxx rather than its numerical value. We’ll use<br>unity to denote a dimensionless quantity, i.e. [x]=1[x] = 1[x]=1 means that xxx is dimensionless. Using<br>this notation, we can express dimensional validity as

[I]=[∫−∞∞e−ax2dx].(2)<br>\left[ I \right] = \left[ \int_{-\infty}^{\infty} e^{-a x^2} dx \right]. \tag{2}<br>[I]=[∫−∞∞​e−ax2dx].(2)

But beyond this, what can we do? The problem is that this problem has no inherent dimensions!<br>If this were a physics problem, perhaps terms would be concretized with<br>units. But here, this is just an abstract mathematical object. However, we can<br>still make progress by assigning an arbitrary dimension to one of<br>the variables and then seeing what that implies. Let’s do this by assigning an<br>arbitrary dimension to xxx, say L\mathbb{L}L (imagine this is<br>“length”). What does this imply about the dimension of aaa?

We can see that xxx and aaa are related via the exponent. Now note that<br>exponents must be dimensionless quantities. That is because xxx raised to nnn is just xxx multiplied<br>by itself that many times:

xn=x⋅x⋅x⋅⋯⋅x⏞n times.(3)<br>x^n = \overbrace{x \cdot x \cdot x \cdot \dots \cdot x}^{\text{$n$ times}}. \tag{3}<br>xn=x⋅x⋅x⋅⋯⋅xn times.(3)

So nnn is just a number here, not a dimensioned quantity. In general, the full term<br>xnx^nxn may have units. If xxx has dimensions feet, and n=2n=2n=2, then x2x^2x2 has<br>dimensions ft2\text{ft}^2ft2 (or area). However, Euler’s number eee is actually<br>dimensionless. It’s a mathematical constant, and much like π\piπ, it doesn’t<br>make sense to talk about the dimension of eee. Thus,<br>we know that −ax2-a x^2−ax2 must have no units, because it’s an exponent, and we know<br>that e−ax2e^{-a x^2}e−ax2 must be dimensionless because it is merely Euler’s number<br>raised to a dimensionless exponent.

For the exponent to be<br>unitless, this implies that aaa must have dimension 1/L21 / \mathbb{L}^21/L2, since

[a][x2]=1L2L2=1.(4)<br>\left[a\right]\left[x^2 \right]= \frac{1}{\mathbb{L}^2} \mathbb{L}^2 = 1. \tag{4}<br>[a][x2]=L21​L2=1.(4)

Now that we have the dimensions of aaa and xxx, let’s analyze the<br>integral. First, observe that the integral is loosely a summation, and it only<br>makes sense to reason about adding two quantities together if they have the same<br>dimensions, e.g. a length plus a length is another length. So we can think of<br>summation and integration as operations that do not change their arguments’<br>dimensions. And dxdxdx is “a little bit of<br>xxx” and thus has the same units as xxx. Here, it has dimension L\mathbb{L}L.

So we can summarize<br>our analysis so far as

[I]=[∫−∞∞e−ax2dx]=[e−ax2]×[dx]=1×L=L.(5)<br>\left[ I \right] = \left[ \int_{-\infty}^{\infty} e^{-ax^2} dx \right] = \left[ e^{-ax^2} \right] \times \left[ dx<br>\right] = 1 \times \mathbb{L} = \mathbb{L}. \tag{5}<br>[I]=[∫−∞∞​e−ax2dx]=[e−ax2]×[dx]=1×L=L.(5)

Now observe that III is a...