# The dimensional analysis tool

#### Prerequisites

Dimensionalities and units provide a powerful tool for working with equations in science. They express a fundamental difference between equations and symbol use in a typical math class and the way they are used in science. As our tool icon, we've chosen a basic tool for measuring distances: a tape measure.

Symbols in science most often represent measurements, not numbers. And because measurements can have different values depending on what unit we choose, we have to be careful in writing our equations so that the dimensionalities of each term (things added or subtracted) in the equation agree, and that the dimensionalities of both sides of the equation are the same.

## Catching errors

So one way dimensionalities can be a useful tool is in showing you when you might have remembered an equation incorrectly or made a typographical error in writing it down— as in the following example.

Suppose you are working a problem in which you need to know the surface area of a sphere. But you remember a bunch of different equations from your high school geometry class. Which is which?

Which equation correctly represents the surface area of a sphere?

1. $2\pi R$

2. $4 \pi R^2$

3. $\frac{4}{3}\pi R^3$

4. $\pi R^2$

You might have memorized this, but, as we know, memory is not always reliable. Can you figure this out without remembering the answer? A good first step is to check the dimensions. The radius of a sphere is a length ([$R$] = L) and an area is the product of two lengths ([area] = L^{2}). Therefore, the answer must have two factors of $R$. We can therefore rule out answers 1 and 3 right away without using any memory of the equations at all. We now only have to choose between 2 and 4. If we remember that 4 is the area of a circle and we realize that this is the circle that cuts through the center of the sphere, it's clear that the area of the surface of the sphere is bigger than that — and since the surfaces of both the top and bottom hemispheres are also bigger than that circle, the answer has to be bigger than $2πR^2$. This tells us the answer must be #2. (Correct)

Here, we combined dimensional analysis with other things we know to generate the correct result. That will often be the case.

## Generating new equations

We'll also find throughout the course that dimensional analysis lets us generate new equations as well as check old ones! This is an extremely powerful tool that we will be using extensively throughout the class.

Here's an example.

The speed, $v$, with which a displacement wave travels on a taut guitar string determines what note it will be play when plucked. It our guitar is out of tune and we want to change its sound, we can change the tensions, $T$, of the string, or replace it by a heavier or lighter string. We therefore can expect that $v$ depends on the mass of the string as well. Can we construct a speed from the tension and mass of the string?

This example gives us a good case where it looks like we are using the same symbol to mean different things and shows us how to be careful! We have that $T$ is then tension in the string, but if we are analyzing the dimensionality of a velocity, we know we will have a time for which we use the notation T! Watch how to be careful!

The dimensionality of a velocity is [$v$] = L/T. Our tension has the dimensionality of a force

[$T$} = [$F$] = [$ma$] = ML/T^{2}

Using a mass, M, and an ML/T^{2}, there's no way we can get a velocity. Dividing $T$ by $m$ gives an (ML/T^{2})/M = L/T^{2}. So we can't do it. We have too many powers of T. But if we think about a guitar, we realize that we don't have a time parameter lying around, but we do have a length — the length of the guitar string. So let's try adding a length to our parameters. Now we have

[$T$] = ML/T^{2} [$m$] = M and [$L$] = L

We want to construct something with dimensions L/T. We can start with $T$. It has a factor of the mass and we need to get rid of it, so let's try $T/m$. That has dimensionality L/T^{2}. We can't get rid of the T, but what if we add another L? Let's try $TL/m$. This has dimensionality L^{2}/T^{2}. But that's just the square of our velocity's dimensions so we can choose

$$v= \sqrt{\frac{TL}{m}}$$

and our result will have the correct dimensions!

This result makes sense since it tells us the speed doesn't depend on the mass of the string along but on $m/L$, it's mass per unit length - its density. That seems reasonable.

We did a lot of thinking and guessing in that solution. And that's a good way to use dimensional analysis as a practical tool. See the associated problem for a discussion of a more formal way that will almost always work.

As you gain experience with the dimensional analysis tool, you'll find it increasingly valuable in many ways, helping you think about what you can do with an equation and what it means.

Joe Redish 7/3/17 and 5/20/19

Last Modified: May 22, 2019