1.5: Dimensional Analysis

You can sometimes figure out something about a physical quantity just by considering its dimensionality. If you know what sorts of things might affect that quantity, and you have good reason to believe that it is just powers of those things multiplied together to give you that quantity, you may be able to figure out (up to a dimensionless constant) the equation that relates that quantity to the things that might affect it just by figuring out what makes the units work.

Consider the example of a simple pendulum: a small mass (the “bob”) hangs at the end of a string. The other end of the string is fixed. The bob may oscillate back and forth. We want to figure out what is the equation for the period \(\ P\) (i.e. the length of time it takes to go through one oscillation). If we think about things that could affect that, there are three obvious possibilities. The first is the mass \(\ m\) of the bob at the end of the pendulum, the second is the length \(\ l\) of the string connecting the bob to the point from which the pendulum hangs, and the third is \(\ g\), the acceleration due to gravity. For each of these quantities, we’ll write down the dimensionality in terms of mass \(\ (M)\), length \(\ (L)\), and time \(\ (T)\). (Note that M here means mass, not meters!)

\(\ \begin{aligned}
\text{[}m\text{]} &=M \\
\text{[}l\text{]} &=L \\
\text{[}P\text{]} &=T \\
\text{[}g\text{]} &=L / T^{2}
\end{aligned}\)

The “bracket” notation, here, means “dimensionality of”. So, the dimensionality of the period is time; the dimensionality of acceleration is length divided by time squared.

If the period is a product of various powers of the different quantities, then we can write:

\(\ [P]=[m]^{a}[l]^{b}[g]^{c}\)

The period itself wouldn’t be equal to this, as there may well be (and, in fact, there is) a dimensionless quantity multiplying everything else. However, even if we don’t get the right formula, we can figure out how the period depends on these other things.

Now, put in the dimensions for each quantity:

\(\ \begin{gathered}
T=M^{a} L^{b}\left(\frac{L}{T^{2}}\right)^{c} \\
T=\frac{M^{a} L^{b+c}}{T^{2 c}}
\end{gathered}\)

Matching up the powers of each dimensionality on the left— which is simple, there is only T to the first power— to the powers on the right, we get these three equations:

\(\ \begin{aligned}
a &=0 \\
b+c &=0 \\
-2 c &=1
\end{aligned}\)

In this case, the equations are easy to solve. The bottom equation gives us \(\ c=-1 / 2\), and that together with the second equation gives us \(\ b=1 / 2\). So, we now know that:

\(\ \begin{gathered}
P \propto l^{1 / 2} g^{-1 / 2} \\
P \propto \sqrt{\frac{l}{g}}
\end{gathered}\)

Without doing any of the actual physics to figure out the period of the pendulum, but only by considering the units on each quantity, we’ve figured out that the period must be proportional to \(\ \sqrt{l / g}\). (If you want to figure out the dimensionless constant in front of \(\ \sqrt{l / g}\), then in fact you do need to consider the full physics.)