# Dimensionless expression for differential equation

I am working through Nonlinear Dynamics and Chaos by Steven H Strogatz. In chapter 3.5 (overdampened beads on a rotating hoop), a differential equation is converted into a dimensionless form. I am trying to work out which dimensions the initial equations had, and why the converted form is dimensionless.

Initial equation:

$$mr \ddot{\phi} = -b \dot{\phi} -mg \sin\phi + mr \omega^2 \sin\phi \cos \phi$$

$$m$$ is mass, $$r$$ is radius, $$\phi$$ is an angle, $$b,g$$ are arbitrary, positive constants, and $$\omega$$ is angular velocity.

Using a characteristic time $$T$$, a dimensionless time $$\tau$$, with $$\tau = \frac{t}{T}$$ is introduced.

$$\dot{\phi}$$ and $$\ddot{\phi}$$ then become $$\frac{1}{T}\frac{d\phi}{d\tau}$$ and $$\frac{1}{T^2}\frac{d^2\phi}{d\tau^2}$$, respectively.

Then the initial equation becomes

$$\frac{mr}{T^2}\frac{d^2\phi}{d\tau^2} = -\frac{b}{T}\frac{d\phi}{d\tau} - m g \sin\phi + mr \omega^2 \sin\phi \cos \phi$$

This is made dimensionless by dividing through a force $$mg$$:

$$(\frac{r}{gT^2})\frac{d^2\phi}{d\tau^2} = (-\frac{b}{mgT})\frac{d\phi}{d\tau} - \sin\phi + (\frac{r \omega^2}{g}) \sin\phi \cos \phi$$

And all the expressions in the brackets are dimensionless.

I understand why the expressions in the brackets are dimensionless, but what about the differentials?

$$\phi$$ is dimensionless.

but would $$\dot{\phi}$$ not have dimension $$\frac{1}{s}$$, and $$\ddot{\phi}$$ $$\frac{1}{s^2}$$?

Because you take the derivative with respect to $$\tau$$. Since $$\tau$$ is dimensionless, the derivative is too.
You substitute $$\dot{\phi} = \frac{1}{T} \frac{d\phi}{d\tau}$$. Now let's consider the dimensions \begin{align} [\dot{\phi}] &= [\frac{d\phi}{dt}] = \frac{rad}{s}\\ [\dot{\phi}] &= [\frac{1}{T} \frac{d\phi}{d\tau}] = \frac{1}{s} \cdot [\frac{d\phi}{d\tau}] \\ \Rightarrow [\frac{d\phi}{d\tau}] &= rad \end{align} and analog for $$\ddot{\phi}$$. Thus, by substituting $$dt \to T d\tau$$ we obtain a dimensionless time $$\tau$$. Just look at the dimension of the last expression \begin{align} s = [dt] = [T d\tau] &= s \; [d\tau] \\ &\Rightarrow [d\tau] = 1 \end{align}