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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 $\ddot{\phi}$ $\frac{1}{s^2}$?

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Because you take the derivative with respect to $\tau$. Since $\tau$ is dimensionless, the derivative is too.

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  • $\begingroup$ OK, that makes sense. $\endgroup$
    – warped
    Dec 29 '19 at 11:22
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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}

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