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I know how to use Buckingham Pi Theorem to, for example derive from the functional equation for a simple pendelum, with the usual methods also described here

$1=fn\left[T_{period}, m, g, L\right]$

$1=fn\left[\frac{g}{L}T_{period}^2\right]=fn\left[\Pi_1\right]$

Everything seemed to work well until I tried to apply the theorem to the governing differential equation:

$(I): m\frac{d^2\Theta}{dt^2}L = - sin(\Theta)mg$

$1 = fn\left[\Theta, m, g, L, t\right]$

$1 = fn\left[\Theta, \frac{g}{L}t^2\right] = fn\left[\Theta, \Pi_1\right]$

Seems to work so far, but when I now try to rewrite $I$ in terms of $\Theta, \Pi_1$, I can't deal with the derivative.


I tried something myself and it seems to work out, but I don't know how rigorous the argumentation is and if the result can be stated in a better way. See the following:

Since we want to take a derivative with respect to time, we define a new dimensional variable $\xi$ with $\xi\bar{t} = t$ with arbitrary $\bar{t} \neq 0$. We also introduce a new function $\Omega\left(\xi\right) = \Theta\left(\xi\bar{t}\right)=\Theta\left(t\right)$. Computing $\frac{d\Omega}{d\xi} = \bar{t}\Theta\left(\xi\bar{t}\right)=\bar{t}\Theta\left(t\right)$ and likewise for higher derivatives.

We now write the functional equation including derivatives like so

$1 = fn\left[\frac{d^2\Theta}{dt^2}, \Theta, m, g, L, t\right]$

substituting $t=\xi\bar{t}, \frac{d^2\Theta}{dt^2}=\frac{1}{\bar{t}^2}\frac{d^2\Omega}{d\xi^2}, \Theta=\Omega$ we have the new functional equation

$1 = fn\left[\frac{1}{\bar{t}^2}\frac{d^2\Omega}{d\xi^2}, \Omega, m, g, L, \bar{t}, \xi\right]$

because the derivative is now in terms of two nondimensional parameters, everything else seems to work great with Buckingham Pi:

$1 = fn\left[\frac{d^2\Omega}{d\xi^2}, \Omega, \frac{g}{L}\bar{t}^2, \xi\right] = fn\left[\frac{d^2\Omega}{d\xi^2}, \Omega, \Pi_2, \xi\right]$

When resubstituting in the actual equation we get

$\frac{d^2\Omega}{d\xi^2} = -sin\left(\Omega\right)\Pi_2$

which indeed yields the correct results.


Although I can't seem to find any mistakes in my reasoning, I am not quite happy with the arbitrary choice of $\bar{t}$. The choice does influence the \Pi-group 2 (although only in magnitude) and also will influence boundary conditions when they are given. As far as I can see, it does not influence the result, when given in natural parameter form (as terms of $m, g, L, t$) because every $\bar{t}$ will get paired again with one $\xi$, yielding just $t$. But I have not been able to prove this point.


I remain with three questions:

  1. What is the standard approach in literature that I can study?
  2. Is there a problem with my derivations, so far?
  3. Can I somehow prove that the choice of $\bar{t}$ does not matter in the final solution?
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    $\begingroup$ Your question consists solely of asking us to check your work. Such questions are off topic on this site. This is not a homework help site. $\endgroup$ – sammy gerbil Aug 11 '16 at 1:02
  • $\begingroup$ Non-dimensionalising is common in physics problems, including those that require ODEs/PDEs. Example: dho.edu.tr/sayfalar/02_Akademik/Egitim_Programlari/… $\endgroup$ – Gert Aug 11 '16 at 1:04
  • $\begingroup$ @sammygerbil this is not a homework problem, this is self-study. Questions about the maths of physics are on-topic. math.stackexchange encourages investigating the problem yourself which I did. I do now see that question (1) was worded wrongly, it was meant to ask about any solution, not just mine. $\endgroup$ – WorldSEnder Aug 11 '16 at 12:25
  • $\begingroup$ The tag is "homework and exercises." This is an exercise within the definition of the policy, a calculation in which you are asking us to check your work. The policy applies to all such questions. See How do I ask homework questions on Physics Stack Exchange? $\endgroup$ – sammy gerbil Aug 11 '16 at 13:15
  • $\begingroup$ @sammygerbil, but aren't there other questions, asking about the idea in general? $\endgroup$ – karatechop Aug 12 '16 at 0:12
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You have done more than half the work yourself. It is convenient to define, $\Pi_1\equiv \sqrt{\frac{g}{L}}t$. There is nothing wrong with the way you have defined it, but my definition reduces work in what follows. Rewrite derivative as:

$\frac{d\theta}{dt}=\frac{d\theta}{d\Pi_1}\frac{d\Pi_1}{dt}=\frac{d\theta}{d\Pi_1}\sqrt{\frac{g}{L}}$

$\frac{d^2\theta}{dt^2}=\frac{d^2\theta}{d\Pi_1^2}\frac{d\Pi_1}{dt}=\frac{d^2\theta}{d\Pi_1^2}\frac{g}{L}$

So your differential equation becomes

$\frac{d^2\theta}{d\Pi_1^2}=-\sin \theta$.

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  • $\begingroup$ Hm, as far as I can understand it, this is more a less chosing $\bar{t}=\sqrt{\frac{L}{g}}$. Thanks for showing me that a good choice of $\bar{t}$ can simplify the equation even more. The remaining problem is then, what if there is no Pi-Group so that $\frac{d\Pi}{dt} \neq fn[t]$ if that can happen. But I think I can work that out myself $\endgroup$ – WorldSEnder Aug 12 '16 at 13:45

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