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In classical physics, when we try to get the force as a function of $r$; Why do we substitue $$r=\frac{1}{u}?$$ I don't get it, what is the point of it? Is the substitution arbitrary?

I am looking for a mathematical explanation of it. I understand it makes the equation solvable, otherwise we would have some messy chain differentiations. But I want it explained in Mathematical term. Perhaps like a proof. That way I wouldn't feel like I am putting in a formula without understanding.

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  • $\begingroup$ You should go into more detail about the problem you have in mind, but assuming this is a two-body problem thing, the answer is "it makes the differential equations easier to solve". In scattering, there's also the fact that annoying boundary conditions at $r=\infty$ become more tractable boundary conditions at $u=0$. $\endgroup$ – eyeballfrog Oct 25 '18 at 18:40
  • $\begingroup$ @eyeballfrog Hi, first off, thank you for your answer. However, I am looking for a mathematical explanation of it. I understand it makes the equation solvable, otherwise we would have some messy chain differentiations. But I want it explained in Mathematical term. Perhaps like a proof. $\endgroup$ – Bertrand Wittgenstein's Ghost Oct 25 '18 at 18:48
  • $\begingroup$ @eyeballfrog Thank you for that explanation; however, I don't understand how that is relevant in 2-particle systems, with $r$ being a finite number. We don't deal with $r\rightarrow \infty$. $\endgroup$ – Bertrand Wittgenstein's Ghost Oct 25 '18 at 18:55
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It has to do with orbit trajectories and angular momentum. The conservation of angular momentum says $$ mr^2\frac{d\theta}{dt} = \ell. $$ Suppose we want to find $r(\theta)$ instead of $r(t)$ in our orbit equations. $dr/dt = (dr/d\theta)(d\theta/dt)$, which means $$ \frac{dr}{dt} = \frac{\ell}{mr^2}\frac{dr}{d\theta} = -\frac{\ell}{m}\frac{d}{d\theta}\left(\frac{1}{r}\right) $$ \begin{multline} \frac{d^2 r}{dt^2} = \frac{d^2 r}{d\theta^2}\left(\frac{d\theta}{dt}\right)^2 + \frac{dr}{d\theta}\frac{d}{dt}\left(\frac{d\theta}{dt}\right) = \frac{\ell^2}{m^2r^4}\frac{d^2r}{d\theta^2}-\frac{2\ell}{mr^3}\frac{dr}{d\theta}\frac{dr}{dt} = \\ = \frac{\ell^2}{m^2r^4}\frac{d^2r}{d\theta^2}-\frac{2\ell^2}{m^2r^5}\left(\frac{dr}{d\theta}\right)^2 = -\frac{\ell^2}{m^2 r^2}\frac{d^2}{d\theta^2}\left(\frac{1}{r}\right) \end{multline} So we see that the orbit trajectories are more easily expressed in terms of the $\theta$ derivatives of $u= 1/r$. In particular, the radial force equation $m\ddot{r}-\ell^2/(mr^3) = F(r)$ becomes $$ \frac{d^2u}{d\theta^2} + u = -\frac{m}{\ell^2u^2}F\left(\frac{1}{u}\right), $$ and the conservation of energy becomes $$ \left(\frac{du}{d\theta}\right)^2 + u^2 - \frac{2m}{\ell^2}V\left(\frac{1}{u}\right) = \frac{2mE}{\ell^2}, $$ where $-dV/dr = F(r)$. In the common case of an inverse-square law force, the force term in the force equation is constant and the potential term in the energy equation is linear, which are easily solved exactly.

Using $u = 1/r$ has the added advantage that it is always bounded, as the bodies never have zero separation, while $r$ can become unbounded in scattering-type problems where the bodies do not form bound orbits.

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