So, I'm following the derivation in D. Morin, Introduction to Classical Mechanics, of the equations for a two-body system. I understand all of it, aside from this one step.

When he's talking about solving Lagrangian for $r(\theta)$, I don't follow the step.

$$\tag{*}L = \frac {1}{2} m\dot{r}^{2}+ \frac{\ell^2}{2mr^2}+u(r)$$

and some how he ends with

$$\tag{7.16}(\frac{1}{r^2} \frac{dr}{d \theta})^2 = \frac{2mE}{\ell^2}- \frac{1}{r^2} - \frac{2mu(r)}{\ell^2}.$$

This is probably a case where my lack of differential equation skills causes trouble.

  • $\begingroup$ Comment to the (*) equation $L = \frac {1}{2} m\dot{r}^{2}+ \frac{\ell^2}{2mr^2}+u(r)$: Either the two plus signs should be minuses, or the LHS should be the energy rather than the Lagrangian. $\endgroup$ – Qmechanic Jun 10 '13 at 11:56

I think there is a sign problem in one of your formulae, however, with your conventions, this is my derivation:

The lagrangian L is $L = \frac {1}{2} m\dot{r}^{2}+ \frac{l^2}{2mr^2}+u(r) = K - V$

where K is the kinetic energy $\frac {1}{2} m\dot{r}^{2} + \frac{l^2}{2mr^2}$ and $V$ is the potential $V(r) = - u(r)$

The constant energy is then $E = K + V = \frac {1}{2} m\dot{r}^{2} + \frac{l^2}{2mr^2} - u(r)$

Multiplying by $\frac{2m}{l^2}$, we get :

$\frac{2m}{l^2} E = \frac{2m}{l^2}(\frac {1}{2} m\dot{r}^{2}) + \frac{1}{r^2} - \frac{2m}{l^2} u(r)$

With $l = m \dot\theta r^2$, you have $\frac{2m}{l^2}(\frac {1}{2} m\dot{r}^{2}) = \frac{2m}{(\large m \dot\theta r^2)^2}(\frac {1}{2} m\dot{r}^{2}) = (\frac{\dot r}{\dot \theta r^2})^2 = (\frac{1}{r^2} \frac{dr}{d \theta})^2$, So we have :

$$\frac{2m}{l^2} E = (\frac{1}{r^2} \frac{dr}{d \theta})^2 + \frac{1}{r^2} - \frac{2m}{l^2} u(r)$$, that is :

$$(\frac{1}{r^2} \frac{dr}{d \theta})^2 = \frac{2m}{l^2} E - \frac{1}{r^2} + \frac{2m}{l^2} u(r)$$

So I do not understand the minus sign for the last term of your second formula.

  • $\begingroup$ I follow your work, thanks. Yes, the minus sign was a typo, sorry. I guess the only part I'm confused about is why multiply by $\frac{2m}{l^2}$? Is this just a trick to get the formula into a workable form, or is there more logic behind it? $\endgroup$ – Astrum Jun 10 '13 at 22:29
  • $\begingroup$ I think it is just a way to get the formula so, as, in the left side term, you have $(\frac{1}{r^2} \frac{dr}{d \theta})^2$. So it is nothing that a different presentation of the equation, while, of corse, using $E = K + V$ instead of $L = K - V$. $\endgroup$ – Trimok Jun 11 '13 at 7:12

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