Assume the two particle problem i.e. two particles with equations of motions as follows:

$\ddot{x_i} = -\frac{1}{m_i}\frac{\partial}{\partial x_i}V(|x_1 - x_2|)$

Since the angular momentum is conserved (in relative coordinates), we know that the two particles have to move in a plane and hence can write the distance between the two particles as follows: $x = r e_r$ and $\dot{x} = \dot{r} e_r + r \dot{e_r} = \dot{r}e_r + r \dot{\phi} e_{\phi}$

Now if I wanna compute the actual value of the angular momentum I seem to run into some trouble. If I do it the following way I get the right value:

$|L| = |\mu x \wedge \dot{x}| = \mu |(r e_r \wedge ( \dot{r}e_r + r \dot{\phi} e_{\phi}))| = \mu | r e_r \wedge r \dot{\phi} e_{\phi} | = \mu r^2 \dot{\phi} $

But if I do it the following way it doesn't work out:

$|L | = |\mu x \wedge \dot{x}| = \mu |x||\dot{x}|\sin(\theta) = \mu |x||\dot{x}| = \mu|r||\sqrt{\dot{r}^2 + r^2 \dot{\phi}^2}| $ where I used the formula for $x$ and $\dot{x}$

Why doesn't that work? I feel like I'm missing something completely trivial here, but can't find it.

  • $\begingroup$ Hint: $|\sin(\theta)|$ is not necessarily equal to $1$. $\endgroup$
    – Qmechanic
    Jul 2 '13 at 13:10
  • $\begingroup$ Ah, we are in an elipse, not a circle. Of course. Thanks! $\endgroup$
    – user17574
    Jul 2 '13 at 13:59

As @Qmechanic points out, you really need to think about the angle between $x$ and $\dot{x}$. In particular, since $x = r e_r$, you can take the inner product with $\dot{x}$ and get a nonzero answer: $x \cdot \dot{x} = r\, \dot{r}$. This contribution needs to be removed when taking the exterior product $x \wedge \dot{x}$, which is what you did in the first version when you eliminated $\lvert r e_r \wedge \dot{r} e_r \rvert$. But setting $\sin\theta = 1$ does not accomplish this.


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