I am working through the classical mechanics textbook SICM[1] and this question is specific to something in the book. I understand in general how conservation of angular momentum leads to orbital motion being confined to a plane in the two-body case. I am just trying to understand the specific way that the textbook justifies it.

I follow till the part where they show using some Lagrangian formalism that the $z$ component of the angular momentum is conserved. And then it says:

The choice of the z-axis is arbitrary, so the conservation of any component of the angular momentum implies the conservation of all components. Thus the total angular momentum is conserved. We can choose the $z$-axis so all of the angular momentum is in the z component. Since $\vec{x}\cdot(\vec{x} \times \vec{v}) = \vec{v}\cdot(\vec{x}\times\vec{x} )=0$, the motion is confined to the plane perpendicular to the angular momentum: $\theta = \pi/2$, and $\dot{\theta}=0$.

I know that angular momentum is defined as $L = \vec{r} \times (m\vec{v})$. How do you go from there to $\vec{x}\cdot(\vec{x} \times \vec{v})$?

The relevant section in the book is linked at [1].


1 Answer 1


Our choice of the $z$-axis was arbitrary, so we can choose other axes and we will find that angular momentum along those axes is also conserved. This implies that the total angular momentum is conserved. In particular, you can choose a coordinate system, and then show that angular momentum is conserved along the $x, y$ and $z$–axes. Now, orient your $z$-axis in the direction of the angular momentum. To see that the orbital motion is perpendicular to the angular momentum, use the dot product: $\vec x\cdot \vec L = \vec x \cdot (\vec x\times m\vec v) = m\vec v\cdot (\vec x\times \vec x) = 0$. Since $\vec L$ is constant, $\vec x$ lies in a plane whose normal is $\vec L$.

  • $\begingroup$ How does $\vec{L}$ equal $(\vec{x} \times m\vec{v})$ here? Is that just considering the particular position where $\vec{r}$ = $\vec{x}$? Or is it that $\vec{x}$ represents position in general and not something relating to the x-axis? $\endgroup$ Oct 14, 2022 at 17:02
  • $\begingroup$ Actually, I think I follow now. $\vec{x}$ here is any position. Therefore, since $\vec{x}\cdot\vec{L}$ os zero, any valid position vector must be perpendicular to $\vec{L}$. Thanks ! $\endgroup$ Oct 14, 2022 at 21:00
  • 1
    $\begingroup$ @ThomasAntony Yes, you are correct. I tried to use the notation used in the reference you provided. I'm not sure myself why it decided to suddenly switch from $\vec r$ to $\vec x$. $\endgroup$
    – Crypton
    Oct 15, 2022 at 3:18

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