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I am having trouble unpacking the following argument from page 72 of Goldstein's classical mechanics book.

We now restrict ourselves to conservative central forces, where the potential is V(r), a function of r only, so that the force is always along r. By the results of the preceding section, we need only consider the problem of a single particle of reduced mass m moving about a fixed center of force, which will be taken as the origin of the coordinate system. Since potential energy involves only the radial distance, the problem has spherical symmetry; i.e., any rotation, about any fixed axis, can have no effect on the solution. Hence, an angle coordinate representing rotation about a fixed axis must be cyclic. These symmetry properties result in a considerable simplification in the problem. Since the problem is spherically symmetric, the total angular momentum vector, $$L = r \times p$$, is conserved. It, therefore, follows that r is always perpendicular to the fixed direction of L in space. This can be true only if r always lies in a plane whose normal is parallel to L. While this reasoning breaks down if L is zero, the motion, in that case, must be along a straight line going through the centre of force, for L = 0 requires r to be parallel to $\dot{r}$, which can be satisfied only in straight-line motion.* Thus, central force motion is always motion in a plane.

Firstly, how does it follow from assuming that the potential in the problem is only dependent on the radial distance between the two bodies, that the system is spherically symmetric? I don't have a problem understanding that the problem IS spherically symmetric, but isn't it more? It has translational symmetry too, no? Is the author only mentioning spherical as we're concerned with the conservation of ang. mom. at the moment?

Secondly, by a combination of this section in Wikipedia and knowledge of Noether's theorem, I can work out that spherical sym -> L=conserved -> r and L are perpendicular. What I'm struggling with is how $\dot{r}$ is ALSO perpendicular to L (and moreover parallel to r) to state that 2 body motion occurs in a plane.

The footnote Goldstein includes is

$\dot{\textbf{r}}= \dot{r}\textbf{e}_r + r\dot{\theta} \textbf{e}_\theta$, hence $\textbf{r}\times \dot{\textbf{r}} =0$ requires $\dot{\theta}=0$

but I'm unsure how this relates back to $L=0$.

The above mentioned Wiki article is more clear to me, but I'm still struggling with the jump from

$$L=r \times p = r\times \mu \dot{r} \text{ is constant}$$

to $\dot{r}$ must be perpendicular to L as well.

Any help with some intuition is appreciated!

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  • $\begingroup$ Symmetry means the potential is invariant under the action of the rotation group, what happens if you rotate the potential by an angle $(\theta,\phi)$? $\endgroup$
    – Triatticus
    Jul 22 '20 at 2:42
  • $\begingroup$ The title looks like set of tags instead of proper title. Perhaps something more descriptive would be better? $\endgroup$
    – Umaxo
    Jul 22 '20 at 5:27
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Is the author only mentioning spherical as we're concerned with conservation of ang. mom. at the moment?

Yes

The above mentioned Wiki article is more clear to me, but I'm still struggling with the jump from $L=r\times p=r\times \mu\dot{r}$ is constant to $\dot{r}$ must be perpendicular to $L$ as well.

The cross product is defined in such a way, that if you make cross product of two vectors, the resulting vector will be by definition perpendicular to both vectors. That is all, there is no magic computation hidden in it, just pure definition of cross product.

The constancy of $L$ is needed only for the fact, that this direction is same for all times and thus the motion happens on a plane which is fixed for all times. Without this, the plane which is perpendicular to $L$ and in which the motion is happening at the instant would be different at different times.

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  • $\begingroup$ Oh my goodness. It is literally the definition of the cross product haha Thank you so much this was embarrassing but very helpful! :) $\endgroup$
    – Lopey Tall
    Jul 22 '20 at 15:00

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