# Conservation of Angular Momentum in Spherically symmetric potential

In Goldstein Book it is given that:

Since the problem is spherically symmetric, the total angular momentum vector $$\boldsymbol{L}=\boldsymbol{r}\times \boldsymbol{p}$$ is conserved.

What does the quote mean and how to prove it?

Classical Mechanics - Goldstein (3rd ed) (Sec 3.2)

• it might help to say where in the book (and possibly also which edition) it says that. (Chapter and section are generally more reliable than page numbers for books that have been printed in multiple formats.) Jun 23, 2019 at 19:59
• Source is already included Jun 23, 2019 at 20:02

A spherical symmetric potential means, the potential $$V$$ depends only on the radius $$r$$, but not on the direction of vector $$\boldsymbol{r}$$. Hence $$V = V(r).$$ From that you get the force $$\boldsymbol{F} = -\boldsymbol{\nabla} V(r) = -\frac{\mathrm d V(r)}{\mathrm d r} \hat{\boldsymbol{r}},$$ which means that the force is parallel to the unit-vector $$\hat{\boldsymbol{r}}$$, and thus parallel to $$\boldsymbol{r}$$.

Using this together with $$\dot{\boldsymbol{p}}=\boldsymbol{F}$$ and $$\dot{\boldsymbol{r}} = \frac{\boldsymbol{p}}{m}$$, you can straight-forward calculate $$\dot{\boldsymbol{L}}$$ (the time-derivative of the angular momentum $$\boldsymbol{L}=\boldsymbol{r}\times \boldsymbol{p}$$) and proove it to be $$\boldsymbol{0}$$. This means that angular momentum $$\boldsymbol{L}$$ doesn't vary with time, i.e. it is conserved.