# Q: Goldstein chapter 1 problem 16: Finding the generalized potential from the force

I have started to work through Herbert Goldstein's, Charles Poole's and John Safko's Classical mechanics, and I am having a bit of trouble with one of the problems (chapter 1 problem 16). The problem is as follows:

A particle moves in a plane under the influence of a force, acting toward a center of force, whose magnitude is

$$F=\frac1{r^2}\left(1-\frac{\dot{r}^2-2\ddot{r}r}{c^2}\right)$$

where $$r$$ is the distance of the particle to the center of force. Find the generalized potential that will result in such a force, and from that the Lagrangian for the motion in a plane.

I was looking through the solution on chegg for this problem, and I can't seem to understand one step they took. I understand how to obtain the lagrangian and equations of motion once I have the generalized potential, but I can't figure out how they got the potential. This is the way that they solve for it: I understand how they obtained the first term in Lagrange's equation (i.e. $$-\frac{\partial}{\partial{r}}\left(\frac{1}{r}+\frac{\dot{r}^2}{c^2r}\right)$$), but I can't figure out how they obtained the second term ($$\frac{d}{dt}\frac{\partial}{\partial{\dot{r}}}\left(\frac{1}{r}+\frac{\dot{r}^2}{c^2r}\right)$$). When I try to extract $$\frac{d}{dt}\frac{\partial}{\partial{\dot{r}}}$$ from $$\frac{2\ddot{r}}{c^2r^2}-\frac{2\dot{r}^2}{c^2r^2}$$, I can't get what they are getting. Does anyone know how they solved for the generalized potential here, or know a better way to do so? I know from what I have done in electromagnetism that the potential of a field can be found through a line integral of the force from $$-\infty$$ to r, but when I approach the problem that way, I get a natural log term at the end. Any help with how to approach this problem would be very appreciated!

• Have you tried differentiating their solution to see how via reverse/backwards engineering it? – Kyle Kanos Jun 12 '15 at 13:21
• Yes I tried that and for the second term I got $\frac{2\ddot{r}}{c^2r^2}-\frac{4\dot{r}^2c^2r}{(c^2r^2)^2}$. It's very close, and I'm sure I'm just making a stupid mistake, but I can't seem to figure out what it is. – Matt Ntiros Jun 12 '15 at 13:36

## 1 Answer

There is a small misprint in the third line (the 3rd term).

If you try to differentiate the second part, you will get exactly part of the force: $\frac{d}{dt}\frac{\partial}{\partial \dot{r}}\left(\frac{1}{r}+\frac{\dot{r}^2}{c^2 r}\right)=\frac{d}{dt}\left(\frac{2\dot{r}^2}{c^2 r}\right)=\frac{2 \ddot{r}}{c^2 r}-\frac{2 \dot{r}^2}{c^2r^2}$

And also your last equation should be like this $F=-\frac{\partial U}{\partial r}+\frac{d}{dt}\frac{\partial U}{\partial \dot{r}}$. Overall this kind of task should be just an exercise where you try to find a potential according to the type of force function.

• Thank you! I can't believe I missed that typo. I assume you mean $\frac{d}{dt}\left(\frac{2\dot{r}}{c^2r}\right)$ right? – Matt Ntiros Jun 12 '15 at 15:47