# Central Force Problem; Need Help Solving Difficult Integral

1. Feb 24, 2008

### logic smogic

Problem
A particle moves in a force field described by

$$F(r)=-\frac{k}{r^{2}}e^{-\frac{r}{a}}$$

where k and a are positive.

Write the equations of motion and reduce them to the equivalent one-dimensional problem. Use the effective potential to discuss the qualitative nature of the orbits for different values of the energy and the angular momentum.

Attempt at Solution
I'm having trouble deriving the potential V. Given the force equation F(r), we should have

$$F(r)=-\nabla V$$

Since the Lagrangian L is defined as $$L=T-V$$, we need to know the potential V. We can derive this by integrating the force equation. But I've searched through many integral tables, and come across nothing for this particular integrand!

Can anyone help with this simple step?

2. Feb 25, 2008

### Fredrik

Staff Emeritus
Why not just use Newton's second law directly? (You know that's what you're going to end up with anyway if you start with the Lagrangian).

$$m\ddot\vec{r}=F(r)\hat{r}$$

To get the equivalent one-dimensional problem, you have to do some polar coordinate magic on the left-hand side.

3. Feb 25, 2008

### logic smogic

The professor would like us to derive the equations of motion using the Lagrangian formulation.

Unless I'm mistaken, you can't arrive at an expression for the potential V by only using F=ma, right? Integrating the expression for force F(r) is the only way to arrive at an expression for the potential V(r).

The possibility of using Newton's laws aside - I can solve the integral via an exponential integral (and, therefore, possible a gamma function?), or by using an infinite sum or Taylor expansion. But I'd be surprised if they gave us a problem that was so difficult to solve analytically.

Thanks Fredrik, I'm still working on it.

4. Feb 25, 2008

### Jezuz

Have you remembered to use the Nabla operator in spherical coordinates when you set up your equations?

5. Feb 25, 2008

### logic smogic

$$V=-\int \int \int \frac{k}{r^{2}} Exp(-\frac{r}{a}) r^{2} sin(\theta) dr d\phi d\theta = -4 \pi k \int Exp(-\frac{r}{a}) dr = 4 \pi k a Exp(-\frac{r}{a})$$

This works. Thanks for the idea.

I'm still confused about one thing, though. For your standard inverse-square law, calculated in spherical, you have

$$F(r)=- \frac{k}{r^{2}}$$

giving (according to the text)...

$$V(r)=- \frac{k}{r}$$

Wouldn't the r^2 cancel in the inverse square, and give a different potential (i.e. V ~ r, not ~1/r)? Just a side thought.

Last edited: Feb 25, 2008
6. Feb 25, 2008

### logic smogic

Alright, going with it, I have used,

$$\frac{d}{dt} \frac{\partial L}{\partial \dot{r}} - \frac{\partial L}{\partial r} = 0$$

for the equation of motion, I get

$$m \ddot{r} - 4 \pi k e^{-\frac{r}{a}} = 0$$

Clearly the $$m \ddot{r}$$ term in the above equation is some additonal force (or is it the same as in the given force law?).

Either way, how do I reduce it to the equivalent one-dimensional problem?

Going on the hunch in parentheses above, would I substitute the force law in...

$$-\frac{k}{r^{2}} e^{-\frac{r}{a}}=4 \pi k e^{-\frac{r}{a}}$$
$$\rightarrow r^{2} = -\frac{1}{4 \pi}$$ ?

Or do I convert from spherical to cartesian?

EDIT: Because of the concern mentioned in the post above, I've decided that this answer must be incorrect. My prof said he used a substitution to do the integral, but I've tried half a dozen substitutions, and none of them made the problem work. I've also tried integrating by parts multiple times, and searching for substitutions then, too. I've checked the errata, and it's not a typo. I'm assuming there's some clever substitution (other than the power, logarithmic, and exponential ones I've been trying) that makes it work.

Last edited: Feb 26, 2008
7. Feb 26, 2008

### logic smogic

Problem solved!

There was a typo in my printing of the textbook, which did not appear in any of the errata available online. The *actual* expression given should be a potential V(r), not force law F(r), and is

$$V(r)=-\frac{k}{r}e^{-\frac{r}{a}}$$

I'm still not sure what they mean by reducing the equation of motion I found,

$$m \ddot{r} + (\frac{k}{ar} + \frac{k}{r^{2}})e^{-\frac{r}{a}}=0$$

to a one-dimensional expression. Any hints?