# Potentital integral

1. Mar 4, 2006

### gulsen

I have a force and need to find the potential, but I'm stuch at a point. Any ideas how to integrate this: $$\frac{(r')^2 - 2 (r'') r}{r^2}$$

2. Mar 4, 2006

### vaishakh

what are r' and r''?

3. Mar 4, 2006

### benorin

Likely unknown functions of an independent variable, say t,

something close is

$$\left( -\frac{r^{\prime}}{r}\right) ^{\prime} = \frac{(r^{\prime})^2-r^{\prime\prime}r}{r^2}$$

but this leaves the term $$-\frac{r^{\prime\prime}}{r}$$ unaccounted for

also close is

$$\left( -\frac{(r^{\prime})^2}{r}\right) ^{\prime} = r^{\prime}\frac{(r^{\prime})^2-2r^{\prime\prime}r}{r^2}$$

but has an extra factor of $$r^{\prime}$$ in it...

Are you sure that is the term to be integrated?

4. Mar 4, 2006

### gulsen

Mathematicians...

Unfortunately, I'm sure this is the force to be integrated. It's a part of an exercise from Classical Mechanics, Goldstein, p32.

5. Mar 8, 2006

### samalkhaiat

[
[/QUOTE]

This is a scalar quantity! Force needs to be represented by a vector field.
You are missing something.

regards

sam

6. Mar 14, 2006

### gulsen

You can put a unit vector $$e_r$$ if you like. But it's so obvious that if we're talking about force, we're talking about a vector. And if you want a more precise definition, r is the distance of the particle to the center of force.

Instead of being so pedantic and saying "hey question wrong, question is incomplete, etc", will someone give a correct answer? Such posts have no use.

Last edited: Mar 14, 2006
7. Mar 14, 2006

### George Jones

Staff Emeritus
I think it works out better if you split the given in the question as

$$\frac{1}{r^2} \left( 1 + \frac{\left( r' \right)^{2}}{c^2} \right) - \frac{2}{c^2} \left( \frac{\left( r' \right)^{2}}{r^2} - \frac{r''}{r} \right)$$

However, I am quite tired right now, so I may have made mistakes while doing the problem

Regards,
George

8. Mar 15, 2006

### samalkhaiat

OK, here is a useful post. The potential you are after is;
$$V= \frac{\dot{r}^2}{2r}$$
and you donot have to carry out any integration**. Just try to write your force in the form;

$$F=-\frac{\partial{V}}{\partial{r}} + \frac{d}{dt}(\frac{\partial{V}}{\partial{\dot{r}}})$$

**Two months ago, I created a thread here called "Integrate This". The purpose of the thread was to share with people some of the tricks that I use in solving complicated integrals without going through a messy process of integration. Unfortunately, the thread was locked (I believe) for psychological reasons .Have a look at that thread, you may find some useful mathematical tricks in there.

regards

sam