# Proving that central force is conservative and getting the energy

1. Jun 1, 2014

### aquance

1. The problem statement, all variables and given/known data
Hey, sorry in advance if something I write is unclear but I am not native English speaker.

I have a central force that is defined as $$F=-f(r) \frac{\vec{r}}{r}$$ where $$f(r)$$ is some function of distance and $$\vec{r}=(x,y,z)$$. I have to calculate potential energy when $$f(r)=\frac{a}{r^{2}}$$ and I have to use $$\vec{r}d\vec{r}=\frac{dr^{2}}{2}$$

2. Relevant equations

3. The attempt at a solution

I wrote Force F as $$\left( -\frac{f(r)x}{r}, -\frac{f(r)y}{r},-\frac{f(r)z}{r} \right)$$ can I or can I not do it?
Calculated the curl, it was 0 so it is conservative. Now I substituted f(r) and got $$\left( -\frac{ax}{r^{3}}, -\frac{ay}{r^{3}},-\frac{az}{r^{3}} \right)$$

To calculate potential energy I have 3 equations:

$$-\frac{ax}{r^{3}} = \frac{dE}{dx}$$
$$-\frac{ay}{r^{3}} = \frac{dE}{dy}$$
$$-\frac{az}{r^{3}} = \frac{dE}{dz}$$

and the result is $$-\frac{a}{2r^{3}}\left(x^{2}+y^{2}+z^{2}\right) + C$$. I didn't use the $$\vec{r}d\vec{r}=\frac{dr^{2}}{2}$$ so I guess I did something wrong

When you reply please tell me where and why did I make a mistake and how to correct it.

Last edited: Jun 1, 2014
2. Jun 1, 2014

### CAF123

I think you missed negatives in your calculations when solving for E. (Since dE/dx = -Fx etc..)

Note also $x^2 + y^2 + z^2 = r^2$.

Did you mean to write $\vec r \cdot d\vec r = dr^2/2$?

Then $$U = - \int \vec F \cdot d\vec r = - \int \frac{f(r)}{r} \vec r \cdot d\vec r \dots$$
And $$\int \frac{dr^2}{r^3} \propto \frac{1}{r}.$$

So yes, your method is fine, but maybe they wanted you to use the approach above.

3. Jun 1, 2014

### aquance

Yes I forgot about the minuses, thanks for noticing that. Yes I meant what you wrote with the r*dr.

Thanks, I thought my method is incorrect :)

4. Jun 1, 2014

### dauto

The only mistake I can find is an extra factor of (1/2) in your final answer that shouldn't be there. It is possible to solve the problem the way you did (after fixing that mistake), but it is easier to solve it using the identity provided in the problem. Try it...

5. Jun 1, 2014

### aquance

Why is 1/2 a mistake?

When I have the $$-\frac{ax}{r^{3}} = \frac{dE}{dx}$$ it turns into

$$-\frac{ax}{r^{3}}dx = dE$$

$$-\int\frac{ax}{r^{3}}dx = \int dE$$

$$E=-\frac{ax^{2}}{2r^{3}} + C$$

6. Jun 1, 2014

### dauto

The integral is incorrect because r is a function of x. It cannot be treated as a constant. r2 = x2 = y2 = z2.

7. Jun 1, 2014

### aquance

but $$\vec{r}$$ is a function of x, $$r$$ is is simply a scalar ? I would have tried the method mentioned here but I have no idea how to even start to be honest ;p I've missed some lectures about how to solve integrals with vectors and such

8. Jun 1, 2014

### dauto

Missing lectures is a bad idea. r is indeed a scalar, but it is not a constant scalar. It is a function of x, y, and z. Try to change the variable of integration from x to r.

9. Jun 1, 2014

### aquance

Okay so I can see how r^2=x^2+y^2+z^2, but I don't know how to change the variable of integration, I can't just substitute sqrt(r^2-y^2-z^2) for x...

10. Jun 1, 2014

### dauto

Yes you can. What do you get for dx?

Tip: It's easier if you use implicit differentiation.

11. Jun 1, 2014

### aquance

Okay, so
$$x=\sqrt{r^{2}-y^{2}-z^{2}}$$

then

$$dx=\frac{r}{\sqrt{r^{2}-y^{2}-z^{2}}} dr$$

to integral now

$$ax∫\frac{1}{r^{2}\sqrt{r^{2}-y^{2}-z^{2}}} dr$$

???

12. Jun 1, 2014

### dauto

You made the same mistake again. You pulled the x out of the integral. You can't do that because x is a function of r. Just substitute the expression for x provided at the top of your post.

13. Jun 1, 2014

### aquance

Ahaaa! Thanks :)

$$a∫\frac{x}{r^{2}\sqrt{r^{2}-y^{2}-z^{2}}} dr =$$

$$a∫\frac{\sqrt{r^{2}-y^{2}-z^{2}}}{r^{2}\sqrt{r^{2}-y^{2}-z^{2}}} dr =$$

$$-\frac{a}{r} + C$$

So that would mean that $$E = -\frac{3a}{r}+ C$$

Is it correct? Or should it be just $$-\frac{a}{r}+ C$$

Last edited: Jun 1, 2014
14. Jun 1, 2014

### dauto

it's just -a/r + C.

15. Jun 1, 2014

### ehild

You can choose the potential zero at infinity, integrate from $\vec r_$ to ∞.

$$V(\vec {r})=\int_{\vec r}^{∞}{\frac{a}{r'^3}\vec r' \cdot \vec{dr'}}$$

You can write r'dr= 0.5 d(r2) and introduce the variable u=r2.

The integral becomes$$\int{0.5 \frac{a}{u^{3/2} }du}$$ What do you get?

Substitute back u=r2 and take the limits.

ehild

Last edited: Jun 1, 2014
16. Jun 1, 2014

### aquance

So it's $$-(0-\frac{a}{r}) = \frac{a}{r}$$ ? Why do we integrate from r1 to r2, and why do we take r2 as infinity?

17. Jun 1, 2014

### ehild

The potential difference is the integral of the negative force between two points, r1 and r2. If you choose r2 at infinity, where the potential is zero, you get the potential at r1.
I edited my post a bit.

ehild