# Force and Potential Energy Coordinates

1. Oct 7, 2015

### teme92

1. The problem statement, all variables and given/known data
Evaluate the force corresponding to the potential energy function $V (r) = \frac{cz}{r^3}$, where $c$ is a constant. Write your answer in vector notation, and also in spherical polars, and verify that it satisfies $∇∧F = 0$.

2. Relevant equations
$F(x)=-\frac{dU}{dx}$

Spherical Coordinates:
$r=\sqrt{x^2+y^2+z^2}$

3. The attempt at a solution
So I'm confused a bit here. I think the vector notation will only have a $z$ component. So differentiate to get $F$ and then as $r$ is a polar coordinate I just change $r^3$ to $\sqrt{x^2+y^2+z^2}^3$. This doesn't look right though. I have a list of relations from cartesian to spherical coordinates but I don't understand how to answer the question with the $\frac{1}{r^3}$. I know how to do the last bit which is the curl. Thanks in advance for any help in the right direction.

2. Oct 7, 2015

### ehild

Your relevant equation is not true for three dimension. So what is the relation between the force and the potential energy function?

3. Oct 7, 2015

### teme92

Ok is it:

$F(x,y,z)=-\Big(\frac{dU}{dx}+\frac{dU}{dy}+\frac{dU}{dz}\Big)$?

4. Oct 7, 2015

### ehild

No. The force is a vector. You wrote a scalar.

5. Oct 7, 2015

### teme92

$F(x,y,z)=-\Big(\frac{dU}{dx}\hat{i}+\frac{dU}{dy}\hat{j}+\frac{dU}{dz}\hat{k}\Big)$?

6. Oct 7, 2015

### ehild

Yes.
Now you can determine the force from the potential function.

7. Oct 7, 2015

### teme92

So $F=\frac{c}{r^3}\hat{k}$?

8. Oct 7, 2015

### ehild

How did you get it?
You have to differentiate the potential energy function with respect to all coordinates, x, y, z.

9. Oct 7, 2015

### teme92

Well there is no x or y component so its only the differentiating with respect to z that matters no?

10. Oct 7, 2015

### ehild

The potential energy is a scalar, it does not have "components". But it depends on all the three coordinates through r $r=\sqrt{x^2+y^2+z^2}$
What are the partial derivatives?

11. Oct 7, 2015

### teme92

So is $r^3 = x^3+y^3+z^3$?

12. Oct 7, 2015

### ehild

NO!!!!!!!!!!!!!!!!!!!!!
$r^3=(x^2+y^2+z^2)^{\frac{3}{2}}$

13. Oct 7, 2015

### teme92

I know that's the obvious answer but I thought it was to do with what the power was. So I have to differentiate:

$\frac{cz}{(x^2+y^2+z^2)^{\frac{3}{2}}}$

With respect to x,y and z?

14. Oct 7, 2015

### ehild

Yes. You need all partial derivatives, $\partial U / \partial x$, $\partial U / \partial y$, $\partial U / \partial z$.

15. Oct 7, 2015

### ehild

Yes.

16. Oct 7, 2015

### teme92

So I have a long mess of an answer that I won't put up cos it seems pointless. I'm fairly sure it was all differentiated correctly though. For the spherical part do I just start again and change $z$ to $rcos\theta$?

17. Oct 7, 2015

Yes.