Calculating equipotential surfaces

1. Jun 22, 2011

Pengwuino

So this may be a bit silly, but one thing I've never really learned in all my years is how one actually goes about calculating equipotential surfaces for arbitrary potentials? Let's say I have a potential that goes like

$$\Phi = A_0e^{-\left({{r}\over{r_0}}\right)^2}\cos^2 (\phi) \sin^2(\theta)$$

where $r, \phi$ are your typical spherical coordinates and $A_0$ is a constant and $r_0$ is there for dimensional purposes and I want to determine even a single equipotential surface, how does one go about doing this? You typically see very simple examples in physics and you're typically finding something trivial like the shells you see around point particles, but this relies on plain ol intuition. How does one determine the surface with arbitrary (but well-behaved) functions such as my example?

Edit: I changed my function a few times so that it's Laplacian has no weird features for what I'm doing.

Last edited: Jun 22, 2011
2. Jun 22, 2011

BruceW

I'd suggest doing the gradient in spherical coordinates on the potential. This would give the normal to the equipotential surface at all points in space. Then just specify a particular value for the potential to get one particular surface.

3. Jun 22, 2011

Pengwuino

Hmm but that doesn't actually give me the surface as far as I can tell. I assume however one can define the surface, it'll be in terms of some parameter or relationship between the other coordinates.

My initial hunch would be you would set $\Phi = \Phi_0$ for generality, then determine the relationship $\phi = \phi(r,\theta)$ and my surface would be defined running through the $0 \le \phi < 2\pi$ and $0 \le \theta < \pi$.

I eventually have to figure out a surface integral which makes me wonder how do I even define a differential area element.

Last edited: Jun 22, 2011
4. Jun 22, 2011

Bill_K

The total differential of Φ is dΦ = ∂Φ/∂x dx + ∂Φ/∂y dy + ∂Φ/∂z dz. To get an equipotential surface, set this equal to zero and solve for dz:

dz = - (∂Φ/∂x dx + ∂Φ/∂y dy)/(∂Φ/∂z)

This gives you a pair of PDEs to solve for the surface expressed as z(x,y):

∂z/∂x = - (∂Φ/∂x)/(∂Φ/∂z)
∂z/∂y = - (∂Φ/∂y)/(∂Φ/∂z)

5. Jun 22, 2011

BruceW

So you want to integrate some other function (lets say f) over a surface defined such that the function $\Phi$ is kept constant?

6. Jun 22, 2011

Pengwuino

Yes, the function 'f' is actually going to be the gradient of $\Phi$.

So what I really need to figure out is given a potential, how can I define equipotential surfaces for it so that I can later integrate the gradient of the potential over those equipotential surfaces.

And to be clear, I would be integrating $\nabla \Phi \cdot d\vec{A}$

7. Jun 22, 2011

BruceW

$$\nabla \Phi \cdot d \vec{A} = \nabla^2 \Phi dV$$
But the Laplacian is zero for your problem , so luckily for you the integral is zero.

If $\Phi$ was not Laplacian, then Bill_K gives the method for finding the change in z when you change x or y (while still keeping on the equipotential surface). So this effectively defines an equipotential surface when you start from a particular value. To use this, you'd need to redefine your problem in cartesian coordinates.
Alternatively, you could do the total differential in spherical coordinates, which is more time-consuming, but you could get r as dependent on polar angle and azimuthal angle, which is more convenient for your problem.
This gives the 'parameter relationship' you were looking for. Then to integrate over this surface, you'd have to define the surface as being equal to the equipotential surface which you have just defined. I'm not sure how you'd do this. Maybe by creating a coordinate system where one variable is constant over the equipotential surface
i.e. you'd need to define one of your variables as equal to $\Phi$, and then integrate over the other two variables?

8. Jun 22, 2011

Born2bwire

I don't think that's quite right. While the value of the Laplacian will be zero along the boundary of the integration (since it's equipotential) that doesn't mean that the interior value so the Laplacian will be zero. In fact, it should just work out to be a constant times the area of the equipotential surface.

Take for example the function,

$$\Phi(r,\theta,\phi) = r$$
This function has spherical shells of equipotential surfaces. Thus,

$$\nabla\Phi = \hat{r}$$
$$\nabla\Phi\cdot d\mathbf{S} = \hat{r}\cdot\hat{r} r^2\sin\theta d\theta d\phi$$
Thus the value of the integral over the equipotential surface comes out to be 4 \pi r^2.

9. Jun 22, 2011

BruceW

Woops, sorry for some reason I thought that Pengwuino said in the OP that his potential $\Phi$ satisfied Laplace's equation. But he was just saying the Laplacian has no weird features. My mistake.

10. Jun 22, 2011

Born2bwire

Yeah, you had me thinking that too for a second.

11. Jun 22, 2011

Pengwuino

I don't think I agree with the first part. The value of the Laplacian doesn't have to be zero along the boundary of the integration. Something trivial like $1/r^2$ has a non-zero laplacian over the spherical shells that define that functions equipotential surfaces.

12. Jun 23, 2011

BruceW

I think I agree with Pengwuino. Surely the Laplacian isn't necessarily zero anywhere for a general potential?
Back to the main problem: since the potential isn't spherical, I would re-write it in terms of the cartesian coordinates. Also, the gradient of the potential will be perpendicular to the equipotential surface, so:
$$\nabla \Phi \ \cdot \ d \vec{A} = \mid \nabla \Phi \mid \ dA$$
Now, to find what the equipotential surface is, we can use Bill K's method:
$$\frac{ \partial z}{ \partial x} = - \frac{ \frac{ \partial \Phi }{ \partial x} }{ \frac{ \partial \Phi }{ \partial z } }$$
This gives us the partial derivative of z with respect to x. And we can use this to define the length integral which is perpendicular to y. ie
$$ds = \sqrt{1 + { \frac{ \partial z}{ \partial x} }^2 } \ dz$$
So the integral becomes:
$$\mid \nabla \Phi \mid \ \sqrt{1 + { \frac{ \partial z}{ \partial x} }^2 } \ dz \ dy$$
Someone please say if this is wrong, its been ages since I did this kind of maths, but I think its right

Edit: After you've done these steps, I think you'll need to put x in terms of y and z before you do the integration, since x doesn't get integrated over? You can get the dependence of x on y and z by using Bill K's method. Basically, calculate the total differential of the potential along the surface, which we know is zero, and then calculate the total differential for x, then use these, along with boundary conditions, to get x(y,z).

Last edited: Jun 24, 2011
13. Jun 23, 2011

Born2bwire

Right. I say we just delete this thread and start over. :shiftyeyes:

14. Jun 24, 2011

BruceW

Haha, we both made unfounded assumptions to begin with. I blame hopeful thinking, trying to make the problem easier.

15. Jun 27, 2011

Pengwuino

Ok so I've worked with a couple of potential functions and determined various surfaces. Now how does one form the surface element $d\vec A$? Gah, I need to grab a good differential geometry text.

I have 2 things I need to do. 1) Determine the differential area element and eventually integrate this function over that surface and 2) determine the limits of integration for a corresponding volume integral. I wonder if Spivak would help in this? His intro text.

Last edited: Jun 27, 2011
16. Jun 28, 2011

Born2bwire

If you are still trying to integrate the gradient of your potential over an equipotential surface, then you can take a few shortcuts. Namely, the gradient of the potential will always be constant and point in the same direction as your surface normal. So the result of the integral is the area of your surface times some constant related to the gradient of your potential. So if you have a simple surface (sphere, cylinder, etc) then you do not really need to work the integral. Only if you have a complicated surface would you need to work the integral by virtue to just get the area. So your surface element depends upon your system of coordinates and the variables that you are integrating over.

For example, in cartesian coordinates then it is always something like $\hat{z}dxdy$. If it is more complicated, then you should try to parameterize your equation for the surface.

17. Jun 28, 2011

Pengwuino

I've decided to use the potential $A_0e^{-{{r}\over{r_0}}^2}rcos(\theta)$

So far, I've determined the equipotential surfaces are given by

$$x^2 + y^2 + {{z^2}\over{2}} = {{r_0^2ln(z)}\over{2}}+C$$

where the C is a constant that will denote the equipotential surface I believe. These surfaces are very complicated so I can't use any symmetry arguments to make things simpler. I wish I knew how to even look at the surface in Mathematica.

18. Jun 29, 2011

BruceW

The gradient of the potential would point in the direction of the surface normal, but why would the gradient of the potential be constant?

19. Jun 29, 2011

Pengwuino

Yah that can't be true for arbitrary surfaces.

20. Jun 29, 2011

Born2bwire

:uhh:

I feel ashamed with myself. I can't believe I've been making these posts. Ok... where is that memory erase and reset button for this thread...