Electrostatics - Finding potential V(r,z) given hyperbolic boundry conditions.

Illeism
Messages
2
Reaction score
0

Homework Statement



I'm trying to derive Equation (1) from the paper: http://idv.sinica.edu.tw/jwang/EP101/Paul-Trap/Winter 91 ajp demo trapping dust.pdf

We are working with a cylindrically symmetric geometry along the z-axis.
r^2 = x^2 + y^2

We have electrodes described by the hyperbolas:
z^2 = z_0^2 + r^2/2
z^2 = r^2/2 - z_0^2

The top and bottom electrode (described by z^2 = z_0^2 + r^2/2) are held at a potential V_0 with respect to our ground ring (described by z^2 = r^2/2 - z_0^2) held at potential 0.

We want to find the potential V(r,z) for the region inside the hyperbolas (the region containing the origin).

The solution is:
V(z,r) = V_0(\frac{1}{4z_0^2})(2z^2+r_0^2-r^2)


Homework Equations



\nabla^2 V = 0
(no charge inside volume)


The Attempt at a Solution



So, for boundary conditions we have:
V(z,r)=V_0 when z^2 = z_0^2 + r^2/2
V(z,r)=0 when z^2 = r^2/2 - z_0^2

I can't think of any appropriate image that would generate hyperbolic potentials, and following Jackson's section on boundary value problems (physics.bu.edu/~pankajm/LN/hankel.pdf) seems to be giving unnecessarily complex solutions, even though the final solution is quite simple.

At this point I'm at a loss. Is there a better way to approach this problem?
 
Physics news on Phys.org
I found a solution, in case anyone is curious:

The symmetry of the problem allows us to assert that we only will have even terms in our potential. So we have something of the form:

V= const + \displaystyle\sum_{\text{even }i}{a_ir^2+b_iz^2}

Hoping that the lowest order term is dominant, we see what happens if we drop the other terms:

V=a r^2 + b z^2 = a(x^2+y^2) + b z^2 + const

And take the laplacian:

\nabla^2 V= 0 = 2a + 2a + 2 b
so
2a=-b

Substituting back into our equation for V, we get:

V=a r^2 - 2a z^2 + const = a(r^2-2z^2) + const

Now, its just a matter of relating our constants (V_0, z_0) for a specific equipotential and we're set!


So, the actual method to generate the potential is starting from a saddle shaped potential (which allows for the trapping of particles; thus the motivation), and then figuring out what electrodes need to be to generate it... which is much easier than starting with the shape of the electrodes!
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top