Type R PowR Suspended 690 Posts user info edit post |
i dont know where to start
1/25/2008 3:42:20 PM |
Aficionado Suspended 22518 Posts user info edit post |
what the hell are you trying to do? 1/25/2008 3:44:40 PM |
Type R PowR Suspended 690 Posts user info edit post |
trying to align my satellite. 1/25/2008 4:00:03 PM |
Charybdisjim All American 5486 Posts user info edit post |
IT would help to show the whole equation. What you have there just looks like part of a differential equation for a wave function Psi in terms of R. Unless the left most "-" is really an "=."
In that case then that's an eigenvalue problem for the wave function of something in a magentic field where the bracketed term is the hamiltonian and E is the energy. Is that an "H" on the bottom as in magnetic field in the original maxwell's equations? Or, is it a misritten M? Or an h or h_bar?
I'm going to guess it's a poorly written "m" because then it looks like a the basic Hamiltonian for a hydrogen atom. It does look like you're missing a constant on the bottom of the potential though. 4*Pi*Epsilon_naught, unless this is just an "electron in a spherical potential" aka simplified hydrogen atom leaving out some of the constants. If that's the case you need to think about separating the equation (like any good multi-variable partial differential equation.) That gradient does suggest you're considering it in 3 dimensions, but the Psi looks only to be a function of R. It will help writing out the laplacian in spherical coordinates and getting rid of the d/dphi and d/dtheta terms. Remember not to drop things just because they have "theta" or "phi" in them though, since when (if) you integrate a wave function in polar coordinates you are still integrating it in 3 dimensions.
Oh yeah, if this is a hydrogen atom remember to use the reduced mass instead of just the particle mass. Assuming the way you wrote the equation (function of r only) is right, you won't have to worry about spherical harmonics. In general, that's not the case. Anyways, have fun.
[Edited on January 25, 2008 at 11:07 PM. Reason : ] 1/25/2008 10:45:52 PM |
skokiaan All American 26447 Posts user info edit post |
http://en.wikipedia.org/wiki/Kepler's_laws_of_planetary_motion 1/25/2008 11:33:51 PM |
Charybdisjim All American 5486 Posts user info edit post |
Haha, now that's just mean. You'll confuse the poor bastard. 1/25/2008 11:42:00 PM |
Type R PowR Suspended 690 Posts user info edit post |
http://youtube.com/watch?v=-xEzGIuY7kw
watch that video and you'll see what i'm talking about.
maybe a few minutes into it, it will pop up. 1/26/2008 12:14:53 AM |
Charybdisjim All American 5486 Posts user info edit post |
Sooo... you're not actually trying to solve it? You just saw it in a wierd al video? 1/26/2008 1:00:16 AM |
Type R PowR Suspended 690 Posts user info edit post |
yup 1/26/2008 11:27:19 AM |
skokiaan All American 26447 Posts user info edit post |
^^ pwnt 1/26/2008 4:18:58 PM |
LS1powered All American 689 Posts user info edit post |
soo easy 1/26/2008 10:55:32 PM |
Walter All American 7760 Posts user info edit post |
1/26/2008 10:56:45 PM |
afripino All American 11425 Posts user info edit post |
= E psi (r)
[Edited on January 26, 2008 at 11:13 PM. Reason : p is the semi-latus rectum....lol] 1/26/2008 11:12:04 PM |
mathman All American 1631 Posts user info edit post |
ye old energy eigenstates. 1/26/2008 11:18:38 PM |
Type R PowR Suspended 690 Posts user info edit post |
on the delta and the e
are those z's or 2? 1/26/2008 11:53:20 PM |
mathman All American 1631 Posts user info edit post |
They are two's.
The (nabla)^2 = ( (partial/partial x)^2 + (partial/partial y)^2 + (partial/ partial z)^2 ) (well at least in Cartesian coordinates, generally it's kinda ugly)
The e^2 comes from the fact it is the potential energy between two charges of strength e. The charge of the electron verses the charge of the proton. Generally for Hydrogen-like atoms I think another factor Z appears to count the number of protons in the nucleus, but I'm kinda rusty on this stuff so feel free to correct me. 1/27/2008 8:57:36 PM |
evan All American 27701 Posts user info edit post |
looks like you need a pro in the area of english language mechanics as well 2/4/2008 6:23:45 AM |