Recent content by Yitzach

I think that analytically will be easier on the processor(s) and so I would like as many analytical solutions as possible. But I suspect that eventually, there will be a need for numerical solutions. Numerical is just a little hard on CPU's if you are lacking with that in your computer.

Let's (mostly me) try this again. I have the following: \Psi_n(x)=\sqrt2\sin(n\pi x)\ x\in[0,1] |\Psi_n(x)|^2=2\sin^2(n\pi x)\ x\in[0,1] \Phi_n(p)=\frac{1}{\sqrt{2\pi\hbar}}\int^\infty_{-\infty}e^{-ipx/\hbar}\Psi(x)dx\ x\in[0,1],p\in\Re3.54 \Phi_n(p)=\frac{\pi...

Perhaps I wasn't quite clear. I have the wavefunction in "configuration" space for the infinite square well which does a very nice job of giving a probability distribution for where the particle should be. I want the probability distribution for the momentum of the particle. Getting the...

I have in mind to build a game to help teach or demonstrate some concepts in QM and I thought it would be nice to be able to measure momentum. So as a proof of concept before I get too many man hours burned on the project I thought it would be good to do the infinite square well. I managed to...

There is nothing I can do to remove the electric permittivity of the "weak conductor." http://en.wikipedia.org/wiki/Permittivity#Classification_of_materials" says that a material is considered a poor conductor when "\frac{\sigma}{\omega\epsilon}<<1" meaning the electric permittivity could be...

I am familiar with it in the form of \vec{J}=\sigma\vec{E} where sigma is the conductivity which is equal to the multiplicative inverse of rho, which was used. I'm not sure what you are purposing as I don't have the current density to find the current with. The current density was found from the...

As noted in the attachment: \vec{J}=\frac{Q\sigma}{4r^2\pi\epsilon} But the current is constant through any sphere between a and b, inclusive. Failure to do so suggests that the current curls. But because the current follows the E-field, which doesn't curl in this case, and there is no...

That works much better. The two currents and resistance are not dependent on position and they are not negative. I decided to assume a charge Q instead of a potential V and found the E-field using Gauss's Law in integral form. 1a) \vec{I}=\frac{\sigma Q}{\epsilon}\hat{r} 1b)...

Homework Statement Two concentric metal spherical shells of radius a and b, respectively, are separated by a weakly conducting material of conductivity sigma. (a) If they are maintained at a potential difference V, what current flows from one to the other? (b) What is the resistance between...

You want to derive the MacLaurin series, not use them to reach the answer. a_(2n) will be a_n of the cosine MacLaurin series. a_(2n+1) will be a_n of the sine Maclarin series. By working it backward you would end up proving that the results of the method work...

You are working the problem backwards. You want to assume a power series expansion is a solution to the problem. y=\Sigma_{i=0}^\infty a_nx^n You can then take the derivative of that twice so you can find y". You can then plug the series into the equation and solve for a_n and you will end up...

Fair enough. So is there a reason for real numbers? Yes I was aware of other methods. I was wondering about the permissibility of what I did for if I wanted to do it again later. Notice that I mentioned four other methods in the initial post.

By what you said here, I take it to mean that once an answer has been arrived at by this method, it would be advisable/required to check 0 and infinity to make sure that they are not the answer. In this case I knew that neither of those could be the answer based on the nature of the question and...

The following came from a step in my EM homework. I came up with the answer that all of my resources (calculator, WolframAlpha, and a friend) were indicating as the correct answer, so it is not about the homework. This is about the permissibility of what was done. 1...

I did drop the pi^-1. GIGO. TI-89 Titanium only knows what I tell it. Nicely caught. And that sigma makes complete sense now that it is less than one whole r.