Recent content by jncarter

To find the number of possible configurations you need to do some counting. There are formulas out there to do this for a given electron configuration (see Introduction to Quantum Mechanics by Griffiths). Let's start with an example. For atomic hydrogen you have one electron in configuration...

The answer to this question isn't really known. It is more of a matter of philosophy. Do you choose to believe that the electron is a particle that we can touch and see that goes through both, neither and one slit all at once? Or do you simply believe that we can only represent to electron as a...

Include a phase factor in your solution. x_{1} = A*e^{i (\omega t +\delta_{1})} Similarly for the second wave vector. Now retain the amplitude relationships and use the phase factor \delta to allow for zero values at time t=0. It may be more useful to use sine or cosine functions at this point...

This type of problem is much easier to do using Lagrangian mechanics. Are you familiar with this method? You could also set this up as an eigenvalue problem. You would use \omega^{2} as your eigenvalue and x_{1}, x_{2} as your eigenvector. You should find that there are two very distinct modes...

You're right about that second equation. That was my bad forgetting parentheses. I also got the chance to talk to my TA yesterday, and you're right about the second derivative. Thanks for all the help. There's just so much algebra with this type of problem, it's easy to lose track of things.

You lost a negative sign \frac{du}{dz} = -\alpha \zeta^{\alpha+1}v- \zeta^{\alpha+2}v' I haven't checked the other part of your work yet. My last post may also be wrong. I found this formula somewhere \frac{d^2}{dx^2}(uv) = \frac{d^2 u}{dx^2}v+2\frac{du}{dx}\frac{dv}{dx}+u\frac{d^2 v}{dx^2}...

Okay, new attempt at the second derivative: \frac{d^2}{dz^2}(z^{-1})=2z^{-3}=2\zeta^{3} \frac{d^2}{d\zeta^2}(\zeta^{\alpha}v) = v\frac{d^2}{d\zeta^2}(\zeta^{\alpha}) + \zeta^{\alpha} \frac{d^2v}{d\zeta^2} \rightarrow = \alpha(\alpha-1)\zeta^{\alpha-2}v+\zeta^{\alpha}\frac{d^2v}{d\zeta^2}...

Oops, u = \zeta^{\alpha}v. \zeta = z^{-1} → d\zeta = -z^{-2}dz=-\zeta^{2} → \frac{d\zeta}{dz}=-\zeta^{2} Note the positive in the \zeta], since z has a negative exponent, \zeta has a positive one. \frac{du}{dz}=\frac{du}{d\zeta} \frac{d\zeta}{dz} = -\zeta^{2}\frac{d}{d\zeta}(\zeta^{\alpha})...

Homework Statement Show that by letting z = \zeta^-1 and u = \zeta^{\alpha}v(\zeta) that the differential equation, z(1-z)\frac{d^{2}u(z)}{d^{2}z}+{\gamma - (\alpha+\beta+1)z}\frac{du(z)}{dz}-\alpha \beta u(z) = 0 can be reduced to \zeta(1-\zeta)\frac{d^{2}v(\zeta)}{d\zeta^{2}} +...

Homework Statement Find the center of mass of a triangle with two equal sides of length a. The triangle's third side is length b and it has a uniform mass of M. Homework Equations R = \frac{1}{M} \int dm \vec{r} dm = \frac{M}{A} A = \frac{1}{2}base*height The Attempt at a Solution...

Homework Statement Use the contour integral \int_{C}\frac{e^{pz}}{1+e^z}dz to evaluate the real integral \int^{\infty}_{- \infty}\frac{e^{px}}{1+e^x}dx 0<p<1 The contour is attached. It is a closed rectangle in the positive half of the complex plane. It height is 2i∏. Homework...

Have you tried solving the problem yet? We need to know where you are having difficulty before trying to help. In other words: Show us your work.

When I took the derivatives more carefully I ended up with the same x-equation, but the y-equation come out as m\ddot{y_{n}} = 2ky_{n} - k(x_{n} + x_{n+1}) My professor pointed out that this is not a restoring force. The yn term needs to be negative, but he says the potential looks like it's...

Precisely.

Remember what a dot product is: \vec{r_{1}}\cdot \vec{r_{2}} = x_{1}x_{2} + y_{1}y_{2} + z_{1}z_{2} = |r_{1}||r_{2}|cos \phi You can either sum of the product of the components of the two vectors or multiply there magnitudes times the angle between the vectors. You can use the above equation...