Recent content by Jufro

When dealing with |x| it often helps to use the piece wise expansion of it: =x, x>0 = -x, x<0. If you were on the interval from [-L,L] you would have 2 integrals one from [-L,0] and one from [0,L]. However, since you are only interested in the function from [0,L] then $|x|^3$= $x^3$. To solve...

Everything seems kosher.

Given a vector v= {vr,vθ,vφ} you can write this in terms of the unit vectors = vr er+ vθ eθ+ vφ eφ. There I am using the notation that ei is the unit vector associated in the i'th direction. If you want to take the time derivative of this vector ∂t v = ∂t (vr er+ vθ eθ+ vφ eφ) =∂t (vr er)+ ∂t...

Given a vector

Looks good to me. Just for ease of clarity, when you list angles either name them as ACB, CAB, CBA, or mention in the order you are writing them down.

In particular I think BvU is trying to draw your attention to the partial derivatives you wrote in your last statements. Check those dimensions to see if they are consistent with the definitions you wrote in the relevant equations section.

I would assume that the ducks are spherical. The best packing fraction for spherical objects is hcp(fcc) at 0.74. While this is a good bound, it has been shown that random packing is between 62 and 64%. By it has been shown I mean I took those numbers off of wikipedia. That being said, if we...

If you look at acting on one of your states |0,1> for example, then you cannot lower both. So these states would not contribute to the splitting.

Two things, when you set up the boundaries of your integral you have to make sure you eliminate a variable at each step. Therefore you need to do the z integral before the r integral. Otherwise, you will integrate r and then introduce another r which you will not be able to get rid of...

The Hall Coefficient is: RH = \frac{1}{nq} For electrons, q= -e, and our hall coefficient is: RH = \frac{-1}{ne} Therefore we know that n-type semiconductors have negative RH.

What you are looking for is something called integration by parts.

Since the differential equation has the form of: (\frac{∂}{∂r})2 u(r) + a(r) \frac{∂}{∂r} u(r) + b(r) u(r) =0 With a(r) having a simple pole at r=0, and b(r) having no more than a pole of order 2 at r=0, it is sufficient to continue with the Frobenius Method: u(r) = rσ \sum knrn You can...

Hope this is what you were looking for. It isn't the final answer because you can still express Cot ø and Exp(iθ) in terms of the spherical harmonics but that part is not the worst. May even be able to find them in a table somewhere.

Isn't the azimuthal angle the argument of the Exp[], in this case θ?

Sorry, a is real and a > 1. Rewriting sin as vela said came to the right answer, thanks.