Recent content by bartek2009

This problem makes me think of the brachistochrone problem from variational calculus. I guess a modified version of it could help with plotting the position of the plane. ---

I think that you lose generality if you put x=(y-z). Probably the proof applies then to that particular case. Anyway, there is my proof: 1. y = z + w 2. (x, z + w) = (x, z) 3. (x, z) + (x, w) = (x, z) 4. (x, w) = 0 now, since x can be any vector, the difference between y and z...

I don't think the second solution satisfies the equation. ---

Do you integrate in a clockwise direction or the usual anti-clockwise manner? If the questions asks you to do it clockwise then I found the answer to be 2*pi*(R^2). ---

If bodies are bound by strong forces the expansion does not affect them. Take an atom for example, the expansion of the universe does not change the mean distance between the nucleus and the electrons. Same principle applies to solar system or even distances between galaxies in local group...

I'm getting tired. ---

One thing to notice is that the expansion only occurs on a cosmological scale (distances of order ~100Mpc) so the stick would not change its initial length. ---

Treating an electron classically, it is in a Coulomb potential well, proportional to 1/distance (similar to the one I plotted in the picture attached). The potential energy is clearly lower at small values of x, i.e. closer to the nucleus. Since total energy must be conserved the decrease in...

In classical consideration, an electron orbits the nucleus and in electron's frame of reference it looks like the nucleus is orbiting the electron. A moving nucleus could essentially be regarded as a current loop and you can then calculate the magnetic field generated by this motion. This is how...

1. If L=0 then J=1/2 only as S=1/2 for an electron and so there is no splitting. 2. In the above case there would be no splitting and magnetic field would be required to cause it (by Zeeman effect) but if L>0 then J can have more than one value and each will have different energy due to...

FD and BE differ by (+/-)1 factor. This can be neglected if exp(e - u) >> 1 which happens in classical regime, that is at high temperatures or at low concentrations. You can try some rough values for semiconductors, Setting zero at the top of valence band, Eg ~ 1eV (energy gap)...

Both distributions, FD: p = 1 / (exp[(e - u)/kT] + 1) and BE: p = 1 / (exp[(e - u)/kT] - 1) approach MB: p = exp[(u - e)/kT] at high temperatures (i.e. in classical limit). So it depends on the temperature. Does it answer your question? :) ---

I see what you mean now. In this case, I think you cannot define the basis. But maybe there is a way to expand functions in an unnormalised basis. ---

You could read about general curvilinear coordinates and that should answer your question (plus it will also explain the factors in other expressions from different coordinate systems like e.g. spherical polar coordinates). Basically the unit vectors change with position in cylindrical...

I'm not familiar with a 2-norm but the basis is normalised "in my book" like so: \int_{-\pi}^{\pi}\phi_{k}^{*}\phi_{k}dx = \frac{1}{2\pi}\int_{-\pi}^{\pi}e^{+ikx}e^{-ikx}dx = \frac{1}{2\pi}\int_{-\pi}^{\pi}dx = 1 ---