NanoComp / meep

free finite-difference time-domain (FDTD) software for electromagnetic simulations
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LED tutorial in cylindrical coordinates #2690

Closed stevengj closed 5 months ago

stevengj commented 8 months ago

It might be nice to extend this tutorial to show how a similar calculation could be performed in cylindrical coordinates, e.g. for random currents distributed uniformly over a disk.

Even if you were doing the calculation in Cartesian coordinates, you would still only need one dipole source per radius r. Given its emitted power P(r), the total emitted power would be ∫P(r) 2πr dr — note the r factor! To do the same thing in cylindrical coordinates, you additionally may need to be careful about the normalization of the cylindrical "dipole" (actually ring) sources.

stevengj commented 7 months ago

In particular, I would:

  1. First place dipole sources at different radii r in vacuum, summing the Fourier series over m. Check that the radiation patterns P(r,θ) are the same (at high enough resolution) up to a scale factor s(r) (probably an r^2 scale factor from the normalization of the delta function). Here, by P(r,θ) I mean the radiated power of a dipole source at radius r, integrated over φ — but this is done implicitly if we evaluate each m's power at φ=0 and multiply by 2π. (In vacuum, the power should be independent of φ, but that won't be the case for other geometries.) You want P(r,θ) s(r) to be the same for dipole sources at different r.
  2. For incoherent sources in some geometry (e.g. a dielectric disc), you just want to integrate P(r,θ) s(r) times 2πr dr over r, to add up the powers of all the dipoles (everywhere in a disk, hence the 2πr factor), where P(r,θ) is recomputed for your geometry of interest but the scale factor s(r) should be the same as for vacuum. This gives the P(θ) [ = ∫ P(r,θ) s(r) r dr ] of all of the incoherent sources in the disk.

Note, as we've discussed previously, that when you compute the contribution from each m, because the different m's are orthogonal when you integrate over φ, and the integral over φ for a single m just gives a 2π factor, you can effectively evaluate P(θ) only at φ=0 for each m and then just sum the powers timse 2π.