Date sön 20 maj 2018

If you read about Mill's Brilliant Light power in the wikipedia article you will notice that it is associated with pseudoscience. And you will get the impression that all are wrong. But is it. We can take one fact that is the base of whole theory, namely that he constructs a set of orbitals that has non radiative properties and hence has all needed mathematical properties to explain the discrete orbitals we today understand the atoms have. And this with only special solutions of Maxwells equations.

$$\rho = A_{lm} Y_{lm}exp(iwt)\delta_{r=r_0}$$

with $$A_{lm}$$ constants and $$Y_{lm}$$ the spherical harmonics and $$delta$$ the uniform measure of a spherical shell at distance $$r_0$$. Now using the Haus theorem we calculate the fourier component:

$$F(\rho) = A_{lm}\int Y_{lm}exp(iwt)\delta_{r=r_0}exp(iut + i k\cdot \hat r)d\hat r$$

The trick is to take express $$exp(ik\cdot \hat r)$$ as an expansion of spherical harmonics. It turns out that we have the identity: expansion

$$exp(i k\cdot \hat r) = \sum_{lm} C_{lm}j_l(|k||r|)Y_{lm}(k)Y_{lm}(r)$$

, with $$C_{lm}$$ nonzero constants.

So putting this in and restrict the integration to the spherical shell and also do the integration in time domain we geta. Let's also interchange the sum and integral:

$$F(\rho) = \sum_{l'm'}C_{l'm'}A_{lm}j_l'(|k|r_0)(\int_{r=r_0}Y_{l'm'}(r)Y_{l'm'}(k)Y_{l,m}dS) \delta_{u=w}$$

No this integral is zero unless $$l=l',m=m'$$ due to the orthogonality of the spherical harmonics and we get:

$$F(\rho) = C_{lm}A_{lm}D_{lm} j_l(|k|r_0) Y_{lm}(k)\delta_{u=w}$$

, with $$D_{lm}$$ the norm of the spherical harmonics.

So the time fourier component $$u$$ is only $$w$$ and hence for a wavelike packet we must have $$|k| = c|w|$$ and if we take $$r_0 c w$$ a zero to the bessel function, we find that all light like wave numbers lead to a zero fourier component, hence according to Haus theorem for specific $$r_0,w$$ the distribution does not radiate and we're done.