Date ons 08 november 2017

The question we now shall analyze is the precession of the orbitsphere. The analysis need to show that we get a factor of two and hence an angular momentum due to precession of the electron orbitsphere. This is needed because the angular momentum of the electron is \(\hbar\) and not \(\hbar / 2\).

In free space we note that a loop precesses at the angular velocity \(\Omega\) depending on the tilt \(\theta\) from the plane (see wikipedia) as,

$$ \Omega = \frac{I_s\sin(\theta)}{I_o \cos(\theta)} w, $$

with \(I_s\) the axisymmetric moment of inerthia and \(I_o\) the orthogonal one.

We know that all those loops precesses with the same angular velocity, hence we will still get a uniform distribution. Using that we have that the total angular momentum of the orbitsphere due to the loops are,

$$ L_w = C \int \sin(\theta) \cos(\theta) w I_s d\theta = \hbar/2. $$

The angular momenta due to \(\Omega\) in the \(\hat z\) direction aligned with \(L_s\) and \(L_\Omega\) is if we take the orthogonal component only

$$ L_\Omega = C \int \Omega_s I_o \cos(\theta) \cos(\theta) d\theta. $$

Now put the relation for \(\Omega\) as a function of \(w\) and we get,

$$ L_\Omega = C \int (\frac{I_s \sin(\theta)}{I_o \cos(\theta)} w)I_o \cos(\theta) \cos(\theta) d\theta = C \int_0^\pi I_s w \cos(\theta) \sin(\theta) d\theta = L_w $$

So in essens the angular momentum of the electron taking into account the precession becomes,

$$ L_tot = L_w + L_\Omega = 2 L_w = \hbar $$

With this and a-deep-question we see that the formulas in GUTCP are fully correct, that we reproduce the electron angular momenta with the model and also at the same time deduces the correct ionisation energy to a good aproximation of hydrogene. But no no no, this can't work. A detailed reading of the wikipedia article shows that one shouled reverse the \(\cos\) and \(\sin\) in the formula for the precession \(\Omega\) darn. Also it shows that the precession rate goes to infinity as the angle goes to 0. Can't work. Please move on.


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