Date mån 15 januari 2018

In GUTCP, we have a special transform between to the circular spped of light frame that changes the wavelength from $$2\pi r$$ to $$r$$. This change of reference frame is mysterious and it is usually a main stumbling block to follow GUTCP. YOu here things like "this is bogus because he magically add a $$2\pi$$". Now I have no clear understanding about the issue, but we should try to find out when there is a natural notion of $$\lambda = 2 \pi r$$ and a notion of $$\lambda = r$$. We will use the link maxwell.pdf where all results is stemming from applying planeWaveExpansion. Here a condition for non radiation is deduced if the angular freqeuncy

$$r w = k\pi, k=1,\ldots$$

or

$$w = k\pi/r, k=1,\ldots$$

Also using composing linear outgoing plane waces in all directions will lead to a electric field of (using $$l=1,c=1$$)

$$\phi = \int exp(i w (e \cdot x) \pm i w t) \, dS(e) = j_1(r|x|)exp(i w t).$$

Note, More generally we have

$$\phi_{lm} = \int exp(i w (e \cdot x) \pm i w t)Y_{lm}(e) \, dS(e) = j_l(r|x|)Y_{lm}(\hat x) exp(i w t).$$

we get the period as $$w x = 2\pi$$ or

$$(k \pi / r) \lambda = 2\pi$$

or

$$\lambda = 2 r / k$$

Now if we argue that $$k=1$$ is instable the ground state would be deduced as

$$\lambda = r$$

As one of Mills expressions for the wavelength. If we on the other hand consider the movement of the wavepackets along the sphere at the speed of light the natural condition will be $$\lambda = 2\pi r$$. My hypothesis is that this duality might be behind the changes of reference system logic seen in GUTCP.

Now, using this correspondance a loop goes from $$2 \pi r \to r$$ so in order for the total chanrge and mass to be constant we need $$m \to 2\pi m$$ and $$e \to 2\pi e$$. That conserves mass and charge. Furthermore the plane wave notion considered as the speed of light frame is interesting to have in the backhead when reading the proof of the g factor and pages 102 and further to transform eq 1.193 into 1.206. Because the formula for B is in the light light frame, and we have for each point on the sphere e.g. in the non light frame, the force:

$$Fmag = e \cdot c x B -> (\frac{e}{2 \pi}) (\frac{1}{\alpha} v x B)$$

From this the formula for the correction due to magnetic field of the nuclear becomes a simple application. The only difficult part remaining is to understand why the formula below is for a light like reference frame??

$$B = \frac{\mu_0 e \hbar}{2 m_p r^3}$$

There should be references for this. Do you know where this may come from? Then applying further special relativistic corrections for v beeing fast will lead to much higher precition of the Ionisation energy just as GUTCP outlines.

There is a mystery why you get the same magnetic force for all loops. This lead to an idea that the electron loop couples with proton loops in an 1 to 1 fashion so that the system - the hydrogen atom is the union of all the coupled loops and this coupling is visible in the speed of light frame. The speed of light frame is remarkable in that the union of loops transforms to one loop on a shell, Now if we assume that the proton loop and the electron loop both have the same normal we would see that a charge element on the electron loop will indeed have a uniform force in the correct direction. This explains that the formula is in the "speed of light" frame.

The derivation of the magnetic field at a distance $$r$$ from the nucleus, which we consdier a loop of small radi $$dr$$ both loops has the same direction of the normal. Then the B field is (Biot-Savart)

$$B=\frac{\mu_0 I}{4\pi}\int_{\theta}\frac{dl \times \hat R}{R^3}$$

Now we consider the half loop from south to north and note $$R(\theta) = r \pm \cos(\theta)dr$$ and $$dl = \cos(\theta)d\theta$$ and using the we can take right side minus the left side:

$$B=\frac{\mu_0 I}{4\pi}\int_{-\pi/2}^{\pi/2}(\frac{1}{(R-dr \cos(\theta))^2} - \frac{1}{(R+dr \cos(\theta))^2})\cos(\theta)$$

$$B=\frac{\mu_0 I}{4\pi}\int_{-\pi/2}^{\pi/2}cos^2(\theta)\frac{2 dr}{r^3}d\theta = \frac{\mu_0 I dr}{2 r^3}$$

Now the angular momenum of the loop is $$\hbar/2$$ due to Larmor precession the total angular momenum is $$\hbar/2$$ without precession, leading to the angualar momentum equation (just as for the electron) to

$$m v dr = \hbar$$

then

$$I = e v = \frac{e \hbar}{m dr}$$

$$B = \frac{\mu_0 e \hbar}{2 m r^3}$$