Anomalous magnetic dipole moment

This is a very short post and it is just a couple of things I've stumbled upon in my latest Physics Journey. Check these out.

The first one is related to the title of this post. This is straight from Wiki

"The one-loop contribution to the anomalous magnetic moment—corresponding to the first and largest quantum mechanical correction—of the electron is found by calculating the vertex function .... The calculation is relatively straightforward and the one-loop result is:

$ a = \frac{\alpha}{2 \pi} \approx 0.0011614 $

where α is the fine structure constant. This result was first found by Julian Schwinger in 1948"

now, how about this
$ \frac {Z_P} {R_k} = $$\frac {\hbar} {q_p^2} / \frac {h} {e^2} = \frac {e^2} {2\pi q_p^2} = \frac{\alpha}{2 \pi}$$ \approx 0.0011614 $

where

$Z_P$ - Planck impedance = $\frac {\hbar} {q_p^2}$
$R_k$ - von Klitzing Constant = $\frac {h} {e^2}$
$q_p$ - Planck Charge
$e$ - charge on an electron
$\alpha$ - fine structure constant

how cool is that, they are the same! Though the calculation for this second version is considerably easier! Is there some sort of equivalence thing going on here? will do more on this later. It's also equal to the classic electron radius divided by the Compton wavelength of the electron!

The second is just a bit of minor number crunching

have a look at this. It is the rest mass of the electron multiplied by the speed of light divided by the Boltzmann constant,

$\frac {m_e c} { 2 \pi^2 k_b}$ $= 1.002 $

$m_e = 9.1093829 10^{-31} kg $
$c = 299 792 458 ms^{-1}$
$k_b$ - Boltzmann constant = $1.3806488 10^{-23} m^2 kg s^{-2} K^{-1}$

That is so close to unity it has to mean something, then again!

Note ,the units of 1.002 is $\frac {K s} {m}$ Temperature / velocity, but this is not as odd as you might think.

Anyway, that is it for now, more on both of these later

intrinsic angular momentum

I was revisiting a couple of my earlier post, particularly one on the momentum of inertia of a photon when I got to thinking about 2 equations I derived in that post, the first equation was

$ E I = L^2$ .... (1)

Which says that energy multiplied by moment of inertia is the angular momentum squared. I also derived a second more specific version of this (originally equation 16), which shows

$ E I =\hbar^2 $   ... (2)

then it occurred to me that when a photon creates matter it has to obey the various conservation laws. A photon that converts to matter will form an electron/positron pair, rather than just a single electron. If it did form a single electron there would be conservation of charge violation and so on.

For two electrons (or an electron positron pair) though, we have

$ E = 2 m_e c^2 $ ...(3)

also the equivalent moment of inertia is

$ I = 2 I_e $ ... (4)

putting these into (2) we have

$ 4 m_e c^2 I_e = \hbar^2  $ ... (5)

so

$ m_e c^2 I_e  = $ $\frac {\hbar^2} {4} $   ... (6)

but from (1) we have

$ E_e I_e = L_e^2 $ for an electron, where $E_e = m_e c^2$ giving

$L_e^2 =$ $\frac {\hbar^2} {4} $  ... (7) , which becomes

$L_e =$ $\frac {\hbar} {2}$ ...(8)

The intrinsic spin of an electron! That's nice eh? There is more to this though. If equation (2) is correct then does it imply all particles have spin. This makes me think we haven't got the Higgs particle sorted yet because it can't have a zero spin version. It's also a bit of a black eye for string theory which, I've been told, likes the idea of spinless particles.

This is not a problem if a particle can have zero moment of inertia, in which case $I=0$ and $L=0$. What would that mean?

After all

$I = m \lambda^2 $ ...(9) or

$I =$ $\frac {\hbar} {\omega}$  ...(10)

so $I$ can't be zero because if m is zero then we still have a moment of inertia because of the frequency of the massless particle given by

$E = \hbar \omega$

 I'll figure what all that in the next post, hopefully!

wrote this while listening to this.