# Fine dipole moment - numbers


## Fine dipole moment

$\alpha_{QED} = \frac{e^2}{\hbar c} = \frac{d_{QED}^2}{a_B^2 \hbar c}$

this is the definition of fine dipole moment depending on the length $a$ between charges $e$.
The lengtha can be taken either as $a_B$ https://en.wikipedia.org/wiki/Bohr_radius or https://en.wikipedia.org/wiki/Classical_electron_radius

The dimensionality of dipole squared is $d^2=erg*cm^3$

## imagine that there are no free charges, but only dipole interactions

- let us compare with fine coupling in low-T glasses

$\alpha_{QSNS} = P \frac{\gamma^2}{\rho c_t^2}$

where $P=\frac{1}{erg*cm^3}$, $\gamma=erg$, and $\rho c^2= erg/cm^3$

To write down explicit dipole moment, we need to introduce length $\xi$

$\frac{\hbar c}{\xi} = \frac{\gamma^2}{\rho c_t^2} \frac{1}{\xi^3}$, therefore $\xi=\gamma \sqrt{\frac{1}{\hbar \rho c_t^3} }$
$\xi \sim 50A \sim 10* a_B$

Therefore $d^2_{QSNS} = P \gamma^2 \xi^6$

## **effective-QSNS charge is about the same as electric charge!!!!!** .. or even larger???
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QED charge is $e^2=\alpha_{QED} \hbar c$, numerically $e^2 \sim 25*10^{-20} erg*cm$
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QSNS charge is $e_{QSNS}^2 = \alpha_{QSNS} \rho c_t^2 \xi^4$, because $e_{QSNS}^2 \xi^2 = P \gamma^2 \xi^6= \alpha_{QSNS} \rho c_t^2 \xi^6$. $d^2_{QSNS}=e_{QSNS}^2 \xi^2 = [erg*cm^3]$
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Numerically, QSNS charge is $e_{QSNS}^2= 10^{-3} 2 (g/cm^3) (4*10^5cm/s)^2 (3*10^{-7}cm)^4 \sim 25*10^{-19} erg*cm$, where $\xi \sim 30A = 3*10^{-7}cm$
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- therefore $e_{QSNS}^2 \sim 10*e^2$
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## compare $d^2_{QSNS}$ and $d^2_{QED}$

- this ratio $d^2_{QSNS}/d^2_{QED}$ implicitly depends on the ratio $a_B/\xi$

- for QSNSs, for electric dipole and elastic dipole moments compare

$e^2 a_B^2 \rightarrow P \gamma^2 \xi^6= \alpha_{QSNS} \rho c_t^2 \xi^6$

$e^2 a_B^2 \sim (5*10^{-10}esu*0.5*10^{-8}cm)^2 \sim 6 *10^{-36} erg*cm^3$
versus
$\alpha_{QSNS} \rho c_t^2 \xi^6 \sim 4*10^{-4} (4*10^5cm/s)^2 (5*10^{-7}cm)^6 \sim 800*125* 10^{-36} erg*cm^3 \sim 10^5 *10^{-36} erg*cm^3$



- **BIGGER $d^2_{QSNS}$**!!!!! ... it feels strange

$d^2_{QSNS}/d^2_{QED} \sim 10^{4}$

do the same for electric dipole moment of LEEs

- electric and elastic energy per volume - just for intuition - comparable!!!

$\frac{e^2}{a_B^4} \rightarrow \alpha_{QSNS} \rho c_t^2$

$\frac{(5*10^{-10}esu)^2}{(5*10^{-8}cm)^4} \sim 4*10^{10} erg/cm^3$ versus $\alpha_{QSNS} \rho c_t^2 \sim 0.6*10^8 erg/cm^3$

elastic energy is about $10^{-3}$ smaller than electric-charge-hydrogen-smoothered energy

## analogies

$a_B^2 \hbar c \rightarrow \rho c_t^2 \xi^6$

speculatively on the dimensionality basis $\hbar \rightarrow \rho c_t \xi^4$

$\hbar=10^{-27} erg*s$
$\rho c_t \xi^4 \rightarrow 1(g/cm^3) *4*10^{5}(cm/s) *(5*10^{-7})^4 \sim 25*10^{-21} erg*s$

just compare dimensionally $\hbar \rightarrow \rho c a_B^4$

- $\hbar_{QSNS} \rightarrow \rho c_t \xi^4 \sim 10^{-20} erg*s$
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$2*4*10^5*(3*10^{-7}cm)^4 \sim 0.6*10^{-20} erg*s$⏎
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# Parents

* Fine structure constant