Electron Structure, Ultra-Dense Hydrogen and Low Energy Nuclear Reactions

Abstract

In this paper, a simple Zitterbewegung electron model, proposed in a previous work, is presented from a different perspective based on the principle of mass-frequency equivalence. A geometric-electromagnetic interpretation of mass, relativistic mass, De Broglie wavelength, Proca, Klein-Gordon, Dirac and Aharonov-Bohm equations in agreement with the model is proposed. A non-relativistic, Zitterbewegung interpretation of the 3.7 keV deep hydrogen level found by J. Naudts is presented. According to this perspective, ultra-dense hydrogen can be conceived as a coherent chain of bosonic electrons with protons or deuterons located in the center of their Zitterbewegung orbits. This approach suggests a possible role of ultra-dense hydrogen in some aneutronic and many-body low energy nuclear reactions.

Full text

1 1,2
1
2
A[V·s·m−1] [eV ]
AM[V·s·m−1] [eV ]
A[V·s·m−1] [eV ]
At[V·s·m−1] [eV ]
m[kg] [eV ]
F[V·s·m−2] [eV 2]
B[V·s·m−2]=[T] [eV 2]
E[V·m−1] [eV 2]
V[J=kg ·m2·s−2] [eV ]
J[A·m−2] [eV 3]
JM[A·m−2] [eV 3]
ρ[A·s·m−3=C·m−3] [eV 3]

x, y, z [m] [eV −1] [1.9732705 ·10−7m'1eV −1]
t[s] [eV −1] [6.5821220 ·10−16 s'1eV −1]
c[2.997 924 58 ·108m·s−1] [1]
~ ~ =h
/2π[1.054 571 726 ·10−34 J·s] [1]
µ04π·10−7V·s·A−1·m−1] [4π]
0[8.854 187 817 ·10−12 A·s·V−1·m−1] [ 1
4π]
e[1.602 176 565 ·10−19 A·s] [0.085 424 546]
α[7.2973525664 ·10−3] [7.2973525664 ·10−3]
me[9.10938356 ·10−31kg] [0.5109989461 ·106eV ]
λc[2.426 310 2389 ·10−12 m] [1.229 588 259 ·10−5eV −1]
KJ[0.4835978525 ·1015 Hz V −1] [2.71914766 ·10−2]
rere=λc
/2π
rcrc=αre
TeTe=2πre
c
ωeωe=2π
Te
γ2
x=γ2
y=γ2
z=−γ2
t= 1 {γx,γy, γz, γt}Cl3,1(R)
γiγj=−γjγiwith i 6=j and i, j ∈ {x, y, z, t}
∂=γx∂
∂x +γy∂
∂y +γz∂
∂z +γt1
c
∂
∂t
I=γxγyγzγt
IM=γxγyγz
ESR

NU
Cl3,1(R)
c~
c= 1
~= 1
T λ
E=~ω=2π~
T=2π~c
λ.
E=ω=2π
T=2π
λNU
.
NU
2π
eV
L(1eV )= 1 eV−1≈1.9732705 ·10−7m≈0.2µm,
T(1eV )= 1 eV−1≈6.582122 ·10−16 s≈0.66 fs.
1 eV ≈1.519268 ·1015 rad s−1.

E=mc2
[E=m]NU .
meωe
reTe
Ee=mec2=~ωe=~c
re
=h
Te
,
Ee=me=ωe=1
re
=2π
TeNU
.
re=λc
/2π≈0.38616 ·10−12 m
rc≈2.8179 ·10−15 mωe
reTe
ΦMh= 2π~e
ΦM=h/e,
[ΦM= 2π/e]N U .
rc
re
e=rrc
re
=√α≈0.0854245NU
.
pc
pc=eA =eΦM
2πre
=~ωe
c=~
re
=mec.
A=~
/ere
pcre
~
pcre=~.

pc
me
pc=eA =Ee=1
re
=me=ωeNU
.
ϕ
AM
ϕ=e
~ˆAM·dl.
ϕ
2π
ϕ=e
~˛AM·dl=e
~ˆ2πre
0
Adl =e
~ˆ2πre
0
~
ere
dl =e
~
~
ere
2πre= 2π,
AMdl
ϕ V
T
ϕ=e
~ˆT
V dt.
ϕ
Te=2π
/ωeϕ= 2π
V=e
4πε0rc
=e
rcNU
Te=2πre
c= [2πre]N U .
ϕ=e
~
Te
ˆ
0
V dt =e
~V Te=e
~V2πre
c=e2
rc
2πreN U
= 2π.
At=V
c=A=|AM|,
[At=V=A=|AM|]NU ,
A2
= (AM+γtAt)2=A2
M−A2
t= 0.
dϕ =e
~V dt
dϕ
dt =ωe=e
~V=e2
4πε0~rc
=cα
rc
=c
re
=mec2
~=ce
~A,
dϕ
dt =ωe=me=eV =eANU
.

∂∂∧A+µ0J= 0
m
∂∂∧A+mc
~2
A= 0
∂∂∧A+m2A= 0NU
µ0J=m2ANU re
2rc.¯
Je
A
¯
Je=Ie
A,
A= 2rerc= 2αr2
e,
¯
Je=Ie
A=Ie
2αr2
e
.
Ie=αAM
2πNU
¯
Je=AM
4πr2
eNU
,
µ0¯
Je= 4π¯
Je=AM
r2
e
=ω2
eAM=m2
eAMNU
.
A=AM+γtAt
A2
= 0
µ0Jet =At
r2
e
=ω2
eAt=m2
eAtNU
,
µ0
¯
Je=µ0¯
Je+γtJet=m2
e(AM+γtAt) = m2
eANU ,
µ0
¯
Je=m2
eANU .
NU.
∂∧
∂∂∧A+m2A= 0
∂F +m2A= 0

∂∧∂F +m2∂∧A= 0
∂∧∂F +m2F= 0.
∂F =−4π
¯
J,
∂·
∂·∂F =−4π∂·
¯
J= 0,
∂∧∂F
∂2F
∂∧∂F =∂2F−∂·∂ F =∂2F.
F
ψ
∂2F+m2F= 0.
∂A=∂∧A
∂2A+m2A= 0
∂2A+ω2A= 0.
i
∂ψ−mψ= 0
∂F −mF = 0.
m ∂F F
mm
rm=1
/r=ωrur,
∂F −ωruF= 0
∂Cl3,1(R)i
∂,
ωF ψ.ru
r2
u= 1
ωr=ru

r·A= 0.
4πJ+ωruF= 0
F=∂A,
ω2A+ωru∂A= 0,
ru∂A+ωA= 0.
ru
∂A+ωruA= 0.
e
e∂A+eωruA= 0.
e
/ω
c−γt
e
ωA2=c−γt,
ru
e
ωruA2=ruc−ruγt.
∂A=F
eF=−ω2(ruc−ruγt).
F= (E+IB)γt
e(E+IB)γt=−ω2(ruc−ruγt).
E
eEγt=ω2ruγt.
eA =ω ω2eAω,
eE=eAωru=−edAM
dt .
m
ω r
ru=ωr=mr
eE=mω2r.
B:
eIBγt=−ω2ruc

eIMB=−ω2ruc
c
eIMBc =−ω2ru.
B c
eIMB∧c=−ω2ru
ec×B=−ω2ru.
p=eAM,−ω2ru
ec×B=edAM
dt =dp
dt
ec×B=−mω2r=−ω2ru.
ωrum
AAM
A=AM+γtAt
A=|AM|=At
ω eA
e
∂2A+e2A2A= 0NU ,
∂2A+αA2A= 0NU .
ν/A
KJ:
v
A=1
2KJNU
.
T
ΦMΦo
AT =ωT
e=h
e= ΦM= 2K−1
JNU
,
ΦM= 2Φo= 4.13566766 ·10−15 V·s
[ΦM= 73.55246018]NU .

Φm=h
/e
z
L=c∆t z l =vz∆t
r NU m =r−1z
r vz
r=rer1−v2
z
c2
m=~ω
c2=me
q1−v2
z
c2
.
pc=eA =~ω
c=~
r
pc=ω=1
r=mNU
.
pcR r
R > r =~
pc
.
pc=eAMp⊥
pkz
pc=p⊥+pk.
p⊥vz
ωexy

p⊥=~ωe
c=mec,
[p⊥=ωe=me]NU ,
pk
ωz=vz
/r
pk=~ωz
c=~vz
cr =~ω
c2vz=mvz
hpk=ωz=vz
r=mvziNU .
ωe=v⊥
r=pc2−v2
z
r=pc2−v2
z
rep1−v2
z
/c2=c
re
.
ω=c
r.
ωz=vz
r
ω2=ω2
e+ω2
z,
p2
c=p2
⊥+p2
k,
m2c2=m2
ec2+m2v2
z.
p pk
p=pk=mvz.
ωz
mem vz=ωzr
p=mvz=~ω
c2vz=~
cr vz=~ωz
c=~2π
λ=~k,
p=mvz=ωvz=vz
r=ωz=2π
λ=kNU
.
p
k=pλ
2π=~.
k=2π
/λλ
T≈8.1·10−21 s
vz
z

~
~
~
µB
τ
ωp~k
~
~
~BE
~k=±1
2~
θ θ ∈π
3,2π
3
~2
k+~2
⊥=~2,
~k=±1
2~.
BE
τ=|µB×BE|=µBBEsin (θ)
ωp=BEµB
~.
EH=~ωpif θ=2π
3
EL=−~ωpif θ=π
3.
νESR
∆E=EH−EL= 2~ωp=~ωESR =hνESR.

νESR = 2BEµB
h.
BE= 1.5 T νESR ≈42 GHz
s µ
νNMR ≈BEµ
hs .
7
3Li s =3
/2µ≈1.645 ·10−26 BE= 1.5 T
νNMR ≈24.8 MHz 11
5B s =3
/2µ≈1.36 ·10−26 J·T−1
νNMR ≈20.5 MHz 87
38Sr s =9
/2µ≈5.52 ·10−27 J·T−1
νNMR ≈278kHz BE= 0.15 T ωp
/2π=1
/2νNMR ≈139kHz
±~
/2
~
m2c2
/~2
ω
m2c2
~2=m2=ω2NU
.
E0≈3.7 keV
r0
E0≈mec2α≈3.7 keV
r0≈~
mec≈0.39 ·10−12 m.
E0
r0re
−E0

E0=1
4π0
e2
re
=~
re
αc =mec2α,
E0=e2
re
=α
re
=ωee2=meαNU
.
630 eV
2.3·10−12 m
74 ·10−12 m
re
−e2
/re≈ −3.7 keVN U
3.3·10−10 m
π dc
T dc=cT =λc≈2.42 ·10−12 m
di
di=sλ2
c−λc
π2
≈2.3·10−12 m.

286 kJ 240 kJ
1.48 eV
3.7 keV
715 MJ ≈198 kWh
αmec2≈3.7 keV
meu reu
meuc2=mec2+αmec2≈514.728 keV.
mec2=~ωe=~c
re
,
meuc2=me(1 + α)c2=~ωeu =~c
reu
,
reu =~
me(1 + α)c=re
1 + α.
4Ep=e2
4πε01
re−1
reu =e2α
4πε0re
=α2
reNU ≈27.2 eV.
27.2 eV
µ B f
B
f=∇(B·µ).
B
σI = 2

7
3Li
7
3Li +H(0) →24
2He +e.
17.34 MeV
8.67 MeV 11
5B
11
5B+H(0) →34
2He +e.
133
55 Cs + 4D(0) →141
59 P r + 4e
88
38Sr + 4D(0) →96
42M o + 4e
138
56 Ba + 6D(0) →150
62 Sm + 6e.
4D(0) 6D(0)

(ESR)

σI=2