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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.
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1 1,2
1
2
A[V·s·m1] [eV ]
AM[V·s·m1] [eV ]
A[V·s·m1] [eV ]
At[V·s·m1] [eV ]
m[kg] [eV ]
F[V·s·m2] [eV 2]
B[V·s·m2]=[T] [eV 2]
E[V·m1] [eV 2]
V[J=kg ·m2·s2] [eV ]
J[A·m2] [eV 3]
JM[A·m2] [eV 3]
ρ[A·s·m3=C·m3] [eV 3]
x, y, z [m] [eV 1] [1.9732705 ·107m'1eV 1]
t[s] [eV 1] [6.5821220 ·1016 s'1eV 1]
c[2.997 924 58 ·108m·s1] [1]
~ ~ =h
/2π[1.054 571 726 ·1034 J·s] [1]
µ04π·107V·s·A1·m1] [4π]
0[8.854 187 817 ·1012 A·s·V1·m1] [ 1
4π]
e[1.602 176 565 ·1019 A·s] [0.085 424 546]
α[7.2973525664 ·103] [7.2973525664 ·103]
me[9.10938356 ·1031kg] [0.5109989461 ·106eV ]
λc[2.426 310 2389 ·1012 m] [1.229 588 259 ·105eV 1]
KJ[0.4835978525 ·1015 Hz V 1] [2.71914766 ·102]
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
λNU
.
NU
2π
eV
L(1eV )= 1 eV11.9732705 ·107m0.2µm,
T(1eV )= 1 eV16.582122 ·1016 s0.66 fs.
1 eV 1.519268 ·1015 rad s1.
E=mc2
[E=m]NU .
meωe
reTe
Ee=mec2=~ωe=~c
re
=h
Te
,
Ee=me=ωe=1
re
=2π
TeNU
.
re=λc
/2π0.38616 ·1012 m
rc2.8179 ·1015 mωe
reTe
ΦMh= 2π~e
ΦM=h/e,
M= 2π/e]N U .
rc
re
e=rrc
re
=α0.0854245NU
.
pc
pc=eA =eΦM
2πre
=~ωe
c=~
re
=mec.
A=~
/ere
pcre
~
pcre=~.
pc
me
pc=eA =Ee=1
re
=me=ωeNU
.
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
rcNU
Te=2πre
c= [2πre]N U .
ϕ=e
~
Te
ˆ
0
V dt =e
~V Te=e
~V2πre
c=e2
rc
2πreN U
= 2π.
At=V
c=A=|AM|,
[At=V=A=|AM|]NU ,
A2
= (AM+γtAt)2=A2
MA2
t= 0.
=e
~V dt
dt =ωe=e
~V=e2
4πε0~rc
=
rc
=c
re
=mec2
~=ce
~A,
dt =ωe=me=eV =eANU
.
A+µ0J= 0
m
A+mc
~2
A= 0
A+m2A= 0NU
µ0J=m2ANU re
2rc.¯
Je
A
¯
Je=Ie
A,
A= 2rerc= 2αr2
e,
¯
Je=Ie
A=Ie
2αr2
e
.
Ie=αAM
2πNU
¯
Je=AM
4πr2
eNU
,
µ0¯
Je= 4π¯
Je=AM
r2
e
=ω2
eAM=m2
eAMNU
.
A=AM+γtAt
A2
= 0
µ0Jet =At
r2
e
=ω2
eAt=m2
eAtNU
,
µ0
¯
Je=µ0¯
Je+γtJet=m2
e(AM+γtAt) = m2
eANU ,
µ0
¯
Je=m2
eANU .
NU.
A+m2A= 0
∂F +m2A= 0
∂F +m2A= 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
ψmψ= 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+ruA= 0.
e
/ω
cγt
e
ωA2=cγt,
ru
e
ωruA2=rucruγt.
∂A=F
eF=ω2(rucruγt).
F= (E+IB)γt
e(E+IB)γt=ω2(rucruγt).
E
eEγt=ω2ruγt.
eA =ω ω2eAω,
eE=eAωru=edAM
dt .
m
ω r
ru=ωr=mr
eE=2r.
B:
eIBγt=ω2ruc
eIMB=ω2ruc
c
eIMBc =ω2ru.
B c
eIMBc=ω2ru
ec×B=ω2ru.
p=eAM,ω2ru
ec×B=edAM
dt =dp
dt
ec×B=2r=ω2ru.
ωrum
AAM
A=AM+γtAt
A=|AM|=At
ω eA
e
2A+e2A2A= 0NU ,
2A+αA2A= 0NU .
ν/A
KJ:
v
A=1
2KJNU
.
T
ΦMΦo
AT =ωT
e=h
e= ΦM= 2K1
JNU
,
ΦM= 2Φo= 4.13566766 ·1015 V·s
M= 73.55246018]NU .
Φm=h
/e
z
L=ct z l =vzt
r NU m =r1z
r vz
r=rer1v2
z
c2
m=~ω
c2=me
q1v2
z
c2
.
pc=eA =~ω
c=~
r
pc=ω=1
r=mNU
.
pcR r
R > r =~
pc
.
pc=eAMp
pkz
pc=p+pk.
pvz
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=pc2v2
z
r=pc2v2
z
rep1v2
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=kNU
.
p
k=pλ
2π=~.
k=2π
/λλ
T8.1·1021 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=EHEL= 2~ωp=~ωESR =ESR.
νESR = 2BEµB
h.
BE= 1.5 T νESR 42 GHz
s µ
νNMR BEµ
hs .
7
3Li s =3
/2µ1.645 ·1026 BE= 1.5 T
νNMR 24.8 MHz 11
5B s =3
/2µ1.36 ·1026 J·T1
νNMR 20.5 MHz 87
38Sr s =9
/2µ5.52 ·1027 J·T1
νNMR 278kHz BE= 0.15 T ωp
/2π=1
/2νNMR 139kHz
±~
/2
~
m2c2
/~2
ω
m2c2
~2=m2=ω2NU
.
E03.7 keV
r0
E0mec2α3.7 keV
r0~
mec0.39 ·1012 m.
E0
r0re
E0
E0=1
4π0
e2
re
=~
re
αc =mec2α,
E0=e2
re
=α
re
=ωee2=meαNU
.
630 eV
2.3·1012 m
74 ·1012 m
re
e2
/re≈ −3.7 keVN U
3.3·1010 m
π dc
T dc=cT =λc2.42 ·1012 m
di
di=sλ2
cλc
π2
2.3·1012 m.
286 kJ 240 kJ
1.48 eV
3.7 keV
715 MJ 198 kWh
αmec23.7 keV
meu reu
meuc2=mec2+αmec2514.728 keV.
mec2=~ωe=~c
re
,
meuc2=me(1 + α)c2=~ωeu =~c
reu
,
reu =~
me(1 + α)c=re
1 + α.
4Ep=e2
4πε01
re1
reu =e2α
4πε0re
=α2
reNU 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
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