- Email: [email protected]
The light controlled fusion
Annals of Nuclear Energy 62 (2013) 57–60
Contents lists available at SciVerse ScienceDirect
Annals of Nuclear Energy journal homepage: www.elsevier.com/locate/anucene
The light controlled fusion BingXin Gong ⇑ 3-708 HuaDu XinCun, No. 57, JianSheBei Road, XinHua HuaDu, Guangzhou, Guangdong 510800, China
a r t i c l e
i n f o
Article history: Received 28 March 2013 Received in revised form 4 June 2013 Accepted 6 June 2013
Keywords: Controlled fusion Nuclei Light Oscillating electric dipoles Kinetic energy Density
a b s t r a c t This is a new technique for controlled fusion. When two nuclei are colliding with each other, light, whose the frequency is higher than the minimal threshold frequency of lithium, will be aimed directly at the two nuclei, the two nuclei will perform the simple harmonic oscillation, the charged particle’s simple harmonic oscillation can be considered as an oscillating electric dipole, and the two oscillating nuclei will radiate the electromagnetic wave. Either of the two oscillating electric dipoles will attract each other, or they will repulse each other. There will be an attraction force between the two oscillating nuclei. When the attraction force is greater than the Coulomb repulsion between the two nuclei, the two nuclei will fuse together. Where the kinetic energy and the density of the two nuclei can be controlled, the electric vector and the frequency of the light can be controlled also and, therefore, the fusion can be controlled. Ó 2013 Elsevier Ltd. All rights reserved.
1. Introduction If the light is directed at the nuclei like 2H and 6Li, the 2H and the 6Li will perform a simple harmonic oscillation and will then radiate the electromagnetic wave. The charged particle’s simple harmonic oscillation can be considered as an oscillating electric dipole. As the 2H and the 6Li have a positive charge, when the 2H and the 6Li are travelling in the same direction along the line, there will be an attraction force between the 2H and the 6Li. The attraction force is inversely proportional to the 4th power of the distance between the two nuclei that perform a simple harmonic oscillation; the repulsion force is inversely proportional to the 2nd power of this distance, the attraction force will be greater than that of the repulsion force. When the attraction force is greater than that of the Coulomb repulsion between 2H and 6Li, the 2H and the 6Li will fuse together. An example shows that if the kinetic energy of the 2H and the 6 Li are greater than 2.15 keV, and the distance between them is shorter than 1012 m, and the frequency of the light is x P 5:14 1017 Hz, the 2H and the 6Li will fuse together.
2. The principle of light controlled fusion Light controlled fusion is based on the following principle:
When a light directs at a charged particle, the charged particle will perform a simple harmonic oscillation and will radiate an electromagnetic wave. The charged particle’s simple harmonic oscillation can be considered as an oscillating electric dipole. When two oscillating electric dipoles are in the same direction along the line, there will be an attraction force between the two oscillating electric dipoles. Suppose the distance between two oscillating nuclei is r, r is also the distance between two oscillating nuclei when they are colliding. Suppose the light which is directed at the nuclei is produced by the oscillating electric dipole. In the far-zone fields of the oscillating electric dipole which has ! charge Q and amplitude a, the electric field intensity EðtÞ is proportional to the 2nd power of the angular frequency x (Cheng, 1989a), !
EðtÞ ¼
Qa x2 cos xt 4pe0 c2 R
ð1Þ
where e0 is the dielectric constant, c is the speed of light, and R is the distance between the point of observation and the centre of the oscillating electric dipole. Let
A¼
Qa 4pe0 c2 R
ð2Þ
The Eq. (1) can then be changed into !
⇑ Tel.: +86 13480212541. E-mail address: [email protected] 0306-4549/$ - see front matter Ó 2013 Elsevier Ltd. All rights reserved. http://dx.doi.org/10.1016/j.anucene.2013.06.007
EðtÞ ¼ Ax2 cos xt
ð3Þ
58
B. Gong / Annals of Nuclear Energy 62 (2013) 57–60 !
This electric field intensity EðtÞ will cause the nuclei to perform a simple harmonic oscillation. The nucleon 1’s simple harmonic oscillation can be considered as an oscillating electric dipole. Suppose the nucleon 1 has a charge of q1, its angular frequency of the simple harmonic oscillation is x, and its amplitude is l1. In the near-zone fields of the oscillating nucleon 1, the ! electric field intensity components in spherical coordinate are Er ðtÞ and ! Eh ðtÞ, themagnetic field intensity component in spherical coordi! nate is H/ ðtÞ (Cheng, 1989b), !
Er ðtÞ ¼ !
Eh ðtÞ ¼
! q1 l1 cos h cos xt r 2pe0 r 3
ð4Þ
! q1 l1 sin h cos xt h 4pe0 r 3
ð5Þ
!
!
P 2 ¼ q2 l2 cos xt r ¼
! q22 1 q l1 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Ax2 þ 1 3 cos xt r 2 m2 2 p e r 0 ðx20 x2 Þ þ x2 c2 !
The electric field intensity EðtÞ has nothing to do with the distance of r, as it will not give the oscillating nucleon 2 the force ! which moves along the direction of r . ! The near-zone electric field intensity Er ðtÞ of the oscillating nucleon 1 will give the oscillating nucleon 2 a force F which moves ! along the direction of r . When two oscillating nuclei are going in the same direction ! along r , as shown in Fig. 1. P1 and P2 are in the same direction ! along r , !
! xq1 l1 sin h p ! H/ ðtÞ ¼ cos xt þ / 2 4pr 2
ð6Þ
where r is the distance between the point of observation and the centre of the oscillating nucleon 1, r l1, r k, and where k is the wavelength of the incident light. Suppose the oscillating nucleon 2 lies at the point of observation, which is in the near-zone fields of the oscillating nucleon 1. r is then the distance between the two oscillating nuclei when they are colliding. ! ! When the electric field intensity Er ðtÞ is in the direction along r , h = 0, therefore ! q l1 Er ðtÞ ¼ 1 3 cos xt r 2pe0 r !
and r ¼ r
2
ð16Þ which shows that F is the attraction force. The Coulomb repulsion between two oscillating nuclei is Ff
Ff ¼
q1 q2
!
! q2 q1 l1 cos xt r 2pe0 r3
ð10Þ
and where x0 is the natural frequency of the nucleon 2, and c is the coefficient of damping (Panofsky and Phillips, 1962), for convenience one can set
c¼
q22 x2 6pe0 m2 c3
ð11Þ
Because c x, one can obtain
x¼
! q2 1 q l1 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Ax2 þ 1 3 cos xt r 2 m2 2pe0 r 2 2 2 2 ðx0 x Þ þ x c !
¼ l2 cos xt r l2 ¼
q2 1 q l qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Ax2 þ 1 1 3 2 m2 2pe0 r ðx20 x2 Þ þ x2 c2
ð17Þ
4pe0 R2f
To fuse together, it needs
The interaction energy between the oscillating nucleon 1 and the oscillating nucleon 2 is W (Jackson, 1998)
The electric field intensity EðtÞ and Er ðtÞ will also cause the nucleon 2 to perform a simple harmonic oscillation, so the nucleon 2 can be considered as a bound oscillator of mass m2 and charge q2, its angular frequency of the simple harmonic oscillation is x, and its amplitude is l2. ! The nucleon 2 will oscillate in the direction of r , according to the equation of motion !
!
ð8Þ ð9Þ
€x þcx_ þ x20 x ¼ q2 Ax2 cos xt r þ
ð15Þ
@ , @r
1 3Aq22 q1 l1 x2 cos2 xt 3q22 q21 l1 cos2 xt F ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ 2 4m2 pe0 r4 8m2 p2 e20 r7 ðx20 x2 Þ þ x2 c2
!
!
!
Rf 6 1015 m
!
H/ ðtÞ ¼ 0
!
ð7Þ
!
Eh ðtÞ ¼ 0
!
F ¼ q2 l2 cos xtð r r Er ðtÞ ¼ P2 r Er ðtÞ !
ð14Þ
ð18Þ
!
W ¼ P2 Er ðtÞ 2
1 Aq22 q1 l1 x2 cos2 xt q2 q2 l cos2 xt ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ 2 1 12 2 6 3 2 2m2 pe0 r 4m2 p e0 r ðx20 x2 Þ þ x2 c2
!
ð19Þ The Coulomb barrier between two oscillating nuclei is Wf
Wf ¼
q1 q2 4pe0 Rf
ð20Þ
When W is greater than the Coulomb barrier Wf between two oscillating nuclei, the two oscillating nuclei will fuse together. Suppose an electron lies at the point of observation, which is in the near-zone fields of the oscillating nucleon 1. For the electron of mass me and negative charge of qe, the Eq. (13) will be changed into
le ¼
qe 1 q l1 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Ax2 þ 1 3 2 me 2pe0 r ðx20 x2 Þ þ x2 c2
ð21Þ
and where le is !the amplitude of the oscillating electron. By defining P e as the oscillating electron’s dipole moment vector ! with magnitude qele cos xt and direction along r such as
ð12Þ ð13Þ
The nucleon 2’s simple harmonic oscillation can be considered also as an oscillating electric dipole. ! By defining P 2 as the nucleon 2’s dipole moment vector with ! magnitude q2l2cos xt and direction along r such that
!
Fig. 1. P1 and P2 are in the same direction along r .
59
B. Gong / Annals of Nuclear Energy 62 (2013) 57–60 !
x ¼ 5:14 1014 Hz; W p ¼ 2:13 eV
!
Pe ¼ qe le cos xt r
! q2 1 q l1 ¼ e qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Ax2 þ 1 3 cos xt r 2 me 2 p e r 0 ðx20 x2 Þ þ x2 c2
ð22Þ
ð7Þ !
The near-zone electric field intensity Er ðtÞ of the oscillating nucleon 1 will give the oscillating electron a force Fe which moves ! along the direction of r , !
!
!
ð23Þ
Because qe is a negative charge, the oscillating electron and the ! oscillating nucleon 1 are going in opposite directions along r , as ! shown in Fig. 2. P1 and Pe are in the opposite directions along r .
3Aq2e q1 l1
2 3q2e q21 l1
x2 cos2 xt þ 4me pe0 r4 8me
cos2 xt p2 e20 r7
!
ð24Þ It shows that Fe is a repulsive force. The interaction energy between the oscillating electron and the oscillating nucleon 1 is We !
1 Aq2e q1 l1 x2 cos2 xt q2 q2 l cos2 xt qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ e 1 12 2 6 3 2 2me pe0 r 4me p e0 r ðx20 x2 Þ þ x2 c2
!
W e ¼ Pe Er ðtÞ
ð25Þ 2
1 Aq2e q1 l1 x2 cos2 xt q2 q2 l cos2 xt W e ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ e 1 12 2 6 3 2 2me pe0 r 4me p e0 r ðx20 x2 Þ þ x2 c2
!
ð26Þ From the Eqs. (2), (19), and (26), one can know that W and We increase with the decrease of r and the increase of A and x, and A increases with the increase of the charge Q and the amplitude a. Resonance will occur when x = x0. One can determine the natural frequency x0 of the nucleon 2; A can be controlled, q1 and q2 are known, so x and r can be known, and r relates to the kinetic energy of the nuclei, therefore one can determine x and the kinetic energy of the nuclei. 3. An example 6
2
Suppose the nucleon 1 is the Li and the nucleon 2 is H. In the photoelectric effect, where the electron has a negative charge, and the nucleon has a positive charge, there is an attraction force between the electron and the nucleon, but when the light’s frequency is high enough it will eject the electron. Apparently, when the light is directed at the electrons and the nuclei, it is the repulsion force between the oscillating electrons and the oscillating nuclei, and so the repulsion force will be greater than the attraction force, so the electrons will be ejected from a metal plate. According to the Eq. (26), the interaction energy between the oscillating electron and the oscillating 6Li is We, ejecting an electron from Lithium which requires a minimal threshold frequency x and minimal threshold energy of Wp,
!
P 2:13 eV ð28Þ 10
F e ¼ qe le cos xtð r r Er ðtÞÞ ¼ Pe r Er ðtÞ
1 F e ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 ðx0 x2 Þ þ x2 c2
However, when W e P W p , the electrons will be ejected from Lithium metal plate. According to the Eq. (26), 2
! q l1 Er ðtÞ ¼ 1 3 cos xt r 2pe0 r !
!
ð27Þ
When the electron is inside the atom, r 10
16
m, l1 10
m,
x = 5.14 1014 Hz, me = 0.91 1030 kg. This means that the electron is a bound electron, x0 – 0, in spite of x x0 or x = x0, 2
1 q2 q2 l cos2 xt qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi e 1 1 2 2 6 2:13 eV 2 4m p e0 r e ðx20 x2 Þ þ x2 c2
ð29Þ
Therefore
1 Aq2 q l1 x2 cos2 xt W e qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi e 1 P 2:13 eV 2 2me pe0 r 3 ðx20 x2 Þ þ x2 c2
ð30Þ
1 Aq2 q l1 cos2 xt 2:13eV r 3 me qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi e 1 P 2 2 p e x2 0 ðx20 x2 Þ þ x2 c2
ð31Þ
1 Aq2 q l1 cos2 xt qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi e 1 2 2pe0 ðx20 x2 Þ þ x2 c2 P
2:13eV 1030 m 0:91 1030 kg
ð32Þ
2
ð5:14 1014 Þ
For 6Li and 2H, similarly, one has 2
1 q22 q21 l1 cos2 xt qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 4m2 p2 e20 r6 ðx20 x2 Þ þ x2 c2
! 2:13 eV
ð33Þ
and where m2 is the mass of the 2H, m2 = 3.34 1027 kg. To let 2H and 6Li fuse together, it needs
1 Aq2 q l1 x2 cos2 xt q q W qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 1 P 1 2 3 2 2m p e r 4 pe0 Rf 2 0 ðx20 x2 Þ þ x2 c2
ð34Þ
where Rf 6 1015 m. Because
1 Aq2 q l1 cos2 xt qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi e 1 2 2pe0 ðx20 x2 Þ þ x2 c2 P
2:13eV 1030 m 0:91 1030 kg ð5:14 1014 Þ
2
ð35Þ
To let 2H and 6Li fuse together, it needs
x P 5:14 1017 Hz
ð36Þ
r 6 1012 m
ð37Þ
but where r = 1012 m,
q2 q1 4pe0 1012
!
Fig. 2. P1 and P2 are in the opposite direction along r .
¼ 6:9 1016 J
ð38Þ
where the Coulomb barrier between the 2H and the 6Li is 6.9 1016 J (4.3 keV). Accordingly, the kinetic energy w of the 2H and the 6Li needs to be
60
B. Gong / Annals of Nuclear Energy 62 (2013) 57–60
w P 3:45 1016 J ð2:15 keVÞ
ð39Þ 17
The light whose the frequency x P 5:14 10 Hz will direct at the 2H and the 6Li, when the kinetic energy of the 2H and the 6Li is w P 2:15 keV, and their distance is r 6 1012 m, the 2H and the 6Li will fuse together, 6
Li þ 2 H ! 24 He þ 22:4 MeV
ð40Þ
From the Eqs. (2), (19), and (26), one can know that W and We will increase with the increase of A and x, as A increases with the increase of the charge Q and the amplitude a. Therefore, if Q increases by a factor of 1000 and a increases 100 times, when the light whose the frequency is x P 5:14 1016 Hz, the kinetic energy of the 2H and the 6Li is w P 0:215 keV, their distance is r 6 1011 m, the 2H and the 6Li will fuse together. 4. Discussion The present model is applicable for heavier projectile-target combinations, because the Eq. (19) is applicable whether there is a light nucleus or a heavy nucleus, as it is the attraction force between two oscillating nuclei which are travelling in the same direction along the line. From the Eqs. (29), (33), and (35), one can see that for it to apply to a projectile-target combination that it is necessary to know the heavy metals’ threshold energy, but as the author is short of data about the thresholds energy of heavy metals, he cannot provide the range of projectile-target combination here. From the Eqs. (2), (19), and (26), one can see that W and We will increase with the increase of A and x, and A will increase with the increase of Q and a. Therefore, with the photoelectric effect, the electrons may be emitted, even where the incident light has a frequency of less than the threshold frequency of the metal. In fact, the two-photon photoelectric emission has shown that the electrons may be emitted, even though the incident photons have energies of less than the threshold energy of the sodium metal (Teich and Wolga, 1968). The increase of A and x can decrease the nucleus’ kinetic energy which the fusion needs.
If the nucleus were considered to be an oscillating electric dipole, it could also explain the characteristics of the nuclear force. For example, it is just the repulsive force between the protons, or the attraction force between the protons which perform the simple harmonic oscillation. The attraction force is inversely proportional to the 4th power of the distance, whereas the repulsive force is inversely proportional to the 2nd power of the distance, so the attraction force will be greater than the repulsive force. The force between the electric dipoles is a non-central force, as is the nuclear force. 5. Conclusion The attraction force between two oscillating nuclei occurs when they are travelling in the same direction along the same line. The repulsive force takes place when the oscillating nucleon and the oscillating electron are travelling in opposite directions along the same line. To have controlled fusion, there needs to be control over the light’s electric vector and the frequency of the light. The kinetic energy and the density of the nucleus are also need to be controlled. Acknowledgements The author thanks the Editor and the Reviewer for their valuable comments and suggestions. The author thanks Dr. Dorothy Middleton and Ms. Huilan Shu and Dr. Joe Ho for their help. References Cheng, D.K., 1989a. Field and Wave Electromagnetics, second ed. Addison-Wesley, p. 605. Cheng, D.K., 1989b. Field and Wave Electromagnetics, second ed. Addison-Wesley, p. 604. Jackson, J.D., 1998. Classical Electrodynamics. John Wiley & Sons Ltd., Third Edition, p. 150. Panofsky, W.K.H., Phillips, M., 1962. Classical Electricity and Magnetism, second ed. Addison-Wesley Publishing Company, Inc., p. 401. Teich, M.C., Wolga, G.J., 1968. Two-quantum volume photoelectric effect in sodium. Phys. Rev. 171, 809–814.
Contents lists available at SciVerse ScienceDirect
Annals of Nuclear Energy journal homepage: www.elsevier.com/locate/anucene
The light controlled fusion BingXin Gong ⇑ 3-708 HuaDu XinCun, No. 57, JianSheBei Road, XinHua HuaDu, Guangzhou, Guangdong 510800, China
a r t i c l e
i n f o
Article history: Received 28 March 2013 Received in revised form 4 June 2013 Accepted 6 June 2013
Keywords: Controlled fusion Nuclei Light Oscillating electric dipoles Kinetic energy Density
a b s t r a c t This is a new technique for controlled fusion. When two nuclei are colliding with each other, light, whose the frequency is higher than the minimal threshold frequency of lithium, will be aimed directly at the two nuclei, the two nuclei will perform the simple harmonic oscillation, the charged particle’s simple harmonic oscillation can be considered as an oscillating electric dipole, and the two oscillating nuclei will radiate the electromagnetic wave. Either of the two oscillating electric dipoles will attract each other, or they will repulse each other. There will be an attraction force between the two oscillating nuclei. When the attraction force is greater than the Coulomb repulsion between the two nuclei, the two nuclei will fuse together. Where the kinetic energy and the density of the two nuclei can be controlled, the electric vector and the frequency of the light can be controlled also and, therefore, the fusion can be controlled. Ó 2013 Elsevier Ltd. All rights reserved.
1. Introduction If the light is directed at the nuclei like 2H and 6Li, the 2H and the 6Li will perform a simple harmonic oscillation and will then radiate the electromagnetic wave. The charged particle’s simple harmonic oscillation can be considered as an oscillating electric dipole. As the 2H and the 6Li have a positive charge, when the 2H and the 6Li are travelling in the same direction along the line, there will be an attraction force between the 2H and the 6Li. The attraction force is inversely proportional to the 4th power of the distance between the two nuclei that perform a simple harmonic oscillation; the repulsion force is inversely proportional to the 2nd power of this distance, the attraction force will be greater than that of the repulsion force. When the attraction force is greater than that of the Coulomb repulsion between 2H and 6Li, the 2H and the 6Li will fuse together. An example shows that if the kinetic energy of the 2H and the 6 Li are greater than 2.15 keV, and the distance between them is shorter than 1012 m, and the frequency of the light is x P 5:14 1017 Hz, the 2H and the 6Li will fuse together.
2. The principle of light controlled fusion Light controlled fusion is based on the following principle:
When a light directs at a charged particle, the charged particle will perform a simple harmonic oscillation and will radiate an electromagnetic wave. The charged particle’s simple harmonic oscillation can be considered as an oscillating electric dipole. When two oscillating electric dipoles are in the same direction along the line, there will be an attraction force between the two oscillating electric dipoles. Suppose the distance between two oscillating nuclei is r, r is also the distance between two oscillating nuclei when they are colliding. Suppose the light which is directed at the nuclei is produced by the oscillating electric dipole. In the far-zone fields of the oscillating electric dipole which has ! charge Q and amplitude a, the electric field intensity EðtÞ is proportional to the 2nd power of the angular frequency x (Cheng, 1989a), !
EðtÞ ¼
Qa x2 cos xt 4pe0 c2 R
ð1Þ
where e0 is the dielectric constant, c is the speed of light, and R is the distance between the point of observation and the centre of the oscillating electric dipole. Let
A¼
Qa 4pe0 c2 R
ð2Þ
The Eq. (1) can then be changed into !
⇑ Tel.: +86 13480212541. E-mail address: [email protected] 0306-4549/$ - see front matter Ó 2013 Elsevier Ltd. All rights reserved. http://dx.doi.org/10.1016/j.anucene.2013.06.007
EðtÞ ¼ Ax2 cos xt
ð3Þ
58
B. Gong / Annals of Nuclear Energy 62 (2013) 57–60 !
This electric field intensity EðtÞ will cause the nuclei to perform a simple harmonic oscillation. The nucleon 1’s simple harmonic oscillation can be considered as an oscillating electric dipole. Suppose the nucleon 1 has a charge of q1, its angular frequency of the simple harmonic oscillation is x, and its amplitude is l1. In the near-zone fields of the oscillating nucleon 1, the ! electric field intensity components in spherical coordinate are Er ðtÞ and ! Eh ðtÞ, themagnetic field intensity component in spherical coordi! nate is H/ ðtÞ (Cheng, 1989b), !
Er ðtÞ ¼ !
Eh ðtÞ ¼
! q1 l1 cos h cos xt r 2pe0 r 3
ð4Þ
! q1 l1 sin h cos xt h 4pe0 r 3
ð5Þ
!
!
P 2 ¼ q2 l2 cos xt r ¼
! q22 1 q l1 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Ax2 þ 1 3 cos xt r 2 m2 2 p e r 0 ðx20 x2 Þ þ x2 c2 !
The electric field intensity EðtÞ has nothing to do with the distance of r, as it will not give the oscillating nucleon 2 the force ! which moves along the direction of r . ! The near-zone electric field intensity Er ðtÞ of the oscillating nucleon 1 will give the oscillating nucleon 2 a force F which moves ! along the direction of r . When two oscillating nuclei are going in the same direction ! along r , as shown in Fig. 1. P1 and P2 are in the same direction ! along r , !
! xq1 l1 sin h p ! H/ ðtÞ ¼ cos xt þ / 2 4pr 2
ð6Þ
where r is the distance between the point of observation and the centre of the oscillating nucleon 1, r l1, r k, and where k is the wavelength of the incident light. Suppose the oscillating nucleon 2 lies at the point of observation, which is in the near-zone fields of the oscillating nucleon 1. r is then the distance between the two oscillating nuclei when they are colliding. ! ! When the electric field intensity Er ðtÞ is in the direction along r , h = 0, therefore ! q l1 Er ðtÞ ¼ 1 3 cos xt r 2pe0 r !
and r ¼ r
2
ð16Þ which shows that F is the attraction force. The Coulomb repulsion between two oscillating nuclei is Ff
Ff ¼
q1 q2
!
! q2 q1 l1 cos xt r 2pe0 r3
ð10Þ
and where x0 is the natural frequency of the nucleon 2, and c is the coefficient of damping (Panofsky and Phillips, 1962), for convenience one can set
c¼
q22 x2 6pe0 m2 c3
ð11Þ
Because c x, one can obtain
x¼
! q2 1 q l1 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Ax2 þ 1 3 cos xt r 2 m2 2pe0 r 2 2 2 2 ðx0 x Þ þ x c !
¼ l2 cos xt r l2 ¼
q2 1 q l qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Ax2 þ 1 1 3 2 m2 2pe0 r ðx20 x2 Þ þ x2 c2
ð17Þ
4pe0 R2f
To fuse together, it needs
The interaction energy between the oscillating nucleon 1 and the oscillating nucleon 2 is W (Jackson, 1998)
The electric field intensity EðtÞ and Er ðtÞ will also cause the nucleon 2 to perform a simple harmonic oscillation, so the nucleon 2 can be considered as a bound oscillator of mass m2 and charge q2, its angular frequency of the simple harmonic oscillation is x, and its amplitude is l2. ! The nucleon 2 will oscillate in the direction of r , according to the equation of motion !
!
ð8Þ ð9Þ
€x þcx_ þ x20 x ¼ q2 Ax2 cos xt r þ
ð15Þ
@ , @r
1 3Aq22 q1 l1 x2 cos2 xt 3q22 q21 l1 cos2 xt F ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ 2 4m2 pe0 r4 8m2 p2 e20 r7 ðx20 x2 Þ þ x2 c2
!
!
!
Rf 6 1015 m
!
H/ ðtÞ ¼ 0
!
ð7Þ
!
Eh ðtÞ ¼ 0
!
F ¼ q2 l2 cos xtð r r Er ðtÞ ¼ P2 r Er ðtÞ !
ð14Þ
ð18Þ
!
W ¼ P2 Er ðtÞ 2
1 Aq22 q1 l1 x2 cos2 xt q2 q2 l cos2 xt ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ 2 1 12 2 6 3 2 2m2 pe0 r 4m2 p e0 r ðx20 x2 Þ þ x2 c2
!
ð19Þ The Coulomb barrier between two oscillating nuclei is Wf
Wf ¼
q1 q2 4pe0 Rf
ð20Þ
When W is greater than the Coulomb barrier Wf between two oscillating nuclei, the two oscillating nuclei will fuse together. Suppose an electron lies at the point of observation, which is in the near-zone fields of the oscillating nucleon 1. For the electron of mass me and negative charge of qe, the Eq. (13) will be changed into
le ¼
qe 1 q l1 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Ax2 þ 1 3 2 me 2pe0 r ðx20 x2 Þ þ x2 c2
ð21Þ
and where le is !the amplitude of the oscillating electron. By defining P e as the oscillating electron’s dipole moment vector ! with magnitude qele cos xt and direction along r such as
ð12Þ ð13Þ
The nucleon 2’s simple harmonic oscillation can be considered also as an oscillating electric dipole. ! By defining P 2 as the nucleon 2’s dipole moment vector with ! magnitude q2l2cos xt and direction along r such that
!
Fig. 1. P1 and P2 are in the same direction along r .
59
B. Gong / Annals of Nuclear Energy 62 (2013) 57–60 !
x ¼ 5:14 1014 Hz; W p ¼ 2:13 eV
!
Pe ¼ qe le cos xt r
! q2 1 q l1 ¼ e qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Ax2 þ 1 3 cos xt r 2 me 2 p e r 0 ðx20 x2 Þ þ x2 c2
ð22Þ
ð7Þ !
The near-zone electric field intensity Er ðtÞ of the oscillating nucleon 1 will give the oscillating electron a force Fe which moves ! along the direction of r , !
!
!
ð23Þ
Because qe is a negative charge, the oscillating electron and the ! oscillating nucleon 1 are going in opposite directions along r , as ! shown in Fig. 2. P1 and Pe are in the opposite directions along r .
3Aq2e q1 l1
2 3q2e q21 l1
x2 cos2 xt þ 4me pe0 r4 8me
cos2 xt p2 e20 r7
!
ð24Þ It shows that Fe is a repulsive force. The interaction energy between the oscillating electron and the oscillating nucleon 1 is We !
1 Aq2e q1 l1 x2 cos2 xt q2 q2 l cos2 xt qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ e 1 12 2 6 3 2 2me pe0 r 4me p e0 r ðx20 x2 Þ þ x2 c2
!
W e ¼ Pe Er ðtÞ
ð25Þ 2
1 Aq2e q1 l1 x2 cos2 xt q2 q2 l cos2 xt W e ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ e 1 12 2 6 3 2 2me pe0 r 4me p e0 r ðx20 x2 Þ þ x2 c2
!
ð26Þ From the Eqs. (2), (19), and (26), one can know that W and We increase with the decrease of r and the increase of A and x, and A increases with the increase of the charge Q and the amplitude a. Resonance will occur when x = x0. One can determine the natural frequency x0 of the nucleon 2; A can be controlled, q1 and q2 are known, so x and r can be known, and r relates to the kinetic energy of the nuclei, therefore one can determine x and the kinetic energy of the nuclei. 3. An example 6
2
Suppose the nucleon 1 is the Li and the nucleon 2 is H. In the photoelectric effect, where the electron has a negative charge, and the nucleon has a positive charge, there is an attraction force between the electron and the nucleon, but when the light’s frequency is high enough it will eject the electron. Apparently, when the light is directed at the electrons and the nuclei, it is the repulsion force between the oscillating electrons and the oscillating nuclei, and so the repulsion force will be greater than the attraction force, so the electrons will be ejected from a metal plate. According to the Eq. (26), the interaction energy between the oscillating electron and the oscillating 6Li is We, ejecting an electron from Lithium which requires a minimal threshold frequency x and minimal threshold energy of Wp,
!
P 2:13 eV ð28Þ 10
F e ¼ qe le cos xtð r r Er ðtÞÞ ¼ Pe r Er ðtÞ
1 F e ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 ðx0 x2 Þ þ x2 c2
However, when W e P W p , the electrons will be ejected from Lithium metal plate. According to the Eq. (26), 2
! q l1 Er ðtÞ ¼ 1 3 cos xt r 2pe0 r !
!
ð27Þ
When the electron is inside the atom, r 10
16
m, l1 10
m,
x = 5.14 1014 Hz, me = 0.91 1030 kg. This means that the electron is a bound electron, x0 – 0, in spite of x x0 or x = x0, 2
1 q2 q2 l cos2 xt qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi e 1 1 2 2 6 2:13 eV 2 4m p e0 r e ðx20 x2 Þ þ x2 c2
ð29Þ
Therefore
1 Aq2 q l1 x2 cos2 xt W e qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi e 1 P 2:13 eV 2 2me pe0 r 3 ðx20 x2 Þ þ x2 c2
ð30Þ
1 Aq2 q l1 cos2 xt 2:13eV r 3 me qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi e 1 P 2 2 p e x2 0 ðx20 x2 Þ þ x2 c2
ð31Þ
1 Aq2 q l1 cos2 xt qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi e 1 2 2pe0 ðx20 x2 Þ þ x2 c2 P
2:13eV 1030 m 0:91 1030 kg
ð32Þ
2
ð5:14 1014 Þ
For 6Li and 2H, similarly, one has 2
1 q22 q21 l1 cos2 xt qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 4m2 p2 e20 r6 ðx20 x2 Þ þ x2 c2
! 2:13 eV
ð33Þ
and where m2 is the mass of the 2H, m2 = 3.34 1027 kg. To let 2H and 6Li fuse together, it needs
1 Aq2 q l1 x2 cos2 xt q q W qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 1 P 1 2 3 2 2m p e r 4 pe0 Rf 2 0 ðx20 x2 Þ þ x2 c2
ð34Þ
where Rf 6 1015 m. Because
1 Aq2 q l1 cos2 xt qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi e 1 2 2pe0 ðx20 x2 Þ þ x2 c2 P
2:13eV 1030 m 0:91 1030 kg ð5:14 1014 Þ
2
ð35Þ
To let 2H and 6Li fuse together, it needs
x P 5:14 1017 Hz
ð36Þ
r 6 1012 m
ð37Þ
but where r = 1012 m,
q2 q1 4pe0 1012
!
Fig. 2. P1 and P2 are in the opposite direction along r .
¼ 6:9 1016 J
ð38Þ
where the Coulomb barrier between the 2H and the 6Li is 6.9 1016 J (4.3 keV). Accordingly, the kinetic energy w of the 2H and the 6Li needs to be
60
B. Gong / Annals of Nuclear Energy 62 (2013) 57–60
w P 3:45 1016 J ð2:15 keVÞ
ð39Þ 17
The light whose the frequency x P 5:14 10 Hz will direct at the 2H and the 6Li, when the kinetic energy of the 2H and the 6Li is w P 2:15 keV, and their distance is r 6 1012 m, the 2H and the 6Li will fuse together, 6
Li þ 2 H ! 24 He þ 22:4 MeV
ð40Þ
From the Eqs. (2), (19), and (26), one can know that W and We will increase with the increase of A and x, as A increases with the increase of the charge Q and the amplitude a. Therefore, if Q increases by a factor of 1000 and a increases 100 times, when the light whose the frequency is x P 5:14 1016 Hz, the kinetic energy of the 2H and the 6Li is w P 0:215 keV, their distance is r 6 1011 m, the 2H and the 6Li will fuse together. 4. Discussion The present model is applicable for heavier projectile-target combinations, because the Eq. (19) is applicable whether there is a light nucleus or a heavy nucleus, as it is the attraction force between two oscillating nuclei which are travelling in the same direction along the line. From the Eqs. (29), (33), and (35), one can see that for it to apply to a projectile-target combination that it is necessary to know the heavy metals’ threshold energy, but as the author is short of data about the thresholds energy of heavy metals, he cannot provide the range of projectile-target combination here. From the Eqs. (2), (19), and (26), one can see that W and We will increase with the increase of A and x, and A will increase with the increase of Q and a. Therefore, with the photoelectric effect, the electrons may be emitted, even where the incident light has a frequency of less than the threshold frequency of the metal. In fact, the two-photon photoelectric emission has shown that the electrons may be emitted, even though the incident photons have energies of less than the threshold energy of the sodium metal (Teich and Wolga, 1968). The increase of A and x can decrease the nucleus’ kinetic energy which the fusion needs.
If the nucleus were considered to be an oscillating electric dipole, it could also explain the characteristics of the nuclear force. For example, it is just the repulsive force between the protons, or the attraction force between the protons which perform the simple harmonic oscillation. The attraction force is inversely proportional to the 4th power of the distance, whereas the repulsive force is inversely proportional to the 2nd power of the distance, so the attraction force will be greater than the repulsive force. The force between the electric dipoles is a non-central force, as is the nuclear force. 5. Conclusion The attraction force between two oscillating nuclei occurs when they are travelling in the same direction along the same line. The repulsive force takes place when the oscillating nucleon and the oscillating electron are travelling in opposite directions along the same line. To have controlled fusion, there needs to be control over the light’s electric vector and the frequency of the light. The kinetic energy and the density of the nucleus are also need to be controlled. Acknowledgements The author thanks the Editor and the Reviewer for their valuable comments and suggestions. The author thanks Dr. Dorothy Middleton and Ms. Huilan Shu and Dr. Joe Ho for their help. References Cheng, D.K., 1989a. Field and Wave Electromagnetics, second ed. Addison-Wesley, p. 605. Cheng, D.K., 1989b. Field and Wave Electromagnetics, second ed. Addison-Wesley, p. 604. Jackson, J.D., 1998. Classical Electrodynamics. John Wiley & Sons Ltd., Third Edition, p. 150. Panofsky, W.K.H., Phillips, M., 1962. Classical Electricity and Magnetism, second ed. Addison-Wesley Publishing Company, Inc., p. 401. Teich, M.C., Wolga, G.J., 1968. Two-quantum volume photoelectric effect in sodium. Phys. Rev. 171, 809–814.