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Narrowing down the candidate of the NAE (nuclear active environment) of t

he nuclear cold fusion.

Tetsuo Sawada (Nihon U.)

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Three important properties of the nuclear CF

(1) Openning of the channel at zero incident energy.

(2) Closing of the channels and ,

which are the main channels of the low energy d+d reaction in vacuum.

(3) Drastic lowering of the repulsive Coulpmb potential between d's, which prevents for the two deuterons to come close and to fuse.

We shall see that the first property is sufficient to narrow down the candidate of the NAE of the nuclear cold fusion uniquely, and so the second and third properties will be derived from the first one.

Hedd 4

ptdd nHedd 3

1

3

energy-momentum conservation in vacuum

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12 2/)'(2/' mpmpQ

4222 cmcpE

• In the d + d → t + p reaction, energy Q is shared by the final state particles of mass m1 and m2 as

in the centr of mass system. Namely trit

on gets 25% whereas p gets 75% of Q.

• However when the mass is zero, we must use the relativistic form of energy

instead. So the energy equation of d+d → + is α γ

, whose solution is

therefore only 0.318% of Q becomes the k

inetic energy of the alpha particle.

42 2/'' MpcpQ

99682.0

)/211/()/211(1/' 44

MQMQQcp

Q = 23.9 MeV. ( Q-value)

M4 = 3732 MeV. ;(mass of alpha particle)

p'

- p'

2

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energy-momentum conservation of nuclear CF

Let us examine the conservation law of d+d → He(4) of the nuclear CF, or generally (two body) → (one body) type. It will be shown that the consevations of energy and momentum are not compatible.

proof: in the center of mass system, the momentum p' of the final state particle is zero from the momentum conservation, whereas from the energy conservation the magnitude of p' must satisfy | p'| ²=(2MQ). qed.

Therefore the reaction of the nuclear CF indicates the existence of the external potential outside of the nuclear system, which receives the momentum transfer.

p'

- pp

p'

- p'

external potential

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5

For our d + d → He(4) reaction, the momentum transfer q is as large as q=422 MeV./c , whose De Broglie wave length is λ/2π = ħ/q =0.56fm.. So we expect that the external potential cannot be soft and spread but it must be sharply localized in order to do the job to receive such large q.

More precisely, the probability of the momentum transfer q by the external potential Vext( r) is the square of the scattering amplitude f(q) given by :

)(]exp[)4/2()( 3 rVrqirdmqf ext

which is essentially the Fourier transformation of the external potential. When the variations of the exponential factor and the potential Vext coincide, f(q) become reasonablly large. Next example will help to understand the situations.

Narrowing down the candidates of the external potential 1 4

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Narrowing down the candidates of the external potential 2

When the electric charge distribution is exponential form of

]exp[)( brCr the external potential has the form:

The Fourier tranformation of the potential φ(R) is :

Since the size of the wawe packet of the electron cloud is the order of 1Å, and ħ/q is 0.56 fm. in our case, q/b ≈ 60000. Therefore | f(q)|² becomes extreamly small. So the electron cloud cannot do the job to absorb such a large momentum transfer q of the d+d → He(4) reaction.

)

21(1

1)(

bRe

RR bR

which reduces to the Coulomb potential at large R.

2222 )/1(

11)(~

bqqq

in which b and √< r ²> relates by < r ²>=12/b

².

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unique candidate of the external field 1

There are two candidates of the external field, to which the nucleus can responds. They are the electric field E and the magnetic field B, because the nucleus has two attributes, namely the (positive) electric charge Ze and the magnetic dipole moment μ. The interaction energy V of these attributes and the external fields are Ze and - φ •Bμ , respectively, where is φthe scalar potential of the electric field: E= - grad . Among them, we alφready observed that E produced by the electron cannot do the job.

On the other hand, since the magnetic Coulomb field produced by the magnetically charged particle (magnetic monopole) is B=*e/r² , the external potential of the nucleon Vm is:

.04.6)2/1(1)(

)(ˆ)/)(*2/()(22

2

araarerF

rFrremerV

with

where

ar

ptotm

, in which F(r) is the form factor of the nucleon.

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8

unique candidate of the external field 2

The Fourier transformation of the external potential Vm(r) is

qqa

qaameeiqV ptotm ˆ

)/1(

)/()/2)(2/(*)(

~222

5

However since, in our d+d → He(4) reaction q = 422MeV./c =3.02μπ and a=6.04μπ , the ratio becomes (a/q)=2.0. Therefore the Fourier transfomation of Vm(r) stays resonablly large.

Contrarily to the case of the electric field produced by the electron cloud, the momentum transfer q from the nuclear system goes smoothly when the external field is the magnetic field produced by the magnetic monopole.

In this way the first mystery of the nuclear CF is solved if the fusion reaction proceeds under the influence of the magnetic monopole. Therefore NAE is the magnetic field produced by magnetic monopole.

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Summary ( How the magnetic monopole is selected as NAE )

(1) In the nuclear CF, the channel of d + d → He(4) reaction opens. However this reaction violates the conservation law of the momentum of the nuclear system. Momentum q must be transfered to the external field. Numerically |q| is as large as 422.0 MeV./c in our case.

(2) Since the nucleus must responds to the external field, we can narrow down the candidates of the fields further. They are the electric field or the magnetic field, which respond to the electric charge Ze and the magnetic dipole moment μ of the nucleus respectively.

(3) Because the magnitude of q is so large, |q|=422MeV./c , whose deBroglie wave length is 0.56fm., the external field must be sharply localized. The electric field produced by the packet of electron , whose size is limitted by the Compton wave length of electron (386fm.) cannot do the job.

(4) There remains only the magnetic field B produced by the magnetic monopole. We shall see that all other mysteries are resolved by the external magnetic field prduced by *e.

We repeat: 8

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Charge quantization condition of Dirac and the super-strong nature of the magnetic Coulomb field

If we remember that the angular momentum density of the electro-magnetic field is r×(E×H)/c , the total angular momentum of the system, where the electric charge Q and the magnetic charge *Q coexist, stored in space is M = - (*QQ/c) e, in which e is the unit vector connecting two charges. Since in the quantum theory, a component of the angular momentum must be the integer multiple of /2 , we arrive at the "charge quantization condition" of ℏDirac : (*QQ/ c)=n/2 with n=0, ±1, ±2, . ℏ ⋯

If we substitute the charges *Q and Q by the smallest non-zero charges *e and e, then we get the relation between the fine structure constant and its magnetic counterpart : (*e²/ c)(e²/ℏ ℏc)= 1/2. Since (e²/ c)= 1/137.04 , we ℏobtain (*e²/ c)= 137.04/4, wich is super-strong. Therefore the external potential ℏVext becomes strong enough to receive the momentum transfer with high probability.

Another important role of the super-strong monopole is that the deuterons are trapped by the monopole deeply enough to be prevented to transit to the continuous t+p or He(3)+n channels. We shall see this in the next page.

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How the main channels d+d → t+p, d+d →He(3)+n are closed ------ derivation of the property (2) of the nuclear CF. -------

If we remember the binding energies of d, t and He(3) are 2.2, 8.5 and 7.7 MeV. respectively, the Q-values of d+d → t+p and d+d → He(3)+n are 4.1 and 3.3 MeV. respectively.

When the nuclear CF proceeds after the two deuterons are trapped by the magnetic monopole *e, the energies of the final states decrease as the binding energies of the *e-d namely B*e-d increase. If B*e-d exceeds Q/2, the energy of the final state becomes negative. Therefore the channel is closed.

To find the binding energy of the ground state of the *e-d system, the variational calculation is carried out with the Hamiltonian:

22000 /)ˆ)(2/(*)(2

1rrmeeAZei

mHwithVHHH ptotN

np

, where A is the vector potential and Z=1 for proton and Z=0 for nuetron. The result is B=2.3 MeV.. Therefore channels are closed.

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Energy levels of the A=4 and Z=2 channels

Since the binding energy of thedeuteron is 2.2 MeV., the threshold of of the d+d channel is E= - 4.4 MeV. and the sky blue belt of the next figure indicates the continuous spectrum of the d+d channel. Likewise, E= - 8.5 MeV. and E= -7.7 MeV. are the threshold of the t+p and He(3)+n channel respectively. Energy of the state where all the nucleons are separated infinitely is chosen as the origin of E (E=0).

In vaccum, if we start the d+d reaction with the zero kinetic energy, its energy level is E= - 4.4 MeV.. From the energy conservation, this energy level means the kinetic energy of the t+p and He(3)+n channel is 4.1 and 3.3 MeV. respectively, and so such reactions can occur. On the other hand, if the d+d reaction starts after d's are trapped by the magnetic monopole *e, and form the bound state d-*e-d, then E= -9.0 MeV.. In such a case, kinetic energy of t+p and He(3)+n becomes negative, which means such reactions cannot occur. However since He(4) is the tightly bound nucleus, kinetic energy for E= - 9.0 MeV. is 19.2 MeV.. So this channel stay open as long as it satisfies the momentum conservation.

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d + d t + p He(3) + n He(4)

-28.2

-7.7-8.5

-4.4

-10.02

-7.95

D=1 monopole

He(4) dominance

-9.0

EMeV.

-5

-10

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Penetration of the repulsive Coulomb barrier 1

The nuclear CF proceeds when the two deuterons are trapped by the same magnetic monopole *e, whose orbital sizes are several fm.. The trapping of the first deuteron goes rather smoothly. However the potential V1(x) felt by the second deuteron is the sum of the repulsive Coulomb and the attractive potential between the magnetic monopole and the magnetic moment of the nucleus when the spin orients properly. The explicit form of V1(x) is

,/)4/(/)( 221 xmxexV ptot

in which the charge quantization condition of Dirac, *ee=1/2 , is used.

Because of the attractive second term, the peak value of V1(x) reduces to 17keV. , which should be compared with the peak value 1MeV. of the pure Coulomb potential VC(x). However this lowering is not sufficient to increase the penetration factor to the reasonable value.

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Penetration of the repulsive Coulomb barrier 2

Further help to increase the penetration factor P comes, when we consider the electron wave function in the magnetic Coulomb potential. Since the binding energy is order of the electron mass me, the relativistic treatment is necessary, so we must solve the Dirac equation in the external magnetic Coulomb field produced by the magnetic monopole. However I skip this.

I shall use only the result. The repulsive Coulomb term is shieled to become

If we substitute V2(x) to the formula of the penetration factor P for E=0 :

We obtain P=6.2×10^{-9} , which should be compared with P=5×10^{ -106} of the case of VC(x) . Thus the third mystery (3) of CF is resolved.

If you are interested in this topic, see " Underlying mechanism of the nuclear CF implied by the energy-momentum consevation" (available at t-sawada@fureai.or.jp)

/)(2]2exp[ dxxVPb

aredwith

222 /)4/(]2exp[)/()( xmxmxexV ptote

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One particle catalyst and the sporadic nature of nuclear CF

We saw that the d+d → He(4) reaction occurs when two deuterons form a bound state with the same magnetic monopole *e. Since such a state is unstable and transits to more stable -particle. Because the spin of is zerα αo, it is not attracted by *e. The magnetic monopole simply emits the alpha particle, and there remains fresh monopole *e, and which starts to attracts surrounding fuel deuterons again. In this way the magnetic monopole plays the roll of the " one-particle catalyst " of the nuclear CF.

At present, since we rely on the floating magnetic monopole in nature and it is a very rare particle, we must wait for long time before the monopole *e comes into and stops in the region where the deuteron density is sufficiently high. Therefore the starting of the nuclear CF is governed by the probability. We may expect to facilitate the nuclear CF, if we can succeed to gather the floating magnetic monopole to the region where the deuteron density is high. In the next we shall consider this possibility.

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Proposal of experiments and devices for the magnetic monopole

• Devices for magnetic monopole:

(A) trapper : necessary to store the magnetic monopole.

(B) detector : super-conducting ring can do the job.

(C) collector : magnetic dipole can collect the floating magntic monopole.

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(A) Magnetic monopole *e trapped in the lattice 1

In order to learn the process of trapping of the magnetic monopole *e by a lattice, by a rare earth for example, let us compute the trapping energy E by using a simple Δmodel. We consider a lattice of atoms with large magnetic moment (e/2mκ e) whose directions are random (see upper figure). When a magnetic monopole *eD comes in, the magnetic moments align spherically (down figure).

Before *e comes in.

After *e comes in and stops.

Because of the thermal agitation, the directions of the spins are random outside of the blue sphare of radius R, and R is determined by:

TkR

eD

m

eB

e 2

1

2 2

*

where kB is the Boltzmann constant.

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19

ee

R

m

R

a

D

r

De

m

edrr

aE

32

*

0

23 2

4

T

cmDR

T

keVDE

63/2 102.97

2,

14.24)

2(

(A) Magnetic monopole *e trapped in the lattice 2

The trapping energy E is the sum of the contributions of the magnetic Δmoment mesured by the unit of Bohr magneton e/2mκ e in the magnetic Coulomb field *eD/r ² . D is 1 and 2 for Dirac and for Schwinger magnetic monopole respectively.

where a is the lattice constant.

When the lattice a is 1Å , the numerical values of E and R are:Δ

It is remarkable that E and R are proportinal to 1/√T. Since at rooΔm temperature T=300˚, E is the order of 1keV., and R is few nm..Δ

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(B) Detector of the magnetic monopole 1

*e D

Total magnetic flux Δϕ produced by the magnetic charge *eD is Δϕ= 4π*eD. Therefore when the magnetic charge *eD passes through the super-conducting ring of the torus shape as indicated by the right figure, the magnetic flux trapped by the ring must increase by Δϕ.

Since the the super-conducting current I circulating the inner surface of the ring is proportional to the total magnetic flux ϕ trapped by the ring, we can determine the change of the flux Δϕ by observing the change of the surface current ΔI.

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(B) Detector of the magnetic monopole 2

C

2 b

I

h

ΔΦ

cmAcmS

D

S

DcI /

)(328.0

2

4 20

ciHrot /4

We change the equation

to the integral form, in which the red line of the figure is the contuor C of Stokes theorem. Because of Meisner effect that the magnetic field inside of the super-conductor is zero, we get the relation between the surface current I and the magnetic field strength ϕ/S where S is the area of the inner radius of the torus. Therefore the change of the current ΔI becomes:

, where ϕ0 is the quantum unit of the magnetic flux : ϕ0=πℏc/eand numerically it is)(1007.2

)(1007.227

0

110

CGSincmG

orMKSinWb

By using the SQUID, the change of the current can be observed.

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(C) Collector of the of the magnetic monopole *e 1

The magnetic dipole or the solenoid of the electric current can serve as the collector of the magnetic monopole, because the magnetically charged particle *e with very low velocity moves on the line of force of the magnetic field.

After Gilbert, we may regared the earth as a huge magnetic dipole, then we can expect to find the magnetic monopole with better probability in the neighborhood of the magnetic north (or south) pole of earth.

In the next page, we shall consider how toslow down the speed of the monopole more to make it be trapped by the potential of the order of 1 keV. in depth.

N

S

trapper

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Trapping of the magnetic monopole *e 1

22 ||)*(3

2aDe

dt

dE

Charged particle looses its energy in the form of radiation, when it enters into the region of the potential. -dE/dt=(2/3)e²a² is Larmor's formula of the radiation loss, in which a is the acceleration. For the case of the magnetic monople, it must become:

Therefore energy loss occurs at the surface of the metals of high permeability, and so in order to slow down *e particle effectively, use of the powder of nm size is desirable. In the figure the green points are the places where the radiation is emitted.

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Energy transfer to lattice and hopping of the magnetic monopole

2124

2max222

22

.307.0)/(4

2

2log

2

11

cmgMeVcmeNKwith

I

Tcm

A

ZKz

dx

dE

eA

e

Contrarily to the stopping power of the case of the parallel beam, in our case the energy deposition is not uniform, but heat is stored in the small neighborhood of the magnetic monopole *e and it becomes the hot spot. As the temperature raises up, the trapping energy of *e by the lattice decreases as 1/√T, and the magnetic monopole must hop to cooler region. Therefore we must observe the discrete hot spots in the cathode. In fact such discrete hot spots spreading with time were observed with the infra-red camera by Boss et. al..

The alpha particle emitted from the magnetic monopole comes to stop with track length 0.1 mm. The mechanism of the stopping power is exactly the same as that of the Bethe-Bloch's formula given below. Namely the energy is transfed to electron first and then to the lattice.

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