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Why Does the Sun Shine? - University of Richmond
Transcript of Why Does the Sun Shine? - University of Richmond
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Why Does the Sun Shine? 1
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Why Does the Sun Shine? 2
1 What is the energy production of the Sun (RSun = 1.36 kW /m2)?
2 Could the reaction H2 + 12O2 → H2O (∆H = 1.25× 106 J/mole)
power the Sun ?
3 How much hydrogen and oxygen would you need to maintain theSun’s power output?
4 How long would the Sun (MSun = 2× 1030 kg) last making H2O?
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Why Does the Sun Shine? 31 Would pp → d + β+ + νe , ∆E = 1.442 MeV work?
2 How close must protons approach each other for fusion to occur?
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Why Does the Sun Shine? 41 Would pp → d + β+ + νe , ∆E = 1.442 MeV work?
2 How close must protons approach each other for fusion to occur?
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The Coulomb Barrier 5
Potential Energy
0 10 20 30 40 50 60 70
-40
-20
0
20
40
r(fm)
V(M
eV)
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Why Does the Sun Shine? 61 Would pp → d + β+ + νe , ∆E = 1.442 MeV work?
2 How close must protons approach each other for fusion to occur?
3 Do the protons in the Sun have enough energy, on average, toovercome the Coulomb barrier? (TSun = 2× 106 K )
4 Would protons in the high-velocity tail of the Maxwellian distributionhave enough energy to overcome the barrier?
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Why Does the Sun Shine? 71 Would pp → d + β+ + νe , ∆E = 1.442 MeV work?
2 How close must protons approach each other for fusion to occur?
3 Do the protons in the Sun have enough energy, on average, toovercome the Coulomb barrier? (TSun = 2× 106 K )
4 Would protons in the high-velocity tail of the Maxwellian distributionhave enough energy to overcome the barrier?
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Why Does the Sun Shine? 81 Would pp → d + β+ + νe , ∆E = 1.442 MeV work?
2 How close must protons approach each other for fusion to occur?
3 Do the protons in the Sun have enough energy, on average, toovercome the Coulomb barrier? (TSun = 2× 106 K )
4 Would protons in the high-velocity tail of the Maxwellian distributionhave enough energy to overcome the barrier?
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The Maxwellian Velocity Distribution 9
P(v)d~v = 4π
(m
2πkBT
)3/2
v2e−mv2/2kBTdv
T=2x106 degrees
T=6x106 degrees
0 2 4 6 80
1
2
3
4
5
v(x10-5 m/s)
P(v)(x
10-
6)
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The Solar Fusion Paradox 10
1 Our Sun generates enormous power PSun =3.8× 1026 J/s.
2 The power source was utterly unknown un-til Einstein discovered his famous resultE = mc2.
3 And Rutherford discovered nuclear physics.
4 Even then, statistical mechanics told us thechances of the reaction p+p → d+β++νeoccurring were small because of the heightof the Coulomb barrier.
5 How can the two protons overcome theirmutual repulsion to fuse and release enoughenergy to power the Sun?
Quantum Tunneling!
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The Solar Fusion Paradox 11
1 Our Sun generates enormous power PSun =3.8× 1026 J/s.
2 The power source was utterly unknown un-til Einstein discovered his famous resultE = mc2.
3 And Rutherford discovered nuclear physics.
4 Even then, statistical mechanics told us thechances of the reaction p+p → d+β++νeoccurring were small because of the heightof the Coulomb barrier.
5 How can the two protons overcome theirmutual repulsion to fuse and release enoughenergy to power the Sun?
Quantum Tunneling!Jerry Gilfoyle Why Does the Sun Shine? 7 / 20
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The Coulomb Barrier 12
Average Energy
Potential Energy
0 10 20 30 40 50 60 70
-40
-20
0
20
40
r(fm)
V(M
eV)
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The Coulomb Barrier? 13
Region 1 Region 2 Region 3
Incident
Wave
Reflected
Wave
Transmitted
Wave
Average
Energy
0 2a0
V0
x
V(x)
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The Plan for Solving the Solar Fusion Paradox 14
1 Approximate the Coulomb barrier with a rectangular barrier.
2 Develop the notion of particle flux or flow.
3 Solve the Schroedinger equation for the rectangular barrier potential.
4 Determine the flux penetrating the barrier.
5 Calculate the probability of the following reaction occurring.
p + p → d + β+ + νe ∆E = 1.442 MeV
6 Compare the results of the previous calculation with the prediction ofclassical physics using the Maxwellian velocity distribution.
P(v)d~v = 4π( m
2πkT
)3/2v2e−mv2/2kTdv
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Particle Flux in a Beam 15
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The Rectangular Barrier 16
Region 1 Region 2 Region 3
Incident
Wave
Reflected
Wave
Transmitted
Wave
Average
Energy
0 2a0
V0
x
V(x)
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The Postulates 17
1 Each physical, measurable quantity, A, has a corresponding operator,A , that satisfies the eigenvalue equation A φ = aφ and measuringthat quantity yields the eigenvalues of A .
2 Measurement of the observable A leaves the system in a state that isan eigenfunction of A .
3 The state of a system is represented by a wave function Ψ which iscontinuous, differentiable and contains all the information about it.
The average value of any observable A is determined by〈A〉 =
∫all space
Ψ∗A Ψd~r .
The ‘intensity’ is proportional to|Ψ|2.
4 The time development of the wave function is determined by
i~∂Ψ(~r , t)
∂t= − ~2
2µ∇2Ψ(~r , t) + V (~r)Ψ(~r , t) µ ≡ reduced mass.
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Summary of TISE Solution 18
The Potential
Eigenfunctions e±ik1x e±ik2x e±ik3x
Wave Numbers k1 =√
2mE~2 k2 =
√2m(E−V0)
~2 k3 = k1
General Solution φ1,2,3 = coeff1 × e ik1,2x + coeff2 × e ik1,2x
Coefficients A, B C , D F , G
Boundary Conditions 1 φ1(0) = φ2(0) φ2(2a) = φ3(2a)
Boundary Conditions 2 ∂φ1(0)∂x = ∂φ2(0)
∂x∂φ2(2a)∂x = ∂φ3(2a)
∂x
1 2 3
E
0 2a0
V0
x
V
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Results So Far 19
Transfer Matrix method
ψ1 = d12p2d21p−11 ψ3
= tψ3
d12 =1
2
(1 + k2
k11− k2
k1
1− k2k1
1 + k2k1
)d21 =
1
2
(1 + k1
k21− k1
k2
1− k1k2
1 + k1k2
)
p2 =
(e−ik22a 0
0 e+ik22a
)p−11 =
(e+ik12a 0
0 e−ik12a
)
Flux
flux = |ψ|2vn where n is the region
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Quantum Tunneling 20
V0 = 10 MeV
a = 10 fm
5 10 15 20 25 300.0
0.2
0.4
0.6
0.8
1.0
Energy (MeV)
T
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Quantum Tunneling 21
V0 = 10 MeV
a = 10 fm
Red Curve - quantumcalculation
Green Curve - classicalexpectation
5 10 15 20 25 300.0
0.2
0.4
0.6
0.8
1.0
Energy (MeV)
T
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Quantum Tunneling 22
V0 = 10 MeV
a = 10 fm
Red Curve - quantum
calculation
Green Curve - classical
expectation
5 10 15 20 25 3010-8
10-6
10-4
0.01
1
Energy (MeV)
T
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The Plan for Solving the Solar Fusion Paradox 23
1 Approximate the Coulomb barrier with a rectangular barrier.
2 Develop the notion of particle flux or flow.
3 Solve the Schroedinger equation for the rectangular barrier potential.
4 Determine the flux penetrating the barrier.
5 Calculate the probability of the following reaction occurring.
p + p → d + β+ + νe ∆E = 1.442 MeV
6 Compare the results of the previous calculation with the prediction ofclassical physics using the Maxwellian velocity distribution.
P(v)d~v = 4π( m
2πkT
)3/2v2e−mv2/2kTdv
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Solving the Solar Fusion Paradox 24
1 Quantum Tunneling
1 Pick reasonable values for the barrier.
2 Get 〈KE〉 in the solar core:
〈KE〉 =3
2kBT whereT ≈ 107 k
V0=0.5 MeV
a=2.0 fm
r=2.0 fm
0 2 4 6 8 100.0
0.2
0.4
0.6
0.8
1.0
r (fm)
V(r)
Potential Energy
3 Calculate T = 1/|t11|2 with:
t11 =1
4
[(1 +
k2
k1
)e−ik22a
(1 +
k1
k2
)+
(1−
k2
k1
)e ik22a
(1−
k1
k2
)]and
kn =√
2mp(〈KE〉−Vn)
~22 Maxwellian velocity
1 Get the Coulomb barrier height and the proton velocity.
Vtop =Z1Z2e2
rso vtop =
√2Vtop
mp
2 Integrate the velocity distribution from vtop .
P =
∫ ∞vtop
4π( m
2πkT
)3/2v2e−mv2/2kTdv
3 Compare.
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