Difference between revisions of "TF DerivationOfCoulombForce"

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<math>\frac{\partial^2 U(r)}{\partial r^2} + \frac{2m}{\hbar^2} E U(r) = 0</math>
 
<math>\frac{\partial^2 U(r)}{\partial r^2} + \frac{2m}{\hbar^2} E U(r) = 0</math>
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[[Image:imageTF_1.jpg]]
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;<math>\psi_{I}</math> :
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<math>
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\frac{\partial^2 U_I (r)}{\partial r^2} + {K_1}^2 U(r) = 0</math> : <math> {K_1}^2 = \frac{2m}{hbar^2}(E + V)</math>
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<math>
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\Longrightarrow</math>  <math>U_I (r) = A sin(k_1 r) + Bcos(k_1 r)</math> : spring simple harmonic motion
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Bounding condition:
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::::<math>\psi_I (r=0)=0</math> <math>\Longrightarrow</math> <math>B=0</math>

Revision as of 03:20, 24 February 2009

Poisson's Equation
2ϕ(ξ)=ρϵ0=eϵ0δ(ξ)

Fourier Transform of Poisson's Equation

1(2π)3/2eikξ2ϕ(ξ)dV=1(2π)3/2eϵ0eikξδ(ξ)dV
1(2π)3/2eikξ(ϕ(ξ))dV=e(2π)3/2ϵ0(1)

Product rule for dervatives

1(2π)3/2{(eikξϕ)(eikξ)(ϕ)}dV=e(2π)3/2ϵ0(1)


Gauss' Theorem:

(eikξϕ)dV=SeikξdA


Definition of derivative:

(eikξ)(ϕ)=(ϕeik)ϕ2eikξ


Substituting

1(2π)3/2{SeikξϕdA(ϕeikξ)dV+ϕ2eikξdV}=e(2π)3/2ϵ0


Gauss' Low:

(ϕeikξ)dV=SϕeikξdA


1(2π)3/2{S{eikξϕϕeikξ}dA+ϕ2eikξdV}=e(2π)3/2ϵ0


1(2π)3/2ϕ(ik)(ik)eikξdV=e(2π)3/2ϵ0


k21(2π)3/2ϕ(ξ)eikξdVxi=e(2π)3/2ϵ0

k2ϕ(k)=e(2π)3/2ϵ0

1.) Coulomb ϕ(k)=e(2π)3/2ϵ01k2 = potential in "k"(momentum) space

To find the potential in "coordinate" (ξ) space just inverse transform

ϕ(ξ)=1(2π)3/2e+ikξϕ(k)dVk
=1(2π)3/2eikξe2π)3/2ϵ01k2dVk
=e(2π)3ϵ0eikξk2dVk


dVk=k2sinθkdθkdϕkdk


=e(2π)3ϵ002πdϕk0πdθk0dk×k2sinθkeikξ
=e(2π)2ϵ00π0sinθkeikξcosθkk2dk

u=cosθ

du=sinθdθ


ϕ(ξ)=e4π2ϵ00infty11eikξuk2duk2dk
=e4π2ϵ00inftyeikξeikξikξdk
=e4π2ϵ01iξ(iπ)=e4π2ϵ01ξ
=\frac{e}{4 {\pi}^2 \epsilon_0} \frac{1}{|\vec{r} - \vec{r}^'|} = Coulomb potential
2) Nuclear potential

Consider the force field generated by a point source (nucleon) at location r from the origin of a coordinate system.

300px

Assume a particle of mass m is e charged to generate the field (In Coulomb force particle was m=o photon).

Definition of relativistic Energy:

E2=(mc2)2+(cp)2

In terms of Hamiltonian

d2dt2ϕ(r)=[(mc2)2+(ci)2]ϕ(r)


In a static case

[(mc2)2c222]ϕ(r)=0

[2(mc)2]ϕ(r)=0

Lets μ=mc=(140MeVc2)c6.6×1016eVs(106eVMeV)

=2.1×1023(3×108m)(1015mfm)=0.7fm(200MeVfm)=140MeV


c=(6.6×1016eVs)(3×108ms)(1015fmm)(1MeV106eV)=198MeVfm200MeVfm
μ=mc2c=mc2200MeVfm if mc2200MeV then interaction length 1fm.


With the source term

[{\nabla}^2 - {\mu}^2]\phi(\xi) = -{{\xi}_0}^' \delta(\xi)

As seen before for Coulomb force


\frac{1}{(2\pi)^{3/2}} \int e^{-ik \xi} [{\nabla}^2 - {\mu}^2] \phi (\xi) dV = -\frac{{{\xi}_0}^'}{(2\pi)^{3/2}} \int e^{-i \vec{k} \cdot \vec{\xi}} \delta (\xi) dV
[-k^2 -{\mu}^2]\phi(\vec{k}) =  -\frac{{{\xi}_0}^'}{(2\pi)^{3/2}}
\Longrightarrow \phi(\vec{k}) = \frac{{{\xi}_0}^'}{(2\pi)^{3/2}} \frac{1}{k^2 + {\mu}^2}
\phi(\vec{\xi}) = \frac{1}{(2 \pi)^{3/2}} \int \frac{e^{+i \vec{k} \cdot \vec{\xi}}}{(2 \pi)^{3/2}} \frac{{{\xi}_0}^'}{k^2 + {\mu}^2} dV_k : inverse fourier transform
= \frac{{{\xi}_0}^'}{(2\pi)^3} \int \frac{e^{+i \vec{k} \cdot \vec{\xi}}}{k^2 + {\mu}^2} k^2 sin\theta d\theta d\phi dk
\phi(\xi) = \frac{{{\xi}_0}^'}{(2\pi)^2} \int \frac{e^{i\vec{k} \cdot \vec{\xi}}}{(k^2 + {\mu}^2)} k^2 d[cos\theta] (2\pi) dk
= \frac{{{\xi}_0}^'}{(2\pi)^2} {{\int}_0}^{\infty} \frac{k^2}{k^2 + {\mu}^2} {{\int}_0}^{\pi} e^{ik\xi cos\theta} d[cos\theta] dk
= \frac{{{\xi}_0}^'}{(2\pi)^2}  {{\int}_0}^{\infty} \frac{k^2}{k^2 + {\mu}^2} dk \frac{e^{ik\xi \mu}}{ik\xi} {|_{-1}}^1
=  \frac{{{\xi}_0}^'}{(2\pi)^2} \frac{1}{i\xi} {{\int}_0}^{\infty} \frac{k}{k^2 + {\mu}^2} (e^{ik\xi} - e^{-ik\xi})
=  \frac{{{\xi}_0}^'}{(2\pi)^2}\frac{1}{i\xi} {{\int}_{-\infty}}^{\infty} \frac{e^{ik\xi}kdk}{k^2 + {\mu}^2}
eikξkdkk+iμkiμ=iπeμξ
\phi(\xi) = \frac{{{\xi}_0}^'}{(2\pi)^2} \frac{e^{-\mu \xi}}{\xi}

\phi(\vec{r}) = \frac{{{\xi}_0}^'}{(2\pi)^2} \frac{1}{|\vec{r} - {\vec{r}}^'|}
{\phi}_{EM}(\vec{r}) = \frac{e}{4\pi \epsilon_0} \frac{1}{|\vec{r} - {\vec{r}}^'|}

Coupling constants are:

??????????????????????????????????
??????????????????????????????????


Summary

There are now at least two forces which act between Nucleons, the Coulomb force and the Nucleon force. We can write the force in terms of a potential

V_{EM} = \frac{e}{4\pi \epsilon_0} \frac{1}{|\vec{r} - {\vec{r}}^'|}


V_{Nuc} =  \frac{{{\xi}_0}^'}{(2\pi)^2} \frac{e^{-\mu |\vec{r} - {\vec{r}}^'|}}{|\vec{r} - {\vec{r}}^'|}
V_{tot} = \frac{1}{4\pi} \left \{ \frac{Q_1 Q_2}{\epsilon_0 e } - \frac{{\xi}_0}{\pi} e^{-\mu |\vec{r} - {\vec{r}}^'|} \right \} \frac{1}{|\vec{r} - {\vec{r}}^'|}
μ=mπc=1R(1.5fm)1
C.) Deuteron
(H12), (D2), (d) = a proton-neutron bound state

General properties: L=0 - orbital angular momentum

Jπ=1+ - (Nuclear spin)

r2=2.095fm - Mean radius.

B.E.=2.2246MeV - Binding energy.

Non-relativistic Schrodiger solution
B.E.=2.2246MeV<<1.8GeVm(H2)c2

weakly bound system

Instead of Dirac equation try 3-D Square Well Schrod. Eq. approximation for Deuteron wavefunction.

350px

V(r)=V0 when rR=2.2

V(r)=0 r>R

Assume ψ=Y00V(r)r : No angular dependence, only radial dependence.

Schrod. Equation

22m2ψ+V(r)ψ=Eψ

2=1r2r(r2r)+1r2sinθθ(sinθθ)+1r2sin2θ2ϕ2


22mY001r2rr2(U(r)/r)r+1r2sinθU(r)r(θsinθY00θ)

+U(r)r3sin2θ2ϕ2Y00+V(r)Y00U(r)r=EU(r)r


1r2rr2rU(r)r=1r2rr2[1rU(r)rU(r)r2]

=1r2[U(r)r+r2U(r)r22rU(r)r2r2r2U(r)rr2U(r)2r3]

=1r2U(r)r2+1r2U(r)r1r2U(r)r2U(r)r2+2U(r)r2

=1r2U(r)r2


U(r)r3sinθ(θsinθθ)Y00=U(r)r3sinθl(l+1)Y00

=U(r)r3sinθ[θsinθθ]14π=0


U(r)r3sinθ2ϕ2Y00=0 : Y00=14π

Schrod. Equation becomes

22mY00r2U(r)r2+V(r)Y00U(r)r=EU(r)Y00r


for rR : V(r)=V0


2U(r)r2+2m2(E+V)U(r)=0

for r>R : V(r)=0


2U(r)r2+2m2EU(r)=0


File:ImageTF 1.jpg

ψI

2UI(r)r2+K12U(r)=0 : K12=2mhbar2(E+V)

UI(r)=Asin(k1r)+Bcos(k1r) : spring simple harmonic motion


Bounding condition:

ψI(r=0)=0 B=0