- Poisson's Equation
- ∇2ϕ(→ξ)=−ρϵ0=−eϵ0δ(→ξ)
Fourier Transform of Poisson's Equation
- 1(2π)3/2∫e−i→k⋅→ξ∇2ϕ(→ξ)dV=−1(2π)3/2eϵ0∫e−i→k⋅→ξδ(→ξ)dV
- 1(2π)3/2∫e−i→k⋅→ξ→∇⋅(→∇ϕ(→ξ))dV=−e(2π)3/2ϵ0(1)
Product rule for dervatives
- 1(2π)3/2∫{→∇⋅(e−i→k⋅→ξ→∇ϕ)−(→∇e−i→k⋅→ξ)⋅(→∇ϕ)}dV=−e(2π)3/2ϵ0(1)
Gauss' Theorem:
- ∫→∇⋅(e−i→k⋅→ξ→∇ϕ)dV=∮Se−i→k⋅→ξ→∇⋅d→A
Definition of derivative:
- (→∇e−i→k⋅→ξ)⋅(→∇ϕ)=→∇⋅(ϕ→∇e−i→k)−ϕ∇2e−i→k⋅→ξ
Substituting
1(2π)3/2{∫e−i→k⋅→ξ→∇ϕ⋅d→A−∫→∇⋅(ϕ→∇e−i→k⋅→ξ)dV+∫ϕ∇2e−i→k⋅→ξdV}=−e(2π)3/2ϵ0
Gauss' Low:
∫→∇⋅(ϕ→∇e−i→k⋅→ξ)dV=∫ϕ→∇e−ikξ⋅d→A
1(2π)3/2{∫{e−i→k⋅→ξ→∇ϕ−ϕ→∇e−i→k⋅→ξ}⋅d→A+∫ϕ∇2e−i→k⋅→ξdV}=−e(2π)3/2ϵ0
1(2π)3/2∫ϕ(−ik)(−ik)e−i→k⋅→ξdV=−e(2π)3/2ϵ0
−k21(2π)3/2∫ϕ(ξ)e−i→k⋅→ξdVxi=−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)3/2∫e+i→k⋅→ξϕ(k)dVk
- =1(2π)3/2∫ei→k⋅→ξe2π)3/2ϵ01k2dVk
- =e(2π)3ϵ0∫ei→k⋅→ξk2dVk
- dVk=k2sinθkdθkdϕkdk
- =e(2π)3ϵ0∫02πdϕk∫0πdθk∫0∞dk×k2sinθkei→k⋅→ξ
- =e(2π)2ϵ0∫0π∫0∞sinθkeikξcosθkk2dk
u=cosθ
du=sinθdθ
- ϕ(ξ)=e4π2ϵ0∫0infty∫−11eikξuk2duk2dk
- =e4π2ϵ0∫0inftyeikξ−e−ikξikξdk
- =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.
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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
−ℏd2dt2ϕ(→r)={(mc2)2+(cℏ→∇i)2}ϕ(r)
In a static case
((mc2)2−c2ℏ2∇2)ϕ(r)=0
[∇2−(mcℏ)2)ϕ(r)=0
