- 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.
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
−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
Lets μ=mcℏ=(140MeVc2)c6.6×10−16eV⋅s(106eVMeV)
- =2.1×1023(3×108m)(10−15mfm)=0.7fm(200MeVfm)=140MeV
- ℏc=(6.6×10−16eV⋅s)(3×108ms)(1015fmm)(1MeV106eV)=198MeV⋅fm≈200MeV⋅fm
- μ=mc2ℏc=mc2200MeV⋅fm ⟹ if mc2≈200MeV 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+iμk−iμ=iπe−μξ
- \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}}^'|}
