TF DerivationOfCoulombForce

Poisson's Equation

[math]\nabla^2 \phi(\vec{\xi}) = - \frac{\rho}{\epsilon_0} =- \frac{e}{\epsilon_0} \delta(\vec{\xi})[/math]

Fourier Transform of Poisson's Equation

[math]\frac{1}{(2 \pi)^{3/2}} \int e^{-i \vec{k} \cdot \vec{\xi}} \nabla^2 \phi(\vec{\xi})dV = - \frac{1}{(2 \pi)^{3/2}} \frac{e}{\epsilon_0} \int e^{-i \vec{k} \cdot \vec{\xi}}\delta(\vec{\xi}) dV [/math]

[math]\frac{1}{(2 \pi)^{3/2}} \int e^{-i \vec{k} \cdot \vec{\xi}} \vec{\nabla} \cdot (\vec{\nabla} \phi(\vec{\xi}))dV = - \frac{e}{(2 \pi)^{3/2}\epsilon_0} (1)[/math]

Product rule for dervatives

[math]\frac{1}{(2 \pi)^{3/2}} \int \left \{ \vec{\nabla} \cdot ( e^{-i \vec{k} \cdot \vec{\xi}} \vec{\nabla} \phi ) - (\vec{\nabla} e^{-i \vec{k} \cdot \vec{\xi}}) \cdot (\vec{\nabla} \phi) \right \} dV = - \frac{e}{(2 \pi)^{3/2}\epsilon_0} (1)[/math]

Gauss' Theorem:

[math]\int \vec{\nabla} \cdot ( e^{-i \vec{k} \cdot \vec{\xi}} \vec{\nabla} \phi ) dV = \oint_S e^{-i \vec{k}\cdot \vec{\xi}} \vec{\nabla}\cdot d\vec{A}[/math]

Definition of derivative:

[math](\vec{\nabla} e^{-i \vec{k} \cdot \vec{\xi}}) \cdot (\vec{\nabla} \phi ) = \vec{\nabla} \cdot (\phi \vec{\nabla} e^{-i \vec{k}}) - \phi {\nabla}^2 e^{-i \vec{k} \cdot \vec{\xi}}[/math]

Substituting

[math]\frac{1}{(2 \pi)^{3/2} } \left \{ \int e^{-i \vec{k} \cdot \vec{\xi}} \vec{\nabla} \phi \cdot d\vec{A} - \int \vec{\nabla} \cdot (\phi \vec{\nabla} e^{-i \vec{k} \cdot \vec{\xi}} ) dV + \int \phi {\nabla}^2 e^{-i \vec{k} \cdot \vec{\xi}} dV \right \} = \frac{-e}{(2 \pi)^{3/2} \epsilon_0}[/math]

Gauss' Low:

[math]\int \vec{\nabla}\cdot (\phi \vec{\nabla} e^{-i \vec{k} \cdot \vec{\xi}} ) dV = \int \phi \vec{\nabla} e^{-i k \xi } \cdot d\vec{A}[/math]

[math]\frac{1}{(2 \pi)^{3/2} } \left \{\int \left \{ e^{-i \vec{k} \cdot \vec{\xi}} \vec{\nabla} \phi - \phi \vec{\nabla} e^{-i \vec{k} \cdot \vec{\xi}} \right \} \cdot d\vec{A} + \int \phi {\nabla}^2 e^{-i \vec{k} \cdot \vec{\xi}} dV \right \} = \frac{-e}{(2 \pi)^{3/2} \epsilon_0}[/math]

[math]\frac{1}{(2 \pi)^{3/2} } \int \phi (-ik) (-ik) e^{-i \vec{k} \cdot \vec{\xi}} dV = \frac{-e}{(2 \pi)^{3/2} \epsilon_0}[/math]

[math]-k^2 \frac{1}{(2 \pi)^{3/2} } \int \phi(\xi) e^{-i \vec{k} \cdot \vec{\xi}} dV_{xi} = \frac{-e}{(2 \pi)^{3/2} \epsilon_0}[/math]

[math]-k^2 \phi(k) = \frac{-e}{(2 \pi)^{3/2} \epsilon_0}[/math]

1.) Coulomb [math]\phi(k) = \frac{e}{(2 \pi)^{3/2} \epsilon_0} \frac{1}{k^2}[/math] = potential in "k"(momentum) space

To find the potential in "coordinate" [math](\xi)[/math] space just inverse transform

[math]\phi (\xi) = \frac{1}{(2 \pi)^{3/2} } \int e^{+ i \vec{k} \cdot \vec{\xi}} \phi (k) dV_k[/math]

[math]= \frac{1}{(2 \pi)^{3/2} } \int e^{i \vec{k} \cdot \vec{\xi}} \frac{e}{2 \pi)^{3/2} \epsilon_0} \frac{1}{k^2} dV_k[/math]

[math]= \frac{e}{(2 \pi)^{3} \epsilon_0} \int \frac{e^{i \vec{k} \cdot \vec{\xi}}}{k^2} dV_k[/math]

[math]dV_k=k^2 sin{\theta}_k d{\theta}_k d{\phi}_k dk[/math]

[math]=\frac{e}{(2 \pi)^{3} \epsilon_0} {{\int}_0}^{2\pi} d{\phi}_k {{\int}_0}^{\pi} d{\theta}_k {{\int}_0}^{\infty} dk \times k^2 sin{\theta}_k e^{i \vec{k} \cdot \vec{\xi}}[/math]

[math]=\frac{e}{(2\pi)^2 \epsilon_0} {{\int}_0}^{\pi} {{\int}_0}^{\infty} sin{\theta}_k e^{ik \xi cos{\theta}_k} k^2 dk[/math]

[math]u=cos\theta[/math]

[math]du=sin\theta d\theta[/math]

[math]\phi(\xi) = \frac{e}{4 {\pi}^2 \epsilon_0} {{\int}_0}^infty {{\int}_{-1}}^1 \frac{e^{ik\xi u}}{k^2} du k^2 dk[/math]

[math]=\frac{e}{4 {\pi}^2 \epsilon_0} {{\int}_0}^infty \frac{e^{ik \xi} - e^{-ik\xi}}{ik\xi} dk[/math]

[math]=\frac{e}{4 {\pi}^2 \epsilon_0} \frac{1}{i\xi} (i\pi) = \frac{e}{4 {\pi}^2 \epsilon_0} \frac{1}{\xi}[/math]

[math]=\frac{e}{4 {\pi}^2 \epsilon_0} \frac{1}{|\vec{r} - \vec{r}^'|} =[/math] Coulomb potential

2) Nuclear potential

Consider the force field generated by a point source (nucleon) at location [math]\vec{r}[/math] 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:

[math]E^2=(mc^2)^2 + (cp)^2[/math]

In terms of Hamiltonian

[math]-\hbar \frac{d^2}{dt^2} \phi(\vec{r}) = \left \{ (mc^2)^2 + (\frac{c\hbar \vec{\nabla}}{i})^2 \right \} \phi (r)[/math]

In a static case

[math]((mc^2)^2 - c^2 {\hbar}^2 {\nabla}^2)\phi(r)=0[/math]

[math][ {\nabla}^2 - (\frac{mc}{\hbar})^2)\phi(r)=0[/math]