Difference between revisions of "TF DerivationOfCoulombForce"

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:<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 = - \frac{e}{(2 \pi)^{3/2}\epsilon_0} (1)</math>
 
:<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 = - \frac{e}{(2 \pi)^{3/2}\epsilon_0} (1)</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>

Revision as of 03:12, 20 February 2009

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 = - \frac{e}{(2 \pi)^{3/2}\epsilon_0} (1)[/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]
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