Difference between revisions of "Electric QuadrupoleMoment Forest NuclPhys I"

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because
 
because
  
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<math>\vec{E} = \int \ro (\vec{r^'}) \frac{(\vec{r} - \vec{r^'})}{|r - r^'|^3} d^3r^' = - \vec{\nabla} \int \frac{\ro (r^')}{|\vec{r} - \vec{r^'}|}</math>
 +
 +
\vec{E} = -\vec{\nabla} \Psi (r)
 +
 +
<math>\Psi (r) =  \int \frac{\ro (r^')}{|\vec{r} - \vec{r^'}|}</math>
  
  
 
[[Forest_NucPhys_I]]
 
[[Forest_NucPhys_I]]

Revision as of 04:42, 7 April 2009

Electric Quadrupole Moment of a Nucleus

Pages 104-111

As in the dipole calculation we assume that the object is in a state such that its maximum total angular momentum is along the z-axis.

or [math]\Psi_{jm} = \Psi_{jj}[/math]


then

[math]Q = \lt \Psi_{jj} |3z^2 - r^2|\Psi_{jj}\gt [/math]

From definition of quadrupole moment for a single charged object/particle.

The origin of this comes from electron-statics.

You expand the electric potential in terms of spherical harmonics.

[math]\Phi(\vec{r}) = {\Sigma_{l=0}}^{\infty} {\Sigma_{m=-l}}^{l} \frac{4\pi}{2l + 1} q_{lm} \frac{Y_{lm}(\theta \psi)}{r^{l+1}}[/math]

because

[math]\vec{E} = \int \ro (\vec{r^'}) \frac{(\vec{r} - \vec{r^'})}{|r - r^'|^3} d^3r^' = - \vec{\nabla} \int \frac{\ro (r^')}{|\vec{r} - \vec{r^'}|}[/math]

\vec{E} = -\vec{\nabla} \Psi (r)

[math]\Psi (r) = \int \frac{\ro (r^')}{|\vec{r} - \vec{r^'}|}[/math]


Forest_NucPhys_I