[tex]V=\frac 1 {4\pi\epsilon_0}\int \frac {\rho(\vec r\;')}{\eta} \;d\;\tau'\;\hbox{ where }\; \vec{\eta} = \vec r - \vec r\;'[/tex](adsbygoogle = window.adsbygoogle || []).push({});

[itex]\vec r\;[/itex] is the position vector of the field point and [itex]\;\vec r\;'\;[/itex] is the position vector of the source point.

Using multiple expansion, the quadrapole term of potential

[tex]V_{QUAD} = \frac 1 {4\pi\epsilon_0} \frac 1 {2r^3}\int(r\;')^n(3\;cos^2\theta'-1)\rho(\vec r\;')d\tau'[/tex]

The book go on to derive [itex]V_{QUAD}[/itex] into coordinate free equation and I am lost. Please explain to me how the steps work:

The book claimed

[tex]V_{QUAD}= \frac 1 {4\pi\epsilon_0} \frac 1 {2\; r^3} \sum ^{3}_{i,j=1}\left [ \hat r_i \hat r_j \int[3r'_i r'_j -(r')^2\delta_{ij}]\rho(\vec r\;')d\;\tau'\right ]\;=\;\frac 1 {4\pi\epsilon_0} \frac 1 {2\; r^3} \left [ 3 \sum ^{3}_{i=1} \hat r_i r'_i \sum^3_{j=1} \sum^3_{j=1} \;=\; (r')^2\sum_{ij}\hat r_i\hat r_j \delta_{ij}\right]\rho(\vec r') \;d \;\tau'[/tex]

[tex]\hbox{ Where }\; \delta _{ij} = \; \begin{array}{cc} 1 & i=j \\ 0 & i \neq j \end{array}[/tex]

[tex]\sum ^{3}_{i=1} \hat r_i r'_i =\hat r \cdot \vec r\;' = r'cos\theta' = \sum ^{3}_{j=1} \hat r_j r'_j \;\hbox{ and }\;\sum_{i,j} \hat r_i \hat r_j \delta_{ij} = \sum \hat r_j \hat r_j = \hat r \cdot \hat r =1 [/tex]

Can anyone explain:

[tex]\sum ^{3}_{i=1} \hat r_i r'_i =\hat r \cdot \vec r\;' = r'cos\theta' = \sum ^{3}_{j=1} \hat r_j r'_j[/tex]

and

[tex] \sum \hat r_j \hat r_j = \hat r \cdot \hat r =1 [/tex]

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# Help me understand the derivation of the quadrapole expansion of charge distribution.

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