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

  1. Jun 15, 2011 #1
    [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]

    [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]
     
    Last edited: Jun 15, 2011
  2. jcsd
  3. Jun 15, 2011 #2

    G01

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    Re: Help me understand the derivation of the quadrapole expansion of charge distribut

    The first is just the definition of the dot product. i.e.:

    [tex]\vec{A}\cdot\vec{B}=A_1B_1+A_2B_2+A_3B_3=\sum_{i=1}^3A_iB_i=ABcos\theta[/tex]

    The components are just numbered instead of lettered. A_1=A_x, and so on.

    The second is just the dot product of a unit vector with itself. By definition, that is 1.
     
  4. Jun 15, 2011 #3
    Re: Help me understand the derivation of the quadrapole expansion of charge distribut

    Thanks for your time.

    Is that just mean both A and B are 3 space vector? Just that simple?!!! And they have to use the [itex]\sum_1^3[/itex] to confuse me?!!!
     
  5. Jun 15, 2011 #4

    G01

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    Re: Help me understand the derivation of the quadrapole expansion of charge distribut

    Yes, that's it. The point is that this notation is "coordinate free." It doesn't matter how you define your three orthogonal axes, the expression for the quadrupole potential in coordinate free notation will always look like this.

    Using summation notation in this case, is just shorthand.

    The Kroneker delta, [itex]\delta_{ij}[/itex] just means that when i and j are not equal, that term does not contribute, since [itex]\delta_{ij}=0[/itex] for [itex]i\not=j[/itex] and [itex]\delta_{ij}=1[/itex] for [itex]i=j[/itex].
     
  6. Jun 16, 2011 #5
    Re: Help me understand the derivation of the quadrapole expansion of charge distribut

    Thanks

    I understand

    [tex]\sum^3_{i=1}\vec r_i \vec r'_i = \vec r \cdot \vec r\;'[/tex]


    What is [tex]\sum ^3_{i,j=1} \hat r_i \hat r_j [/tex]?
     
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