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Potential of a Quadrupole is given as,

$$V(r) = \frac{1}{4\pi e_0}\left(\frac{1}{R}q+\frac{1}{R^2}\sum_{i=x,y,z}\hat{R_i}\vec{p_i} + \frac{1}{R^3}\sum_{i,j=x,y,z}\hat{R_i}\hat{R_j}Q_{ij}\right)$$

where $$Q_{ij} = \frac{1}{2}\sum^N_{n = 1}q_n(3r_{n,i}r_{n,j}-(r_n)^2\delta_{ij})$$

I am kind of confused with the last term

$$\frac{1}{R^3}\sum_{i,j=x,y,z}\hat{R_i}\hat{R_j}Q_{ij}$$

So if we have 4 particles, How can I expand this term ?

$$\frac{1}{R^5}[x^2Q_{xx}+y^2Q_{yy}+z^2Q_{zz}+2xyQ_{xy}+2xzQ_{xz}+2yzQ_{yz}]$$

Then $$Q_{xx} = \frac{1}{2}[q_1(3r_{1,x}r_{1,x}-r_1^2)+q_2(3r_{2,x}r_{2,x}-r_2^2)+q_3(3r_{3,x}r_{3,x}-r_3^2)+q_4(3r_{4,x}r_{4,x}-r_4^2)]$$ $$Q_{yy} = \frac{1}{2}[q_1(3r_{1,y}r_{1,y}-r_1^2)+q_2(3r_{2,y}r_{2,y}-r_2^2)+q_3(3r_{3,y}r_{3,y}-r_3^2)+q_4(3r_{4,y}r_{4,y}-r_4^2)]$$ and so on.

Is these expansions true ?

I thought that $\hat{R_i}\hat{R_j}=\delta_{ij}$ and what is exactly $\hat{R_i}$ ? I am kind of lost.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – tpg2114 Nov 18 '19 at 17:23
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You've mis-transcribed the source you stated, arXiv:1112.3376, for this formula. The correct expression would be $$ V(\boldsymbol R) = \frac{1}{4\pi e_0}\left(\frac{1}{R}q+\frac{1}{R^2}\sum_{i=x,y,z}\hat{R_i}\boldsymbol{p}_i + \frac{1}{R^3}\sum_{i,j=x,y,z}\hat{R_i}\hat{R_j}Q_{ij}\right), $$ with the main difference being the argument in the function call on the left-hand side, the vector-valued $\boldsymbol R$, which is the point at which the potential is being evaluated.

(The formula you've given also takes the unorthodox choice to notate the cartesian components of the dipole moment vector $\boldsymbol p$ as $\boldsymbol{p}_i$ instead of $p_i$, which is misleading notation and should not be used. The error, however, comes from the original.)

In this formula, as usual in electromagnetism, the vector $\hat{\boldsymbol R} = \boldsymbol R/R$ represents the normalized vector $\boldsymbol R$, whose components are $\hat R_i$. If you want to examine the components directly it is generally easier to multiply each term in the multipole expansion by a factor of $R^\ell/R^\ell$, so that it reads $$ V(\boldsymbol R) = \frac{1}{4\pi e_0}\left(\frac{1}{R}q+\frac{1}{R^3}\sum_{i=x,y,z}{R_i}\boldsymbol{p}_i + \frac{1}{R^5}\sum_{i,j=x,y,z}{R_i}{R_j}Q_{ij}\right), $$ without the hats, and then $R_i$ and $R_j$ are just the cartesian components of $\boldsymbol R$.

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  • $\begingroup$ Thanks for the clafirication $\endgroup$ – Reign Nov 18 '19 at 9:48
  • $\begingroup$ I am confused is it something like this $$\frac{1}{2R^5}[x^2Q_{xx}+y^2Q_{yy}+z^2Q_{zz}+2xyQ_{xy}+2xzQ_{xz}+2yzQ_{yz}]$$ or $$\frac{1}{R^5}[x^2Q_{xx}+y^2Q_{yy}+z^2Q_{zz}+xyQ_{xy}+xzQ_{xz}+yzQ_{yz}]$$ I mean theres 1/2. is it there because of the $i=j$ ? in this case the expansion should be $$\frac{1}{R^5}[x^2Q_{xx}+y^2Q_{yy}+z^2Q_{zz}+xyQ_{xy}+xzQ_{xz}+yzQ_{yz}]$$ $\endgroup$ – Reign Nov 18 '19 at 16:25

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