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The potential due to the electric dipole moment is given

$$ \Phi_{dipole}(\vec{x}) = \frac{1}{4 \pi \epsilon_0} \frac{(\vec{x} - \vec{x'}) \cdot \iiint_V \rho(\vec{x'}) \vec{x'}\, d^3 x'}{|\vec{x} - \vec{x'}|^3} $$

$\rho(\vec{x'})$ is charge density at $x'$, and $\vec{p} \equiv \iiint_V \rho(\vec{x'}) \vec{x'}\, d^3 x'$ is the definition of the dipole moment. I want to prove its discrete form:

$$ \vec{p} = \sum_i q_i \vec{x'_i} $$

$$ \Phi_{dipole}(\vec{x}) = \frac{1}{4 \pi \epsilon_0} \frac{(\vec{x} - \vec{x'}) \cdot \sum_i q_i \vec{x'_i}}{|\vec{x} - \vec{x'}|^3} $$

I started from the diagram below.

enter image description here

However, I got a term that can only be cancelled when $q_+ = -q_-$. The definition of dipole moment is general even continuous. I don't know why I need the extra condition, $q_+ = -q_-$ to get the general dipole moment formula.

$$ - \frac{1}{2} \frac{ (\vec{x} - \vec{x'}) \cdot (q_+\vec{x_-} + q_-\vec{x_+}) }{ |\vec{x} - \vec{x'}|^3 } $$

Here is my process.

$$\begin{align*} \Phi(\vec{x}) &= \frac{1}{4 \pi \epsilon_0} \left[ \frac{q_+}{|\vec{x} - \vec{x_+}|} + \frac{q_-}{|\vec{x} - \vec{x_-}|} \right] \\ &= \frac{1}{4 \pi \epsilon_0} \left[ \frac{q_+}{|\vec{x} - \vec{x_+}|} + \frac{q_-}{|\vec{x} - \vec{x'} + \vec{d}|} \right] \\ &= \frac{1}{4 \pi \epsilon_0} \left[ \frac{q_+}{|\vec{x} - \vec{x_+}|} + \frac{q_-}{ ( |\vec{x} - \vec{x'}|^2 + |\vec{d}|^2 + 2 |\vec{x} - \vec{x'}| |\vec{d}| \cos\theta )^{1/2} } \right] \\ &= \frac{1}{4 \pi \epsilon_0} \left[ \frac{q_+}{|\vec{x} - \vec{x_+}|} + \frac{q_-}{ |\vec{x} - \vec{x'}| ( 1 + \frac{ |\vec{d}|^2 }{ |\vec{x} - \vec{x'}|^2 } + 2 \frac{ |\vec{d}| }{ |\vec{x} - \vec{x'}| } \cos\theta )^{1/2} } \right] \\ &\approx \frac{1}{4 \pi \epsilon_0} \left[ \frac{q_+}{|\vec{x} - \vec{x_+}|} + \frac{q_-}{ |\vec{x} - \vec{x'}| ( 1 + 2 \frac{ |\vec{d}| }{ |\vec{x} - \vec{x'}| } \cos\theta )^{1/2} } \right] \quad (|\vec{d}| \ll |\vec{x} - \vec{x'}|) \\ &\approx \frac{1}{4 \pi \epsilon_0} \left[ \frac{q_+}{|\vec{x} - \vec{x_+}|} + \frac{q_-}{ |\vec{x} - \vec{x'}| ( 1 + \frac{1}{2} \times2 \frac{ |\vec{d}| }{ |\vec{x} - \vec{x'}| } \cos\theta ) } \right] \\ &\quad ( (1 + k)^n \approx 1 + nk \text{, Taylor expansion} ) \\ &= \frac{1}{4 \pi \epsilon_0} \left[ \frac{q_+}{|\vec{x} - \vec{x_+}|} + \frac{q_-}{ |\vec{x} - \vec{x'}| + |\vec{d}| \cos\theta } \right] \\ &\approx \frac{1}{4 \pi \epsilon_0} \left[ \frac{q_+}{ |\vec{x} - \vec{x'}| - |\vec{d}| \cos\theta } + \frac{q_-}{ |\vec{x} - \vec{x'}| + |\vec{d}| \cos\theta } \right] \\ &\quad ( \text{apply the same steps on the } q_+ \text{ term} )\\ &= \frac{1}{4 \pi \epsilon_0} \left[ \frac{ q_+ |\vec{x} - \vec{x'}| + q_- |\vec{x} - \vec{x'}| + q_+ |\vec{d}| \cos\theta - q_- |\vec{d}| \cos\theta }{ |\vec{x} - \vec{x'}|^2 - |\vec{d}|^2 \cos^2\theta } \right] \\ &\approx \frac{1}{4 \pi \epsilon_0} \left[ \frac{ q_+ |\vec{x} - \vec{x'}| + q_- |\vec{x} - \vec{x'}| + q_+ |\vec{d}| \cos\theta - q_- |\vec{d}| \cos\theta }{ |\vec{x} - \vec{x'}|^2 } \right] \quad (|\vec{d}| \ll |\vec{x} - \vec{x'}|) \\ &= \frac{1}{4 \pi \epsilon_0} \left[ \frac{q_+}{|\vec{x} - \vec{x'}|} + \frac{q_-}{|\vec{x} - \vec{x'}|} + (q_+ - q_-) \frac{ (\vec{x} - \vec{x'}) \cdot \vec{d} }{ |\vec{x} - \vec{x'}|^3 } \right] \\ &= \frac{1}{4 \pi \epsilon_0} \left[ \frac{q_+}{|\vec{x} - \vec{x'}|} + \frac{q_-}{|\vec{x} - \vec{x'}|} + (q_+ - q_-) \frac{1}{2} \frac{ (\vec{x} - \vec{x'}) \cdot (\vec{x_+} - \vec{x_-}) }{ |\vec{x} - \vec{x'}|^3 } \right] \\ &= \frac{1}{4 \pi \epsilon_0} \left[ \frac{q_+}{|\vec{x} - \vec{x'}|} + \frac{q_-}{|\vec{x} - \vec{x'}|} + \frac{1}{2} \frac{ (\vec{x} - \vec{x'}) \cdot (q_+\vec{x_+} + q_-\vec{x_-}) }{ |\vec{x} - \vec{x'}|^3 } - \frac{1}{2} \frac{ (\vec{x} - \vec{x'}) \cdot (q_+\vec{x_-} + q_-\vec{x_+}) }{ |\vec{x} - \vec{x'}|^3 } \right] \\ &= \frac{1}{4 \pi \epsilon_0} \frac{ (\vec{x} - \vec{x'}) \cdot [q \vec{x_+} + (-q) \vec{x_-}] }{ |\vec{x} - \vec{x'}|^3 } \quad (\text{let } q_+ = -q_- = q) \\ \end{align*}$$

The first and the second term in the second last statement are the contributions of the monopole moment. That is reasonable. But the terms, which are listed below, in the second last statement can only be merged together to get the dipole moment fomula if $q_+ = -q_-$.

$$ \frac{1}{2} \frac{ (\vec{x} - \vec{x'}) \cdot (q_+\vec{x_+} + q_-\vec{x_-}) }{ |\vec{x} - \vec{x'}|^3 } - \frac{1}{2} \frac{ (\vec{x} - \vec{x'}) \cdot (q_+\vec{x_-} + q_-\vec{x_+}) }{ |\vec{x} - \vec{x'}|^3 } $$

Did I do something wrong?

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  • $\begingroup$ @lpz I know these two terms are contributed by the monopoles. $$ \frac{1}{4 \pi \epsilon_0} \left[ \frac{q_+}{|\vec{x} - \vec{x'}|} + \frac{q_-}{|\vec{x} - \vec{x'}|}\right] $$ What I cannot understand are these two terms. $$ \frac{1}{4 \pi \epsilon_0} \left[ \frac{1}{2} \frac{ (\vec{x} - \vec{x'}) \cdot (q_+\vec{x_+} + q_-\vec{x_-}) }{ |\vec{x} - \vec{x'}|^3 } - \frac{1}{2} \frac{ (\vec{x} - \vec{x'}) \cdot (q_+\vec{x_-} + q_-\vec{x_+}) }{ |\vec{x} - \vec{x'}|^3 }\right] $$ $\endgroup$
    – IvanaGyro
    Commented Nov 26, 2022 at 13:36
  • $\begingroup$ @Ghoster Thank you! I have updated my question. $\endgroup$
    – IvanaGyro
    Commented Nov 26, 2022 at 13:38
  • $\begingroup$ Sorry, didn’t read well your question $\endgroup$
    – LPZ
    Commented Nov 26, 2022 at 16:28

1 Answer 1

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This is normal since you are doing a multipole expansion about: $$ x’=\frac{1}{2}x_++\frac{1}{2}x_- $$ The monopole is: $$ Q=q_++q_- $$ and the dipole moment is: $$ P=q_+(x_+-x’)+q_-(x_- -x’) \\ =(q_+-q_-)d $$ with your notation: $$ d=\frac{1}{2}x_+-\frac{1}{2}x_- $$ This is the term you obtained after using the large distance limit ($|d|\ll|x-x’|$).

To get the dipole moment: $$ P_0=q_+x_++q_-x_- $$ you’ll need to perform the multipole expansion about the origin, not about $x’$.

In general, the dipole moment $P=\sum q_i(x_i-r)$ depends on the choice of the reference point $r$ iff the total charge $Q=\sum q_i$ is not zero. Indeed changing $r$ to $r’$ changes $P$ to: $$ P’=P+Q(r-r’) $$ and this is why by choosing $r$ to be the position of the center of charge, $P’$ is the effective dipole moment of an offset monopole at $r$.

This explains why you only get your expected dipole moment $P_0$ when $q_++q_-=0$ since it is the only case where the origin of the dipole is irrelevant.

Hope this helps.

Answer to comment

This is because you are expanding in distances with respect to $x'$. Equivalently, you get about the origin: $$ \phi(x) = \frac{Q}{4\pi\epsilon_0|x|}+\frac{P_0\cdot x}{4\pi\epsilon_0|x|^3}+O\left(\frac{1}{|x|^3}\right) $$ Or about $x'$: $$ \phi(x) = \frac{Q}{4\pi\epsilon_0|x-x'|}+\frac{P\cdot (x-x')}{4\pi\epsilon_0|x-x'|^3}+O\left(\frac{1}{|x-x'|^3}\right) $$ In either case, and in higher multipolar terms, you need to stay consistent with your reference pole.

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  • $\begingroup$ You are right. What I got is the dipole moment about $x'$. But I am confused. The origin in my processes always keep at the same position. Why I got the dipole moment about $x'$ instead of the origin? $\endgroup$
    – IvanaGyro
    Commented Nov 27, 2022 at 18:41

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