Multipole Expansion of Potential and Dipole Moments
📂Electrodynamics Multipole Expansion of Potential and Dipole Moments Multiple Expansion When a distribution of charges is viewed from sufficiently far away, it appears almost as though it were a point charge. In other words, if the total charge of the charge distribution is Q Q Q Q Q Q 1 4 π ϵ 0 Q r \dfrac{1}{4\pi\epsilon_{0}} \dfrac{Q}{r} 4 π ϵ 0 1 r Q
But if the total charge is 0 0 0 0 0 0 Multiple expansion is the answer to how to express the potential as an approximate formula when the total charge is 0 0 0 V ( r ) V(\mathbf{r}) V ( r ) 1 r n \dfrac{1}{r^{n}} r n 1 multipole expansion .
The potential at position r \mathbf{r} r
V ( r ) = 1 4 π ϵ 0 ∫ 1 ρ ( r ′ ) d τ ′
\begin{equation}
V(\mathbf{r}) = \dfrac{1}{4\pi\epsilon_{0}} \int \dfrac{1}{\cR}\rho (\mathbf{r}^{\prime})d\tau^{\prime}
\label{1}
\end{equation}
V ( r ) = 4 π ϵ 0 1 ∫ 1 ρ ( r ′ ) d τ ′
Where = r − r ′ ( = ∣ ∣ ) \bcR = \mathbf{r} - \mathbf{r}^{\prime} (\cR = \left| \bcR \right|) = r − r ′ ( = ∣ ∣ )
By using the law of cosines to find the magnitude of \cR
2 = r 2 + ( r ′ ) 2 − 2 r r ′ cos α = r 2 [ 1 + ( r ′ r ) 2 − 2 r ′ r cos α ] = r 2 [ 1 + r ′ r ( r ′ r − 2 cos α ) ]
\begin{align*}
\cR ^2 =&\ r^2+(r^{\prime})^2-2rr^{\prime}\cos\alpha
\\ =&\ r^2\left[1+\left(\dfrac{r^{\prime}}{r}\right)^2-2\dfrac{r^{\prime}}{r}\cos\alpha\right]
\\ =&\ r^2\left[1+\dfrac{r^{\prime}}{r}\left( \dfrac{r^{\prime}}{r}-2\cos\alpha\right)\right]
\end{align*}
2 = = = r 2 + ( r ′ ) 2 − 2 r r ′ cos α r 2 [ 1 + ( r r ′ ) 2 − 2 r r ′ cos α ] r 2 [ 1 + r r ′ ( r r ′ − 2 cos α ) ]
For convenience, let’s completely substitute the second term in the angle brackets with ϵ = r ′ r ( r ′ r − 2 cos α ) \epsilon=\dfrac{r^{\prime}}{r}\left( \dfrac{r^{\prime}}{r}-2\cos\alpha \right) ϵ = r r ′ ( r r ′ − 2 cos α )
= r 1 + ϵ ⟹ 1 = 1 r ( 1 + ϵ ) − 1 / 2
\cR=r\sqrt{1+\epsilon} \implies \dfrac{1}{\cR} = \dfrac{1}{r}(1+\epsilon)^{-1/2}
= r 1 + ϵ ⟹ 1 = r 1 ( 1 + ϵ ) − 1/2
If r \mathbf{r} r r ′ r \dfrac{r^{\prime}}{r} r r ′ ϵ ≪ 1 \epsilon \ll 1 ϵ ≪ 1
Binomial Series
If ∣ x ∣ < 1 |x| < 1 ∣ x ∣ < 1
( 1 + x ) α = 1 + α x + α ( α − 1 ) 2 ! x 2 + α ( α − 1 ) ( α − 2 ) 3 ! x 3 + ⋯
(1 + x )^{\alpha} = 1 + \alpha x + \dfrac{\alpha (\alpha-1)}{2!}x^{2} + \dfrac{\alpha (\alpha-1)(\alpha-2)}{3!}x^{3} + \cdots
( 1 + x ) α = 1 + αx + 2 ! α ( α − 1 ) x 2 + 3 ! α ( α − 1 ) ( α − 2 ) x 3 + ⋯
Therefore, ( 1 + ϵ ) − 1 / 2 (1+\epsilon)^{-1/2} ( 1 + ϵ ) − 1/2
1 = 1 r ( 1 + ϵ ) − 1 / 2 = 1 r ( 1 − 1 2 ϵ + 3 8 ϵ 2 − 5 16 ϵ 3 + ⋯ )
\dfrac{1}{\cR} = \dfrac{1}{r}(1+\epsilon)^{-1/2} = \dfrac{1}{r}\left( 1- \dfrac{1}{2}\epsilon+\dfrac{3}{8}\epsilon ^2 -\dfrac{5}{16}\epsilon ^3 +\cdots \right)
1 = r 1 ( 1 + ϵ ) − 1/2 = r 1 ( 1 − 2 1 ϵ + 8 3 ϵ 2 − 16 5 ϵ 3 + ⋯ )
Substituting ϵ \epsilon ϵ
1 = 1 r [ 1 − 1 2 r ′ r ( r ′ r − 2 cos α ) + 3 8 ( r ′ r ) 2 ( r ′ r − 2 cos α ) 2 − 5 16 ( r ′ r ) 3 ( r ′ r − 2 cos α ) 3 + ⋯ ]
\dfrac{1}{\cR}=\dfrac{1}{r}\left[ 1- \dfrac{1}{2}\dfrac{r^{\prime}}{r}\left( \dfrac{r^{\prime}}{r}-2\cos\alpha \right) +\dfrac{3}{8}\left( \dfrac{r^{\prime}}{r} \right)^2 \left( \dfrac{r^{\prime}}{r}-2\cos\alpha \right) ^2 -\dfrac{5}{16}\left( \dfrac{r^{\prime}}{r}\right)^3 \left( \dfrac{r^{\prime}}{r}-2\cos\alpha \right) ^3 +\cdots \right]
1 = r 1 [ 1 − 2 1 r r ′ ( r r ′ − 2 cos α ) + 8 3 ( r r ′ ) 2 ( r r ′ − 2 cos α ) 2 − 16 5 ( r r ′ ) 3 ( r r ′ − 2 cos α ) 3 + ⋯ ]
Organizing according to each order of r ′ r \dfrac{r^{\prime}}{r} r r ′ Appendix .
1 = 1 r [ 1 + ( r ′ r ) ( cos α ) + ( r ′ r ) 2 ( 3 cos 2 α − 1 2 ) + ( r ′ r ) 3 ( 5 cos 2 α − 3 cos α 2 ) + ⋯ ]
\dfrac{1}{\cR}=\dfrac{1}{r}\left[ 1+\left(\dfrac{r^{\prime}}{r}\right)\left( \cos\alpha \right) +\left( \dfrac{r^{\prime}}{r} \right)^2 \left( \dfrac{3\cos^2\alpha -1 }{2}\right) +\left( \dfrac{r^{\prime}}{r}\right)^3 \left( \dfrac{5\cos^2\alpha-3\cos\alpha}{2} \right) +\cdots \right]
1 = r 1 [ 1 + ( r r ′ ) ( cos α ) + ( r r ′ ) 2 ( 2 3 cos 2 α − 1 ) + ( r r ′ ) 3 ( 2 5 cos 2 α − 3 cos α ) + ⋯ ]
Here, each bracket can be expressed as the series in the form of ∑ n = 0 ∞ a n ( r ′ r ) n \sum \limits_{n=0}^{\infty} a_{n}\left( \dfrac{r^{\prime}}{r}\right)^n n = 0 ∑ ∞ a n ( r r ′ ) n a n a_{n} a n
a 0 = 1 a 1 = cos α a 2 = 3 cos 2 α − 1 2 a 3 = ( 5 cos 2 α − 3 cos α 2 ) ⋮
\begin{align*}
a_{0} =&\ 1
\\ a_{1} =&\ \cos\alpha
\\ a_{2} =&\ \dfrac{3\cos^2\alpha-1}{2}
\\ a_{3} =&\ \left( \dfrac{5\cos^2\alpha-3\cos\alpha}{2} \right)
\\ \vdots &
\end{align*}
a 0 = a 1 = a 2 = a 3 = ⋮ 1 cos α 2 3 cos 2 α − 1 ( 2 5 cos 2 α − 3 cos α )
This is the same as the Legendre polynomial P n ( cos α ) P_{n}(\cos \alpha) P n ( cos α ) cos α \cos\alpha cos α
1 = 1 r ∑ n = 0 ∞ ( r ′ r ) n P n ( cos α )
\dfrac{1}{\cR}=\dfrac{1}{r}\sum\limits_{n=0}^{\infty}\left( \dfrac{r^{\prime}}{r}\right)^n P_{n}(\cos\alpha)
1 = r 1 n = 0 ∑ ∞ ( r r ′ ) n P n ( cos α )
Substituting this into the potential formula ( 1 ) \eqref{1} ( 1 ) r r r
V ( r ) = 1 4 π ϵ 0 ∑ n = 0 ∞ 1 r n + 1 ∫ ( r ′ ) n P n ( cos α ) ρ ( r ′ ) d τ ′
V(\mathbf{r})=\dfrac{1}{4\pi\epsilon_{0}}\sum \limits_{n=0}^{\infty} \dfrac{1}{r^{n+1}} \int (r^{\prime})^nP_{n}(\cos\alpha) \rho (\mathbf{r}^{\prime}) d\tau^{\prime}
V ( r ) = 4 π ϵ 0 1 n = 0 ∑ ∞ r n + 1 1 ∫ ( r ′ ) n P n ( cos α ) ρ ( r ′ ) d τ ′
Expanding this series again results in
V ( r ) = 1 4 π ϵ 0 [ 1 r ∫ r ′ cos α ρ ( r ′ ) d τ ′ + 1 r 2 ∫ r ′ cos α ρ ( r ′ ) d τ ′ + 1 r 3 ∫ ( r ′ ) 2 3 cos 2 α − 1 2 ρ ( r ′ ) d τ ′ + ⋯ ]
\begin{align*}
V(\mathbf{r}) = \dfrac{1}{4\pi\epsilon_{0}} \bigg[ &\dfrac{1}{r} \int r^{\prime}\cos\alpha \rho (\mathbf{r}^{\prime})d\tau^{\prime}
\\ &+ \dfrac{1}{r^2}\int r^{\prime}\cos\alpha \rho (\mathbf{r}^{\prime})d\tau^{\prime} + \dfrac{1}{r^3}\int(r^{\prime})^2\dfrac{3\cos^2\alpha -1 }{2}\rho (\mathbf{r}^{\prime})d\tau^{\prime} + \cdots \bigg]
\end{align*}
V ( r ) = 4 π ϵ 0 1 [ r 1 ∫ r ′ cos α ρ ( r ′ ) d τ ′ + r 2 1 ∫ r ′ cos α ρ ( r ′ ) d τ ′ + r 3 1 ∫ ( r ′ ) 2 2 3 cos 2 α − 1 ρ ( r ′ ) d τ ′ + ⋯ ]
The first term is the potential created by a monopole, the second term is by a dipole, and the third term is by a quadrupole. The n n n 2 n − 1 2^{n-1} 2 n − 1
Dipole Moment and Dipole Term Since the multipole expansion is a series with respect to the reciprocal of r r r r r r monopole .
V mono ( r ) = 1 4 π ϵ 0 Q r
V_{\text{mono}}(\mathbf{r}) = \dfrac{1}{4\pi\epsilon_{0}}\dfrac{Q}{r}
V mono ( r ) = 4 π ϵ 0 1 r Q
If the total charge of the gathered charges is 0 0 0 0 0 0 + + + − - − 0 0 0 0 0 0 dipole .
V dip ( r ) = 1 4 π ϵ 0 1 r 2 ∫ r ′ cos α ρ ( r ′ ) d τ ′
V_{\text{dip}}(\mathbf{r})=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{1}{r^2}\int r^{\prime} \cos \alpha \rho (\mathbf{r}^{\prime})d\tau^{\prime}
V dip ( r ) = 4 π ϵ 0 1 r 2 1 ∫ r ′ cos α ρ ( r ′ ) d τ ′
Here, since r ^ ⋅ r ′ = r ′ cos α \hat{\mathbf{r}}\cdot\mathbf{r}^{\prime}=r^{\prime}\cos\alpha r ^ ⋅ r ′ = r ′ cos α
V dip ( r ) = 1 4 π ϵ 0 1 r 2 r ^ ⋅ ∫ r ′ ρ ( r ′ ) d τ ′
V_{\text{dip}}(\mathbf{r})=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{1}{r^2}\hat{\mathbf{r}}\cdot\int \mathbf{r}^{\prime} \rho (\mathbf{r}^{\prime})d\tau^{\prime}
V dip ( r ) = 4 π ϵ 0 1 r 2 1 r ^ ⋅ ∫ r ′ ρ ( r ′ ) d τ ′
This integral value is independent of r \mathbf{r} r dipole moment of the charge distribution, denoted as p \mathbf{p} p
p = ∫ r ′ ρ ( r ′ ) d τ ′
\mathbf{p}=\int\mathbf{r}^{\prime}\rho (\mathbf{r}^{\prime})d\tau^{\prime}
p = ∫ r ′ ρ ( r ′ ) d τ ′
The dipole moment allows for a simple representation of the dipole’s potential.
V dip ( r ) = 1 4 π ϵ 0 p ⋅ r ^ r 2
V_{\text{dip}}(\mathbf{r})=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{\mathbf{p}\cdot\hat{\mathbf{r}} } {r^2}
V dip ( r ) = 4 π ϵ 0 1 r 2 p ⋅ r ^
Appendix 1 − 1 2 r ′ r ( r ′ r − 2 cos α ) + 3 8 ( r ′ r ) 2 ( r ′ r − 2 cos α ) 2 − 5 16 ( r ′ r ) 3 ( r ′ r − 2 cos α ) 3 + ⋯
1- \dfrac{1}{2}\dfrac{r^{\prime}}{r}\left( \dfrac{r^{\prime}}{r}-2\cos\alpha \right) +\dfrac{3}{8}\left( \dfrac{r^{\prime}}{r} \right)^{2} \left( \dfrac{r^{\prime}}{r}-2\cos\alpha \right)^2 -\dfrac{5}{16}\left( \dfrac{r^{\prime}}{r}\right)^3 \left( \dfrac{r^{\prime}}{r}-2\cos\alpha \right)^3 +\cdots
1 − 2 1 r r ′ ( r r ′ − 2 cos α ) + 8 3 ( r r ′ ) 2 ( r r ′ − 2 cos α ) 2 − 16 5 ( r r ′ ) 3 ( r r ′ − 2 cos α ) 3 + ⋯
Expanding the square and cubic terms from the above equation yields
1 − 1 2 r ′ r ( r ′ r − 2 cos α ) + 3 8 ( r ′ r ) 2 [ ( r ′ r ) 2 − 4 r ′ cos α r + 4 cos 2 α ]
1- \dfrac{1}{2}\dfrac{r^{\prime}}{r}\left( \dfrac{r^{\prime}}{r}-2\cos\alpha \right) +\dfrac{3}{8}\left( \dfrac{r^{\prime}}{r} \right)^2 \left[ \left( \dfrac{r^{\prime}}{r}\right)^2-\dfrac{4r^{\prime}\cos\alpha}{r}+4\cos^2\alpha \right]
1 − 2 1 r r ′ ( r r ′ − 2 cos α ) + 8 3 ( r r ′ ) 2 [ ( r r ′ ) 2 − r 4 r ′ cos α + 4 cos 2 α ]
− 5 16 ( r ′ r ) 3 [ ( r ′ r ) 3 − 3 ( r ′ r ) 2 2 cos α + 3 ( r ′ r ) 4 cos 2 α − 8 cos 3 α ] + ⋯
-\dfrac{5}{16}\left( \dfrac{r^{\prime}}{r}\right)^3 \left[ \left( \dfrac{r^{\prime}}{r}\right)^3 -3\left(\dfrac{r^{\prime}}{r}\right)^22\cos\alpha + 3\left( \dfrac{r^{\prime}}{r}\right)4\cos^2\alpha-8\cos^3\alpha \right] +\cdots
− 16 5 ( r r ′ ) 3 [ ( r r ′ ) 3 − 3 ( r r ′ ) 2 2 cos α + 3 ( r r ′ ) 4 cos 2 α − 8 cos 3 α ] + ⋯
Now, organizing according to the order of r ′ r \dfrac{r^{\prime}}{r} r r ′
1 + ( r ′ r ) cos α + ( r ′ r ) 2 ( − 1 2 + 3 2 cos 2 α ) + ( r ′ r ) 3 ( − 3 2 cos α + 5 2 cos 3 α ) + ⋯
1+\left(\dfrac{r^{\prime}}{r}\right)\cos\alpha +\left(\dfrac{r^{\prime}}{r}\right)^2\left(-\dfrac{1}{2} +\dfrac{3}{2}\cos^2\alpha \right)+\left( \dfrac{r^{\prime}}{r} \right)^3 \left( -\dfrac{3}{2}\cos\alpha +\dfrac{5}{2}\cos^3\alpha \right) + \cdots
1 + ( r r ′ ) cos α + ( r r ′ ) 2 ( − 2 1 + 2 3 cos 2 α ) + ( r r ′ ) 3 ( − 2 3 cos α + 2 5 cos 3 α ) + ⋯
Upon organizing this gives
1 + ( r ′ r ) ( cos α ) + ( r ′ r ) 2 ( 3 cos 2 α − 1 2 ) + ( r ′ r ) 3 ( 5 cos 2 α − 3 cos α 2 ) + ⋯
1+\left(\dfrac{r^{\prime}}{r}\right)\left( \cos\alpha \right) +\left( \dfrac{r^{\prime}}{r} \right)^2 \left( \dfrac{3\cos^2\alpha -1 }{2}\right) +\left( \dfrac{r^{\prime}}{r}\right)^3 \left( \dfrac{5\cos^2\alpha-3\cos\alpha}{2} \right) +\cdots
1 + ( r r ′ ) ( cos α ) + ( r r ′ ) 2 ( 2 3 cos 2 α − 1 ) + ( r r ′ ) 3 ( 2 5 cos 2 α − 3 cos α ) + ⋯
