Derivation of the expansion of the potential in rectangular coordinates

[MathematicalPhysicist]

Homework Statement

I want to derive the expansion of $\Phi(x)$ in rectangular coordinates:

$$ \Phi(\vec{x}) = \frac{1}{4\pi \epsilon_0} \bigg[ \frac{q}{r}+\frac{\vec{p}\cdot \vec{x}}{r^3}+\frac{1}{2}\sum_{i,j} Q_{ij} \frac{x_ix_j}{r^5}+\ldots\bigg]$$

Homework Equations

$$\vec{p}= \int \vec{x}' \rho (\vec{x}') d^3 x'$$
$$Q_{ij}=\int (3x_i'x_j'-r'^2\delta_{ij})\rho (\vec{x}')d^3x'$$
$$q=\int \rho(\vec{x}') dx'$$
$$\Phi(\vec{x})=\frac{1}{4\pi\epsilon_0} \int \frac{\rho (\vec{x}')}{|\vec{x}-\vec{x}'|}d^3 x'$$

The Attempt at a Solution

I don't have time to make the calculations by myself, if you know some derivation in the net of this identity, one needs to expand the $1/|\vec{x}-\vec{x}'|$ by Taylor series.

I tried googling for the answer but didn't find a derivation in the notation of Jackson's.

Thanks in advance!

 

[NFuller]

I don't have time to make the calculations by myself

Sorry but, to be frank, you need to make time. The point of homework is to learn how to solve the problems on your own.

one needs to expand the 1/|⃗x−⃗x′|1/|x→−x→′|1/|\vec{x}-\vec{x}'| by Taylor series.

If you are using Jackon's ED text, he shows the expansion in equation 3.70.

 

[MathematicalPhysicist]

Sorry but, to be frank, you need to make time. The point of homework is to learn how to solve the problems on your own.

If you are using Jackon's ED text, he shows the expansion in equation 3.70.

Strictly speaking it's not HW, I finished my EM coursework this last semester; It's just for my own knowledge.
Someone already has done the calculation I assume, just need to look it out.

 

[NFuller]

Strictly speaking it's not HW, I finished my EM coursework this last semester; It's just for my own knowledge.

Alright, I assumed it was since this is posted in the homework forum.

Someone already has done the calculation I assume, just need to look it out.

The expansion is
$$\frac{1}{|\mathbf{x}-\mathbf{x}^{\prime}|}=4\pi\sum_{l=0}^{\infty}\sum_{m=-l}^{l}\frac{1}{2l+1}\frac{r_{<}^{l}}{r_{>}^{l+1}}Y^{*}_{lm}(\theta^{\prime},\phi^{\prime})Y_{lm}(\theta,\phi)$$
You just need to insert this into
$$\Phi(\vec{x})=\frac{1}{4\pi\epsilon_0} \int \frac{\rho (\vec{x}')}{|\vec{x}-\vec{x}'|}d^3 x'$$
and evaluate the first three terms. Then, use the definitions of $q$, $\mathbf{p}$, and $Q_{ij}$ to replace those terms which will give you the final expansion for $\Phi$.

 

[MathematicalPhysicist]

Alright, I assumed it was since this is posted in the homework forum.

The expansion is
$$\frac{1}{|\mathbf{x}-\mathbf{x}^{\prime}|}=4\pi\sum_{l=0}^{\infty}\sum_{m=-l}^{l}\frac{1}{2l+1}\frac{r_{<}^{l}}{r_{>}^{l+1}}Y^{*}_{lm}(\theta^{\prime},\phi^{\prime})Y_{lm}(\theta,\phi)$$
You just need to insert this into
$$\Phi(\vec{x})=\frac{1}{4\pi\epsilon_0} \int \frac{\rho (\vec{x}')}{|\vec{x}-\vec{x}'|}d^3 x'$$
and evaluate the first three terms. Then, use the definitions of $q$, $\mathbf{p}$, and $Q_{ij}$ to replace those terms which will give you the final expansion for $\Phi$.

If it were a simple plug and chug I would have done it by myself; but there are some terms that aren't so simple.

 

[NFuller]

If it were a simple plug and chug I would have done it by myself; but there are some terms that aren't so simple.

It is that simple. Which terms are you having trouble with? How far have you got?

 

[MathematicalPhysicist]

It is that simple. Which terms are you having trouble with? How far have you got?

I can see how you get the term $q/r$, but not the other two terms.

And it's not like I don't have other tasks at hand besides this, I am quite tired... :-(

 

[NFuller]

I can see how you get the term q/rq/rq/r, but not the other two terms.

To get the dipole term $\Phi(\mathbf{x})_{d}$ for example, we evaluate
$$\int Y_{lm}^{*}(\theta^{\prime},\phi^{\prime})r^{\prime l}\rho(\mathbf{x}^{\prime})d^{3}x^{\prime}$$
for $l=1$ and $m=0,1$. For $m=0$,
$$\int Y_{1,0}^{*}(\theta^{\prime},\phi^{\prime})r^{\prime}\rho(\mathbf{x}^{\prime})d^{3}x^{\prime}=\sqrt{\frac{3}{4\pi}}\int z^{\prime}\rho(\mathbf{x}^{\prime})d^{3}x^{\prime}=\sqrt{\frac{3}{4\pi}}p_{z}$$
For $m=1$,
$$\int Y_{1,1}^{*}(\theta^{\prime},\phi^{\prime})r^{\prime}\rho(\mathbf{x}^{\prime})d^{3}x^{\prime}=-\sqrt{\frac{3}{8\pi}}\int (x^{\prime}-iy^{\prime})\rho(\mathbf{x}^{\prime})d^{3}x^{\prime}=-\sqrt{\frac{3}{8\pi}}(p_{x}-ip_{y})$$
Plugging this in for $\Phi$
$$\Phi(\mathbf{x})_{d}=\frac{1}{\epsilon}\left[\frac{1}{3}\sqrt{\frac{3}{4\pi}}p_{z}\sqrt{\frac{3}{4\pi}}\frac{\text{cos}(\theta)}{r^{2}}+\frac{1}{3}\sqrt{\frac{3}{8\pi}}(p_{x}-ip_{y})\sqrt{\frac{3}{8\pi}}\frac{\text{sin}(\theta)e^{-i\phi}}{r^{2}}-\frac{1}{3}\sqrt{\frac{3}{8\pi}}(p_{x}-ip_{y})\sqrt{\frac{3}{8\pi}}\frac{\text{sin}(\theta)e^{i\phi}}{r^{2}}\right]$$
This is simply
$$\Phi(\mathbf{x})_{d}=\frac{1}{4\pi\epsilon}\frac{\mathbf{p}\cdot\mathbf{x}}{r^{3}}$$
Can you work out the quadrupole terms?

 

[MathematicalPhysicist]

I should have known it would be that much tedious, I'll continue from there, thanks.

 

[MathematicalPhysicist]

To get the dipole term $\Phi(\mathbf{x})_{d}$ for example, we evaluate
$$\int Y_{lm}^{*}(\theta^{\prime},\phi^{\prime})r^{\prime l}\rho(\mathbf{x}^{\prime})d^{3}x^{\prime}$$
for $l=1$ and $m=0,1$. For $m=0$,
$$\int Y_{1,0}^{*}(\theta^{\prime},\phi^{\prime})r^{\prime}\rho(\mathbf{x}^{\prime})d^{3}x^{\prime}=\sqrt{\frac{3}{4\pi}}\int z^{\prime}\rho(\mathbf{x}^{\prime})d^{3}x^{\prime}=\sqrt{\frac{3}{4\pi}}p_{z}$$
For $m=1$,
$$\int Y_{1,1}^{*}(\theta^{\prime},\phi^{\prime})r^{\prime}\rho(\mathbf{x}^{\prime})d^{3}x^{\prime}=-\sqrt{\frac{3}{8\pi}}\int (x^{\prime}-iy^{\prime})\rho(\mathbf{x}^{\prime})d^{3}x^{\prime}=-\sqrt{\frac{3}{8\pi}}(p_{x}-ip_{y})$$
Plugging this in for $\Phi$
$$\Phi(\mathbf{x})_{d}=\frac{1}{\epsilon}\left[\frac{1}{3}\sqrt{\frac{3}{4\pi}}p_{z}\sqrt{\frac{3}{4\pi}}\frac{\text{cos}(\theta)}{r^{2}}+\frac{1}{3}\sqrt{\frac{3}{8\pi}}(p_{x}-ip_{y})\sqrt{\frac{3}{8\pi}}\frac{\text{sin}(\theta)e^{-i\phi}}{r^{2}}-\frac{1}{3}\sqrt{\frac{3}{8\pi}}(p_{x}-ip_{y})\sqrt{\frac{3}{8\pi}}\frac{\text{sin}(\theta)e^{i\phi}}{r^{2}}\right]$$
This is simply
$$\Phi(\mathbf{x})_{d}=\frac{1}{4\pi\epsilon}\frac{\mathbf{p}\cdot\mathbf{x}}{r^{3}}$$
Can you work out the quadrupole terms?

You should also have a $m=-1$ since the inner sum goes from $-l$ to $l$.

 

[NFuller]

You should also have a $m=-1$ since the inner sum goes from $-l$ to $l$.

Yes, it's in there. I didn't explicitly show it since
$$Y_{l,-m}(\theta,\phi)=(-1)^{m}Y_{l,m}^{*}(\theta,\phi)$$