# Verifying a macroscopic Maxwell equation

1. Mar 22, 2017

### JulienB

1. The problem statement, all variables and given/known data

Hi everybody! I have trouble understanding the following problem, hopefully somebody can help!

Show that the electrostatic potential

$\phi(\vec{x}) = \phi_\text{free} (\vec{x}) + \phi_\text{ind} (\vec{x}) = \frac{1}{4 \pi \epsilon_0} \bigg( \int d^3 x'\ \frac{\rho_\text{free} (\vec{x}')}{|\vec{x} - \vec{x}'|} + \int d^3x'\ \frac{\vec{P} (\vec{x}') \cdot (\vec{x} - \vec{x}')}{|\vec{x} - \vec{x}'|^3} \bigg)$

solves the macroscopic Maxwell equation $\vec{\nabla} \cdot \vec{D} = \vec{\nabla} \cdot (\epsilon_0 \vec{E} + \vec{P}) = \rho_\text{free}$.

2. Relevant equations

$\vec{E} = - \vec{\nabla} \phi$
$\Delta_x \bigg(\frac{1}{|\vec{x} - \vec{x}'|} \bigg) = -4 \pi \delta^{(3)} (\vec{x} - \vec{x}')$
$\frac{\vec{x} - \vec{x}'}{|\vec{x} - \vec{x}'|^3} = \vec{\nabla} \bigg(\frac{1}{|\vec{x} - \vec{x}'|} \bigg)$

3. The attempt at a solution

I have seen the solution already, but there is one step I don't understand. Here is the beginning of the proof:

$\vec{\nabla} \cdot \vec{D} = - \epsilon_0 \Delta \phi + \vec{\nabla} \cdot \vec{P}$

which means I am expecting to find $\Delta \phi = -\frac{1}{\epsilon_0} \rho_\text{free} (\vec{x}) + \frac{1}{\epsilon_0} \vec{\nabla} \cdot \vec{P}$. First I apply the Laplace-operator on $\phi_\text{free}$:

$\Delta \phi_\text{free} (\vec{x}) = \frac{1}{4 \pi \epsilon_0} \int d^3x'\ \rho_\text{free} (\vec{x}') \Delta_x \bigg(\frac{1}{|\vec{x} - \vec{x}'|} \bigg)$
$= \frac{1}{4 \pi \epsilon_0} \int d^3x'\ \rho_\text{free} (\vec{x}') (-4 \pi \delta^{(3)} (\vec{x} - \vec{x}'))$
$= -\frac{1}{\epsilon_0} \rho_\text{free} (\vec{x})$

Here I have used the property of the Dirac delta function in the integral, and I got my first term pretty easily. The second term is actually precisely what is bothering me:

$\Delta \phi_\text{ind} (\vec{x}) = \frac{1}{4 \pi \epsilon_0} \int d^3x'\ \vec{P} (\vec{x}') \cdot \Delta_x \bigg( \frac{(\vec{x} - \vec{x}')}{|\vec{x} - \vec{x}'|^3} \bigg)$
$= \frac{1}{4 \pi \epsilon_0} \int d^3x'\ \vec{P} (\vec{x}') \cdot \Delta_x \bigg( \vec{\nabla}_x \frac{1}{|\vec{x} - \vec{x}'|} \bigg)$
$= \frac{1}{4 \pi \epsilon_0} \int d^3x'\ \vec{P} (\vec{x}') \cdot \vec{\nabla}_x \bigg( \Delta_x \frac{1}{|\vec{x} - \vec{x}'|} \bigg)$
$= \frac{-1}{\epsilon_0} \int d^3x'\ \vec{P} (\vec{x}') \cdot \vec{\nabla}_x \delta^{(3)} (\vec{x} - \vec{x}')$

Now I am not very happy with this expression. I guess I could put $\vec{\nabla}_x$ in front of $\vec{P}$ since it doesn't apply on $x'$, but then I would get the wrong sign. And if I were to apply the Dirac delta function's property first, the result would be different and senseless. I think that this is anyway not the right way to proceed. In the solution, my teacher just indicates that he has used integration by parts but I do not really understand how! Here is what he wrote:

$\phi_\text{ind} (\vec{x}) = \frac{1}{4 \pi \epsilon_0} \int d^3x'\ \vec{P} (\vec{x}') \cdot \vec{\nabla}_x \bigg(\frac{1}{|\vec{x} - \vec{x}'|} \bigg)$
$\underbrace{=}_{P.I.} - \frac{1}{4 \pi \epsilon_0} \int d^3x'\ \vec{\nabla} \cdot \vec{P} (\vec{x}') \frac{1}{|\vec{x} - \vec{x}'|}$

(P.I. refers to integration by parts in German)

I assume that he performed the integration by parts by defining $u=\vec{P} (\vec{x}')$ and $v' = \vec{\nabla}_{x'} \bigg(\frac{1}{|\vec{x} - \vec{x}'|} \bigg)$ (since $\vec{\nabla}_{x} \bigg(\frac{1}{|\vec{x} - \vec{x}'|} \bigg) = \vec{\nabla}_{x'} \bigg(\frac{1}{|\vec{x} - \vec{x}'|} \bigg)$ I believe), but then I get:

$\phi_\text{ind} (\vec{x}) = \frac{1}{4 \pi \epsilon_0} \vec{P} (\vec{x}) \frac{1}{|\vec{x} - \vec{x}'|} - \frac{1}{4 \pi \epsilon_0} \int d^3x'\ \vec{\nabla}_{x'} \cdot \vec{P} (\vec{x}') \frac{1}{|\vec{x} - \vec{x}'|}$

which looks pretty close apart from the fact that his first term is gone. How is that? Is the polarization disappearing at the boundaries? Or do we assume that the boundaries are $\pm \infty$, in which case it is the term that $\frac{1}{|\vec{x} - \vec{x}'|}$ is equal to 0. I feel like it is important that I do not misunderstand this step.

Julien.

2. Mar 22, 2017

### qnach

This is not introductory physics, for sure.

3. Mar 23, 2017

### JulienB

@qnach Really? Last time I posted about non relativistic electrodynamics in the advanced physics forum, it got moved here

4. Mar 23, 2017

### qnach

Yes
You are RIGHT
People in this forum is sick.
I simply mimic their tone.

5. Mar 23, 2017

### JulienB

@qnach but more seriously, can I move it to the other forum then? I don't see a button for doing so.

6. Mar 23, 2017

### SammyS

Staff Emeritus
You can use the "Report" in the lower left of any post in the thread, and ask a Mentor to place this thread in the appropriate forum, if Introductory Physics is not the correct place. (I'll do that.)

7. Mar 23, 2017

### JulienB

@sammy Thanks a lot for the info and for moving the thread of course. Hopefully somebody will have an answer now :)

8. Mar 25, 2017

### JulienB

Problem solved, integration by parts was just never used. The solution is simply that:

$\vec{\nabla}_{x} \bigg(\frac{1}{|\vec{x} - \vec{x}'|} \bigg) = \begin{bmatrix} \partial_x \\ \partial_y \\ \partial_z \end{bmatrix} \frac{1}{\sqrt{(x-x')^2 + (y-y')^2 + (z-z')^2}}$
$= \frac{(\vec{x} - \vec{x}')}{|\vec{x} - \vec{x}'|^3} = - \vec{\nabla}_{x'} \left(\frac{1}{|\vec{x} - \vec{x}'|} \right)$

since

$\vec{\nabla}_{x'} \bigg(\frac{1}{|\vec{x} - \vec{x}'|} \bigg) = \begin{bmatrix} \partial_{x'} \\ \partial_{y'} \\ \partial_{z'} \end{bmatrix} \frac{1}{\sqrt{(x-x')^2 + (y-y')^2 + (z-z')^2}}$
$=\frac{(\vec{x}' - \vec{x})}{|\vec{x} - \vec{x}'|^3}$.

I hope this is useful to anyone having their teacher believe that this comes from integration by parts. XD

Julien.

9. Mar 25, 2017

### TSny

This looks OK except it appears to me that the $\vec{\nabla}_x$ operator in the first integral should be with respect to the primed variables; that is, $\vec{\nabla}_{x'}$.

Yes
OK, except for a sign error. $\vec{\nabla}_{x} \bigg(\frac{1}{|\vec{x} - \vec{x}'|} \bigg) = -\vec{\nabla}_{x'} \bigg(\frac{1}{|\vec{x} - \vec{x}'|} \bigg)$

Yes, that all looks good. You can assume the integration over $d^3x'$ is over all of space. So, the boundary term is at infinity.

10. Mar 25, 2017

### TSny

Is the overall sign of the last expression correct?

Similarly, is the sign of the last expression correct?

Don't you need to do integration by parts in order to get the $\vec{\nabla}_{x'}$ operator to act on $\vec{P}$?