- #1
JulienB
- 408
- 12
Homework Statement
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}##.
Homework 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)##
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.Thank you very much in advance for your answers, I'm looking forward to reading you!Julien.