Solving Griffiths E&M Integral with Polarization in z-Direction

ehrenfest
Messages
2,001
Reaction score
1

Homework Statement


This question refers to Griffiths E and M book.

Of course I am stuck with the integral.

I chose z to be in the same direction as the polarization.

The integral is then
P \int d\Omega r'^2 dr'\frac{\left(\mathbf{r} -\mathbf{r'}\right)\cdot \hat{\mathbf{z}}}{\left(\mathbf{r} -\mathbf{r'}\right)^2}

BTW, how do you get the script r that Griffiths uses in latex?


Homework Equations





The Attempt at a Solution

 
Physics news on Phys.org
Your denominator should be ({\bf r-r')}^3.
 
Yes, the integral is a pain in the neck, but you should notice that it is the same integral as a uniformly charged sphere. I would probably just use mathematica, but if you don't have that luxury then I believe you can use trig substitutions and partial fractions to get it done.
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top