Potential Energy of a 2D Crystal Lattice

[SquidgyGuff]

Homework Statement

Use a computer to calculate numerically the potential energy, per ion, for an ifinite 2D square, ionic crystal with separation a; that is, a plane of equally spaced charges of magnitude e and alternating sign.

Homework Equations

The Attempt at a Solution

The closest I can come to a solution is by calculating the potential energy of the center charge and adding the contributions of the other charges manually and it was far too inaccurate. I tried to use a sum to find E and then use it in the formula for potential energy, but it is so terribly wrong.
http://www.sciweavers.org/upload/Tex2Img_1440630831/render.png
In the back of the book it lists this as the answer, but I have no clue how they arrived at it. The book says that you have to find the potential energy contributed from the top half and the line to the right and multiply it by two, but I'm not sure how they got their sums.
http://www.sciweavers.org/upload/Tex2Img_1440631169/render.png

 

[haruspex]

In the back of the book it lists this as the answer, but I have no clue how they arrived at it. The book says that you have to find the potential energy contributed from the top half and the line to the right and multiply it by two, but I'm not sure how they got their sums.
http://www.sciweavers.org/upload/Tex2Img_1440631169/render.png

Do you understand what they mean by "top half" and "line to the right"? Can you see which of the two sums refers to which of those?

 

[SquidgyGuff]

Do you understand what they mean by "top half" and "line to the right"? Can you see which of the two sums refers to which of those?

I know the sum on the left represents the line of alternating charges to the right of the center and I can see how it is derived, but no the sum of the charges above it.

 

[haruspex]

I know the sum on the left represents the line of alternating charges to the right of the center and I can see how it is derived, but no the sum of the charges above it.

The second is a double sum, over m and n. m goes from minus infinity to plus infinity, while n goes from 1 to infinity. In terms of the relative positions of charges in the half plane above the centre, which of m and n would you say corresponds to the x co-ordinate?

 

[SquidgyGuff]

The second is a double sum, over m and n. m goes from minus infinity to plus infinity, while n goes from 1 to infinity. In terms of the relative positions of charges in the half plane above the centre, which of m and n would you say corresponds to the x co-ordinate?

m is definetley represents x because it is infinite in both directions, where as n is only infinite in one direction. I think I understand now! It's essentially the same as the first, but in 2 dimensions with the distance represented as sqrt(m^2 + n^2) times a. Is this a particularly hard question for 3rd semester of college?

 

[SquidgyGuff]

The second is a double sum, over m and n. m goes from minus infinity to plus infinity, while n goes from 1 to infinity. In terms of the relative positions of charges in the half plane above the centre, which of m and n would you say corresponds to the x co-ordinate?

The (-1)^m in the first term makes sense to me, but how would one know to use (-1)^(m+n) for the second term?

 

[haruspex]

The (-1)^m in the first term makes sense to me, but how would one know to use (-1)^(m+n) for the second term?

The sign alternates in both directions, so a step of 1 in either m or n will switch the sign. That tells you the sign will go like -1^m+n. (You could equally write -1^m-n etc.). All that is left to decide is the sign at the start of the sum.
I've no idea how this rates as appropriate for third semester in your academic culture, and besides that was a very long time ago for me.
As you continue with this question, there are opportunities for it to become quite a bit more involved yet. Those sums only converge quite slowly. You can solve the half line sum analytically, but not the half plane sum. There are ways to speed up the convergence, and ways to break the sum up into a finite sum that you can do in software, plus an infinite sum in which you can make approximations and obtain an analytic solution.