Exploring the Quantum Theory of Magnetism: From Sum to Integral

[LagrangeEuler]

http://books.google.rs/books?id=vrc...e&q=Quantum theory of magnetism Ising&f=false

In page 266 why we are alowed to go from double sum to double integral?

\lim_{N\to \infty}\frac{1}{N}\sum_{q_1,q_2}F(q_1,q_2) \rightarrow \frac{1}{4\pi^2}\int^{2\pi}_0\int^{2\pi}_0dq_1dq_2F(q_1,q_2)

 

[danielu13]

For some reason, there was no page 266 for me. But anyway, it's because of the limit definition of an integral, which works for double integrals just as it does a single integral. A nice overview of this can be found here.

 

[LagrangeEuler]

For some reason, there was no page 266 for me. But anyway, it's because of the limit definition of an integral, which works for double integrals just as it does a single integral. A nice overview of this can be found here.

Why integrals are from zero to $2\pi$. Why not
$\rightarrow \frac{1}{16\pi^2}\int^{4\pi}_0\int^{4\pi}_0...$

 

[danielu13]

I believe both of those should be equal; if you take the integral at 4\pi and then divide by 16\pi^2, it should be equal to taking the integral at 2\pi and dividing by 4\pi^2, correct?

 

[LagrangeEuler]

I believe both of those should be equal; if you take the integral at 4\pi and then divide by 16\pi^2, it should be equal to taking the integral at 2\pi and dividing by 4\pi^2, correct?

So you said that I can go from sum to integral in the way

\rightarrow \frac{1}{a^2}\int^{a}_0\int^{a}_0...?

 

[LagrangeEuler]

I believe both of those should be equal; if you take the integral at 4\pi and then divide by 16\pi^2, it should be equal to taking the integral at 2\pi and dividing by 4\pi^2, correct?

Well look for example $lim_{N\to \infty}\frac{1}{N}\sum_{q_1,q_2)ln(q_1-2\pi)$. Is it the same? How I go from that sum to integral?