Casimir Effect, no idea

1. Oct 24, 2013

epislon58

Hello,

I am attempting to repeat the math found on page 4 of this paper using the Euler-maclaurin summation formula. How would I incorporate the conversion factor because I can not figure it out for the life in me!

Thank you!

http://www.hep.caltech.edu/~phys199/lectures/lect5_6_cas.pdf

Last edited: Oct 24, 2013
2. Oct 25, 2013

fzero

Using the conversion factor means that we can write

$$\nu = -\lim_{\epsilon\rightarrow 0} \frac{d}{d\epsilon} e^{-\epsilon \nu}.$$

Note that I take $\epsilon\rightarrow 0$ instead of $\infty$ like in the text. I suspect the $\infty$ is a typo.

This conversion factor is useful, because we can then write the sum as

$$\sum_{\nu = 1}^\infty \nu = -\lim_{\epsilon\rightarrow 0} \frac{d}{d\epsilon} \sum_{\nu = 1}^\infty e^{-\epsilon \nu}.$$

Since $e^{-\epsilon} < 1$, we can recognize this as a geometric series and do the sum. Expanding the result in $\epsilon$ will leave a divergent term, a finite term, and terms that vanish as $\epsilon\rightarrow 0$. Introducing $e^{-\epsilon \nu}$ into the integral gives a result that cancels the divergent term in the sum. We are then left with a finite result when we compute the difference between the sum and integral.

Alternatively, one can use the Euler-Maclaurin formula, which is also referred to in the text.

3. Oct 25, 2013

epislon58

Thank you, but the text states that they used the Euler-Macluarin and the conversion factor. Could this have been a typo?

4. Oct 25, 2013

fzero

You only need one of the methods. Give it a try.

5. Oct 25, 2013

epislon58

oh ok thank you. I will let you know how it turns out asap!

6. Oct 25, 2013

TSny

For using the Euler-Maclaurin formula, the hint might be that you should use $e^{-\epsilon \nu}$ as a "convergence" factor (rather than a "conversion" factor). Then let $\epsilon \rightarrow 0$ (not ∞, as fzero has already noted).

Thus, consider the argument of the sum or integral to be $\nu e^{-\epsilon \nu}$. Without the convergence factor, you run into trouble for $\nu \rightarrow \infty$ in the Euler-Maclaurin formula.