Following the section 2.3 of the book "Advances in the Casimir Effect" by M. Bordag, the Casimir energy is given, for a maseless field, by $$E(a) = E_0 (a) - E_{0M} (a)$$ where $E_0 (a)$ is the vacuum energy in a circle of circumference $a$, and $E_{0M}(a)$ is the vacuum energy in the free Minkowski space within the length of the circumference, $a$. This substraction is done to avoid the infinite that arises in the vacuum energy.

With the antiperiodic conditions in the circle, one gets a wavefunction with a frequency of: $$\omega_n = \frac{2\pi c}{a}\left(n+\frac{1}{2}\right), n=0, \pm 1, \pm2...$$ So we can write: $$E_0(a)=\frac{\hbar}{2} \sum_{n=-\infty}^{\infty} \omega_n= \frac{\pi \hbar c}{a} \sum_{n=-\infty}^{\infty} \left(n+\frac{1}{2}\right)$$ For the free Minkowski space, the frequency for a maseless field is $$\omega_k = ck$$ and its energy in an interval of length $a$ is: $$E_{0M} = \frac{\hbar}{2} \int^{\infty}_{-\infty} \frac{dk}{2\pi}\omega_k a = \frac{\hbar a c}{2\pi}\int^{\infty}_0 k dk \overset{\text{$2 \pi t = ak$}}{=} \frac{2\pi \hbar c}{a} \int_0^{\infty} t dt$$ In order to use the Abel-Plana formula for odd half integers, given by: $$\sum_{n=0}^{\infty} F\left(n+\frac{1}{2}\right) - \int_0^{\infty} F(t) dt =-i\int^{\infty}_0 \frac{dt}{e^{2\pi t}+1} [F(it)-F(-it)]$$ We must write the series in $E_0(a)$ beginning with $0$:

$$E_0(a) =\frac{\pi \hbar c}{2a} + \frac{ \pi \hbar c }{a} \sum_{n= \pm 1, \pm2...}^{\infty} \left(n+\frac{1}{2}\right) = \frac{\pi \hbar c}{2a} + \frac{2 \pi \hbar c}{a} \sum_{n=1}^{\infty} \left(n+\frac{1}{2} \right) = -\frac{\pi \hbar c }{a} +\frac{2 \pi \hbar c}{a} \sum_{n=0}^{\infty} \left(n+\frac{1}{2} \right)$$

So we have: $$E(a)= E_0(a)-E_{0M}(a) = -\frac{\pi \hbar c }{a} +\frac{2 \pi \hbar c}{a} \left(\sum_{n=0}^{\infty} \left(n+\frac{1}{2} \right) - \int_0^{\infty} t dt \right)$$ And picking $F(t)=t$ in the Abel-Plana formula, I get that the parenthesis above is $\frac{1}{24}$ (I've done it explicitly, but you can check the result here https://www.calculadora-de-integrales.com/#expr=2x%2F%28e%5E%282pix%29%2B1%29&lbound=0&ubound=inf), so:

$$E(a)= -\frac{\pi \hbar c }{a} +\frac{ \pi \hbar c}{12a}$$

But in the book they get:

$$E(a)= \frac{ \pi \hbar c}{12a}$$

So I think I must be doing something wrong, perhaps related with expressing the infinite series beggining with $0$, because the term arises from there... Any ideas?


1 Answer 1


I've solved it, in case anyone wants to see the solution. The problem is that the expression $\omega_n$ arises from: $$\omega_n^2 = c^2 k_n^2$$ with $$k_n = \frac{2 \pi }{a}\left(n+\frac{1}{2}\right), n= 0, \pm 1, \pm 2...$$ so the expression I wrote for $\omega_n$ has an absolute value.

This means that when writing $E_0 (a)$ beginning with $0$:

$$E_0(a) =\frac{ \pi \hbar c }{a} \sum_{n=0, \pm 1, \pm2...}^{\infty} \left|n+\frac{1}{2}\right| = \frac{ \pi \hbar c}{a} \left(\sum_{n=0, 1, 2...}^{\infty} \left|n+\frac{1}{2}\right| + \sum_{n= -1, -2...}^{\infty} \left|n+\frac{1}{2}\right| \right) = $$ $$=\frac{2 \pi \hbar c}{a}\sum_{n=0, 1, 2...}^{\infty} \left(n+\frac{1}{2}\right)$$

Where it's easy to check that the infinite series inside the big parenthesis are the same. So that solves the extra term I got when calculating this.


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