Ultralight Dark Photon and Casimir Effect

Abstract

We investigate the influence of a dark photon on the Casimir effect and calculate the corresponding leading contribution to the Casimir energy. For expected magnitudes of the photon - dark photon mixing parameter, the influence turns out to be negligible. The plasmon dispersion relation is also not noticeably modified by the presence of a dark photon.

Appendices

Appendix A: Euler-Maclaurin Summation Formula

The Euler-Maclaurin summation formula was originally obtained in 1782 by Euler and independently and almost simultaneously by Maclaurin [61]. A rigorous treatment of this very important tool of numerical analysis with diverse applications can be found, for example, in [62]. There exist several elementary derivations [63,64,65] of this remarkable formula. However, we prefer a formal heuristic derivation [66], which in the simplest way demystifies the appearance of Bernoulli numbers in it. Note that using Banach spaces of entire functions of exponential type, the formal derivation of the Euler-Maclaurin formula can be made mathematically rigorous [67].

Let $\hat D=\frac {d}{dx}$ be a differential operator, so that

$$ \hat D f(x)=\frac{d f}{dx}, f(x+n)=e^{n\hat D}f(x), $$

(A.1)

where the second equation is a formal expression of the Taylor formula. Then

$$ \sum\limits_{n=0}^{N-1}f(x+n)=\left (\sum\limits_{n=0}^{N-1} e^{n\hat D}\right )f(x)= \frac{e^{N\hat D}-1}{e^{n\hat D}-1}f(x)=\frac{1}{e^{n\hat D}-1}\left[f(x+N)-f(x) \right ]. $$

(A.2)

The Bernoulli numbers B_n are defined by the power series expansion of their exponential generating function:

$$ \begin{aligned} &\frac{x}{e^{x}-1}=\sum\limits_{n=0}^{\infty} \frac{B_{n}}{n!} x^{n}, B_{0}=1, B_{1}=-\frac{1}{2}, B_{2}=\frac{1}{6}, \\ &B_{3}=0, B_{4}=-\frac{1}{30}, B_{5}=0, B_{6}=\frac{1}{42} \ldots \end{aligned} $$

(A.3)

Comparing (A.3) and (A.2), we can write

$$ \sum\limits_{n=0}^{N-1}f(x+n)=\left (\hat D^{-1}+\sum\limits_{n=1}^{\infty} \frac{B_{n}}{n!} \hat D^{n-1}\right )\left [f(x+N)-f(x)\right ]. $$

(A.4)

On the other hand,

$$ {\int\limits_{0}^{N}} f(x+t) dt={\int\limits_{0}^{N}} e^{t\hat D}f(x) dt= \hat D^{-1}\left (e^{N\hat D}-1\right )f(x)=\hat D^{-1}\left [f(x+N)- f(x)\right ]. $$

(A.5)

Therefore

$$ \sum\limits_{n=0}^{N-1}f(x+n)-{\int\limits_{0}^{N}} f(x+t) dt=\sum\limits_{n=1}^{\infty} \frac{B_{n}}{n!} \hat D^{n-1}\left [f(x+N)-f(x)\right ]. $$

(A.6)

When $N\to \infty$, for convergence we need $\hat D^{k}f(\infty )=0$, k = 0, 1,… and (A.6) in this limit takes the form

$$ \sum\limits_{n=0}^{\infty}f(x+n)-\int\limits_{0}^{\infty} f(x+t) dt= -\sum\limits_{n=1}^{\infty} \frac{B_{n}}{n!} \hat D^{n-1} f(x)=\frac{1}{2} f(x)- \sum\limits_{n=2}^{\infty} \frac{B_{n}}{n!} \frac{d^{n-1}f(x)}{dx^{n-1}} . $$

(A.7)

The form of the Euler-Maclaurin Summation Formula used in the main text corresponds to x = 0 in (A.7).

Appendix B: Evaluation of the Integrals

First we evaluate the integral from entry 3.411.1 in [48]. From the definition of the gamma function

$$ {\Gamma}(\nu)=\int\limits_{0}^{\infty} x^{\nu-1}e^{-x}dx, $$

(B.1)

and using

$$ \frac{1}{1-e^{-x}}=\sum\limits_{k=0}^{\infty} e^{-kx}, $$

(B.2)

we get after interchanging the orders of the integration and summation

$$ \begin{array}{@{}rcl@{}} \int\limits_{0}^{\infty} \frac{x^{\nu-1}}{e^{x}-1}dx&=&\int\limits_{0}^{\infty} \frac{x^{\nu-1}e^{-x}}{1-e^{-x}}dx=\int\limits_{0}^{\infty}\sum\limits_{k=0}^{\infty} x^{\nu-1}e^{-(k+1)x}dx \\ &=& \sum\limits_{k=0}^{\infty}\int\limits_{0}^{\infty} x^{\nu-1}e^{-(k+1)x}dx={\Gamma}(\nu)\sum\limits_{k=0}^{\infty} \frac{1} {(k+1)^{n}}={\Gamma}(\nu) \zeta(\nu). \end{array} $$

(B.3)

Now, if we differentiate the just proved identity

$$ \zeta(\nu)=\frac{1}{\Gamma(\nu)}\int\limits_{0}^{\infty} \frac{x^{\nu-1}}{e^{x}-1} dx $$

(B.4)

with respect to ν and take into account $\frac {dx^{\nu -1}}{d\nu }=\ln {x} x^{\nu -1}$, we get

$$ \begin{array}{@{}rcl@{}} \zeta^{\prime}(\nu)&=&\frac{1}{\Gamma(\nu)}\int\limits_{0}^{\infty}\frac{x^{\nu-1} \ln{x}}{e^{x}-1}dx-\frac{{\Gamma}^{\prime}(\nu)}{{\Gamma}^{2}(\nu)} \int\limits_{0}^{\infty}\frac{x^{\nu-1}}{e^{x}-1}dx \\&=& \frac{1}{\Gamma(\nu)}\int\limits_{0}^{\infty}\frac{x^{\nu-1} \ln{x}}{e^{x}-1}dx-\psi(\nu) \zeta(\nu), \end{array} $$

(B.5)

where $\psi (\nu )={\Gamma }^{\prime }(\nu )/{\Gamma }(\nu )$. Therefore,

$$ \int\limits_{0}^{\infty}\frac{x^{\nu-1} \ln{x}}{e^{x}-1}dx= {\Gamma}(\nu) \zeta^{\prime}(\nu)+\psi(\nu) {\Gamma}(\nu) \zeta(\nu). $$

(B.6)

Note that from $\psi (\nu +1)=\frac {1}{\nu }+\psi (\nu )$ and ψ(1) = −γ, it follows that

$$ \psi(3)=\frac{1}{2}+1+\psi(1)=\frac{1}{2}\left (3-2\gamma\right). $$

(B.7)