Monte Carlo study of the BMV vacuum linear magnetic birefringence experiment

Abstract

QED vacuum can be polarized and magnetized by an external electromagnetic field, therefore acting as a birefringent medium. This effect has not yet been measured. In this paper, after having recalled the main facts concerning vacuum magnetic birefringence polarimetry detection method and the related noise sources, we detail our Monte Carlo simulation of a pulsed magnetic field data run. Our Monte Carlo results are optimized to match BMV experiment 2014 data. We show that our Monte Carlo approach can reproduce experimental results giving an important insight into the systematic effects limiting experiment sensitivity.

Graphic abstract

Appendix: discussion on the intensity ratio formula

The formula used in 2014 to determine the ellipticity from the intensities is in our notation

$$\begin{aligned} P_e(t) = \hat{C}[P_t(t)]\left( \sigma ^{2}+(\varGamma +\hat{C}[\varPsi (t)])^{2}\right) , \end{aligned}$$

(30)

in this “Appendix” we will show that the error performed from the exact expression Eq. 11 is negligible. We recall that we have defined the operator $\hat{C}$ associated to the cavity transfer function $H(\nu )$ with a cutoff frequency $\nu _c$ so that $\varPsi _{f}(t)=\hat{C}[\varPsi (t)]=\mathscr {F}^{-1}[H(\nu )\mathscr {F}[\varPsi (t)]]$ where $\mathscr {F}$ is the operator of the Fourier transform.

We consider that the ordinary intensity shows a small modulation around a DC value as $P_{t}(t)=P_{t}^{\mathrm{DC}}\left( 1+\beta \cos \omega t\right)$ with $\beta \ll 1$. To simplify we will also consider that the magnetic field is oscillating as $B(t)=A\cos \omega _{\mathrm{B}}t$; therefore, the ellipticity $\varPsi (t)$ can be written as

$$\begin{aligned} \varPsi (t)&=\alpha B^{2}(t)=\frac{\alpha A}{2}\left( 1+\cos 2\omega _{{B}}t\right) \nonumber \\&=\varPsi ^{{DC}}\left( 1+\cos 2\omega _{{B}}t\right) . \end{aligned}$$

(31)

We calculate $P_{e}(t)$ using Eq. 11 taking the case $\varPsi (t)\ll \varGamma$ so

$$\begin{aligned} P_{e}(t)\simeq \sigma ^{2}P_{t}(t)+\hat{C}\left( P_{t}(t)[\varGamma ^{2}+2\varGamma \varPsi (t)]\right) ; \end{aligned}$$

(32)

injecting our expressions for $P_{t}(t)$ and $\varPsi (t)$, we obtain

$$\begin{aligned} P_{e}(t)&=\sigma ^{2}P_{t}(t)+\hat{C}\left( P_{t}^{{DC}}\left( \varGamma ^{2}+2\varGamma \varPsi (t)\right) \right) \nonumber \\&\quad + \hat{C}\left( P_{t}^{{DC}}\beta \cos \omega t\left( \varGamma ^{2}+2\varGamma \varPsi (t)\right) \right) \nonumber \\&=\sigma ^{2}P_{t}(t)+P_{t}^{{DC}}\varGamma ^{2}+2\varGamma P_{t}^{{DC}}\varPsi _{f}(t)\nonumber \\&\quad +P_{t}^{{DC}}\varGamma ^{2}\beta \hat{C}\left( \cos \omega t\right) +2\varGamma P_{t}^{{DC}}\beta \hat{C}\left( \varPsi (t)\cos \omega t\right) \nonumber \\&=\sigma ^{2}P_{t}^{{DC}}+\sigma ^{2}P_{t}^{{DC}}\beta \cos \omega t +P_{t}^{{DC}}\varGamma ^{2}+2\varGamma P_{t}^{{DC}}\varPsi _{f}(t)\nonumber \\&\quad +P_{t}^{{DC}}\varGamma ^{2}\beta \hat{C}\left( \cos \omega t\right) +2\varGamma P_{t}^{{DC}}\beta \varPsi ^{{DC}}\hat{C}\left( \cos \omega t\right) \nonumber \\&\quad +2\varGamma P_{t}^{{DC}}\beta \varPsi ^{{DC}}\hat{C}\left( \cos \omega t\cos 2\omega _{{B}} t\right) . \end{aligned}$$

(33)

In order to calculate the ellipticity $\varPsi ^{2014}(t)$ of 2014, we have to determine the inverse of $P_{t,f}(t)=\hat{C}\left( P_{t}(t)\right)$

$$\begin{aligned} \frac{1}{P_{t,f}(t)}=\frac{1}{P_{t}^{{DC}}\left( 1+\beta C\left( \cos \omega t\right) \right) } \simeq \frac{1}{P_{t}^{{DC}}}\left( 1-\beta C\left( \cos \omega t\right) \right) , \end{aligned}$$

because $\beta \ll 1$. We calculate the ratio $P_{e}(t)/P_{t,f}(t)$, neglecting of terms in $\beta ^{2}$

$$\begin{aligned} \frac{P_{e}(t)}{P_{t,f}(t)}&\simeq \sigma ^{2}+\varGamma ^{2}+2\varGamma \varPsi _{f}(t)+\sigma ^{2}\beta \left( \cos \omega t-\hat{C}\left( \cos \omega t\right) \right) \nonumber \\&\quad -2\varGamma \beta \varPsi _{f}(t)\hat{C}\left( \cos \omega t\right) +2\varGamma \beta \varPsi ^{\mathrm{DC}}\hat{C} \left( \cos \omega t\right) \nonumber \\&\quad +2\varGamma \beta \varPsi ^{{DC}}\hat{C}\left( \cos \omega t\cos 2\omega _{{B}}t\right) \nonumber \\&=\sigma ^{2}+\varGamma ^{2}+2\varGamma \varPsi _{f}(t)+\sigma ^{2}\beta \left( \cos \omega t-\hat{C}\left( \cos \omega t\right) \right) \nonumber \\&\quad +2\varGamma \beta \varPsi ^{{DC}}\left( \hat{C}\left( \cos \omega t\cos 2\omega _{{B}}t\right) \right. \nonumber \\&\quad -\left. \hat{C}\left( \cos \omega t\right) \hat{C}\left( \cos 2\omega _{{B}}\right) \right) . \end{aligned}$$

(34)

We finally obtain the ellipticity as it was computed in 2014, that is $\varPsi ^{2014}(t)=\left( \frac{P_{e}(t)}{P_{t,f}(t)}-\sigma ^{2}-\varGamma ^{2}\right) /2\varGamma$ and we calculate the error as $\delta (t)=\varPsi ^{2014}(t)-\varPsi _{f}(t)$

$$\begin{aligned} \delta (t)&=\frac{\sigma ^{2}\beta }{2\varGamma }\left( \cos \omega t-\hat{C}\left( \cos \omega t\right) \right) \nonumber \\&\quad +\beta \varPsi ^{{DC}}\left( \hat{C}\left( \cos \omega t\cos 2\omega _{{B}}t\right) \right. \nonumber \\&\quad -\left. \hat{C}\left( \cos \omega t\right) \hat{C}\left( \cos 2\omega _{{B}}\right) \right) \nonumber \\&=\frac{\sigma ^{2}\beta }{2\varGamma }\left( \cos \omega t-\hat{C}\left( \cos \omega t\right) \right) \nonumber \\&\quad +\beta \varPsi ^{{DC}}\left( \frac{1}{2}\hat{C}\left( \cos (\omega +2\omega _{{B}})t\right) \right. \nonumber \\&\quad +\left. \frac{1}{2}\hat{C}\left( \cos (\omega - 2\omega _{{B}})t\right) -\hat{C}\left( \cos \omega t\right) \hat{C}\left( \cos 2\omega _{{B}}\right) \right) . \end{aligned}$$

(35)

We first study the case where $\omega \ll \omega _{c}$, because a magnetic shot typically last about ten milliseconds we can consider that during the duration of the shot we have $\omega t\ll 1$, so

$$\begin{aligned} \delta (t)\simeq \beta \varPsi ^{{DC}}\left( \hat{C} \left( \cos 2\omega _{{B}}t\right) -\hat{C}\left( \cos 2\omega _{{B}}t\right) \right) =0. \end{aligned}$$

(36)

In the limit $\omega \gg \omega _{c}$, because the cavity filtering on a cosinus is

$$\begin{aligned} \hat{C}\left( \cos \omega t\right) =\frac{1}{\sqrt{1+(\omega /\omega _{c})^{2}}} \cos \left( \omega t -\arctan (\omega /\omega _{c})\right) , \end{aligned}$$

(37)

so for this limit we have

$$\begin{aligned} \delta (t)\simeq \frac{\sigma ^{2}\beta }{2\varGamma }\cos \omega t. \end{aligned}$$

(38)

If we use the typical parameters of an experiment like the one in Ref. [6], we found that the amplitude of the error is of the order of $10^{-9}$.

We can therefore consider that the formula used in 2014, although incorrect, presents a negligible error.