5 Interferometric detection of the fluctuations.

The fluctuations introduced in the previous sections may - at least in principle - be detectable in interferometers built to search for gravitational waves [11,12] or higher order QED effects [13]. To estimate how much the fluctuations influence the propagation of light we proceed as follows: if we take a photon with frequency $\nu _{0}$ and wavelength $\lambda _{0}$ in a region where the gravitational potential vanishes, then its wavelength in a region with a (negative) gravitational potential $\Phi$ is approximately

$$\lambda ({\vec{x}},t)\approx \lambda _{0}\left( {1+{\frac{{\Phi ({\vec{x}},t)}% }{{c^{2}}}}}\right) .$$

while the frequency does not change (see [14,15]). The gravitationally loaded vacuum behaves as a refractive medium with refractive index

$$n({\vec{x}},t)\approx \left( {1-{\frac{{\Phi ({\vec{x}},t)}}{{c^{2}}}}}\right)$$

the optical path length changes, and the resulting phase change along a path ${\mathfrak{L}}$ is

$$\varphi _{L}(t) \approx {\frac{2\pi }{\lambda _{0}}}\int\limits_{{% \mathfrak{L}}}n{\left[ {{\vec{x}}(t+\ell /c-L/c),t+\ell /c-L/c}\right] \mathrm{d} % \ell }$$

$$\approx {\frac{2\pi }{\lambda _{0}}}\int\limits_{{\mathfrak{L}}}\left( {1... ...\vec{x}}(t+\ell /c-L/c),t+\ell /c-L/c)}}{{c^{2}}}}}\right) {% \mathrm{d} \ell }$$ (9)

where $\ell$ is the path length, $L$ is the total path length and 9 is the phase change at the end of the path at time $t$ (notice that the wave enters the path at an earlier time $t-L/c$). Thus in a two-arm interferometer with distinct paths ${\mathfrak{L}}_{1}$ and ${\mathfrak{L}}_{2}$ and a zero-field phase difference $\phi_{0}$ between the arms, the phase difference due quantum fluctuations is

$$\Delta \varphi (t) = (\varphi _{2}(t)-\varphi _{1}(t))-\varphi _{0}$$

$$= {\frac{{2 \pi}}{\lambda _{0}}}\left[\,\int\limits_{{\mathfrak{L}}... ...l /c-L/c),t+\ell /c-L/c}\right] }}{{c^{2}}}}}\right) \mathrm{d} \ell } \right .$$

$$\qquad \left . -\int\limits_{{\mathfrak{L}}_{1}}{\left( {1-{\frac... ...+\ell /c-L/c}\right] }}{{c^{2}}}}}\right) \mathrm{d} \ell }\right]-\varphi _{0}$$

$$= {\frac{{2 \pi}}{{\lambda _{0} c^{2}}}}\left\{ {\int\limits_{{\mat... ...}\right) ,t+{\frac{ \ell }{c}}-{\frac{L}{c}}}\right] \mathrm{d} \ell }} \right.$$

$$\qquad \left. -\int\limits_{{\mathfrak{L}}_{1}}{\Phi \left[ {{\ve... ...\right) ,t+{\frac{ \ell }{c}}-{\frac{L}{c}}}\right] \mathrm{d} \ell }\right\} ;$$ (10)

then we find the total irradiance at the end of the two paths:

$$I(t) = I_{1}+I_{2}+2\sqrt{I_{1}I_{2}}\cos \left[ {\varphi _{0}+\Delta \varphi (t)}\right]$$

$$\approx I_{1}+I_{2}+2\sqrt{I_{1}I_{2}}\cos \varphi _{0}-2\Delta \varphi (t)\sqrt{I_{1}I_{2}}\sin \varphi _{0}.$$

Therefore the variance of the irradiance noise is (we set $\int \int _{1,2} \mathrm{d} \ell _{1} \mathrm{d} \ell _{2} \equiv \int \int _{... ...mathfrak{L}}_{2},{\mathfrak{L}}_{2}} \mathrm{d} \ell _{1} \mathrm{d} \ell _{2}$)

$$\left\langle \left\vert I\right\vert ^{2}\right\rangle = 4I_{1}I_{2}\sin ^{2}\varphi _{0}\left\langle {\ \left\vert \Delta \varphi \right\vert ^{2}}\right\rangle$$

$$= {\frac{ {16 \pi ^{2} I_{1}I_{2} \sin ^{2}\varphi _{0}}}{{\lambda ... ...c{{\ell _{2}}}{c}}}\right) , {\frac{{ \ell _{2}}}{c}}}\right] }\right\rangle }.$$ (11)

The variance 11 does not depend on $t$ because of the stationarity of the phase noise 10, so that we can set $t=L/c$ and simplify the notation; likewise the correlation function of the irradiance is given by

$$R_{I}(\tau ) = \left\langle {\Delta I(t)\Delta I(t+\tau )}\right\rangle$$

$$= 4I_{1}I_{2}\sin ^{2}\varphi _{0}\left\langle {\Delta \varphi (t)\Delta \varphi (t+\tau )}\right\rangle$$

$$= {\frac{{16 \pi ^{2} I_{1}I_{2} \sin ^{2} \varphi _{0}}}{{\lambda ... ...}}}{c}}+\tau } \right) ,{\frac{{\ell _{2}}}{c}}+\tau }\right] }\right\rangle },$$

and, finally, the spectral density of the irradiance fluctuations, from the Wiener-Kintchine theorem, is

$$\left\langle {\left\vert \Delta I(f)\right\vert ^{2}}\right\rangle = \int\limits_{-\infty }^{+\infty }{R_{I}(\tau )\mathrm{e} ^{-2\pi \mathrm{i} f \tau }\mathrm{d} \tau}$$

$$= 2\int_{0}^{\infty }{R_{I}(\tau )\cos (2\pi f\tau )\mathrm{d} \tau }$$

$$= {\frac{{16 \pi ^{2} I_{1}I_{2}\sin ^{2}\varphi _{0}}}{{\lambda _{... ...ht),{\frac{{\ell _{2}}}{c}}+\tau }\right] }% \right\rangle \cos (2\pi f\tau ) .$$