Thermal States and black holes II

Following from 

Hong Liu's Lectures on Holography and,

Gustavo Cesar Valdivia Mera's https://arxiv.org/pdf/2001.09869.pdf

Considering Rindler Space:

   Starting with the accelerated observer:

        $x^\mu = (x^0;x^1$

        $u^\mu = (\frac{dx^0}{d\tau}; \frac{dx^1}{d \tau})$

                   $= (\gamma;\gamma v)$

              Instantaneous co-moving frame

        $a^\mu = (0;g)$ 

        $a^\mu a_nu = a^0a_0 + a^1 a_1 = g^2$

 $t(\tau ) = \frac{1}{g} \sinh({g \tau})$

 $x(\tau) = \frac{1}{g} \cosh ({g\tau})$

. . . . . ,  and so on from the arxiv article

then in exponential terms  

  $x = \frac{ \frac{e^{g \tau}}{g} + \frac{e^{g \tau}}{g}}{2}$

  $t = \frac{  \frac{e^{g \tau}}{g} -  \frac{e^{g \tau}}{g}}{2}$

$\bar{v} = t + x$

$\bar{u} = t - x$

$\bar{v} = \frac{e^{g \tau}}{g}$

$\bar{u} = - \frac{e^{g \tau}}{g}$

setting spatial coordinate in Rindler space to $\xi$

. . . . .

$x = \frac{e^{g\xi}}{g}\cosh(g \tau)$

$t = \frac{e^{g\xi}}{g}\sinh(g \tau)$

$ds^2 = e^{2g \xi}(-d\tau^2 + d\xi^2)$

Now one can consider massless scalar field theory in Rindler space.

             Following  (97) through (103) one arrives at:

                 $\Box \phi(t;x) = \Box \phi(\tau;\xi) = 0$

. . . . . .

 From (107)   

$\phi(x^\mu) = \int_0^\infty \frac{dk}{\sqrt{4 \pi \omega_k}} (a(k) e^{ik^\mu x_\mu} + a^{\dagger}(k) e^{-ik^\mu x_\mu})  + \int_{-\infty}^0\frac{dk}{\sqrt{4 \pi \omega_k}} (a(k) e^{ik^\mu x_\mu} + a^{\dagger}(k) e^{-ik^\mu x_\mu})$

 From (108)

$\phi(x^\mu) = \int_0^\infty \frac{dk}{\sqrt{4 \pi \omega_k}} (a(k) e^{i{- \omega_k t + kx}} + a^{\dagger}(k) e^{i{- \omega_k t + kx}})  + (b(k) e^{i{- \omega_k t - kx}} + b^{\dagger}(k) e^{i{- \omega_k t - kx}})$

From (109) through (115) one gets the Right modes and Left modes

$\phi(\tau,\xi) = \int_0 ^{\infty} d\omega (c(\omega) h_\omega + c^{\dagger}(\omega)h^* _\omega + d(\omega) j_\omega + d^\dagger (\omega) j^*_\omega)$

$h_\omega = \frac{e^{- i \omega(\tau - \xi)}}{\sqrt{4 \pi \omega}}$

$j_\omega = \frac{e^{- i \omega(\tau + \xi)}}{\sqrt{4 \pi \omega}}$