# Solving Klein-Gordon equation in the Rindler coordinates - the Unruh effect

I am reading 't Hooft's notes on Black Holes. I want to find the solutions of the Klein-Gordon equation $(\tilde{x},\tilde{y}, \rho, \tau)$ in the Rindler coordinates which are $$x=\tilde{x}\,\,\,\,\ y= \tilde{y}\,\,\,\,\,\,z=\rho \cosh{\tau}\,\,\,\,\,t=\rho \sinh{\tau}$$.  Firstly looking at the solution at first glance, why are there two parameters $\omega$, $\mu$, when the Klein-Gordon solution in Minkowski space has a solution with a parameter $\omega$ satisfying the mass on-shell condition.

Secondly, how do you get that solution? Following 't Hooft's advice, I wrote the usual solution with $k_3=0, k_0=\mu$ in terms of the Rindler coordinates and fourier transformed with time.

$$\Phi_{\mu}=\int d{\tau}\, e^{i{\tilde{k} \cdot \tilde{x}-\mu \rho \frac{e^{\tau}-e^{-\tau}}{2}}} e^{i\omega \tau}$$

Setting $s=e^{i\tau}$, I get the first expression of equation 17.7, but without the $e^{\tau}$ factors for $\alpha, \beta$. I am obviously, going wrong somewhere, I am Fourier transforming to frequency space, so why is $\tau$ still there? Shouldn't it be just $\mu$ and $\omega$ or $\mu$ and $\tau$?

Since you have Fourier transformed to extract the component with the single frequency $\omega$, its time dependence is given by $e^{-i\omega \tau}$. You multiply this factor after Fourier transformation to express the behavior of this single-frequency component in the time domain.
What you obtain from the above is the last expression in 17.7 (i.e., the one without $e^{\tau}$ factors). Then, using 17.9, you can derive the expression with the $e^{\tau}$ factors.
• Sorry for the late reply. Somehow i didn't get the notification for a answer on this question. I have no clue what you are talking about, I have $\Phi$, which is a function of time, and I am fourier transforming to the frequency domain. What does it mean when you say the transformed expression is time-independant, hasn't the time variable been changed to $\omega$? – user7757 Jul 1 '13 at 12:49
• What you really do here is to extract a solution corresponding to the single frequency $\omega$. Technically, you do this by Fourier transforming and then multiplying $e^{-i\omega \tau}$. – higgsss Jul 3 '13 at 12:07