- #1
Haorong Wu
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- TL;DR Summary
- I derive two different frequencies in Rindler coordinates in 2 and 4 dimensions. Why they are different?
I consider a scalar massless field obeying the Klein-Gordon equation ##\Box \psi=0 ##.
First, in Minkowski spacetime, a solution of the equation is $$ u_{\mathbf k}(x^\mu)=((2\pi)^3 2 \omega)^{-1/2} e^{ik_\mu x^\mu}$$ where ##\mathbf k=(\omega, \vec k)##. So this mode has a frequency of ##\omega##.
Next, in the right part of the Rindler coordinates in 2-dimensions, where ##t=\frac 1 a e^{a\xi} \sinh (a\eta)##, and ##z=\frac 1 a e^{a\xi} \cosh (a\eta)##, the metric is ##ds^2=e^{2a\xi}(-d\eta^2+d\xi^2)##. The solution to the Klein-Gordon equation will be $$g_k=(4 \pi \omega)^{-1/2} e^{-i\omega \eta+i k \xi}.$$ So this mode has a frequency of ##\omega##. I have read several papers, all of which consider this scenario. One of them says that the derivation in 4-dimensions will be similar.
Now, I am considering in 4-dimensional Rindler coordinates, where the metric is ## ds^2=e^{2a\xi}(-d\eta^2+d\xi^2)+dx^2+dy^2##. I solve the Klein-Gordon equation##\Box \psi=[e^{-2a\xi} (-\partial^2_\eta+\partial^2_\xi)+\partial^2_x+\partial^2_y]\psi=0 ## yielding $$g_k=(4 \pi \omega)^{-1/2} \exp [i(-e^{2a\xi} \omega \eta +e^{2a\xi} k_z \xi +k_x x+k_y y)] $$ where I have assumed the wavevector is ##\mathbf k=(\omega, k_1, k_2, k_3)##. Since the ##x## and ##y## axes are orthogonal to the acceleration direction, I choose ##k_x## and ##k_y## to be unaltered. I am not sure what the frequency of this mode is. I have defined its frequency to be ##\omega##, but from its expression, it should have a frequency of ##e^{2a\xi} \omega##. Why it is enlarged by a factor ##e^{2a\xi}## compared to the previous 2-dimensional case? If the accelerating observer has a proper acceleration ##a##, then ##\xi=0## and ##e^{2a\xi}=1##. But I am not sure about this argument because some following analyse, such as find the Bogoliubov transform between ##u_{\mathbf k} ## and ##g_k##, will involve the coordinate ##\xi##.
Also, I consider an accelerating observer, whose 4-velocity in Rindler coordinates will be ##\mathbf u=(e^{-a\xi}, 0, 0, 0)## since its coordinates in ##x##, ##y##, ##\xi## coordinates are constant and ##\mathbf u \cdot \mathbf u=-1##. So the frequency observed by this observer should be ##-\mathbf u \cdot \mathbf k=e^{a \xi} \omega=\omega## since ##\xi=0## for the observer.
So should I take ##\omega## to be the frequency of this mode? Then why its wave function displays a different frequency? Also, I do not see any Doppler effects. Should I worry about that?
Many thanks.
First, in Minkowski spacetime, a solution of the equation is $$ u_{\mathbf k}(x^\mu)=((2\pi)^3 2 \omega)^{-1/2} e^{ik_\mu x^\mu}$$ where ##\mathbf k=(\omega, \vec k)##. So this mode has a frequency of ##\omega##.
Next, in the right part of the Rindler coordinates in 2-dimensions, where ##t=\frac 1 a e^{a\xi} \sinh (a\eta)##, and ##z=\frac 1 a e^{a\xi} \cosh (a\eta)##, the metric is ##ds^2=e^{2a\xi}(-d\eta^2+d\xi^2)##. The solution to the Klein-Gordon equation will be $$g_k=(4 \pi \omega)^{-1/2} e^{-i\omega \eta+i k \xi}.$$ So this mode has a frequency of ##\omega##. I have read several papers, all of which consider this scenario. One of them says that the derivation in 4-dimensions will be similar.
Now, I am considering in 4-dimensional Rindler coordinates, where the metric is ## ds^2=e^{2a\xi}(-d\eta^2+d\xi^2)+dx^2+dy^2##. I solve the Klein-Gordon equation##\Box \psi=[e^{-2a\xi} (-\partial^2_\eta+\partial^2_\xi)+\partial^2_x+\partial^2_y]\psi=0 ## yielding $$g_k=(4 \pi \omega)^{-1/2} \exp [i(-e^{2a\xi} \omega \eta +e^{2a\xi} k_z \xi +k_x x+k_y y)] $$ where I have assumed the wavevector is ##\mathbf k=(\omega, k_1, k_2, k_3)##. Since the ##x## and ##y## axes are orthogonal to the acceleration direction, I choose ##k_x## and ##k_y## to be unaltered. I am not sure what the frequency of this mode is. I have defined its frequency to be ##\omega##, but from its expression, it should have a frequency of ##e^{2a\xi} \omega##. Why it is enlarged by a factor ##e^{2a\xi}## compared to the previous 2-dimensional case? If the accelerating observer has a proper acceleration ##a##, then ##\xi=0## and ##e^{2a\xi}=1##. But I am not sure about this argument because some following analyse, such as find the Bogoliubov transform between ##u_{\mathbf k} ## and ##g_k##, will involve the coordinate ##\xi##.
Also, I consider an accelerating observer, whose 4-velocity in Rindler coordinates will be ##\mathbf u=(e^{-a\xi}, 0, 0, 0)## since its coordinates in ##x##, ##y##, ##\xi## coordinates are constant and ##\mathbf u \cdot \mathbf u=-1##. So the frequency observed by this observer should be ##-\mathbf u \cdot \mathbf k=e^{a \xi} \omega=\omega## since ##\xi=0## for the observer.
So should I take ##\omega## to be the frequency of this mode? Then why its wave function displays a different frequency? Also, I do not see any Doppler effects. Should I worry about that?
Many thanks.