I Understanding Frequency in Rindler Coordinates for a Scalar Massless Field

Haorong Wu
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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.
 
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Hint 1: Take the 4-dimensional solution in Rindler coordinates and choose ##k_x=k_y=0##. Does it solve the 2-dimensional Klein-Gordon equation in Rindler coordinates?

Hint 2: Take the 4-dimensional Klein-Gordon equation in Rindler coordinates and multiply it (from the left) with ##e^{2a\xi}##. Can you solve this new equation?

Hint 3: Take the solution in Minkowski spacetime and express it in Rindler coordinates. Do you see something interesting? Take the solution in Rindler spacetime (in both 2 and 4 dimensions) and express it in Minkowski coordinates. Do you see something interesting?

Hint 4: When talking about solutions of the Klein-Gordon equation, sometimes you said "a solution", sometimes you said "the solution". Does the difference matter? If it does, then what's correct: "a" or "the"?

Hint 5: For any solution ##\psi(x)## in any system of coordinates ##x##, define frequency as
$$\Omega(x)=\frac{i\partial_0\psi(x)}{\psi(x)}$$
and apply the definition to various solutions. What do you obtain? Does this definition make sense?

Hint 6: For a moment forget relativity theory, both general and special. What does the word "frequency" mean? For what kind of phenomena is it defined?

Hint 7: Consider a table at rest in an inertial frame. At the table there is a classical pendulum swinging with the frequency ##\omega##, as seen by the observer not moving with respect to the table. What's the frequency of swinging as seen by an observer moving with velocity ##v## with respect to the table? What if ##v## is not constant, what's the frequency of swinging as seen by an observer moving with acceleration ##a##?

Conclusion: Can you give qualitative answers to all the questions above, without actual calculations? If you can, can you now answer your own question: how to understand frequency in Rindler coordinates?
 
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Thank you very much, @Demystifier for those hints.

A1: Yes, they must solve the 2-dimensional Klein-Gordon equation. They just scale the "frequencies" ##\omega## and ##k_z## by a factor ##e^{2a\xi}##.

A2: I do not understand this change. Multiplying a factor should not change the equation since the factor can be safely removed from both side. Anyway, it can also give equivalent solutions $$g_k=(4 \pi \omega)^{-1/2} \exp [i(- \omega \eta + k_z \xi +e^{-2a\xi}k_x x+e^{-2a\xi}k_y y)].$$ The transverse frequencies are shrank by a factor of ##e^{-2a\xi}##.

A3: The Minkowski solutions in Rindler coordinates read $$u_k=C \exp [i(- \omega \frac 1 a e^{a\xi} \sinh (a\eta) + k_z \frac 1 a e^{a\xi} \cosh (a\eta) +k_x x+k_y y)] .$$
The two-dimensional Rindler solutions in Minkowski coordinates read $$g_k=C \exp [i(-\omega arctanh (t/z) /a + k_z \ln (a^2(z^2-t^2)) /2a ] . $$
The four-dimensional Rindler solutions in Minkowski coordinates read $$g_k=C \exp [i(-a(z^2-t^2)\omega arctanh (t/z) + a(z^2-t^2)k_z \ln (a^2(z^2-t^2)) /2+k_x x+k_y y)] .$$
But I could not see their relations, other than that the Minkowski solutions are linear combinations of Rindler solutions, and vice versa.

A4: Maybe I should always stick to "a" because frequencies can take an arbitrary value.

A5: This equation will give the number before the time coordinate. In Rindler coordinates, the two-dimensional solutions will give ##\omega ##, while the four-dimensional solutions will give ##e^{2a\xi} \omega ##. In a way, they are both right for frequencies. I think the difference is for the four-dimensional solutions, the frequencies are bounded by the transverse frequency ##k_x##, and ##k_y##, while for the two-dimensional solutions, the frequencies have no restrictions from them.

A6: A frequency is the reciprocal of a period, and a period is a time after which the wave will be the same. So I think the definition in hint 5 is well-defined because after substituting ##\eta \rightarrow \eta+2\pi/ \omega## or ##\eta \rightarrow \eta+2\pi/ e^{2a\xi}\omega## in 2- or 4- dimensional solutions, respectively, they all induce a factor of ##e^{i2\pi}=1##.

A7: This can be solved by the doppler shift equations. Whether ##v## is constant or not should be the same. If the observer is accelerating, we may use his or her velocity at a given time.So, the frequency is given by the number before the time coordinate. If we let the parameter ##a## in Rindler metric be the proper acceleration ##\alpha## of a Rindler observer, then ##\xi=0## and ##e^{2a\xi} \omega = \omega##. So the Doppler effect is not taken into consideration.
 
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