Understanding Frequency in Rindler Coordinates for a Scalar Massless Field

[Haorong Wu]

TL;DR

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.

 

[Demystifier]

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?

 

[Haorong Wu]

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.