- #1
Wise Owl
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I have been reading these notes on Rindler coordinates for an accelerated observer. In Rindler coordinates, the hyperbolic motion of the observer is expressed through the coordinate transformation
$$t=a^{-1}e^{a{{\xi}}}\sinh a{\eta}\\
{}x=a^{-1}e^{a{{\xi}}}\cosh a{\eta}.$$On a space-time diagram, the null light rays act as a horizon for the observer. This is because light sent from outside the observer's "Rindler wedge" can never catch up.
Now let's consider the wave equation. In regular Minkowski space, the equation reads $$\square\,\varphi = \bigg(\frac{{\partial}^2}{{\partial}t^2}-\frac{{\partial}^2}{{\partial}x^2}\bigg)\,\varphi=0$$ with a general solution corresponding to plane waves $$\varphi = e^{\pm ikx-i{\omega}t}.$$ In Rindler coordinates, the wave equation is $$\square\,\varphi = e^{-2a \xi}\bigg(\frac{{\partial}^2}{{\partial}\eta^2}-\frac{{\partial}^2}{{\partial}\xi^2}\bigg)\,\varphi=0$$ Since this equation has the same form of that for Minkowski space, I would expect the solutions to be the same. However, in the notes, the solution depends on what region of space-time is being considered. Specifically, the given solution is (see eqs. (17), (18) )
$$^R\varphi =
\begin{cases}
e^{ik\xi -i{\omega}{\eta}} & \text{in }R \\
0 & \text{in }L
\end{cases}\\
^L\varphi =
\begin{cases}
0 & \text{in }R \\
e^{ik\xi +i{\omega}{\eta}} & \text{in }L
\end{cases}
$$where $$^R\varphi$$ and $$^L\varphi$$ correspond to the solutions in R and L, the right and left "Rindler wedges", respectively. The sum of these solutions is the general solution to the wave equation over the entire space-time.
My question:
1. Why does the solution need to be broken down into these L and R wedges? Since the wave equation is identical to that for regular Minkowski space, why wouldn't the solution be identical as well?
2. How are the signs chosen for each wedge? For instance, why does the right wedge have $$e^{ik\xi \textbf{-}i{\omega}{\eta}}$$ while the left wedge has $$e^{ik\xi \textbf{+}i{\omega}{\eta}}\,\,?$$
$$t=a^{-1}e^{a{{\xi}}}\sinh a{\eta}\\
{}x=a^{-1}e^{a{{\xi}}}\cosh a{\eta}.$$On a space-time diagram, the null light rays act as a horizon for the observer. This is because light sent from outside the observer's "Rindler wedge" can never catch up.
Now let's consider the wave equation. In regular Minkowski space, the equation reads $$\square\,\varphi = \bigg(\frac{{\partial}^2}{{\partial}t^2}-\frac{{\partial}^2}{{\partial}x^2}\bigg)\,\varphi=0$$ with a general solution corresponding to plane waves $$\varphi = e^{\pm ikx-i{\omega}t}.$$ In Rindler coordinates, the wave equation is $$\square\,\varphi = e^{-2a \xi}\bigg(\frac{{\partial}^2}{{\partial}\eta^2}-\frac{{\partial}^2}{{\partial}\xi^2}\bigg)\,\varphi=0$$ Since this equation has the same form of that for Minkowski space, I would expect the solutions to be the same. However, in the notes, the solution depends on what region of space-time is being considered. Specifically, the given solution is (see eqs. (17), (18) )
$$^R\varphi =
\begin{cases}
e^{ik\xi -i{\omega}{\eta}} & \text{in }R \\
0 & \text{in }L
\end{cases}\\
^L\varphi =
\begin{cases}
0 & \text{in }R \\
e^{ik\xi +i{\omega}{\eta}} & \text{in }L
\end{cases}
$$where $$^R\varphi$$ and $$^L\varphi$$ correspond to the solutions in R and L, the right and left "Rindler wedges", respectively. The sum of these solutions is the general solution to the wave equation over the entire space-time.
My question:
1. Why does the solution need to be broken down into these L and R wedges? Since the wave equation is identical to that for regular Minkowski space, why wouldn't the solution be identical as well?
2. How are the signs chosen for each wedge? For instance, why does the right wedge have $$e^{ik\xi \textbf{-}i{\omega}{\eta}}$$ while the left wedge has $$e^{ik\xi \textbf{+}i{\omega}{\eta}}\,\,?$$
