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A Solution to the wave equation in Rindler coordinates

  1. Aug 12, 2016 #1
    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 $$e^{ik\xi \textbf{+}i{\omega}{\eta}}\,\,?$$
     
  2. jcsd
  3. Aug 12, 2016 #2

    Bill_K

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    In both cases, you haven't written the general solution, what you've written is a complete set of particular solutions, two different sets in Minkowski and Rindler coordinates. The general solution of the two-dimensional wave equation can be written f(x - t) + g(x + t) where f and g are arbitrary functions. It's easy to show that x - t is a function of ξ - η alone, and x + t is a function of ξ + η alone, so an equivalent form for the general solution is F(ξ - η) + G(ξ + η ). The particular solutions R and L result from making suitable choices for F and G. The value of R and L is that they may be used to expand arbitrary right- and left-going waves.
     
  4. Aug 12, 2016 #3

    dextercioby

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    Welcome back to PF, Bill_K. I've definitely missed your valuable contributions here.
     
  5. Aug 12, 2016 #4

    strangerep

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    +1.

    @WannabeNewton cried himself to sleep for a week when you (Bill_K) disappeared. :biggrin:
     
  6. Aug 13, 2016 #5

    atyy

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    Welcome back Bill_K!
     
  7. Aug 13, 2016 #6

    vanhees71

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    2016 Award

    Yeah! Welcome back!
     
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