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IrishBoy123

  • one year ago

Relativistic Doppler Effect

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  1. IrishBoy123
    • one year ago
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    Does anyone have a simple proof for this, ideally just a Lorentz transform or a few lines of calculus, and not a whole load of algebra.... \(f_{obs \ receding} = \sqrt{\frac{1-v/c}{1+v/c}}f_{source}\)

  2. Michele_Laino
    • one year ago
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    here we have to keep in mind that the wave quadrivector: \[\Large \left( {{k_0},{\mathbf{k}}} \right)\] will transfor itself like the position quadrivector: \[\Large \left( {{x_0},{\mathbf{x}}} \right) = \left( {ct,{\mathbf{x}}} \right)\]

  3. Michele_Laino
    • one year ago
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    transform*

  4. Michele_Laino
    • one year ago
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    so we can write: \[\Large \left\{ \begin{gathered} {k_0}' = \gamma \left( {{k_0} - {\mathbf{\beta }} \cdot {\mathbf{k}}} \right) \hfill \\ k{'_{||}} = \gamma \left( {{k_{||}} - {\mathbf{\beta }} \cdot {\mathbf{k}}} \right) \hfill \\ {\mathbf{k}}{'_ \bot } = {\mathbf{k}} \bot \hfill \\ \end{gathered} \right.\]

  5. Michele_Laino
    • one year ago
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    now if we consider the case in which \[\Large {\mathbf{\beta }},\;{\mathbf{k}}\] have the same direction, we can rewrite the first equation, as below:: \[\Large \omega ' = \gamma \omega \left( {1 - \beta } \right)\] being: \[\Large {k_0} = \frac{\omega }{c}\] Now keeping mind that: \[\Large \beta = \frac{V}{c},\;\gamma = \frac{1}{{\sqrt {1 - \frac{{{V^2}}}{{{c^2}}}} }}\] where V is the relative speed between those inertial system to which our Lorentz transforms are referring to, we got your relationship

  6. Michele_Laino
    • one year ago
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    oops.. the second equation, is wrong, here is the right one:

  7. Michele_Laino
    • one year ago
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    \[\Large \left\{ \begin{gathered} {k_0}' = \gamma \left( {{k_0} - {\mathbf{\beta }} \cdot {\mathbf{k}}} \right) \hfill \\ k{'_{||}} = \gamma \left( {{k_{||}} - \beta {k_0}} \right) \hfill \\ {\mathbf{k}}{'_ \bot } = {{\mathbf{k}}_ \bot } \hfill \\ \end{gathered} \right.\]

  8. Michele_Laino
    • one year ago
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    nevertheless the Doppler effect comes from the first equation

  9. IrishBoy123
    • one year ago
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    that looks wonderful and exactly what i was looking for. i only need to do it along x so i can probably simplify even more, eg draw it on some frames. i've been going round in circles on this - i tried just transforming the wavelength in the hope it would "just work out" but it doesn't - so thank you very much!

  10. Michele_Laino
    • one year ago
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    :)

  11. Astrophysics
    • one year ago
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    .

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