IrishBoy123
  • IrishBoy123
Relativistic Doppler Effect
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
IrishBoy123
  • IrishBoy123
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.... \(f_{obs \ receding} = \sqrt{\frac{1-v/c}{1+v/c}}f_{source}\)
Michele_Laino
  • Michele_Laino
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: \[\Large \left( {{x_0},{\mathbf{x}}} \right) = \left( {ct,{\mathbf{x}}} \right)\]
Michele_Laino
  • Michele_Laino
transform*

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Michele_Laino
  • Michele_Laino
so we can write: \[\Large \left\{ \begin{gathered} {k_0}' = \gamma \left( {{k_0} - {\mathbf{\beta }} \cdot {\mathbf{k}}} \right) \hfill \\ k{'_{||}} = \gamma \left( {{k_{||}} - {\mathbf{\beta }} \cdot {\mathbf{k}}} \right) \hfill \\ {\mathbf{k}}{'_ \bot } = {\mathbf{k}} \bot \hfill \\ \end{gathered} \right.\]
Michele_Laino
  • Michele_Laino
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
Michele_Laino
  • Michele_Laino
oops.. the second equation, is wrong, here is the right one:
Michele_Laino
  • Michele_Laino
\[\Large \left\{ \begin{gathered} {k_0}' = \gamma \left( {{k_0} - {\mathbf{\beta }} \cdot {\mathbf{k}}} \right) \hfill \\ k{'_{||}} = \gamma \left( {{k_{||}} - \beta {k_0}} \right) \hfill \\ {\mathbf{k}}{'_ \bot } = {{\mathbf{k}}_ \bot } \hfill \\ \end{gathered} \right.\]
Michele_Laino
  • Michele_Laino
nevertheless the Doppler effect comes from the first equation
IrishBoy123
  • IrishBoy123
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!
Michele_Laino
  • Michele_Laino
:)
Astrophysics
  • Astrophysics
.

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