Relativistic Doppler Effect

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Relativistic Doppler Effect

Physics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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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.... \(f_{obs \ receding} = \sqrt{\frac{1-v/c}{1+v/c}}f_{source}\)
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: \[\Large \left( {{x_0},{\mathbf{x}}} \right) = \left( {ct,{\mathbf{x}}} \right)\]
transform*

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