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IrishBoy123
 one year ago
Relativistic Doppler Effect
IrishBoy123
 one year ago
Relativistic Doppler Effect

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IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0Does 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.... \(f_{obs \ receding} = \sqrt{\frac{1v/c}{1+v/c}}f_{source}\)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4here 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
 one year ago
Best ResponseYou've already chosen the best response.4so 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
 one year ago
Best ResponseYou've already chosen the best response.4now 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
 one year ago
Best ResponseYou've already chosen the best response.4oops.. the second equation, is wrong, here is the right one:

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4\[\Large \left\{ \begin{gathered} {k_0}' = \gamma \left( {{k_0}  {\mathbf{\beta }} \cdot {\mathbf{k}}} \right) \hfill \\ k{'_{}} = \gamma \left( {{k_{}}  \beta {k_0}} \right) \hfill \\ {\mathbf{k}}{'_ \bot } = {{\mathbf{k}}_ \bot } \hfill \\ \end{gathered} \right.\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4nevertheless the Doppler effect comes from the first equation

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0that 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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