## IrishBoy123 one year ago Relativistic Doppler Effect

1. 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.... $$f_{obs \ receding} = \sqrt{\frac{1-v/c}{1+v/c}}f_{source}$$

2. 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: $\Large \left( {{x_0},{\mathbf{x}}} \right) = \left( {ct,{\mathbf{x}}} \right)$

3. Michele_Laino

transform*

4. Michele_Laino

so we can write: $\Large \left\{ \begin{gathered} {k_0}' = \gamma \left( {{k_0} - {\mathbf{\beta }} \cdot {\mathbf{k}}} \right) \hfill \\ k{'_{||}} = \gamma \left( {{k_{||}} - {\mathbf{\beta }} \cdot {\mathbf{k}}} \right) \hfill \\ {\mathbf{k}}{'_ \bot } = {\mathbf{k}} \bot \hfill \\ \end{gathered} \right.$

5. 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

6. Michele_Laino

oops.. the second equation, is wrong, here is the right one:

7. Michele_Laino

$\Large \left\{ \begin{gathered} {k_0}' = \gamma \left( {{k_0} - {\mathbf{\beta }} \cdot {\mathbf{k}}} \right) \hfill \\ k{'_{||}} = \gamma \left( {{k_{||}} - \beta {k_0}} \right) \hfill \\ {\mathbf{k}}{'_ \bot } = {{\mathbf{k}}_ \bot } \hfill \\ \end{gathered} \right.$

8. Michele_Laino

nevertheless the Doppler effect comes from the first equation

9. 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!

10. Michele_Laino

:)

11. Astrophysics

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