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When rearranging the doppler shift equation in terms of wavelength to get in terms of $\beta$ we get: $\beta = \cfrac{(\lambda_{source})^2-(\lambda_{obs})^2}{(\lambda_{source})^2 + (\lambda_{obs})^2}$. My question is since $\beta>0$ when the observed wavelength is greater than the source e.g. redshifted, we would get a negative beta so do we have to rearrange the equation another way or would the answer just be the modulus of this equation. Hope this makes sense, thanks!

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Note that for the longitudinal Doppler shift \begin{equation} \left. \beta\boldsymbol{=} \begin{cases} \dfrac{v}{c}\boldsymbol{>}0\,,\quad \texttt{if observer moves away from the source}\vphantom{\dfrac{a}{\dfrac{a}{b}}}\\ \dfrac{v}{c}\boldsymbol{<}0\,,\quad \texttt{if observer moves towards the source} \end{cases} \right\} \tag{01}\label{01} \end{equation}

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    $\begingroup$ Ah okay that makes sense, not sure why I was so fixed on beta>0. Thanks! $\endgroup$ Commented Aug 16, 2021 at 22:18
  • $\begingroup$ Welcome @Dirac Delta Yeah. $\endgroup$
    – Frobenius
    Commented Aug 16, 2021 at 22:19
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    $\begingroup$ Why not exist your excellent images 😀? $\endgroup$
    – Sebastiano
    Commented Aug 16, 2021 at 23:22
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    $\begingroup$ @Frobenius I was joking 😄😄😄😄 $\endgroup$
    – Sebastiano
    Commented Aug 17, 2021 at 9:36

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