Hello,(adsbygoogle = window.adsbygoogle || []).push({});

I am well and truly stumped on Doppler shift. The complete expression is

[tex]f'=f \gamma (1-\beta cos(\theta))[/tex]

where:

[tex]\beta=v/c[/tex]

[tex]v=source\ velocity[/tex]

[tex]\gamma=\frac{1}{\sqrt{1-\beta^{2}}}[/tex]

[tex]\theta[/tex] is the angle between received photons (lab frame) and the direction of motion of the star.

Now for a directly receding source, where [tex]\theta=180^{0}[/tex] this gives the shifted frequency to be, after moving around,

[tex]f'=f*\sqrt{\frac{1+\beta}{1-\beta}}[/tex].

Now in a lot of text books this is the other way round ie,

[tex]f'=f*\sqrt{\frac{1-\beta}{1+\beta}}[/tex].

Why is this is so?? It's been giving me a headache for ages trying to figure it out.

Please help!

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Doppler shift clarification

Loading...

Similar Threads for Doppler shift clarification |
---|

B Why Light Experienced a Doppler Shift? |

I Redshift and atomic oscillations |

B Visual Effects of Light Travel |

B Why do photons allow Doppler shift |

I Could emission theory produce Doppler-shift formula for moving mirrors? |

**Physics Forums | Science Articles, Homework Help, Discussion**