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## Main Question or Discussion Point

Hello,

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!

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!