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## Main Question or Discussion Point

Hi,

I was looking up the formula for this on wikipedia and it said that the frequency shift is:

[itex]\frac{\sqrt{1 - v^2/c^2}}{1+\frac{v}{c}\cos{\theta}}[/itex]

In the case where the the emitter is directly above the observer when the photon arrives it says it simplifies to:

[itex]\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}[/itex]

Now, I can't get that to work out.

In this special case we have that:

[itex]\cos{\theta}=\frac{v}{c}[/itex]

So the doppler shift becomes:

[itex]\frac{\sqrt{1 - v^2/c^2}}{1+v^2/c^2}[/itex]

which doesn't cancel the numerator because there's a minus sign missing.

If you take the same situation but reverse the direction of v nothing changes because the new angle becomes [itex]\pi-\theta[/itex], so the sign of the cosine term cancels the minus sign from v, which is clearly wrong since it should go from a blueshift to a redshift. Is there supposed to be a +/- in front of the cosine term?

I was looking up the formula for this on wikipedia and it said that the frequency shift is:

[itex]\frac{\sqrt{1 - v^2/c^2}}{1+\frac{v}{c}\cos{\theta}}[/itex]

In the case where the the emitter is directly above the observer when the photon arrives it says it simplifies to:

[itex]\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}[/itex]

Now, I can't get that to work out.

In this special case we have that:

[itex]\cos{\theta}=\frac{v}{c}[/itex]

So the doppler shift becomes:

[itex]\frac{\sqrt{1 - v^2/c^2}}{1+v^2/c^2}[/itex]

which doesn't cancel the numerator because there's a minus sign missing.

If you take the same situation but reverse the direction of v nothing changes because the new angle becomes [itex]\pi-\theta[/itex], so the sign of the cosine term cancels the minus sign from v, which is clearly wrong since it should go from a blueshift to a redshift. Is there supposed to be a +/- in front of the cosine term?