- #1
fulis
- 6
- 0
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?