- #1
fisicist
- 46
- 7
Hi all!
I've got a question about the cosmological redshift. We're given the metric
[itex]ds^2 = c^2\,dt^2 - a(t)^2 \left[ dr^2 + r^2\,d\theta^2 + r^2\sin^2 \theta\,d\varphi^2 \right][/itex]
Now light moves on null geodesics, so [itex]c^2\,dt^2 - a(t)^2\,dr^2[/itex] for radially moving light. For a GR exercise, we are asked to calculate the redshift. In order to do so, we are given the clue that [itex]\lambda = c / f[/itex]. Why is that true? Why don't we use c/a instead of c? By using c, I get, indeed, the correct result, whereas by using c/a I would get that the observed wavelength and the emitted wavelength are equal.
Thank you!
I've got a question about the cosmological redshift. We're given the metric
[itex]ds^2 = c^2\,dt^2 - a(t)^2 \left[ dr^2 + r^2\,d\theta^2 + r^2\sin^2 \theta\,d\varphi^2 \right][/itex]
Now light moves on null geodesics, so [itex]c^2\,dt^2 - a(t)^2\,dr^2[/itex] for radially moving light. For a GR exercise, we are asked to calculate the redshift. In order to do so, we are given the clue that [itex]\lambda = c / f[/itex]. Why is that true? Why don't we use c/a instead of c? By using c, I get, indeed, the correct result, whereas by using c/a I would get that the observed wavelength and the emitted wavelength are equal.
Thank you!