# Proof of expression to combine different redshifts

I am trying to arrive to the following expression:

$1+z = (1+z_C)(1+z_G)(1+z_D)$

so the total redshift is the product of the cosmological redshift, the gravitational redshift and the Doppler redshift. I know that

$1+z = \frac{\lambda_{observed}}{\lambda_{emitted}}$

but I don't know why would you make the product of them in order to get the total redshift. How do you derive the formula of the total redshift?

• Add some context. What are $z,z_C,z_G,z_D$? – glS Jan 4 '15 at 11:46
• $z_C$ is the cosmological redshift, $z_G$ is the gravitational redshift, $z_D$ is the Doppler redshift, and $z$ is the total redshift. – Cosmo_1 Jan 4 '15 at 12:03

Suppose light starts out with wavelength $\lambda_\mathrm{emit}$. If you were not moving with respect to the source in either a proper sense (Doppler shift) or comoving sense (cosmological redshift), but the light has to climb out of a potential well, it would have wavelength $\lambda_\mathrm{G} = (1 + z_\mathrm{G}) \lambda_\mathrm{emit}$.
Now consider a photon emitted with $\lambda_\mathrm{G}$, but cosmologically far away. You would observe the wavelength $\lambda_\mathrm{CG} = (1 + z_\mathrm{C}) \lambda_\mathrm{G}$.
Finally, suppose you are receiving photons of wavelength $\lambda_\mathrm{CG}$, but then you start moving very fast. Now those same photons will appear to you to have wavelength $\lambda_\mathrm{CGD} = (1 + z_\mathrm{D}) \lambda_\mathrm{CG}$.
With all three effects at work, you simply multiply the factors: $$\lambda_\mathrm{CGD} = (1 + z_\mathrm{C}) (1 + z_\mathrm{G}) (1 + z_\mathrm{D}) \lambda_\mathrm{emit}.$$ The same argument holds for any combination of these effects in any order: $$\lambda_\mathrm{obs} = \lambda_\mathrm{emit} \prod_i (1 + z_i).$$