When observing the Doppler effect of a particular galaxy, is the wavelength change greater for long wavelengths? According to the simple Doppler effect formula, the wavelength change is proportional to the wavelength at rest and the observed speed.


1 Answer 1


Yes, cosmological redshift, and gravitational redshift in general, is generally proportional to wavelength. Within general relativity, light rays in vacuum are modeled as propagating along null geodesics, and the only difference in this description as applied to one wavelength of light (as seen by some observer) versus another is a rescaling of the geodesic's affine parameter, which uniformly rescales its associated frequency everywhere along the light ray's trajectory.

To be more explicit, if a light ray propagates along a null geodesic $\gamma(s)$ (here $\gamma$ is a map describing the null geodesic curve; it maps some real interval into the spacetime manifold) between observer $A$ at $\gamma(s_A)$ and observer $B$ at $\gamma(s_B)$ whose four-velocity vectors are $u_A$ and $u_B$, then the frequencies of the light ray as seen by these observers are

$$\omega_A = \langle \gamma'(s_A), u_A\rangle \quad \quad \omega_B = \langle \gamma'(s_B), u_B \rangle,$$

where the inner product is evaluated according to the spacetime metric.

To describe the same procedure for a light ray with initial frequency $\tilde \omega_A$ as measured by $A$, all that changes is that we describe the null geodesic via a reparameterization $\tilde \gamma(r)$ of $\gamma$ given by $\tilde \gamma(r) := \gamma(\frac{\tilde \omega_A}{\omega_A} r)$. Setting $r_A := \frac{\omega_A}{\tilde \omega_A} s_A$ and $r_B = \frac{\omega_B}{\tilde \omega_B} s_B$, this yields $$\tilde \gamma'(r_A) = \frac{\tilde \omega_A}{\omega_A} \gamma'(s_A) \quad \quad \tilde \gamma'(r_B) = \frac{\tilde \omega_A}{\omega_A} \gamma'(s_B),$$ so that $$\tilde \omega_B = \langle \tilde \gamma'(r_B), u_B \rangle = \frac{\tilde \omega_A}{\omega_A} \langle \gamma'(s_B), u_B \rangle = \tilde \omega_A \frac{\omega_B}{\omega_A},$$ and hence $\frac{\tilde \omega_B}{\tilde \omega_A} = \frac{\omega_B}{\omega_A}$. That is, the ratio between the the frequencies (and therefore wavelengths) observed by $A$ and $B$ is the same, irrespective of the initial frequency observed by $A$. Notice that precisely the same formalism and argument applies no matter the four-velocity of $A$ (so cosmologically with $A$ a galaxy, whether it is comoving or not), meaning the "Doppler" component of redshift thought of as due to relative motion is indeed captured here.

A caveat to the conclusion is that, in reality, light rays don't quite travel on null geodesics when they pass through matter due to nontrivial indices of refraction, and dispersion within media can mean that different frequencies would travel slightly different paths (not just differently parameterized paths), which can result in differing relative redshifts for different initial frequencies. I rather expect this is utterly negligible for cosmological observations, however.

  • $\begingroup$ I want to fully understand what you wrote, but first, I want to understand the relationship between the wavelength change caused by the Doppler effect of the short wavelength and the wavelength change of the long wavelength. The more blue the light, the smaller the wavelength change? $\endgroup$ Jun 2, 2021 at 5:26
  • $\begingroup$ If the observed speed is constant, is it right that the size of the red-light red-shift should be greater than the size of the blue-light red-shift? $\endgroup$ Jun 2, 2021 at 5:32
  • $\begingroup$ @빛나는밤 Indeed, the bluer the light, the smaller the wavelength change. The conclusion of my discussion was that the ratio $\frac{\lambda_B}{\lambda_A}$ is constant, no matter the initial wavelength $\lambda_A$. This implies what you say. That ratio depends only on the path the light travels through spacetime and the four-velocities of the observers $A$ and $B$. $\endgroup$
    – jawheele
    Jun 2, 2021 at 5:51

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