Cosmological Redshift: Why Does k Remain the Same?

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SUMMARY

The discussion centers on the behavior of the wavenumber (k) of photons in an expanding universe, specifically addressing why k remains unchanged while the wavelength (λ) increases and frequency (f) decreases due to cosmic redshift. Participants argue that k should not remain constant, as it is defined as 1/λ, suggesting that if λ increases, k must decrease correspondingly. The relationship between k and f is also examined, with references to the four-vector representation of light waves and the implications of propagation in vacuum versus other mediums.

PREREQUISITES
  • Understanding of cosmological redshift and its implications
  • Familiarity with the concepts of wavenumber (k), wavelength (λ), and frequency (f)
  • Knowledge of four-vectors in physics, particularly in the context of light waves
  • Basic grasp of phase velocity and its relationship to light propagation
NEXT STEPS
  • Research the mathematical relationship between wavenumber (k) and wavelength (λ) in cosmology
  • Explore the implications of cosmic redshift on light propagation in different mediums
  • Study the properties of four-vectors and their applications in relativistic physics
  • Investigate the concept of phase velocity and its effects on wave behavior in various contexts
USEFUL FOR

Astronomy students, physicists, and anyone interested in the fundamental principles of light behavior in an expanding universe.

karlzr
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condidering a photon in expanding cosmos, it 's said that the wavenumber k remains unchanged, the wavelength \lambda increases, proportional to the scale factor a(t) of the universe, and the frequency w decreases in the opposite way, that is the cosmic redshift.

so, why does k remain the same?
 
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Hi, karlzr -- By the way, to make Greek letters show up, you have to surround them in itex tags. To see how to do that, click on the Quote button for this post, and look at how I did this one: \lambda.

I don't think it's true that k stays the same. The four-vector (f,k) is a lightlike vector for a light wave, so f=k (in units where c=1). Since f experiences a cosmological red-shift, k must go down as well. Also, as far as I know k is simply defined as 1/\lambda, so 1/\lambda can't vary independently of k.

Conceivably the answer could be different if you weren't talking about propagation in vacuum, since the k and f are related by the phase velocity, which doesn't have to equal c. So then I could imagine that it would be possible to have f and k not related in the usual way, although I think you'd still have k=1/\lambda as a matter of definition.
 

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