In multiple articles I have seen the age of a star, within the milky way, referred to as its redshift (typically denoted by $z$). I know that $z$ can be calculated as $z=\frac{\lambda_{obsv} - \lambda_{emit}}{\lambda_{emit}}$ but what I can't find is how this is then used to calculate the age of a star within our galaxy? What characteristic does the redshift show us (spin, velocity, temperature, etc.) and how is this used mathematically to calculate age? Or is it just a method to determine where it falls on the HR diagram?

  • $\begingroup$ Can you give some references? $\endgroup$ – pela May 9 '16 at 5:46
  • 1
    $\begingroup$ @RobJeffries ncbi.nlm.nih.gov/pubmed/26560034 "The typical redshift of formation for stars in the bulge with $[Fe/H]<-1$ is $z \approx 10$...half of them have binding energies $E<-1 \times 10^{-4} km^{2}s^{-2}$, which is consistent with a formation redshift of $z>-15$..." While there is reference to metallicity, the author(s) here are clearly using redshift in a manner to reference age. I know how metallicity can be used to estimate age, but here they repeatedly reference a redshift value that correlates to age. $\endgroup$ – NotSoSN May 9 '16 at 12:48
  • 3
    $\begingroup$ @RobJeffries Please don't confuse your lack of understanding of my question for a lack knowledge or integrity on my part. If you want references, ask, but it is completely unnecessary, rude, and insulting to make such remarks. If you can't answer, don't respond, if you need more info, ask, but don't ever speak to someone you don't know in a professional setting in a disrespectful manner, it only serves to show your own ignorance. $\endgroup$ – NotSoSN May 9 '16 at 12:52
  • 1
    $\begingroup$ @NotSoSN My comment was accurate. Your sources did not say what you claim in your question. I was going to answer your question, since you provided some context and revealed what the question is actually about. Now I won't bother. Cheerio. $\endgroup$ – ProfRob May 9 '16 at 12:57
  • 2
    $\begingroup$ @RobJeffries If you can't answer my question without being condescending I'm more than happy to find answers elsewhere. If my question was poorly phrased based on a misunderstanding of the reference it would be a simple task to point that out. A little respect and courtesy goes a long way. $\endgroup$ – NotSoSN May 9 '16 at 13:06

The redshift that is referred to is not a Doppler redshift, but a cosmological redshift. The difference is that the former is caused by the source moving through space, while the latter is caused by the "stretching" of the wavelength of the light as it travels through space.

The cosmological redshift happens gradually along its journey, and hence it is a measure of the distance to the source. But since traveling through space takes time, it is also a measure of the lookback time to the source — that is, the time that has passed since its emission. This notion of the term can thus be used as a timeline for phenomena in the Universe, and perhaps somewhat confusingly, it is sometimes used this way even when referring to local phenomena.

For instance, Earth was formed 4.54 billion years ("Gyr") ago. If some unrelated galaxy emitted light at the same time, and if that light reaches us today, then that galaxy must be at a particular distance$^\dagger$. During its journey it has been redshifted to $z \simeq 0.42$. Thus we may say that Earth was formed at redshift $0.42$.

Similarly, stars (in the Milky Way or elsewhere) that are said to have formed at redshift $z=10$ and $z=15$, can also be said to have formed 13.3 Gyr and 13.5 Gyr ago, respectively.

In the figure below, I plotted the relation between cosmological redshift and lookback time (assuming a Planck 2015 cosmology). Now hug each other, Rob and NotSoSN.


$^\dagger$5.4 billion lightyears, actually. Not just 4.54 billion lightyears, because space has expanded in the meantime.

  • $\begingroup$ Not to forget that redshift (and blueshift) can also be used to calculate the spin of a star. $\endgroup$ – Tanenthor May 12 '16 at 17:17
  • $\begingroup$ @Tanenthor: That effect is just a regular Doppler shift. $\endgroup$ – pela May 12 '16 at 19:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.