7
$\begingroup$

How the blueshift from Andromeda galaxy was measured? Since the measurement of blueshift/redshift has to do with the identification of relative distance of spectral lines how this identification is possible under the influence of such a big number of stars with very different characteristics?

$\endgroup$
  • $\begingroup$ You mean like this astrosurf.com/buil/redshift/demo.htm? It looks like even hobby astronomy equipment can do this reasonably well today. $\endgroup$ – CuriousOne Jan 6 '15 at 8:06
  • $\begingroup$ @CuriousOne If this is the or one method to see the blueshift then yes. My question is what represent this spectral lines? All stars together? $\endgroup$ – HolgerFiedler Jan 6 '15 at 8:11
  • $\begingroup$ I see. Given the aperture of this size telescope, yes, it's the combined spectra of a large number of stars in the galaxy. As you can see, the slit is a lot narrower than the galaxy, so it's by far not the entire galactic light that is being used. Andromeda is close enough to see individual giant stars in it, but those have a significant relative motion with regards to its center of mass themselves, so the individual blueshift measurement is probably not going to return much more information than the combined spectrum. $\endgroup$ – CuriousOne Jan 6 '15 at 8:27
4
$\begingroup$

The measurement of the velocity vector of a nearby, large galaxy is not a simple matter. If you just dropped a spectrograph slit down randomly somewhere in Andromeda then you could get a wide variety of answers, since it rotates with velocities of $\sim +/- 200$ km/s in different parts of its disk.

To estimate the centre of mass redshift one must use different tracers at different positions in the Galaxy. One of the more accurate techniques is using Hydrogen 21cm radiation. An early study is shown in van de Hulst et al. (1957). Fig.5 in that paper shows the change in radial velocity as one moves along the major axis of the galaxy from one side to another. One then models these with some density weighted rotation curve to estimate what the centre of mass velocity is (obviously quite close the the velocity measured in the brightest part of Andromeda). van de Hulst got -296 km/s (towards us) with an uncertainty of less than 10 km/s. A more modern value is $-301 \pm 1$ km/s (van der Marel & Guhathakurta 2008).

This line of sight velocity of the centre of mass of Andromeda is incredibly precise, but this includes the motion of our Sun around our Galactic centre. Correcting for this puts Andromeda on a collision course with the Galaxy with a closing speed of around 100 km/s.

$\endgroup$
  • $\begingroup$ Good to see you chose to stick around. :-) $\endgroup$ – Jim Apr 2 '15 at 13:03
7
$\begingroup$

It's true that stars can be very different but the spectral lines are the same in all of them, though the relative intensities of the lines will differ from star to star. So if for example you're measuring the Lyman $\alpha$ line then every star in Andromeda will have this line at the same frequency. This allows you to measure an average position for this line for all of Andromeda and get an average velocity for all the stars in Andromeda.

This is what Slipher did in 1913 (I'm not sure what lines he used) to get the relative velocity of 300km/sec.

Because Andromeda is rotating the relative velocity of individual stars depends on their distance from the centre of Andromeda and which side they're on (relative to us). So the 300 km/sec can only be an average over all stars - the lines would be broadened by the motion with the galaxy. For nearby galaxies we can resolve the rotational motion and measure the galaxy rotation curve.

$\endgroup$
  • $\begingroup$ Slipher's paper says only that he used "the region of spectrum from F to H." I wonder if those are references to Fraunhofer lines, which Wikipedia tells me would be 486-397 nm and would include Hβ-Hδ and some iron and calcium. This makes some amount of sense given he used a spectrum of (sunlight reflected off) Saturn for reference. $\endgroup$ – user10851 Jan 6 '15 at 11:37

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.