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Distance Measures

  1. Dec 27, 2012 #1

    I have a question dealing with distances of, lets say galaxies. I researched a little and saw that there are plenty of coordinates. Mostly used are spherical (celestial) ones, RA α and DEC δ. To have a three dimensional distance measure, one uses the redshift and from there the line of sight distance.
    According to Hoog arXiv:astro-ph/9905116v4, this computes like:

    [tex] D_C = D_H \int_0^z \frac {dz'}{E(z')} [/tex]

    where D_H is a constant and E(z) is
    [tex] E(z)=\sqrt{\Omega_M (1 + z)^3 +\Omega_k (1 + z)^2 +\Omega_M } [/tex].
    So if I got it correctly I some up little line elements along dz, depending on the cosmology i consider ([tex] \Omega_k=0 [/tex]). But this is just the line of sight distance right? What about the position on the skymap α and δ? How can I compute the distances of objects that differ in redshift z and in celestial coordinates? Does it even make sense as the universe expanded in between?
  2. jcsd
  3. Dec 27, 2012 #2
    Hi Madster:

    Yes, that is what the author calls it.

    Do you understand that this is a theoretical construct; in other words, no single observer can make this observation....This relates to the fact that cosmological 'distance' measures are really four dimensional measures, not a 'three dimensional distance measure' as you state. Some convention for TIME must be made.

    If you already understand this, then we must await a post from an expert. If not, try
    studying this first diagram: http://en.wikipedia.org/wiki/Metric_expansion_of_space

    Two objects: Earth moves along the brown line of the left, a quasar [to serve as a distant object] moves along the yellow worldline on the right and they are moving apart as time increase upwards in the illustration.

    The red line is the path of a light beam emitted by the quasar about 13 billion years ago and reaching the Earth in the present day. That is all we can SEE, or observe. The orange line shows the present-day distance between the quasar and the Earth, about 28 billion light years. THAT curve is one of instantaneous time....where we calculate the quasar to be NOW. It is not directly 'observable'.
  4. Dec 27, 2012 #3


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    More specifically, to make this sort of measurement, you need a chain of observers, all of whom measure the distance to the next observer in the chain "at the same time".

    The results in general depend on the details of the chain, the usual chain in cosmology are called "co-moving" observers, ones that see the cosmic microwave background radiation as isotropic.

    The results also depend on how the chain defines "at the same time", usually once you have the chain specified you can assume Einstein clock synchorinzation as the method.
  5. Dec 27, 2012 #4

    Hi. So in comoving coorinates mentioned by pervect this quasar will be 13 billion light years away no matter how much the scale factor evolves, no? What then is the correct calculus of two randomly selected objects in comoving coordinates?
  6. Dec 27, 2012 #5


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    As far as I know, the current labmda-cdm models hold the spatial slices to be flat, so you could use ordinary trig if you knew the two distances (which cosmologists confusingly call proper distance. even though it requires the use of comoving observers).

    But I haven't been able to find a good clear reference to confirm my unfortunately fallible memory about this topic.

    I'll refrain from working out the trig unless asked, esp. in light of the other uncertanties.
  7. Dec 28, 2012 #6
    So I use the formula above to get the "proper distant" (comoving) between the observer, us, and the distant object because I integrate from 0, that's again us, to the measured redshift z of the object. OK then but how to get this comoving distance between objects that are at z1=0.1 and z2=0.3 e.g.?
  8. Dec 28, 2012 #7
    Madster: Cosmological distances, [recession] velocities and the conventions used to determine them are about the most the most confusing subject I've come across in these forums....

    from my notes: a combination of discussions in these forums and Wikipedia:

    Putting these together with pervect's post 'flat special slices' he recalls makes sense because the FLRW model assumes homogeneous and isotropic space; if you freeze your measurement time than sounds reasonable that space portion would be virtually flat.

    here is another piece.....which may help...

    Marcus, an expert of these forums, often refers people to an online cosmological calculator

    and he explains some aspects of the model in this loooooong discussion:


    Another calculator he uses is:

    Jorrie's calculatorhttp://www.einsteins-theory-of-relativity-4engineers.com/cosmocalc_2010.htm

    In the models, you need to utilize standard LCDM parameters, like dark energy fraction of 0.73, and one 'signpost would be that a galaxy with a redshift of 1.8 is at the CEH .

    I believe the standard parameters are already plugged in the NedWright calculator.

    Marcus used the calculator and answered an interesting question which puts some pieces in perspective:

    Hope this helps.....if you do determine an answer, please post the explanation of what you think you found....sooner or later an expert may stumble here and help.
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