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Redshift > 1.46 means Recessional Velocity over C?

  1. Mar 19, 2006 #1
    I read this recently, but did not understand how one calculates this, since I'm used to:

    Distance = Recessional Velociy / Hubble's Constant
    = Beta x C / Hubble's Constant


    I thought that as Z approaches infinity, Recessional Velociy approaches C. But clearly this cannot be true if a Red Shift of z =1.47 implies superluminal recessional velocity. Where are I going wrong?

    Thanks much.

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  2. jcsd
  3. Mar 19, 2006 #2


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  4. Mar 19, 2006 #3
    And here is a paper you really need to read:
    http://arxiv.org/abs/astro-ph/0310808" [Broken]

    Expanding Confusion: common misconceptions of cosmological horizons and the superluminal expansion of the Universe
    Authors: Tamara M. Davis, Charles H. Lineweaver
    Comments: To appear in Publications of the Astronomical Society of Australia, 26 pages (preprint format), 6 figures. Version 2: Section 4.1 revised

    We use standard general relativity to illustrate and clarify several common misconceptions about the expansion of the Universe. To show the abundance of these misconceptions we cite numerous misleading, or easily misinterpreted, statements in the literature. In the context of the new standard Lambda-CDM cosmology we point out confusions regarding the particle horizon, the event horizon, the ``observable universe'' and the Hubble sphere (distance at which recession velocity = c). We show that we can observe galaxies that have, and always have had, recession velocities greater than the speed of light. We explain why this does not violate special relativity and we link these concepts to observational tests. Attempts to restrict recession velocities to less than the speed of light require a special relativistic interpretation of cosmological redshifts. We analyze apparent magnitudes of supernovae and observationally rule out the special relativistic Doppler interpretation of cosmological redshifts at a confidence level of 23 sigma.
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  5. Mar 19, 2006 #4
    Ok, I see my problem, how then does one calculate recessional velocity from redshift in Cosmology without making the SR mistake I made?
  6. Mar 19, 2006 #5
    If you look at equation 1. in the reference I linked, it gives the appropriate relation between recessional velocity and redshift for general relativity.
  7. Mar 20, 2006 #6


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    I think I mentioned Ned Wright cosmology tutorial in another page on another forum, but it appears I edited out the references to his cosmological distance calculator.

    This will calculate comoving radial distance from the redshift (z), and Hubble's law will give the recessional velocity from the comoving distance (this is mentioned on the calculator output).

    The javascript calculator itself is at

    the background information on it is contained in
    http://www.astro.ucla.edu/~wright/cosmo_02.htm, including a link to the derivation (which won't be intelligible without knowledge of GR).

    The conversion from redshifts to distances (and recessional velocities) is highly model dependent, though. Ned Wright's calculator defaults to H0 = 71 (he doesne't give units, presumably this is in (km/sec)/Mpc, [itex]\Omega_m = .27[/itex], and [itex]\Omega_v = .73[/itex], which I believe are the Lambda-CDM values.
  8. Mar 20, 2006 #7
    Thanks Pervect, great post, clear as day.
  9. Mar 21, 2006 #8


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    Excellent reference, matt.o. That is my personal favorite for explaining superluminal recession velocities. Ned Wright also gives a very good explanation for this, as well as numerous other issues in cosmology. His FAQ is a great read, IMO, for anyone wishing to learn the fundamentals.
  10. Mar 21, 2006 #9


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    Here's my favorite example:

    The text in


    associated with the attached image evaluates a special case where both the standard SR distances and velocities and the cosmoloigcal distances (comoving distance) and velocities (Hubble velocity) are defined.

    More advanced readers might recognize this as the "Milne" cosmology, other less advanced readers may not recognize it from this description - Ned Wright doesn't use this particular phrase in his text.

    The point of this image though is to point out that the cosmologists are measuring something that's different from the standard SR notion of distance and velocity. This can be seen by this specific example, which applies to a cosmology where both concepts exist.

    The SR notions of distance and velocity cannot, unfortunately, be generalized to situations where the Riemann tensor is non-zero (i.e. the spacetime is curved). SR notions of distance and velocity require a Minkowskian metric. The example was specifically chosen so that the space-time in question has both a Minkowskian metric (a static flat space-time description) and a Milne metric (a non-static expanding space-time description) which are mathematically totally equivalent, representing the same physical spacetime (a spacetime with a very low matter density).

    The cosmological defintions of distance and velocity are more general than the SR defintions, but they do NOT reduce to the SR defintions of distance and velocity in this limit of low matter density.

    As I went on at some length in another thread, the important point here is "many distances, many velocities". One should be aware that the definitions of distance (of which there are many) used by cosmologists are not the same as those used in SR, and that this remark also applies to velocities (though in this case only the Hubble recessional velocity is widely used AFAIK).

    This should not be terribly surprising to someone who is familiar with SR - one of the reasons for this difference in defintions is the usual problems of defining "simultaneous" events, which is a necessary prerequisite to be able to measure distance.

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