Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

General redshift formula

  1. Apr 5, 2013 #1

    tom.stoer

    User Avatar
    Science Advisor

    Quick question: is there a general formula for the redshift zC[g] for a photon travelled in a spacetime with arbitrary metric g along a light-like geodesic C?

    I do neither want to use a special symmetry for g (spherical, homogeneous and isotropy) nor do I want to use a specific expression for g, like FRW metric or whatever.
     
  2. jcsd
  3. Apr 5, 2013 #2

    marcus

    User Avatar
    Science Advisor
    Gold Member
    Dearly Missed

    That's a nice question! The Bunn Hogg paper would not exactly address it because I think they were specializing to the FRW case. But they might have some expression that could be usefully adapted.

    http://arxiv.org/abs/0808.1081
    The kinematic origin of the cosmological redshift
    Emory F. Bunn, David W. Hogg
    (Submitted on 7 Aug 2008 (v1), last revised 14 Apr 2009 (this version, v2))
    A common belief about big-bang cosmology is that the cosmological redshift cannot be properly viewed as a Doppler shift (that is, as evidence for a recession velocity), but must be viewed in terms of the stretching of space. We argue that, contrary to this view, the most natural interpretation of the redshift is as a Doppler shift, or rather as the accumulation of many infinitesimal Doppler shifts. The stretching-of-space interpretation obscures a central idea of relativity, namely that it is always valid to choose a coordinate system that is locally Minkowskian. We show that an observed frequency shift in any spacetime can be interpreted either as a kinematic (Doppler) shift or a gravitational shift by imagining a suitable family of observers along the photon's path. In the context of the expanding universe the kinematic interpretation corresponds to a family of comoving observers and hence is more natural.
    6 pages. Am.J.Phys.77:688-694,2009

    One equivalent way to analyze the redshift is as the cumulative effect of a large (infinite) number of Doppler shifts along the path.
     
    Last edited: Apr 5, 2013
  4. Apr 6, 2013 #3

    tom.stoer

    User Avatar
    Science Advisor

    Hi marcus, thanks, but the paper doesn't mention anything like that.

    The most general formula I know is for the Robertson-Walker line element

    [tex]d\tau^2 = dt^2 - a^2(t) \, \left( \frac{dr^2}{1-kr^2} + r^2\,d\Omega^2 \right)[/tex]

    where spherical symmetry and isotropy has been used, but where the scale factor a(t) is still an arbitrary function of time.

    It says that for light emitted from a co-moving source and received by a co-moving observer the redshift z is given by scale factors

    [tex]1+z = a_0 / a_1[/tex]
     
    Last edited: Apr 6, 2013
  5. Apr 6, 2013 #4

    tom.stoer

    User Avatar
    Science Advisor

    What I am looking for is something like

    [tex]z_C[g] = \int_C ds \, f[g][/tex]
    which can be applied to arbitrary metrics g and arbitrary (light-like) geodesics C.

    EDIT: wait, this can't be fully correct, b/c this eq. looks covariant, whereas frequencies and redshifts are source- and observer-dependent; that means it must be something like

    [tex]z_{C,S,O}[g] [/tex]
    but now I am even more confused about the r.h.s.
     
    Last edited: Apr 6, 2013
  6. Apr 6, 2013 #5

    fzero

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    It's worthwhile to look back over the derivation of the redshift for FRW and see why you could get such an explicit formula in the first place. First, the null geodesic equation ##ds^2=0## is separable. As a result, we can show that the scaled time-elapsed between emission and detection of successive crests of a wave is independent of positions. Then you use the approximation that the scale factor is constant over the period of the wave.

    The FRW derivation does not require a solution of the geodesic equation, but it's clear that in the general case, you will need one. Obviously you're not going to get too far by keeping the metric arbitrary. You might get a bit further if you restrict to a class of metrics for which the null geodesic equation is separable, but these probably all be FRW-like, where we don't restrict to constant spatial curvature.
     
  7. Apr 6, 2013 #6

    Chalnoth

    User Avatar
    Science Advisor

    I'm not sure it's possible to write down something quite that simple, unfortunately. There are a few steps that you need to take to do this with a general metric:
    1. Find the path the light ray follows. In the simplest case, this can be simply by solving [itex]ds^2 = 0[/itex] (because light must follow null geodesics). If the metric is sufficiently complicated, this won't be enough, and you'll have to solve the geodesic equation, which is far more complicated. See here:
    http://en.wikipedia.org/wiki/Geodesics_in_general_relativity

    2. Once you've found the path the light ray follows, you can pick a suitable set of observers along that path in the way that Bunn & Hogg did in the paper Marcus linked above.
     
  8. Apr 6, 2013 #7

    tom.stoer

    User Avatar
    Science Advisor

    Chalnoth, thanks for the comment, but what you are explaining is much more than I had in mind.

    I do not want to find the geodesic C, I only want to have an expression which works for arbitrary, unspecified, light-like geodesics.

    Simple example: Suppose I am asking for the general expression for the length of a geodesic C. The answer is simply

    [tex]S = \int_C ds[/tex]

    where C remains unspecified! That's all I am looking for.
     
  9. Apr 6, 2013 #8

    fzero

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    If you look at Wald (5.3.6) and the preceeding discussion, you'll find an expression for the redshift in the case that there exists a Killing vector. This is quite an improvement on the heuristic way the redshift is usually computed for FRW.
     
  10. Apr 7, 2013 #9

    tom.stoer

    User Avatar
    Science Advisor

    Thanks, great, this is still not fully generic b/c it relies on the Killing vector, but it's an improvement compared to the usual derivation
     
  11. Apr 20, 2013 #10

    tom.stoer

    User Avatar
    Science Advisor

    I thought about it, and I can't believe my results. Applying the idea of Bunn and Hogg the infinitesimal Doppler shift at two spacetime points separates by dr, dt is always

    df/f = - H dt

    But H is always

    H = d ln(a) / dt

    and therefore the diffential equation can always be integrated w/o ever solving for a geodesic. That means that given two spacetime points connected by a light-like geodesic and with co-moving time t the formula

    1 + z = a(t0) / a(te)

    is always valid, regardless what a(t) is.

    But this can't be correct.

    The problem is that the definition of H(t) and a(t) cold make sense locally i.e. for infinitesimal Doppler shifts, but not globally b/c there we expect dependence on spatial coordinates.

    Therefore the first equation has to be modified

    df / dt = -H(r) dt

    were we assume that for each infinitesimal Doppler shift the approximation of an RW line element is still valid locally. But I am afraid that this starting point with H(r) defined locally does not really make sense.

    Any ideas?
     
    Last edited: Apr 20, 2013
  12. Apr 20, 2013 #11

    Chalnoth

    User Avatar
    Science Advisor

    Yes, it is correct. The wavelength of photons is increased by the same factor as the expansion.

    In order to have dependence upon spatial coordinates, you have to have a metric which changes in space. The FRW metric does not.
     
  13. Apr 20, 2013 #12

    Haelfix

    User Avatar
    Science Advisor

    Tom is asking about a Doppler shift formula for a general metric.

    It seems to me the problem is really one of choice. Namely, how do you want to compare two distant tangent spaces. There are many different geodesics that could be chosen, and hence there is no canonical choice. But it seems to me that you would want one that reproduces standard intuition. Namely you want a choice of geodesic that yields something that looks like the standard decomposition into cosmological redshift terms + Doppler shift terms.

    I believe I worked this out a long time ago in a problem set or somesuch, but I think the general gist is at the very minimum, you need a globally hyperbolic spacetime that can be foliated by hypersurfaces of constant time. This fixes one of the degrees of freedom, and you can then parralel transport the four vectors along the induced geodesic (eg the mutual points of intersection along the worldlines of a chosen test particle).

    I do not remember if there was additional arbitrariness (but most likely yes, eg you probably want some symmetry to kill off diagonal terms or you will have to make some choices about shear terms)
     
  14. Apr 21, 2013 #13

    tom.stoer

    User Avatar
    Science Advisor

    Agreed.

    But what are intesting spacetimes with Killing vectors?
     
  15. Apr 21, 2013 #14

    martinbn

    User Avatar
    Science Advisor

    What do you mean? Kerr, FRW, aren't these interesting?
     
  16. Apr 21, 2013 #15

    tom.stoer

    User Avatar
    Science Advisor

    No, they aren't interesting ;-) what I am really interested in are metrics w/o symmetries, generalizations of the derivation using Killing vector fields etc.
     
  17. Apr 21, 2013 #16

    Chalnoth

    User Avatar
    Science Advisor

    Right. That is horribly ugly and incredibly difficult to do properly. For reference, nobody has solved the Einstein equations for a metric that does not obey spherical symmetry. Granted, finding null geodesics is a simpler problem than that, but it is in no way simple for a general metric.
     
  18. Apr 21, 2013 #17

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    :confused: There are loads of exact solutions to Einstein's equation that aren't spherically symmetric.
     
  19. Apr 21, 2013 #18

    fzero

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    In principle, given a sufficiently nice 3-manifold ##M_3##, we can construct an Einstein 4-manifold as the total space of a circle bundle over ##M_3##. This includes various gravitational instanton solutions, so I would say that they are "interesting." I don't have any more insight into the problem w/o a Killing vector.
     
  20. Apr 22, 2013 #19

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Sorry, I'm having trouble understanding this. In the context of a general metric and coordinate systems (as in your original post) what are: H; a; co-moving t?

    The shift depends not just on the locations of the emission and reception events, but also on the worldlines (i.e., 4-velocities) of the observers at emission and reception 4-velocities, as you said in a previous post.

    Later this week I will try and post a calculation for an FRLW example (possibly using non-zero peculiar velocities), but instead of using a usual method like

    https://www.physicsforums.com/showthread.php?p=3978731#post3978731

    I will try and use a general method that applies to all metrics and coordinates systems.

    I haven't done this, so I don't know if I will be successful, but, when I get some time, I would love to have a go at this.
     
  21. Apr 22, 2013 #20

    tom.stoer

    User Avatar
    Science Advisor

    agreed - I only wanted to express my cluelessness.

    I think the best way i to try to generalize Wald's method, i.e. the parallel transport of the projection of a wave vector k on a velocity u.

    Another idea could be to use the Greens function for a general metric and to apply a reduction to geometric optics.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: General redshift formula
  1. Blueshift and Redshift (Replies: 3)

  2. Cosmological redshift (Replies: 8)

Loading...