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Do clocks speed up in an expanding Universe?

  1. May 13, 2014 #1
    For simplicity I assume a flat radial FRW metric (with [itex]c=1[/itex]):
    ds^2 = -dt^2 + a^2(t)\ dr^2
    Now let us consider the path of a light ray, a null geodesic, with [itex]ds=0[/itex] so that we have:
    dt = a(t)\ dr
    Now at the present time [itex]t_0[/itex] we define [itex]a(t_0)=1[/itex] so that we have:
    dt_0 = dr
    The interval of radial co-ordinate [itex]dr[/itex] does not depend on time so that it is the same in both equations.

    We combine the two equations to eliminate [itex]dr[/itex] giving:
    dt = a(t)\ dt_0
    If [itex]dt[/itex] is always a fixed interval of cosmological time [itex]t[/itex] then from our perspective at the present time [itex]t_0[/itex] it is represented by the time interval [itex]dt_0[/itex] given by:
    dt_0 = \frac{dt}{a(t)}
    Thus one second measured in the future at time [itex]t[/itex] is equivalent to [itex]1/a(t)[/itex] seconds of our present time [itex]t_0[/itex].

    Therefore in an expanding Universe clocks speed up from the perspective of our present time.

    Is this effect real or apparent?

    Are there any observations that could decide between the two?

    The conventional view is that this effect is only apparent and is the cause of the cosmological redshift.
    Last edited: May 13, 2014
  2. jcsd
  3. May 13, 2014 #2
    I don't quite follow. Why are we considering a null-geodesic? A comoving observer will measure the proper time, that is, the "proper distance" in the time direction:

    $$\mathrm{d}\tau \equiv \sqrt{-g_{00}}\mathrm{d}t \qquad (\mathrm{d}s^2 = g_{\mu\nu}\mathrm{d}x^{\mu}\mathrm{d}x^{\nu})$$
    Since for RW metric ##g_{00} = -1## the clocks always run at the same rate. If we had a metric for which ##g_{00} = -b(t)## is some function of t then clocks would run at different rates at different times. Compare for example to the Schwarchild metric where the time component of the metric depends on the distance from the center and so clocks run differently for different observers.
  4. May 13, 2014 #3
    I use a null-geodesic in order to relate time intervals measured by observers at different times.

    As the observers are comoving then their proper times [itex]\tau[/itex] are indeed the same as their cosmological times [itex]t[/itex]. However I want to know how the rates of their proper/cosmological times compare with each other.

    I think the situation is analogous to the case of special relativity. Imagine that two observers in different inertial frames observe the same light beam. Since the speed of the light remains the same in each frame, the observers' time and distance measures must change.
    Last edited: May 13, 2014
  5. May 13, 2014 #4
    OK, I guess you are thinking of an expanding clock. So a light ray travels a fixed comoving distance and the time it takes for it to do that is used as a time standard. Do I understand you correctly? In that case I would agree that by that clock the rates would be different. However, most clocks we use are not like that, they are held together by EM forces and so do not expand with the universe. I think by a conventional clock the rates would be the same.
  6. May 13, 2014 #5
    No - the time is the same time as measured by standard rigid clocks.

    The analysis I use above is a short-hand version of the argument normally used to derive the cosmological redshift.
  7. May 13, 2014 #6
    To me your analysis seems very much to be saying that it will take longer for a light ray to travel a fixed comoving distance in the future than now. That is of course true, but only has relevance to the rate at which clocks run if one chooses to identify the travelling of a comoving distance by light as a clock.

    The derivation of the cosmological redshift is indeed the consequence of the fact that it will take light more time to travel a fixed comoving distance at later times (e.g. for two consecutive wave crests). The time measured by the observer between two crests is longer than by the emitter is because the second crest had a longer way to travel because the universe has expanded. To try to bring this back to clocks, for a "rigid clock" both crests would have the same distance to travel, hence no time difference.

    But if I'm wrong, perhaps someone else will have better insight.
  8. May 13, 2014 #7
  9. May 13, 2014 #8
  10. May 13, 2014 #9


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    I think Lubos answered your question in that other thread. Why don't we start with why you think his answer is missing the mark?
  11. May 13, 2014 #10
    Last edited: May 13, 2014
  12. May 13, 2014 #11
    the descriptive above is conformal time.

    d=ax where d is the physical proper distance x is the comoving distance. with convention c=1
    [itex]\eta[/itex] is the conformal time.

    [tex] d\eta=\frac{dx}{a}=\frac{ctd}{a}[/tex]

    gives particle horizon

    [tex]\eta = \int_{0}^{t} \frac{dt'}{a(t')}[/tex]


    edit forgot for flat geometry (euclidean)
    Last edited: May 13, 2014
  13. May 13, 2014 #12
    Just a side not question, do we even use conformal distance and conformal time anymore. At one time the particle horizon used to be called the comoving horizon ( I have some pretty old textbooks and article:P) This at one time used to be thought of as the region of causal connection, (in older articles).? The reason I ask in my newer textbooks no longer include it.
    Last edited: May 13, 2014
  14. May 14, 2014 #13
    I think my argument I gave in the "conformal time or cosmological time?" post was wrong. Therefore Lubos's criticism refers to a false argument.

    Contrary to what I stated in that post observers do not measure intervals of conformal time [itex]dt/a(t)[/itex], that vary with time, instead they simply measure fixed intervals of cosmological time [itex]dt[/itex]. I now agree with Lubos that all observers will measure the speed of light to be c.

    Let me summarise my new improved argument as follows:

    The equation that I derive from the null geodesic in this thread:
    dt_0 = \frac{dt}{a(t)},
    gives the time interval [itex]dt_0[/itex], at the present time [itex]t_0[/itex], that corresponds to a fixed time interval [itex]dt[/itex] at some future time [itex]t[/itex].

    Thus from our perspective today clocks in the future speed up. I think this effect is real rather than simply apparent. It implies that the energy scale of all physical processes is increasing with the scale factor [itex]a(t)[/itex]. I think this means that the Planck mass is proportional to [itex]a(t)[/itex].
    Last edited: May 14, 2014
  15. May 14, 2014 #14
    For the case where light travels a fixed comoving (coordinate) distance.

    I think the question you need to answer is why a fixed comoving (i.e. coordinate) distance traveled by light would have something to do with the rate of a clock.

    In contrast, consider a rigid clock comprised of two parallel mirrors with a photon bouncing between them. The time it takes for the photon to travel between the mirrors is taken as a time unit.

    Since the clock is rigid, the proper distance between two mirrors is fixed, let's call it ##d_M##. Suppose Alice, living today (##t_0##), observers one bounce of the photon. The photon travels on null-geodesics so it will tavel the proper distance ##2d_M## in cosmic time interval ##\mathrm dt_{\mathrm{Alice}} = a(t_0)\mathrm dr_0 = 2d_M##, where ##\mathrm dr_0## is the comoving distance between the plates.

    Suppose a million years pass and Bob finds this same clock in the future (##t_1##). Bob also observes one bounce of the photon which takes cosmic time ##\mathrm dt_{\mathrm{Bob}} = a(t_1)\mathrm dr_1 = 2d_M##. The proper distance between the mirrors is fixed while the comoving (coordinate) distances at different times scale with the expansion of the universe, $$2d_M = a(t_1)\mathrm dr_1 = a(t_0)\mathrm dr_0 \quad \Leftrightarrow \quad \mathrm dr_1 = \frac{a(t_0)}{a(t_1)}\mathrm dr_0$$ Hence, the cosmic time corresponding to one bounce of the photon (one tick of the clock) is ##\mathrm dt_{\mathrm{Bob}} = \mathrm dt_{\mathrm{Alice}}## so the clocks run at the same rate.
  16. May 14, 2014 #15
    The light ray connects the time interval measured by Bob, [itex]dt_{\mathrm{Bob}}[/itex], with a corresponding time interval in Alice's present [itex]dt_0[/itex]:
    dt_0 = \frac{dt_{\mathrm{Bob}}}{a(t_{\mathrm{Bob}})}.
    Last edited: May 14, 2014
  17. May 14, 2014 #16
    Let me put it this way: what is ##dr## for you and why do you think that it should be constant? Like I've been saying, all you have given is the different times it takes a light ray to travel a fixed coordinate distance (remember that coordinate distance is not a physical distance measured by observers). What exactly is the connection of this to clocks and how different observers measure time is entirely unclear (in my opinion there is none).
  18. May 14, 2014 #17
    Let me ask you a simple question, if conformal time was real and not simply a consequence of the coordinate system. Then how do you explain that when we measure a standard candle at some faraway location. Lets say in a galaxy near the CMB, we measure a type 1a supernova.
    Type Ia supernovae have a characteristic light curve, their graph of luminosity as a function of time after the explosion. see the graph on this page

    http://en.wikipedia.org/wiki/Type_Ia_supernova. regardless of how far away we measure the luminosity to time relation. The ratio of luminosity to time remains the same. Keep in mind the further we look the further back in time we look.

    If time was running slower as your trying to prove, then interactions we observe would be running slower. (our rate of time would be faster) Keep in mind scientists do monitor reaction rates all the time in the past.

    Why do they not see a time dilation???

    Simple reason is there is no time dilation involved, conformal time is a mathematical construct of the coordinate system being used, nothing more. Conformal time and conformal distance has been known to scientists since the FLRW metric was developed. Do you honestly believe your the first one to misconstrue conformal time as a time dilation? Trust me your not its a well argued and well proven fact that conformal time does not = time dilation.

    This is easily proven as any form of known reaction rate can serve as our rigid clock
    Last edited: May 14, 2014
  19. May 14, 2014 #18
    Ned Wright reports that time dilation in supernova light curves has been observed:


    He thinks it's an "apparent" effect - I think it's a real effect.
    Last edited: May 14, 2014
  20. May 14, 2014 #19
    Ok - Here is my argument in terms of light clocks.

    I start from the null geodesic equation relating an interval of cosmological time ##dt## to an interval of radial co-ordinate ##dr##:
    c\ dt = a(t)\ dr
    Now a rigid clock always has a proper length of ##dr## as it does not expand with the Universe.
    The velocity of light is always ##c##.
    Therefore the time interval ##d\tau## measured by this rigid lightclock is:
    d\tau = \frac{dr}{c}
    I substitute this expression into the null geodesic equation above to obtain:
    d\tau = \frac{dt}{a(t)}
    Therefore lightclocks (in fact all types of physical clock) measure conformal time.
  21. May 14, 2014 #20
    all those papers he referenced are 14+ years old lol. This is one of the reasons why you don't see conformal distances used anymore. that and those papers don't account for the effects of dark energy
    Last edited: May 14, 2014
  22. May 14, 2014 #21

    if you won't listen to the professionial cosmologists when they tell you its Apparent. Then we certainly cannot convince you otherwise
  23. May 14, 2014 #22
    No, the proper (physical) distance is ##\mathrm ds = a(t)dr##. This is what is constant for a rigid clock. ##\mathrm dr## is the coordinate distance.
  24. May 14, 2014 #23
    At the present time, ##t_0##, with ##a(t_0)=1##, imagine a rigid rod adjacent to an equal length of space each with the same proper length ##ds_0=a(t_0)dr=dr##.

    Now at a later time, ##t##, the space will have expanded to a proper length ##ds=a(t)dr## but the rigid rod will continue to have its original proper length ##ds_0=dr##.
  25. May 14, 2014 #24
    Ok I am done hinting....you obviously aren't catching the hints. This above does not work as it does not work beyond the particle horizon. Nor does it consider the cosmological constant

    the correct formula you should be using is


    in the notation from Barbera Ryden's "Introductory to Cosmology"

    the particle horizon used to be the comoving horizon or the region of causality. Now our region of causality is the cosmic event horizon. Thanks to the cosmological constant


    edit noticed the previous article was copyrighted so I removed it, note none of the calculations this far, either posted or referenced include [itex]\Lambda[/itex]. In particular conformal distance or time
    Last edited: May 15, 2014
  26. May 15, 2014 #25
    The point is that the coordinate distance corresponding to the distance between the two ends of the rod will change. If one end of the rod is fixed at the origin and the other end is at coordinate distance ##\mathrm dr_0## from it today then the proper length of the rod is ##\mathrm ds_0 = a(t_0)\mathrm dr_0 = \mathrm dr_0## (for ##a(t_0) = 1##). It will take light the time ##\mathrm dt_0 = a(t_0)\mathrm dr_0 = \mathrm dr_0## to travel from one end of the rod to the other.

    At some later time ##t_1##, the other end of the rod will be at coordinate distance ##\mathrm dr_1## from the origin (where the other end is fixed), and the proper length of the rod is ##ds_1 = a(t_1)\mathrm dr_1##. It will take light a time interval ##\mathrm dt_1 = a(t_1)\mathrm dr_1## to travel from one end to the other.

    Now we require that the rod is rigid, i.e., that its proper length remains constant (##\mathrm ds_0 = \mathrm ds_1##), from which it follows that

    $$a(t_1)\mathrm dr_1 = a(t_0)\mathrm dr_0 = dr_0 \quad \Leftrightarrow \quad \mathrm dr_1 = a^{-1}(t_1) \mathrm dr_0.$$

    Thus at time ##t_1## it will take light to travel from one end of the rod to the other a time interval

    $$\mathrm dt_1 = a(t_1)\mathrm dr_1 = a(t_1) a^{-1}(t_1) \mathrm dr_0 = \mathrm dr_0 = \mathrm dt_0,$$

    the same as at time ##t_0##. Therefore if you build a clock which measures time by counting how many times a light ray travels the length of the rigid rod, for example, this clock will be always running at the same rate.
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