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Is there a redshift in a conformally flat metric space?

  1. Aug 12, 2013 #1
    Hello PF:

    I noticed a thread on PF in which TOM STOER and others were discussing how to calculate the redshift for an arbitrary metric. I need to talk to Tom if he is still on this list.
    The question has arisen in an applied physics field whether the following conformally flat metric will exhibit a Hubble shift or not:

    ds2=a(t)2[-dt2+dr2] (cartesian coordinates)

    This is a much simpler problem than Tom's "arbitrary" metric, and since this metric is well known (indeed common) in Relativity literature, someone must know the answer to this question off the top of his head.
    I presume that there is no Hubble shift for this metric b/c while the scale factor stretches the wavelengths, the scale factor also multiplies dt producing a "cosmic time dilation" which blue shifts the distant light and exactly offsets the wavelength stretching redshift. However, I am unable to produce a simple mathematical demonstration of this.
    I have been told that Wald, Weinberg and other texts outline how to calculate the Hubble shift but I find most texts only treat the FLRW metric which is a very special case b/c dt is actually the proper time in FLRW and objects at a fixed comoving coordinate are actually at rest (have a geodesic worldline)). This is not so in the conformally flat metric, so the simple calculation is not applicable.
    In the discussion Tom Stoer posted he noted that if the emitting star had 4 momentum ue and the observer at the origin had 4 momentum uo that you can deduce the following:

    fo/fe=[itex]\tau[/itex]e/[itex]\tau[/itex]o

    for a train of light pulses.
    Given that light travels in straight 45 degree lines at constant speed c=1 in the conformal metric, it seems to me that it can't be more than a few lines of math from the above equation to the result that:

    fo/fe=1 (i.e. no Hubble shift)

    which is what I think is the case for the conformally flat metric.

    So, if anyone can simply tell me, or indicate how I can prove this, we would be very grateful, b/c it turns out that this metric is of fundamental importance in a field only remotely related to Relativity, and the question of whether a Hubble shift exists or does not exist for this metric turns out to be an immediate proof or disproof of the theory being investigated.

    best wishes, tardy
     
  2. jcsd
  3. Aug 13, 2013 #2

    atyy

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  4. Aug 13, 2013 #3
    Yes, but to bring them into that form the scale factor is not a simple function of the time only... it is generally a highly complex function of the time and the space coordinates, specifically the radial coordinate in spherically symmetric solutions. The FLRW expressed in "conformal time" does not describe a different physical system from the nonconformal form. The metric I have written does describe a physically different system I.e., there is no "cosic time dilation" in FLRW space. There is in the space described by the conformally flat metric. Correct me if I'm wrong.
    tardy
     
  5. Aug 13, 2013 #4

    Mentz114

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    Certainly the FLRW's are conformally flat ( ie Weyl tensor is zero ). Tardy's metric is also CF but it's different in that the coordinate speed of light is always 1 (c in old money) but is 1/a(t) in FLRW. I haven't done the calculation but it looks like there is no frequency shift between the (non-geodesic) co-moving worldlines.
     
  6. Aug 13, 2013 #5
    Yes, that's what it looks like to me too. But frankly so far I've been unable to do the calculation. Im wondering if the entire calculation has to be done in proper time or if the straight line diagonal light lines of conformal time vs. comoving coordinates makes the calculation trivial?
    tardy
     
  7. Aug 13, 2013 #6

    atyy

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  8. Aug 13, 2013 #7
    It is mathematically the same, but it is not physically the same. Below eqn. 8 the authors say:

    "Hence the parametric (conformal) time is shown
    on clocks that slow down with the expansion
    of the universe relative to the clocks showing
    cosmic time, and stops in the limit a → ∞.

    Notice the difference between "conformal time clocks" and "cosmic time clocks" Conformal time is only a mathematical device in their analysis. Physical reality runs in accordance with comic time.

    This is distinctly different from the "tardy-metric case" where conformal time is actually taken to be real physical time, not a "slow clock". IOW all motion in the universe appears to be "speed up" as a(t) increases.

    Obviously the tardy-metric does not obey the EFE's and doesn't even obey Newton's first law... but that happens to be irrelevant in the research application we are working on.

    The only thing physically required of the tardy-metric is that it obey the (kinematical) laws of SR.
    It does not have to obey even the dynamical laws of SR because it is in an empty (massless)
    Lorentzian metrical space. There are only light rays in the space, no mass particles.
    tardy
     
  9. Aug 13, 2013 #8

    Mentz114

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    I don't think so.

    The frequency shift between worldlines is given by ( I've slipped into spherical polar coords [itex]t,r,\theta,\phi[/itex])

    ##\frac{\nu_1}{\nu_2} = \frac{(g_{ab}k^a u^b)_1}{(g_{ab}k^a u^b)_2}##
    where ##u^a## is a worldline and ##k^a## is a null geodesic connecting two worldlines. The subscripts 1,2 refer to the emission and absorption events.

    In this case the null geodesics are just ##k^a=n\partial_t+n\partial_r##. Suppose ##u^a = u_t\partial_t + u_r \partial_r##. Then

    ##g_{ab}k^a u^b= g_{tt}n u_t+g_{rr}n u_r=a(t)^2n(-u_t+u_r)##.

    If the emission takes place at ##t=t_1## and the absorption at ##t_2## then the frequency ratio is ##\frac{a(t_1)^2n(u_r-u_t)_1}{a(t_2)^2n(u_r-u_t)_2}##

    The normalization ##g_{ab}u^au^b=-1## gives ##u_t=\frac{\sqrt{{a\left( t\right) }^{2}\,{u_r}^{2}+1}}{a\left( t\right) }##.

    If ##u_r=0## and ##n=1/a(t)## we no get frequency change, if ##n## is something else we get ##a(t_1)n(t_1)/a(t_2)n(t_2)##.

    I can't work out if ##n=1/a(t)## makes ##k^a## a null geodesic (##k^b\nabla_b k_a = 0## ?)

    [Later]
    If we make the transformation ##(x')^a = a(t)x^a## then ##(dx'^a)^2=(dx^a)^2 a(t)^2## and the metric is just the Minkowski metric. The null geodesics in these coords is ##{k'}^a=\partial_{t'}+\partial_{r'}##. Transforming back to our original coords gives ##n=1/a(t)##. This is just the value which gives complete cancellation of ##a(t)## from the numerator and denominator of the frequency ratio.

    Unless I've made an error, it looks like there is no redshift between worldlines where only ##dt/d\tau \ne 0##.
     
    Last edited: Aug 13, 2013
  10. Aug 13, 2013 #9

    Mentz114

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    Actually, what you're calling the 'Tardy metric' is a solution of the EFE. Not a vacuum solution but a fluid, and the pressure can be made zero ( to give a dust solution) by setting ##a(t)=Ht^2## where H is a constant.
     
  11. Aug 13, 2013 #10
    I'm only an M.S. in Physics (1967). It will take me days to follow through your calculation.
    It does occur to me that ur=0 since the emitter and receiver are both at fixed comoving coordinates and have no radial velocity.
    My guess is that yes, n=1/a(t) is likely correct since the worldlines of the emitter and absorber only differ from a normal fixed coordinate worldline with ut=c by having a wordline with ut= c/a(t)
    I won't bother you with questions now, but "eureka", I think you've found it!
    tardy
     
  12. Aug 13, 2013 #11

    Mentz114

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    I blundered with the coordinate transformation, which should be

    ##t'=\int a(t) dt \rightarrow dt'=a(t)dt##, and ##r'=a(t)r \rightarrow dr'=a(t)dr##. The integral goes away when we apply the ##\partial_{t}## operator to the vector, so I think it still works.

    But don't be hasty, there's probably a much simpler way to work this out.
     
  13. Aug 13, 2013 #12
    Dear Mentz114:
    I am greatly indebted to you for your elegant demonstration that there is no Hubble shift in the conformally flat metric ds2=a(t)2[-dt2+dr2]

    Because of that I guess I owe it to you to tell you how this metrical question arose in the first place. I'm sure its more than you will want to know, however I am writing a book about it and want to give you credit for the mathematical proof and get your permission to publish it.

    It turns out that this metric comes from the field of Psychology, and describes what we might call "seen reality"; as opposed to "absolute physical reality" which is what laboratory rulers and clocks record.

    A newborn person is about 1 foot tall and grows to 6 feet in 18 years. Concurrently, his mental speed (cognitive speed aka Intelligence) rises from about 3 bits/sec to 16 bits/sec at adulthood (for an average IQ of 100). As a consequence of this fact, it appears nominally to the average person that the world "gets smaller" and "gets slower" as he grows up.

    The mathematical description of this turns out to be a "conformal" metric. IOW, we can look at the problem as if the world was undergoing a "Hubble contraction" and at the same time that time was slowing down. Hence the metric ds2=a(t)2[-dt2+dr2]

    IOW, what we have done is take the ordinary Lorentz metric and substituted our "foot size" for the laboratory ruler, and our "mental speed" for the laboratory clock, and then said, pretend that we are not actually growing bigger and faster... but that it is the world that is "shrinking" and "getting slower".

    Okay, so far so good.... and believe it or not that turns out to be a conformal metric! But the first time I told this idea to a group of physicists, one of them raised his hand and said "if the world looks like it's shrinking" why wouldn't all of the stars in the sky turn blue at night... a Hubble blue shift in the head! Well I was floored by the question, until I suddenly realized that the conformally flat metric probably doesn't exhibit a Hubble shift at all.

    Well, that was months ago, and I have been struggling ever since to fid a mathematical proof of that fact......... the proof that you just found !

    And now, as if all that isn't enough to get me thrown off PF, I will finally mention the punch line to all this. a'(t) turns out to be the human growth curve... you know, those yearly pencil marks on the kitchen door frame your mother used to make on your birthday recording how tall you were. a'(t) starts out small and increases ... so the metric uses 1/a'(t) =a(t) because the world does just the inverse of what our body does... the world starts out very large and comes down to a=1 at full growth at age 18. But there in likes the punch line.

    a(t) for the human race never actually falls all the way to 1. No one ever reaches his full height and weight or full intelligence. In fact actuall biometric data (Secular Trend) shows that in the world population a(t) only falls to about 1.2 worldwide. IOW, on average, as much as 20% of our body and our brain is missing.... simply ungrown. This means that the world appears, to the average person, to be about 20% bigger and 20% faster, than it actually is.

    Long story short... this 20% of the world is actually invisble to us (because 20% of what we should be seeing is now 20% above the Fourier cutoff frequency of our visual system. Bottom line, this "invisble world" that we can't see is called by the theologians "Heaven", and you guessed it, the perfect 100% fully grown man (who dies not exist on Earth) is known to history as "God".

    Well, like I said, this is probably more than you wanted to know. But we are indebted to you for proving that there is no Hubble shift in the metric.... for after all... if it predicted a Hubble blueshift... the entire theory would have crashed and burned right there... since anyone can see there is no blueshift visible in the stars at night.

    The name of the book is the Scientific Proof of God by the way, and your mathematical proof will appear in the last chapter on the Relativistic connection. I'll cite you as Mentz114.... not to worry, if the book sells a billion copies biographers are sure to find out your real name so you don't have to disclose it!

    regards, tardy
     
  14. Aug 13, 2013 #13

    Mentz114

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    Dear Tardy,

    I wish to disassociate myself from the views expressed in your last post. Under no circumstances mention my name, ever.

    You are a loony crackpot and you will shortly be banned from this forum.

    Why do I get all the batty ones ?
     
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