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Worldline of photon in comoving coordinates.

  1. Jun 22, 2008 #1
    Can someone provide an equation that would enable me to plot the red "pear shaped" line shown in this diagram? http://www.astro.ucla.edu/~wright/omega0.gif

    The red line represents the path of a photon as the universe expands in comoving coordinates. Can the path be derived from the Friedmann-Robertson-Walker (FWR) metric? I do not really need the derivation. Just a simple equation that I can plot on a graph, like the parametric equation for a circle, x^2+t^2=1

    Thanks :)
     
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  3. Jun 22, 2008 #2

    George Jones

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    Can you give a link that puts this in context.

    My guess is that numerical integration is involved, but I won't know for sure until I do the derivation.
     
  4. Jun 22, 2008 #3
    This link is the context from Ned Wright's cosmology page. http://www.astro.ucla.edu/~wright/cosmo_02.htm and
    http://www.astro.ucla.edu/~wright/cosmo_03.htm

    I am not sure if there is enough mathematical context there to help you there.

    and this is link that gives some of the maths of the FRW metric.

    http://www.phys.washington.edu/users/dbkaplan/555/lecture_03.pdf
     
  5. Jun 22, 2008 #4

    George Jones

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    I don't need much :biggrin:; I just want to make I know what is plotted versus what.
     
  6. Jun 22, 2008 #5

    Mentz114

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    I'm assuming t is the vertical axis, r ( space) horizontal.
    The plot you cite looks like a 'big-bang/big-crunch', because the photon ends up back where it started. So it might be a FLRW for the k=1 case. The expansion factor for this case is

    [tex] \frac{\kappa M}{6}( 1 - cos(ct))[/tex]

    Getting the expansion factor requires solving the EFE for the model, which actually turns out to be possible.
     
    Last edited: Jun 22, 2008
  7. Jun 22, 2008 #6
    What would it be without the big crunch bit? I was hoping for the case where there is no acceleration or deceleration in the balanced on a knife edge case :p

    The return of the photon does not necessarily mean a big crunch. The departing galaxies are moving superluminaly relative to Earth in comoving coordinates and they do not return like the photon. The photon represents light we now now from the big bang (the CMB).

    The superluminal bit should not be a cause for alarm as the nothing is moving faster than light relative to local spacetime/comoving observers :)
     
  8. Jun 22, 2008 #7

    Mentz114

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    Hi Kev,
    that diagran cannot be the worldline of a CMBR photon. They originate when the universe is quite large already and full of plasma. When matter stopped aborbing/emitting photons they were free to travel and become the CMBR. So they were emitted from everywhere in everybody's past light cone at a certain time.

    Superluminal ? Perhaps the plot is a case where we emit light, then the expansion allows us to catch it up. Yes, that must be it. I'll see if I can come up with a formula.
     
    Last edited: Jun 22, 2008
  9. Jun 22, 2008 #8
    I tend to agree. I can only assume there is an initial offest not visible in the scale of the diagram that represents the radid inflation phase so the photon from the CMB has a head start displacement.
     
  10. Jun 22, 2008 #9

    Mentz114

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    Sorry, Kev, I was editing while you were typing, so there's a discrepancy. I'm not so sure the expansion can result in catching up a photon.

    Anyhow, the expansion factor in the case of a critical universe is proportional to t^2. So the co-moving velocity is proportional to t.

    Can you tell me exactly the link is to the picure ? What do they say it is ?
     
  11. Jun 22, 2008 #10
    This is the exact link: http://www.astro.ucla.edu/~wright/cosmo_02.htm

    It says:

    The light cones for distant galaxies in the diagram above are tipped over past the vertical, indicating v > c. (Superluminal) The space-time diagram below shows a "zero" (really very low) density cosmological model plotted using the [tex]D_{now}{/tex] and t of the Hubble law.

    It also says it is the past light cone in the comoving model.

    My guess is that a given time the photon is moving at the local superluminal velocity minus one so that initially it is moving away from us but eventually at the hublle horixon it breaks even and then starts coming towards us. THe FRW metric should tell us the distance the comoving object is at any cosmological time t and possibly its velocity relative to us. It should just be a case of subtracting c from that velocity to give the velocity of the photon at any given time and then integrating the result (I think).
     
    Last edited: Jun 22, 2008
  12. Jun 22, 2008 #11

    Mentz114

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    Hi Kev,

    it says "The red pear-shaped object is our past light cone". To plot the curve you need the transformation between ('D_now', hubble t) and (x,t) of SR. There might be enough information there. Although they don't say which model ( extremely low density ?) they are using.

    [edit] That is a an excellent tutorial but they leave out the details.

    This is very good for the Friedman bit,

    http://en.wikipedia.org/wiki/Friedmann_equations
     
    Last edited: Jun 22, 2008
  13. Jun 22, 2008 #12

    George Jones

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    From the Friedmann-Robertson-Walker metric, I get

    [tex]
    \left( x , y \right) = \left( \pm 3 \left (t^{\frac{2}{3}} - t \right) , t \right),
    [/tex]

    for [itex]0 \le t \le 1[/itex], which seems to reproduce the curve.

    I'll post my assumptions and derivation later today (if my daughter takes a long enough nap) or tomorrow. The derivation isn't that long, and is very intuitive.
     
    Last edited: Jun 22, 2008
  14. Jun 22, 2008 #13

    Mentz114

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    Hi George,

    I got

    [tex] x = a_0(t-t_0)^{\frac{1}{3}}t^{\frac{1}{3}}, 0\le t\le t_0[/tex]

    which does not quite reproduce the pear-shape. Yours is no doubt correct.

    M
     
    Last edited: Jun 22, 2008
  15. Jun 22, 2008 #14

    George Jones

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    I'm not sure about that!
     
  16. Jun 22, 2008 #15

    George Jones

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    Think of events on the past light cone as being emission events (by hypothetical stars). At each time, there is a star that is at just the right distance so that light reaches us now. For each time in the past, there is a unique (up to rotational symmetry) such star.

    Where, at the time of emission, was the the star located? I take "where" to mean proper distance from us, as measured at the time of emission.

    Assumptions:

    1) the universe contains only matter (no radiation or dark energy);

    2) the universe has flat spatial sections.

    Then, the Friedmann-Robertson-Walker metric is

    [tex]
    ds^2 = dt^2 - a \left( t \right)^2 dr^2,
    [/tex]

    with [itex]a \left( t \right) = t^{2/3}[/itex]. [itex]r[/itex] is a comoving coordinate.

    A lightlike worldline has [itex]ds^2 = 0[/itex], so

    [tex]
    dr = - \frac{dt}{ a \left( t \right)} = - t^{-\frac{2}{3}} dt.
    [/tex]

    The negative square root is taken because, as the light approaches us, its comoving distance from us decreases.

    Now, integrate this between appropriate limits. Since all points are equivalent, take our comoving spatial coordinate to be [itex]r = 0[/itex]. Arbitrarily, take now to be [itex]t = 1[/itex]. These are the upper limits of the integrations. Suppose that light was emitted at [itex]t = t[/itex] and [itex]r = r[/itex]. These are the lower limits of the integrations.

    [tex]
    \int^{0}_{r} dr' = - \int^{1}_{t} t'^{-\frac{2}{3}} dt'
    [/tex]

    [tex]
    r = 3 \left( 1 - t^{\frac{1}{3}} \right).
    [/tex]

    The proper distance [itex]D[/itex] is the scale factor multiplied by the comoving distance, i.e.,

    [tex]
    D \left( t \right) = a \left( t \right) r \left( t \right) = 3 t^{\frac{2}{3}} \left( 1 - t^{\frac{1}{3}} \right) = 3 \left (t^{\frac{2}{3}} - t \right).
    [/tex]
     
    Last edited: Jun 22, 2008
  17. Jun 22, 2008 #16
    Hi George,

    I think you may have cracked it. Good work! Thanks :smile:
    I superimposed a plot of your equation over the Ned Wright diagram and got a near perfect match ... but only where I set n=100 rather than 3 where:

    [tex]D \left( t \right) = n \left (t^{\frac{n-1}{n}} - t \right)[/tex]

    Is there anything sacrosanct about 3 or is it an arbitary variable depeding on the age/scale of the universe chosen? What is the significance of n or (n-1)/n ?

    Fast work by the way. You really know your stuff!
     
  18. Jun 23, 2008 #17

    George Jones

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    I didn't really look closely at first, but now I see that Ned Wright made the same two assumptions that I did.

    So, unless I made a mistake (quite possible), the [itex]t^{2/3}[/itex] cannot be changed. The [itex]3[/itex] out front can, however, be changed, as this will just change the relative scales of the [itex]x[/itex] and [itex]y[/itex] axes, or, in computer-speak, will change the aspect ratio of the diagram.
     
  19. Jun 23, 2008 #18

    Mentz114

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    Nice work, George.
     
  20. Jun 23, 2008 #19

    OK. I see where the discrepancy comes from now. :biggrin: Your curve with [itex]t^{2/3}[/itex] matches the curve given by Ned Wright in this model http://www.astro.ucla.edu/~wright/cosmo_03.htm which is for a matter dominated universe where the worldlines of galaxies are curving back towards each other.

    The diagram I asked about in post#1 was from the model with almost zero mass density and straight worldline for the receding galaxies here http://www.astro.ucla.edu/~wright/cosmo_02.htm and in that case an equation with [itex]t^{\frac{n-1}{n}}[/itex] with a value of n that tends towards infinity, matches the data. However a value of n = 100 produces a reasonable match and the curve does not change much from n=100 to n aproaching infinity.

    In either case, your equation matches the model when a suitable choice of n is used, so as Mentz said, good work!
     
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