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Mistake in cosmology chapter in "Reflections on Relativity"?

  1. Oct 31, 2014 #1
    http://mathpages.com/rr/s7-01/7-01.htm

    I am completely unable to follow the following sequence of working between equations 2 and 3. AFAIK the final answer is correct, but the intermediate steps seem to be a "casserole of nonsense". I would appreciate feedback from anyone who can follow this, or who can correct it if it is wrong . . .
    --------------------------------------------------------------------------------------------------------------------------------------------

    Now suppose we embed a Euclidean three-dimensional space (x,y,z) in a four-dimensional space (w,x,y,z) whose metric is


    http://mathpages.com/rr/s7-01/7-01_files/image004.gif [Broken]

    where k is a fixed constant equal to either +1 or -1. If k = +1 the four-dimensional space is Euclidean, whereas if k = -1 it is pseudo-Euclidean (like the Minkowski metric). In either case the four-dimensional space is "flat", i.e., has zero Riemannian curvature. Now suppose we consider a three-dimensional subspace comprising a sphere (or pseudo-sphere), i.e., the locus of points satisfying the condition


    http://mathpages.com/rr/s7-01/7-01_files/image005.gif [Broken]


    From this we have w2 = (1 - r2)/k = k - kr2, and therefore


    http://mathpages.com/rr/s7-01/7-01_files/image006.gif [Broken]


    Substituting this into the four-dimensional line element above gives the metric for the three-dimensional sphere (or pseudo-sphere)


    http://mathpages.com/rr/s7-01/7-01_files/image007.gif [Broken]


    Taking this as the spatial part of our overall spacetime metric (2) that satisfies the Cosmological Principle, we arrive at


    http://mathpages.com/rr/s7-01/7-01_files/image008.gif [Broken]
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Oct 31, 2014 #2

    WannabeNewton

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    What exactly in the derivation is confusing you? It looks perfectly fine and sensible to me.
     
  4. Oct 31, 2014 #3
    OK, firstly "From this we have w2 = (1 - r2)/k = k - kr2". I'm OK with the first equality but the second makes me want to gag. Is he invoking some power series/binomial approximation without using an approximation sign?

    Secondly, "and therefore[PLAIN]http://mathpages.com/rr/s7-01/7-01_files/image006.gif [Broken] [Broken] " is OK from the first equality, but does not follow from the second.

    Thirdly, I can't get the final answer by squaring [PLAIN]http://mathpages.com/rr/s7-01/7-01_files/image006.gif [Broken] [Broken] and substituting into the polar metric at the top. I get $$(dw)^2 = k^2r^2 / (1 - kr^2)(dr)^2$$ so $$k(dw)^2 + (dr)^2 = (k^3r^2/(1 - kr^2) + 1) (dr)^2$$ and not $$k(dw)^2 + (dr)^2 = (1 /(1 - kr^2)) (dr)^2$$

    I must be missing some invisible steps I suppose, can you tell me where?
     
    Last edited by a moderator: May 7, 2017
  5. Oct 31, 2014 #4

    Fredrik

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    It's much simpler than that. :) Since ##k=\pm1##, we have ##1/k=k##.
     
  6. Oct 31, 2014 #5

    PAllen

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    k is +1 or -1, by stipulation.

    [edit: cross posted with Fredrik].
     
  7. Oct 31, 2014 #6
    Good catch guys, I need to revisit my calculations . . .
     
  8. Oct 31, 2014 #7

    Fredrik

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    Maybe I'm making some mistake too, but I keep getting
    $$dw=\frac{-kr}{\sqrt{k(1-r^2)}}dr$$ and therefore
    $$dw^2=\frac{r^2}{k(1-r^2)}dr^2.$$ How are you guys getting a factor of ##k## in the ##r^2## term in the denominator (and not in the other term as well)?
     
  9. Oct 31, 2014 #8
    OK, so ##dw^2 = k - kr^2## because ##k = 1/k##. Therefore ##w = \sqrt(k - kr^2)## and $$\frac{dw}{dr} = \frac{1}{2 \sqrt(k - kr^2)} . -2kr = \frac{-kr}{\sqrt(k -kr^2)}$$
    So I still haven't quite got there. I'm not going to attempt substituting back into the metric until I get this right . . . any clues?
     
  10. Oct 31, 2014 #9

    PAllen

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    I get the same as you. Previously, I answered only the initial question. Further, if you plug this back into the metric, k drops out altogether.
     
  11. Oct 31, 2014 #10
    Hmm, I'm not entirely sure this can be saved, that would be a shame as it's nice & concise. I've attempted working backwards from the metric but not come up with a sensible intermediate step as yet . . .

    I think we need $$\frac{dw}{dr} = \frac {\pm r} {\sqrt(1 - kr^2)}$$
     
    Last edited: Oct 31, 2014
  12. Oct 31, 2014 #11

    PAllen

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    Well, everything works out fine if one changes the definition of [pseudo-]sphere to have the k in front of the r2 term to start. Maybe that is the mistake?
     
  13. Oct 31, 2014 #12
    You mean as in ##w^2 + kr^2 = 1## ?
     
  14. Oct 31, 2014 #13

    PAllen

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    Yes. If I work out everything from there, I get the right intermediate expressions, and the right final metric.
     
  15. Oct 31, 2014 #14
    Cool, thanks for that, I think I'll check it all out tomorrow, my brain hurts!
     
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