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I'd like to know your interpretation.

  1. Dec 18, 2011 #1
    With Hobson's notation:

    H=(da/dt)/a

    dH/dt = ((d2a/dt2)/a) - ((da/dt)/a)2

    If ((d2a/dt2)/a) is negative, (dH/dt) is certainly negative.
    But if ((d2a/dt2)/a) is positive, (dH/dt) can be positive or negative.

    What is your interpretation?
     
  2. jcsd
  3. Dec 18, 2011 #2

    mathman

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    dH/dt = ((d2a/dt2)/a) - ((da/dt)/a)2

    Above is incorrect. Should be:

    dH/dt = (d2a/dt2)/a - (da/dt)/a2

    Your second term is wrong.
     
  4. Dec 18, 2011 #3

    I like Serena

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    No, it is correct.
    I think you're forgetting the application of the chain rule.

    I also think EhsanZ's interpretation is correct, although I do not know what "Hobson's notation" is.
    I couldn't find it with google, except for in this thread. ;)
     
  5. Dec 18, 2011 #4

    BillSaltLake

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    Actually I think the only thing wrong with second eq. is a lack of superscripting of the 2s:

    dH/dt = ((d2a/dt2)/a) - ((da/dt)/a)2
     
  6. Dec 18, 2011 #5

    George Jones

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    [itex]d^2 a/dt^2 > 0[/itex] and [itex]dH/dt < 0[/itex] means that the expansion of the universe is accelerating while the Hubble constant is decreasing. We think that this is happening today.
    This (I presume) refers to the notation used in the book General Relativity: An Introduction for Physicists by Hobson, Efstathiou, and Lasenby.

    EhsanZ, play around with the spatially flat, matter-only Lemaitre model given on page 406 and in problem 15.23. This a good analytic approximation to our universe.
     
  7. Dec 18, 2011 #6

    Chronos

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    George is correct. If you look at the equations dispassionately, it is obvious the result can never be negative.
     
  8. Dec 19, 2011 #7

    mathman

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    You're right. My bad.
     
  9. Dec 20, 2011 #8
    No my friend! U made a mistake.
     
  10. Dec 20, 2011 #9
    Doesn't matter! Everybody makes mistakes.:smile:
     
  11. Dec 20, 2011 #10
    I meant the notation that Hobson had used in his book named "General Relativity: An Introduction for Physicists ".
     
  12. Dec 20, 2011 #11
    Yes, you're right my friend. I should've written it more carefully.
    Thanks
     
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