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Question regarding Friedmann equation

  1. Jan 16, 2016 #1
    It has occurred to me that the Friedmann equation

    Friedmann.png
    allows for a solution
    H = H0 = 0 .
    This seems to say that independently of the value of the scale factor a, and the various Ωs, a static universe is possible. I am guessing that this solution is spurious and is a side effect of the derivation of the equation.

    I am curious about how the derivation introduced this spurious solution of the cosmological form of the GR equations, and would much appreciate someone posting an explanation.

    Regards,
    Buzz
     
  2. jcsd
  3. Jan 16, 2016 #2

    PeterDonis

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    What makes you think that? You do realize that, if ##H_0 = 0##, the LHS is undefined, right?
     
  4. Jan 16, 2016 #3

    ChrisVer

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    how js H=0 a solution inependently of the value of a??? Only if the Omegas are zero this can be true...
     
  5. Jan 17, 2016 #4
    Hi Chris:

    Thanks for your post.

    If H0 = 0, then H = 0 no matter what values a or the Ωs have. I understand that H can also be zero with H0 not zero for some combinations of value for a and the Ωs. Such a solution would correspond to a mathematically possible configuration of a GR universe, but H = H0 = 0 does not.

    Regards,
    Buzz
     
  6. Jan 17, 2016 #5
    Hi Peter:

    Got it. Thank you. I have also seen the equation somewhere in the form
    H = H0 √ . . .
    This form, together with another of my occasional senior moments, are the sources of the spurious solution.

    Regards,
    Buzz
     
  7. Jan 17, 2016 #6

    Orodruin

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    If H is zero the critical density is zero and so there is nothing in the universe. You recover Minkowski space.
     
  8. Jan 17, 2016 #7

    ChrisVer

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    Well a's are not constant. In some point they can give an overall zero result and it's when the H=0 (acceleration halts for a moment).
    The thing then is that if you want it to be always zero indepent of the value of a, the omegas to be zero. As for example the only way for:
    ax^2+ bx+c =0 (always 0 independently of the value of x) would need a=b=c=0.
     
  9. Jan 17, 2016 #8

    PeterDonis

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    More precisely, if H is zero at all times then the critical density is zero.
     
  10. Jan 17, 2016 #9

    Chalnoth

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    If you permit this kind of division by zero, you can prove 1 = 2. All you've shown is that if you divide by zero, you can prove anything.
     
  11. Jan 18, 2016 #10
    If you permit this kind of division by zero, you can prove 1 = 2. All you've shown is that if you divide by zero, you can prove anything.
    Hi Chalnth:

    Sorry I wasn't clearer in my post #5.

    What you said above is what I meant by "another of my occasional senior moments".

    Regards,
    Buzz
     
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