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Homework Help: General Relativity - FRW Metric - FRW Equations show that ..

  1. May 16, 2017 #1
    1. The problem statement, all variables and given/known data
    cosmobaby.png

    2. Relevant equations
    see above
    3. The attempt at a solution
    Using the conservation equation for ##p=0##

    I find: ##\rho =\frac{ \rho_0}{a^3}##; (I am told this is ##\geq0## , is ##a\geq0## so here I can conclude that ##\rho_0 \geq =0 ## or not?)

    Plugging this and ##p=0## into the first Einstein equation I get:

    ##\dot{a^2}+k-\Lambda a^2=\frac{8\pi G \rho_0}{a}##

    So a stationary solution is to solve for ##a## and get no time independence, so don't we need something of the form:

    ##\frac{da}{dt} a^k =0## or can I find a more general expression to this?

    This is ofc not possible to get since ##\Lambda## and ##k## are constants and can not depend on ##a##?

    Many thanks
     
  2. jcsd
  3. May 16, 2017 #2

    phyzguy

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    In a time independent solution, a is constant, so Λ and k can depend on a and still be constant.
     
  4. May 16, 2017 #3
    mmm

    so ##k=8 \pi G \rho_0 / a ## and ##\Lambda =0 ## ?
     
  5. May 16, 2017 #4

    phyzguy

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    How did you conclude Λ=0? Remember k = 0, +/-1.
     
  6. May 16, 2017 #5
    ##k=0## and ##\Lambda = \frac{-8\pi G \rho_0}{a}## ?

    Basically need to solve for ##8\pi G \rho_0/a +\Lambda a^2 -k = 0## subject to the above values of ##k## only allowed ?
     
  7. May 16, 2017 #6

    phyzguy

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    Aren't there two equations?
     
  8. May 16, 2017 #7
    equating ##a## coefficients, but baring in mind that ##\Lambda## can depend on ##a## and ##k## can not, so ##k## for the 0th order equation?
     
  9. Apr 10, 2018 #8
    apologies to re-bump but
    re-looking at this , ##\rho =\frac{ \rho_0}{a^3} \geq 0 ##
    ## \rho=\frac{\rho_0}{a} \frac{1}{a^2} \geq 0 ## and so since ##\frac{1}{a^2}>0 ## it must be that ##\rho=\frac{\rho_0}{a} \geq 0 ##

    so from the other equation ##0+k-\Lambda a^2 \geq 0## is the required constraint ?
     
  10. Apr 11, 2018 #9
    I'm getting a bit confused since it says ##k=-1,0,1## , but since the question doesn't specify whether ##\Lamda## is ##>0## or ##<0## there is no need to/no way to narrow down this further
     
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