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Lecture 23. Relativistic Friedman Equation _FE_ - B. Relativistic

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Lecture 23. Relativistic Friedman Equation _FE_ - B. Relativistic Powered By Docstoc
					  A. Polnarev. (MTH6123). 2010. Part IV. Lecture 23.
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Lecture 23.                  Last updated 16.03.10


  B.     Relativistic Friedman Equation (FE)



   The Energy Conservation Equation in GR IV B 1
   The FE for FLRW metric. Great surprise IV B 2



  1.   The Energy Conservation Equation in GR

The acceleration equation plus the equation of state contain three variables to find: the scale factor R (or a), the
energy density ε (ε = ρc2 ) and pressure P (when P < 0 it is better to consider P as a tension). Hence we need one
more equation. Let us take as the third equation

                                                           i
                                                          T0;i = 0,                                             (B.1)

which, as was shown in the previous lecture is the consequence of the EFEs and the Bianchi identity. Using the rules
of the covariant differentiation we have

                                i      i         n
                               T0;i = T0,i + Γi T0 − Γn Tn = T0,0 + Γα T0 − Γα Tα =
                                              in      i0
                                                         i    0
                                                                     α0
                                                                        0
                                                                             β0
                                                                                β




                     1     ˙
                          R α      ˙
                                  R α β    1               ˙
                                                          3R   ˙
                                                              R α β                 1     ˙
                                                                                         3R
                 =     ˙+    δα −    δ T =           ˙+      + δβ δα P          =     ˙+    ( + P ) = 0.        (B.2)
                     c    cR      cR β α   c              R   R                     c    R

Hence,
                                                           ˙
                                                          3R           P
                                                ρ=−
                                                ˙                 ρ+        .                                   (B.3)
                                                          R            c2

This is the energy conservation equation. If P = 0 this equation gives the equation of mass conservation for dust. It
is interesting to mention that this equation is identical to the first law of thermodynamics which says that the change
of energy ,E = ρc2 V , in volume, V , surrounded by the surface of area S is equal to the work done by the pressure
forces, FP = SP ,
                                                                  ˙      ˙
                                 dE = −FP dx = −P Sdx = −P dV, or E = −P V ,                                    (B.4)

Let us consider a sphere of radius r, in this case
                                          4π 3                    ·        P 3 ·
                                    V =      r , and       r3 ρ       =−      r , then                          (B.5)
                                           3                               c2


                                            ˙     P                   ˙
                                                                     3r              P
                                  ρr3 + 3r2 Rρ = − 2 3r2 , i.e.ρ = −
                                  ˙                            ˙                ρ+        .                     (B.6)
                                                  c                  r               c2



                                                               69
  A. Polnarev. (MTH6123). 2010. Part IV. Lecture 24.
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Taking into account that in the homogeneously expanding Universe
                                                         r
                                                         ˙  R˙
                                                           = ,                                                  (B.7)
                                                         r  R
we obtain exactly the energy conservation equation derived from the EFEs and the pure geometrical Bianchi identity.
It means that the first law of thermodynamics is a consequence of the EFEs! Actually we completed the required set
of equations describing the relativistic cosmological models. Now we are going to derive from these two equations,
the acceleration equation and the energy conservation equation, the relativistic version of the Friedman equation.


  2.    The FE for FLRW metric. Great surprise

From the energy conservation equation we obtain
                                                  P         ˙
                                                            ρR
                                                     = −ρ −     .                                               (B.8)
                                                  c2        3R˙
Putting this expression for p into the acceleration equation, we have


                               ¨   4πGR             3P           4πGR                    ˙
                                                                                         ρR
                               R=−            ρ+            =−                ρ − 3ρ −         =
                                     3              c2             3                      R˙

                                          4πG    ˙         4πG     ·
                                      =       2ρRR + ρR2 =
                                                     ˙         ρR2 .                                            (B.9)
                                            ˙
                                           3R                ˙
                                                            3R
                                                 ˙
then multiplying both sides of this equation by 2R and taking into account that
                                                            ·
                                                    ˙¨
                                                   2RR = R2 ,                                                  (B.10)
we obtain
                                                    ·       8πG     ·
                                               ˙
                                               R2       =       ρR2 ,                                          (B.11)
                                                             3
hence
                                              ˙    8πG 2
                                              R2 =    ρR − kc2 ,                                               (B.12)
                                                    3
This is the Friedmann equation in the relativistic Cosmology [k appears here as a dimensionless integration constant,
exactly as in the Newtonian cosmological models, but as one can easily show is related now with 3-curvature].
This is rather surprising result: the Friedman equation is identical to the equation obtained in the Newtonian theory
and does not contain the pressure term at all. However, the pressure is really extremely important. Indeed, if we put
the equation of state
                                                     P = αρc2 ,                                                (B.13)
into the energy conservation equation we obtain
          ˙
          ρ             ˙
                        R                                                                           ·
            = −3(1 + α)   and (ln ρ)· + 3(1 + α)(ln R)· = [ln ρ + 3(1 + α) ln R]· = ln[ρR3(1+α) ]       = 0,   (B.14)
          ρ             R
thus
                                      ln[ρR3(1+α) ] = C and ρR3(1+α) = C .                                     (B.15)
Finally we obtain that
                                                                 3(1+α)
                                                            R0
                                               ρ = ρ0                     .                                    (B.16)
                                                            R



                                                            70
  A. Polnarev. (MTH6123). 2010. Part IV. Lecture 23.
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  3.   When the space curvature is negligible

If k = 0 the FE is reduced to
                                                      ˙
                                                      R2   8πGρ
                                                         =      .                                              (B.17)
                                                      R2     3
Substituting into this equation the expression for ρ obtained above, we have
                                                                           3(1+α)
                                              ˙              1/2
                                              R     8πGρ            R0        2
                                                =                                   ;                          (B.18)
                                              R       3             R

we can solve this equation by the separation of variables. For that let us introduce the following notations
                                                                                        1/2
                                       R            3(1 + α)                   8πGρ
                                  x=      ,   β=             and A =                          .                (B.19)
                                       R0               2                        3

In terms of x, β and A the above equation can be written as
                                                      1 β
                                xxβ−1 = A, hence,
                                ˙                       dx = Adt, and xβ = βAt + C.                            (B.20)
                                                      β
Since x = 0 at t = 0 we obtain that C = 0. Thus
                                                                       1        2
                                         xβ ∼ t, and R ∼ x ∼ t β = t 3(1+α) .                                  (B.21)

In the case k = ±1 the solution (B.21) is still valid if

                                                    8πGρR2
                                                                   |k| = 1.                                    (B.22)
                                                      3c2
As follows from Eqs.(B.16)the LHS of (B.22) goes as Rβ , where β = 2 − 3(1 + α) = −(1 + 3α). Hence if α > −1/3
the solution (B.21) is valid for small R, if α < −1/3 (B.21)is valid for large R. WE will see during the next lecture
that the latter case corresponds to the expansion with acceleration.
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