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I Energy-momentum tensor and Friedmann Equations

  1. Sep 26, 2016 #1
    Hi everyone,

    I want to derive the Friedmann equations from Einstein Field Equations. However, I have a problem that stems from the energy-momentum tensor. I am also trying to keep track of ## c^2 ## terms.

    FRW Metric:
    $$ ds^2= -c^2dt^2 + a^2(t) \left( {\frac{dr^2}{1-kr^2} + r^2 d\theta^2 + r^2 \sin^2\theta d\phi^2} \right)$$

    EFE:
    $$ G_{\mu \nu} \equiv R_{\mu \nu }- \frac{1}{2} R g_{\mu \nu} = \frac{8 \pi G}{c^4} T_{\mu \nu} $$

    I calculated metric dependent components which are Christoffel symbols, Riemann Curvature Tensor, Ricci Tensor & Scalar. What I've found is

    $$ R_{tt} = -3 \frac{\ddot{a}}{a} $$

    $$ R_{ii} = \dfrac{g_{ii}}{c^2a^2}( a \ddot{a} +2 \dot{a}^2 +2kc^2 ) $$

    $$ R=g^{\mu \nu}R_{\mu \nu} = 6\left[\frac{\ddot{a}}{ac^2}+ \left(\frac{\dot{a}}{ac}\right)^2 +\frac{k}{a^2}\right] $$

    For time-time components I want to obtain:
    $$ \boxed{\left(\frac{\dot{a}}{a}\right)^2 = \frac{8 \pi G}{3c^2} \rho - \frac{kc^2}{a^2(t)}} $$
    and for the space-space components:
    $$ \boxed{\frac{2\ddot{a}}{a} + \left(\frac{\dot{a}}{a}\right)^2 = -\frac{8 \pi G }{c^2}p - \frac{kc^2}{a^2(t)}} $$
    here ## \rho ## is the energy density, not the mass density.

    So in order to obtain the boxed equations, energy-momentum tensor ## T_{\mu \nu} ## must be ## diag(\rho c^2, p g_{ij}) ## . However from the expression ## T_{\mu \nu}=\left(\rho + \frac{p}{c^2}\right)u_\mu u_\nu +pg_{\mu \nu} ## I can't get what I want for ## u^\alpha=(c,0,0,0) ## . Am I using wrong form of ## T_{\mu \nu} ## or 4-velocity ## u^\alpha ## ?

    Thanks for your help,

    K.
     
  2. jcsd
  3. Sep 26, 2016 #2

    Chalnoth

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    The correct form of ## T_{\mu \nu} ## is certainly ## diag(\rho c^2, p g_{ij}) ## (though possibly with a negative sign somewhere depending upon the sign convention you're using), and that is the correct 4-velocity. My guess is that you've got a sign error.
     
  4. Sep 26, 2016 #3
    Thanks for the answer, I am using (-,+,+,+) convention. The problem is that for 00 component ## (\rho + p/c^2) \times c^2 + p(-c^2) ## does not give ##\rho c^2 ##
     
  5. Sep 26, 2016 #4

    Chalnoth

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    Doesn't it?

    Edit: Ahh, you have a ##c^2## discrepancy. I believe it should be ## (\rho + p/c^2) \times c^2 + p(-1) ##. This is one reason why physicists usually leave off the factors of ##c##: being only unit conversion factors, they add no information and can cause errors.
     
  6. Sep 26, 2016 #5
    Yes I agree with you, however in our case we kept ## c ## in any term, so we should also keep ## c ## in the metric.

    The other possibility is that
    $$ T_{\mu \nu} = (\rho + p)u_\mu u_\nu + pg_{\mu \nu} $$
    by choosing the above expression as our energy-momentum tensor, then we can obtain the intended result.
     
  7. Sep 27, 2016 #6

    Chalnoth

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    Yes, but the fact that you've got that extra ##c^2## in there means that you made a mistake. Probably some conflict in the convention of where to place the factors of ##c##.

    You can see that it's a mistake because it's not possible to add (or subtract) two parameters with different units. Imagine trying to add a position to a velocity, or a position to a time. It makes no sense.

    In the above equations, ##\rho## has units of matter density, so ##\rho c^2## has units of energy density. The pressure ##p## also has units of energy density.

    My guess is you made a mistake when raising/lowering the indices of the metric.

    That only works if you use units where ##c=1##. With your convention of using units where ##c \neq 1##, you can't add a matter density to an energy density.
     
  8. Sep 27, 2016 #7
    I highlighted that "## \rho ## is the energy density, not the mass density" in my main post. Thus, there is no problem with ## T_{\mu \nu} = (\rho + p)u_\mu u_\nu + pg_{\mu \nu} ## . On the other hand, this time ## T_{00} ## is in the unit of energy density x c2 = mass density x c4 which is meaningless.
     
  9. Sep 27, 2016 #8

    Chalnoth

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    With that convention, sure, that works fine.
     
  10. Apr 16, 2017 #9
    High to everybody!

    With just one and half year delay (sorry), here is a way to derive the Friedmann equations keeping track of the “c” term.

    Starting point is the fully contravariant energy-stress tensor of a perfect fluid in a Local Inertial Frame:

    (1) T_Hij = diag[ ρ * c^2, p, p, p ]

    where Hij means “High ij indexes”, ρ is the mass density, p is the pressure of the fluid.

    In the LIF the metric tensor is:

    (2) g_Hij = g_Lij = diag[ +1, -1, -1, -1 ]

    and the contravariant 4-velocity tensor is:

    (3) u_Hi = d/dτ [ c*t, x, y, z ] = γ [ c, dx/dt, dy/dt, dz/dt ],

    where τ is the proper time and γ the relativistic contraction factor.

    So, if the fluid and the LIF are at rest:

    (4) u_Hi = [ c, 0, 0, 0 ]

    If we leave the LIF and enter a generic frame, the energy-stress tensor becomes:

    (6) T_Hij = (ρ – p/c^2)*(u_H0)^2 + p*g_Hij

    Note that eq.6 becomes eq.1 if u_Hi is as in eq.4 and g_Hij as in eq.2 .

    Now, if the new reference frame is the Friedmann Walker Robertson one

    the fully covariant metric tensor is:

    (7) g_Lij = diag[ + c^2, - a^2/(1- k * x^2), - (a^2 * x^2), - (a^2 * x^2 * sinθ^2) ]

    the fully contravariant metric tensor is:

    (8) g_Hij = diag[ +1/c^2, - (1- k * x^2)/a^2, - 1/(a^2 * x^2), - 1/(a^2 * x^2 * sinθ^2) ]

    and the contravariant 4-velocity tensor is:

    (9) u_Hi = d/dτ [ t, x, θ, φ] = γ [ 1, dx/dt, dθ/dt, dφ/dt ]

    So, if the fluid and the FWR frame are at rest:

    (10) u_Hi = [ 1, 0, 0, 0 ]

    Thus in the FRW frame, from eq.6, by eq.10 and eq.8, the fully contravariant energy-stress tensor is:

    (11) T_Hij = diag[ + ρ, + p*(1- k * x^2)/a^2, + p/(a^2 * x^2), + p/(a^2 * x^2 * sinθ^2)]

    The trick to remove the distorting effect of the metric on the fully contravariant energy-stress tensor as in eq.11 is to consider its Mixed indexes tensor. Since:

    (12) T_Mij = T_Hik * g_Lkj

    from eq.12, eq.11 and eq.7 we obtain:

    (13) T_Mij = diag[ + ρ*c^2, - p, - p , -p ]

    Therefore the Einstein Field Equations to be solved (keeping track of the “c” term) are:

    (14) G_Mij = (8 π G/ c^4)*T_Mij

    The mixed Einstein tensor G_Mij in eq.14 may be computed using the open source wxMaxima algebra system app by, for instance, the following (.wmx) program:

    kill(all)$
    dim: 4$
    array(g,dim,dim)$
    g[1,1]: c^2$
    g[1,2]: 0$
    g[1,3]: 0$
    g[1,4]: 0$
    g[2,1]: 0$
    g[2,2]: -(a(t)^2)/(1-k*x^2)$
    g[2,3]: 0$
    g[2,4]: 0$
    g[3,1]: 0$
    g[3,2]: 0$
    g[3,3]: -(a(t)*x)^2$
    g[3,4]: 0$
    g[4,1]: 0$
    g[4,2]: 0$
    g[4,3]: 0$
    g[4,4]: -(a(t)*x*sin(theta))^2$
    gg: genmatrix(g,dim,dim)$
    kill(ctensor)$
    load(ctensor)$
    lg: gg$
    cmetric(false) $
    ct_coords: [t,x,theta,phi]$
    christof(false)$
    ricci(false)$
    R: scurvature()$
    einstein(true)$

    The output is:
    (15) G_M00 = 3*k / a^2 + 3*(da/dt)^2 / a^2
    and
    (16) G_M11 = k / a^2 + (da/dt)^2 / (c^2 * a^2) + 2 (d^2 a / dt^2) / (c^2 * a)

    From eq.15, eq.14 and eq.13, we obtain the equations:
    (17) 3*k / a^2 + 3*(da/dt)^2/ a^2 = (8 π G/ c^4)* ρ*c^2
    and
    (18) k / a^2 + (da/dt)^2 / (c^2 * a^2) + 2 (d^2 a / dt^2) / (c^2 * a) = - (8 π G/ c^4)* p

    which are the basis for the deduction of the Friedmann equations by keeping track of the “c” term.
     
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