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B Metric tensor of a perfect fluid in its rest frame

  1. Mar 27, 2017 #1
    The stress-energy tensor of a perfect fluid in its rest frame is:

    (1) Tij= diag [ρc2, P, P, P]

    where ρc2 is the energy density and P the pressure of the fluid.

    If Tij is as stated in eq.(1), the metric tensor gij of the system composed by an indefinitely extended perfect fluid in its rest frame must be the solution of the corresponding Einstein’s equation:

    (2) Rij –1/2 gij R = (8πG/c4) Tij

    The metric tensor

    (3) gij= diag [1, -1, -1, -1]

    cannot be the solution of eq.(1) because from eq.(3) it follows that:

    (4) Rij –1/2 gij R = 0


    If the foregoing argument is true, man cannot say that in a generic frame, where the fluid’s four-velocity is Ui and the metric tensor is gij, the stress-energy tensor is:

    (5) Tij= (ρ +P/c2) Ut Ut + P gij


    Where am I wrong?
     
  2. jcsd
  3. Mar 27, 2017 #2

    PeterDonis

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    You have misstated this. The correct statement is that the stress-energy tensor of a perfect fluid in local inertial coordinates at some chosen event takes the form you give. But that does not mean there is any global coordinate chart in which the SET at every event takes that form.

    ...Minkowski spacetime is a vacuum solution, and a perfect fluid is not a vacuum. But that is a global statement, not a local one.

    This does not follow from anything you have said. You are simply misapplying a local equation as if it had to apply globally.

    You might be confused by the fact that, at a single event, you can put the metric tensor into the Minkowski form at that event. So at that event, you can obviously write the stress-energy tensor in the form (5) in local inertial coordinates at that event. But equation (5) is a tensor equation, so it will be valid in any coordinate chart, even one in which the metric tensor ##g_{ij}## (and also the fluid 4-velocity ##U_i##) do not take the forms they take in local inertial coordinates at a given event. But to apply equation (5) in a general, global coordinate chart, you need to find the appropriate expressions for the metric tensor and the fluid 4-velocity in that global chart.
     
  4. Mar 27, 2017 #3

    pervect

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    It might help to study the interior Schwarzschild metric. Wiki has a page on it, <<link>>, I haven't checked it though.

    Note that this is for a spherically symmetric ball of perfect fluid, and that said ball must be larger than it's Schwarzschild radius to have a stable solution.

    Things to do would be to confirm that the metric satisfies Einstein's field equations, and to check that the density in the appropriate frame field is constant. The pressure won't be constant of course - it'll increase as you get deeper.

    In order to do the later step, to physically interpret the tensor as a local density and local pressure, you'll need to understand how to change basis, from the coordinate basis to a local orthonormal basis. Some helpful keywords would be vierbein and tetrad, wiki's article on frame fields in GR might be of some help <<link>>.

    You should suspect on physical grounds that a perfect fluid will self gravitate, that it cannot be of infinite extent (or it would become a black hole), and that if it is non-rotating and in equilibrium, the ball of perfect fluid will be spherically symmetrical. But from your post, you don't seem to suspect any of this :(.
     
  5. Mar 28, 2017 #4
    PeterDonis
    Thanks a lot: now I understand where I was wrong.
    Actually eq.(1) has been deduced (as I see in Physicspages) in a local orthonormal inertial frame, where gij = ηij, so that it makes no sense to look for the solution of eq.(2) with Tij as stated in eq.(1).
    It does make sense with Tij as stated in eq.(5) only, as they do , for instance, when deducing the Friedmann equations (that’s where my doubt first rose).

    Pervect
    Thanks for the link about the Schwarzschild metric inside the Schwarzschild radius. I have not jet dared to enter into that range.
     
  6. Mar 28, 2017 #5

    PeterDonis

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    You're welcome! I'm glad the discussion was helpful.
     
  7. Apr 21, 2017 #6
    Hi everyone!
    The foregoing discussion about the metric tensor of a perfect fluid in its rest frame has helped me very much in solving the problem from which my doubt arose: the deduction of the Friedmann equations from the Einstein field equations, keeping track of the “c” term.
    If someone is interested in this topic, here is the solution I elaborated.

    The 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.
    Thus, 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
    Please 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 ]
    Thus, if the fluid and the FWR frame are at rest:
    (10) u_Hi = [ 1, 0, 0, 0 ]

    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.
     
  8. Apr 21, 2017 #7

    pervect

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    I'm finding this very hard to read due to the formatting.

    You can use the Latex language here on PF to format upper and lower indices.

    \#\# T^{ij}\#\#, without the backslashes, becomes ##T^{ij}##. To do subscripts, put T_{ij} between a pair of #'s and you get ##T_{ij}##. You can also use a pair of $ signs to do non-inline latex. It can look a bit better if you're not trying to mix latex with normal text (i.e. not inline).

    Some other useful tidbits: \Gamma^i{}_{jk} becomes ##\Gamma^i{}_{jk}## , while \Gamma^i_{jk} becomes ##\Gamma^i_{jk}##

    This only scratches the surface, for instance one can do matrices:

    $$\begin{bmatrix} +1 & 0 & 0 & 0 \\
    0 & -1 & 0 & 0 \\
    0 & 0 & -1 & 0 \\
    0 & 0 & 0 & -1 \\
    \end{bmatrix}
    $$

    is generated by

    \\begin{bmatrix} +1 & 0 & 0 & 0 \\
    0 & -1 & 0 & 0 \\
    0 & 0 & -1 & 0 \\
    0 & 0 & 0 & -1 \\
    \\end{bmatrix}

    (use a single backslash, not a pair).
     
  9. Apr 22, 2017 #8
    Thanks Pervect for the suggestion.
    I try it again by Latex

    The starting point is the fully contravariant energy-stress tensor of a perfect fluid in a Local Inertial Frame:
    (1) ##T^{ij}##= \begin{bmatrix} + ρ c^2& 0 & 0 & 0 \\
    0 & +p& 0 & 0 \\
    0 & 0 & +p & 0 \\
    0 & 0 & 0 & +p \\
    \end{bmatrix}
    In the LIF the metric tensor is:
    (2) ##g^{ij}##=##g_{ij}##=\begin{bmatrix} +1& 0 & 0 & 0 \\
    0 & -1& 0 & 0 \\
    0 & 0 & -1 & 0 \\
    0 & 0 & 0 & -1 \\
    \end{bmatrix}
    and the contravariant 4-velocity tensor is:
    (3) ## u^{i} ##= d/dτ [ ct, x, y, z ] = γ [ c, dx/dt, dy/dt, dz/dt ],
    where τ is the proper time and γ the relativistic contraction factor.
    Thus, if the fluid and the LIF are at rest:
    (4) ## u^{i} ## = [ c, 0, 0, 0 ]
    If we leave the LIF and enter a generic reference frame, the energy-stress tensor becomes:
    (6) ##T^{ij}##=(ρ – p/c^2)##u^{0}## ##u^{0}## + p ##g^{ij}##
    Please note that eq.6 becomes eq.1 if ## u^{i} ## is as in eq.4 and ##g^{ij}## as in eq.2 .
    Now, if the new reference frame is the Friedmann-WalkerRobertson's one
    the fully covariant metric tensor is:
    (7) ##g_{ij}##=\begin{bmatrix} + c^2& 0 & 0 & 0 \\
    0 & - a^2 /(1- k x^2) & 0 & 0 \\
    0 & 0 & - a^2 x^2 & 0 \\
    0 & 0 & 0 & -a^2 x^2 sin^2 θ\\
    \end{bmatrix}
    the fully contravariant metric tensor is:
    (8) ##g^{ij}## =\begin{bmatrix} +1/ c^2& 0 & 0 & 0 \\
    0 & - (1- k x^2)/a^2& 0 & 0 \\
    0 & 0 & - 1/a^2 x^2 & 0 \\
    0 & 0 & 0 & - 1/ a^2 x^2 sin^2 θ \\
    \end{bmatrix}
    and the contravariant 4-velocity tensor is:
    (9) ## u^{i} ## = d/dτ [ t, x, θ, φ] = γ [ 1, dx/dt, dθ/dt, dφ/dt ]
    Thus, if the fluid and the FWR frame are at rest:
    (10) ## u^{i} ##= [ 1, 0, 0, 0 ]
    In the FRW frame, from eq.6, by eq.10 and eq.8, the fully contravariant energy-stress tensor is:
    (11) ##T^{ij}## =\begin{bmatrix} +ρ & 0 & 0 & 0 \\
    0 & + p a^2/(1- k x^2)& 0 & 0 \\
    0 & 0 & + p a^2 x^2 & 0 \\
    0 & 0 & 0 & + p a^2 x^2 sin^2 θ \\
    \end{bmatrix}
    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^{i}_{j}##=##T^{ik}####g_{kj}##
    from eq.12, eq.11 and eq.7 we obtain:
    (13) ##T^{i}_{j}##= \begin{bmatrix} + ρ c^2& 0 & 0 & 0 \\
    0 & -p& 0 & 0 \\
    0 & 0 & -p & 0 \\
    0 & 0 & 0 & -p \\
    \end{bmatrix}
    Therefore the Einstein field equations to be solved, keeping track of the “c” term, are:
    (14) ##G^{i}_{j}## = (8πG/c^4) ##T^{i}_{j}##
    The mixed Einstein tensor ##G^{i}_{j}## 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 of this wxMaxima program is:
    (15) ##G^{0}_{0}## = 3 k/a^2 + 3 (##\frac {da} {dt}##)^2 / a^2
    and
    (16) ##G^{1}_{1}## = ##G^{2}_{2}## =##G^{3}_{3}## =k / a^2 + (##\frac {da} {dt}##)^2 / (c^2 a^2) + 2 ##\frac {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 (##\frac {da} {dt}##)^2 / a^2 = +(8 π G/ c^4) ρ c^2
    and
    (18) k / a^2 + (##\frac {da} {dt}##)^2 / (c^2 a^2) + 2 ##\frac {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.

    If You find some error please let me know
    Thanks
     
  10. Apr 23, 2017 #9
    Sorry
    I made a mistake in the signs of eq. 6. The right equation is:
    (6) ##T^{ij}##=##(ρ + p/c^2)## ##u^{0}## ##u^{0}## ##- p ## ##g^{ij}##
     
  11. Apr 23, 2017 #10

    vanhees71

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    The general ideal-fluid energy-momentum tensor tensor is
    $$T^{\mu \nu}=(\epsilon+P) u^{\mu} u^{\nu}-g^{\mu \nu} P.$$
    I use the convention, where ##u_{\mu} u^{\mu}=1##. ##\epsilon## is the proper energy density and ##P## the pressure.

    Unfortunately many of your formulae are very hard to read, let alone the code. Attached you find a Mathematica notebook as pdf (I've no clue, how to upload the Mathematica nb file itself:frown:). with the corresponding calculation with Robertson-Walker metric.
     

    Attached Files:

  12. Apr 23, 2017 #11

    Ibix

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    Do you have a zip utility (WinZip etc)? Zip files can be uploaded.
     
  13. Apr 23, 2017 #12

    vanhees71

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    Good idea. So here it is.
     

    Attached Files:

  14. Apr 23, 2017 #13
    to Vahnees71

    I just wanted to deduce the Friedmann equations keeping track of the “c” term, i.e. without setting equal to one the velocity of light.
    I never understood the utility of the natural units. By natural units it is not possible to appreciate, for instance, the difference in the 4-velocity ## u^{i} ## of a perfect fluid at rest in a LIF (where ## u^{i} ## = [ c, 0, 0, 0 ] ) and in the FWR frame ( where ## u^{i} ##= [ 1, 0, 0, 0 ] ).

    About the computing app to solve the Einstein equations, Maxima has the advantage (meiner Meinung nach) of being open source with GR dedicated tensor tools.
     
  15. Apr 23, 2017 #14

    vanhees71

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    I don't know maxima. Obviously it works well. If I understand your first posting right, it looks very similar to what Mathematica gives, so it should be fine.

    The merit of natural units is that I don't have to keep track of these ennoying conversion factors ##c##. It's like in the US, where you measure distances in miles and heights in feet as if these lengths should be measured in different units.

    Anyway, it's convenient to define ##u_{\mu}## as dimensionless, i.e., as ##u^{\mu}=(1,0,0,0)## in the local rest frame, because usually you need it in relativistic fluid dynamics in terms of a projector to invariant temporal and spatial components of vectors and tensors.
     
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