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Deriving the Navier-Stokes equation from energy-momentum tensor

  1. Dec 5, 2011 #1
    The energy-momentum tensor for a perfect fluid is [itex]T^{ab}=(\rho_0+p)u^au^b-pg^{ab}[/itex] (using the +--- Minkowski metric).

    Using the conservation law [itex]\partial_bT^{ab}=0[/itex], I'm coming up with [itex](\rho+\gamma^2p) [\frac{\partial\mathbb{u}}{{\partial}t}+ (\mathbb{u}\cdot\mathbb{\nabla})\mathbb{u}]= -{\nabla}p[/itex] instead of [itex]\rho[\frac{\partial\mathbb{u}}{{\partial}t}+ (\mathbb{u}\cdot\mathbb{\nabla})\mathbb{u}]= -{\nabla}p[/itex] (disregarding the body force part of the equation). Why is there a term for the part in brackets multiplied by [itex]\rho[/itex], but not for [itex]\gamma^2p[/itex]?
     
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  3. Dec 5, 2011 #2

    dextercioby

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    Where did the gamma come from ? Can you post your derivation/scaned copy of it ?
     
  4. Dec 6, 2011 #3
    [itex]u^a=\gamma(1,\mathbf{u})[/itex]
    Keeping b constant and cycling a from x to z, I get [itex]\partial_tT^{at}= \partial_t[\gamma^2(\rho_0+p)\mathbf{u}][/itex]
    [itex]\partial_xT^{ax}= \partial_x[\gamma^2(\rho_0+p)u_x\mathbf{u}]+\partial_xp[/itex]
    [itex]\partial_yT^{ay}= \partial_y[\gamma^2(\rho_0+p)u_y\mathbf{u}]+\partial_yp[/itex]
    [itex]\partial_zT^{az}= \partial_z[\gamma^2(\rho_0+p)u_z\mathbf{u}]+\partial_zp[/itex]
    [itex]\rho=\gamma^2\rho_0[/itex], so that is why you see a [itex]\gamma^2[/itex] coefficient for p but not for [itex]\rho[/itex]. When I solve for [itex]\partial_bT^{tb}=0[/itex], I get [itex]\partial_t(\rho+\gamma^2p)+ \partial_x[(\rho+\gamma^2p)u_x]+ \partial_y[(\rho+\gamma^2p)u_y]+ \partial_z[(\rho+\gamma^2p)u_z]=0[/itex], so summing the four equations for [itex]T^{ab}[/itex] above and simplifying using [itex]\partial_bT^{tb}=0[/itex], I get [tex](\rho+\gamma^2p)\partial_t\mathbf{u}+ (\rho+\gamma^2p)u_x\partial_x\mathbf{u}+ (\rho+\gamma^2p)u_y\partial_y\mathbf{u}+ (\rho+\gamma^2p)u_z\partial_z\mathbf{u}+ {\nabla}p=\mathbf{0}[/tex]
    This simplifies to what I put in the first post: [itex](\rho+\gamma^2p)[\partial_t\mathbf{u}+ (\mathbf{u}\cdot\mathbf{\nabla})\mathbf{u}]= -\mathbf{\nabla}p[/itex]
     
  5. Dec 6, 2011 #4

    dextercioby

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    There must be a specially relativistic correction to the Galilei invariant Euler equations. You may wish to check L&L <Fluid Mechanics>, pp. 505 to 508.
     
  6. Dec 6, 2011 #5
    It looks like the reason the [itex]\gamma^2p[/itex] term is not in the Navier-Stokes equation is because a nonrelativistic limit is applied where [itex]p<<\rho[/itex], so [itex]\rho+\gamma^2p\approx\rho[/itex].
     
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