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Homework Help: Continuity in weak field approx.

  1. Jul 31, 2006 #1
    Given c=1, weak field approx for g: g(/mu,mu)=eta(/mu,mu)-2phi
    Derive eqn contiuity: d(rho)/dt+u(j)rho,j=-rho u(j/),j (all der. partial)
    Given T(mu,nu/)=(rho+p)u(mu/)cross u(nu/)+pg(mu,nu/)
    using divergenceT =0 i.e.T(mu,nu/;nu)=0:

    Step in the proof is Gamma(0/mu,j)u(mu)=-phi,j

    I get phi,j instead of -phi,j for this.

    Steps:Gamma(0/alpha,j)u(alpha/)=Gamma(0/0j)u(0/)+Gamma(0/jj)u(j/)
    The last term is zero since phi,0=0 and Gamma(/0jj)=phi,0
    The first term, Gamma(0/0j)u(0/)=Gamma(0/0j)=g(00/)Gamma(/00j)
    =(-1)(-phi,0)=phi,0.
    For the proof to work, this term must be -phi,j
    Anyone see how?
     
    Last edited: Aug 1, 2006
  2. jcsd
  3. Jul 31, 2006 #2

    quasar987

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    Suggestion: re-write that in LaTeX.

    Learn LaTeX here: https://www.physicsforums.com/showthread.php?t=8997

    It's easy, it takes a second. click on the desired latex image to see how it was written. ex: click on this to see the code I used,

    [tex]\frac{d\rho}{dt}[/tex]
     
  4. Aug 1, 2006 #3
    latex of problem

    Given:
    c:=1
    Weak field approximation: [itex]g^{\mu\mu}=\eta^{\mu\mu}-2\phi[/itex]
    Non-diagonal terms zero; [itex]-1<<\phi<0[/itex]
    [tex]T^{\mu\nu}=(\rho+p)u^\mu\otimes u^\nu+pg^{\mu\nu}[/tex] [tex]u^j=v^j\equiv\frac{dx^j}{dt}<<1[/tex];[tex]p<<\rho[/tex];[tex] u^0\approx1[/tex];[tex] u^0_,\alpha\approx 0[/tex]
    Then one readily derives the connection-coefficents:
    [tex]\Gamma_{\alpha\alpha\alpha}=-\phi,_\alpha[/tex]
    [tex]\Gamma_{\alpha\alpha\beta}=\Gamma_{\alpha\beta\alpha}=-\phi_,\alpha[/tex]
    [tex]\Gamma_{alpha\beta\beta}=phi_,\alpha[/tex]
    [tex]\Gamma^0{ }_{\alpha\beta}=g^{00}\Gamma_{0alpha\beta}\approx-\Gamma_{0\alpha\beta}[/tex]
    [tex]\Gamma^j{}_{\alpha\beta}=g^{jj}\Gamma_{j\alpha\beta}=\Gamma_{j\alpha\beta}[/tex]

    In applying the eqn [itex]T^{\mu\nu}_{;\nu}=0[/itex] one gets 4 eqns for [itex]\mu\nu=[/itex]00 0j j0 jj

    In one of these eqns is the term: [itex]\Gamma^0{ }_{\alpha j}u^\alpha[/itex]
    This=[itex]\Gamma^0{ }_{0j}u^0+\Gamma^0{}_{jj}u^j[/itex]
    The last term is zero since [itex]\phi_,0=0[/itex]
    The first term I get as [itex]\Gamma^0{ }_{0j}=g^{00}\Gamma_{00j}\approx(-1)(-\phi_,j)=+\phi,j/[/itex]
    Whereas to get the eqn continuity, the whole point of the exercise,
    [tex]\frac{\partial\rho}{\partial t}+v^j\frac{\partial \rho}{\partial x^j}=-\rho\frac{\partial v^j}{\partial x^j}[/tex]
    the above term must be negative
    Anyone see where the sign mistake is?
     
  5. Aug 5, 2006 #4
    Latex Problem(tidied up)

    Given:
    c:=1
    Weak field approximation: [itex]g^{\mu\mu}=\eta^{\mu\mu}-2\phi[/itex]
    Non-diagonal terms zero; [itex]-1<<\phi<0[/itex]
    [tex]T^{\mu\nu}=(\rho+p)u^\mu\otimes u^\nu+pg^{\mu\nu}[/tex] [tex]u^j=v^j\equiv\frac{dx^j}{dt}<<1[/tex]
    [tex]p<<\rho[/tex];[tex] u^0\approx1[/tex]
    [tex] u^0,_\alpha\approx 0[/tex]
    Then one readily derives the connection-coefficents:
    [tex]\Gamma_{\alpha\alpha\alpha}=-\phi,_\alpha[/tex]
    [tex]\Gamma_{\alpha\alpha\beta}=\Gamma_{\alpha\beta\alpha}=-\phi,_\alpha[/tex]
    [tex]\Gamma_{alpha\beta\beta}=\phi_,\alpha[/tex]
    [tex]\Gamma^0{ }_{\alpha\beta}=g^{00}\Gamma_{0alpha\beta}\approx-\Gamma_{0\alpha\beta}[/tex]
    [tex]\Gamma^j{}_{\alpha\beta}=g^{jj}\Gamma_{j\alpha\beta}=\Gamma_{j\alpha\beta}[/tex]

    In applying the eqn [itex]T^{\mu\nu}_{ ;\nu}=0[/itex] one gets 4 eqns for [itex]\mu\nu=[/itex]00 0j j0 jj

    In one of these eqns is the term: [itex]\Gamma^0{ }_{\alpha j}u^\alpha[/itex]
    This=[itex]\Gamma^0{ }_{0j}u^0+\Gamma^0{}_{jj}u^j[/itex]
    The last term is zero since [itex]\phi,_0=0[/itex]
    The first term I get as [itex]\Gamma^0{ }_{0j}=g^{00}\Gamma_{00j}\approx(-1)(-\phi,_j)=+\phi,_j/[/itex]
    Whereas to get the eqn continuity, the whole point of the exercise,
    [tex]\frac{\partial\rho}{\partial t}+v^j\frac{\partial \rho}{\partial x^j}=-\rho\frac{\partial v^j}{\partial x^j}[/tex]
    the above term must be negative
    Anyone see where the sign mistake is?
     
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