GenRel:deriv'n eq'n continuity in weak field approx.

[adamdunne]

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?

[quasar987]

Suggestion: re-write that in LaTeX.

Learn LaTeX here: http://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,

\frac{d\rho}{dt}

[adamdunne]

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

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

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

[adamdunne]

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

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

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