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: 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,

$$\frac{d\rho}{dt}$$

 

[adamdunne]

latex of problem

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:
$$\Gamma_{\alpha\alpha\alpha}=-\phi,_\alpha$$
$$\Gamma_{\alpha\alpha\beta}=\Gamma_{\alpha\beta\alpha}=-\phi_,\alpha$$
$$\Gamma_{alpha\beta\beta}=phi_,\alpha$$
$$\Gamma^0{ }_{\alpha\beta}=g^{00}\Gamma_{0alpha\beta}\approx-\Gamma_{0\alpha\beta}$$
$$\Gamma^j{}_{\alpha\beta}=g^{jj}\Gamma_{j\alpha\beta}=\Gamma_{j\alpha\beta}$$

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]

Latex Problem(tidied up)

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:
$$\Gamma_{\alpha\alpha\alpha}=-\phi,_\alpha$$
$$\Gamma_{\alpha\alpha\beta}=\Gamma_{\alpha\beta\alpha}=-\phi,_\alpha$$
$$\Gamma_{alpha\beta\beta}=\phi_,\alpha$$
$$\Gamma^0{ }_{\alpha\beta}=g^{00}\Gamma_{0alpha\beta}\approx-\Gamma_{0\alpha\beta}$$
$$\Gamma^j{}_{\alpha\beta}=g^{jj}\Gamma_{j\alpha\beta}=\Gamma_{j\alpha\beta}$$

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?