## Diffeomorphisms

 Quote by fzero Not necessarily, though you should be a bit more careful about setting things up. For example $$\xi^{a;b}$$ is not itself an antisymmetric 2-form. Well you do use the fact that it's a Killing vector. The fact that it's spacelike should also be tied to the fact that you're using a spacelike hypersurface.
Surely from Killing's equation, $\nabla_a \xi_b = - \nabla_b \xi_a$ will be antisymmetric? Do I need to do anything about the two surfaces then?

 Quote by fzero The stress tensor for a perfect fluid is (175).
I have been trying to take the $\nu=0$ component of $T^{\mu \nu}{}_{; \mu}=0$ but it simply will not work! Is this the way I should go about it?

 Quote by fzero The solutions for $$a(\eta)$$ are (427). What you claim clearly doesn't happen for $$\eta>0$$ in the flat and open universes.
But just from a plot of sinh, it looks like a smiley face (i.e. positive second derivatie), no?

 Quote by fzero Yes, the horizon is composed of the boundary of the present positions of particles which were in casual contact with us at some time in the past. Again, I don't have the figures, but the one on p. 130 is probably relevant.
So the particle horizon is the boundary of the past light cone?

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 Quote by latentcorpse Surely from Killing's equation, $\nabla_a \xi_b = - \nabla_b \xi_a$ will be antisymmetric? Do I need to do anything about the two surfaces then?
OK that'll do. As for the two surfaces, you need to figure out what your surface $$\Omega$$ is. Is $$\partial \Omega$$ necessarily connected?

 I have been trying to take the $\nu=0$ component of $T^{\mu \nu}{}_{; \mu}=0$ but it simply will not work! Is this the way I should go about it?
It looks like you need to compute a trace of one of the Christoffel symbols.

 But just from a plot of sinh, it looks like a smiley face (i.e. positive second derivatie), no?
$$\ddot{a}$$ is the 2nd derivative w.r.t. coordinate time, not $$\eta$$. I can't see the figures, so I don't know if they should be helpful to you.

 So the particle horizon is the boundary of the past light cone?
The figure is probably clearer than I would put into words. The horizon is a boundary on spacetime at our present expansion parameter. To determine the points that make up the horizon, you need to follow the worldlines of the particles that crossed our past light cone at some earlier time.

 Quote by fzero OK that'll do. As for the two surfaces, you need to figure out what your surface $$\Omega$$ is. Is $$\partial \Omega$$ necessarily connected?
I thought S is the surface given in the formula for J_S. Then $\partial \Omega=S$ and $\Omega$ would just be the enclosed 3-sphere?

 Quote by fzero The figure is probably clearer than I would put into words. The horizon is a boundary on spacetime at our present expansion parameter. To determine the points that make up the horizon, you need to follow the worldlines of the particles that crossed our past light cone at some earlier time.
Well from the figure, it appears to me that the particle horizon actually is the boundary of the past light cone. But if this is so, why don't they just define it as this? This is why I'm having doubts about my interpretation. But I can't think of an example where they wouldn't be equal.

Also, for the $\nabla_\mu T^{\mu \nu}$ question, can I take $u^\mu=(1,0,0,0)$ as we are assuming a comoving observer. And apparently they have $u^\alpha=\delta^\alpha{}_0$.
However, even though our notes say we can do this for a comoving observer, I find that
$\nabla_\mu T^{\mu \nu}=0$
$\nabla_\mu ( \rho u^\mu u^\nu ) + \nabla_\mu ( p u^\mu u^\nu) - \nabla_\mu (pg^{\mu \nu})=0$
Taking $\nu=0$ we get
$\nabla_0 ( \rho u^0) + \nabla_i ( \rho u^i) + \nabla_0 ( pu^0) + \nabla_i ( p u^i) - \nabla_0p$
$\frac{\partial \rho}{\partial t}=0$
which is clearly incomplete so something isn't right...

Thanks.

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 Quote by latentcorpse Also, for the $\nabla_\mu T^{\mu \nu}$ question, can I take $u^\mu=(1,0,0,0)$ as we are assuming a comoving observer. And apparently they have $u^\alpha=\delta^\alpha{}_0$. However, even though our notes say we can do this for a comoving observer, I find that $\nabla_\mu T^{\mu \nu}=0$ $\nabla_\mu ( \rho u^\mu u^\nu ) + \nabla_\mu ( p u^\mu u^\nu) - \nabla_\mu (pg^{\mu \nu})=0$ Taking $\nu=0$ we get $\nabla_0 ( \rho u^0) + \nabla_i ( \rho u^i) + \nabla_0 ( pu^0) + \nabla_i ( p u^i) - \nabla_0p$ $\frac{\partial \rho}{\partial t}=0$ which is clearly incomplete so something isn't right... Thanks.
If you take the calculation another line you should find something like $$\dot{\rho} + (\rho +p){\Gamma^0}_{00}=0$$.
 Quote by fzero If you take the calculation another line you should find something like $$\dot{\rho} + (\rho +p){\Gamma^0}_{00}=0$$.
How did you manage to get any $p$ terms surviving? All of mine cancel.