 Quote by fzero
OK that'll do. As for the two surfaces, you need to figure out what your surface [tex]\Omega[/tex] is. Is [tex]\partial \Omega[/tex] necessarily connected?
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I thought S is the surface given in the formula for J_S. Then [latex]\partial \Omega=S[/latex] and [latex]\Omega[/latex] 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.
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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 [latex]\nabla_\mu T^{\mu \nu}[/latex] question, can I take [latex]u^\mu=(1,0,0,0)[/latex] as we are assuming a comoving observer. And apparently they have [latex]u^\alpha=\delta^\alpha{}_0[/latex].
However, even though our notes say we can do this for a comoving observer, I find that
[latex]\nabla_\mu T^{\mu \nu}=0[/latex]
[latex]\nabla_\mu ( \rho u^\mu u^\nu ) + \nabla_\mu ( p u^\mu u^\nu) - \nabla_\mu (pg^{\mu \nu})=0[/latex]
Taking [latex]\nu=0[/latex] we get
[latex]\nabla_0 ( \rho u^0) + \nabla_i ( \rho u^i) + \nabla_0 ( pu^0) + \nabla_i ( p u^i) - \nabla_0p[/latex]
[latex]\frac{\partial \rho}{\partial t}=0[/latex]
which is clearly incomplete so something isn't right...
Thanks.