Can someone explain why p terms are not canceling in this Diffeomorphisms proof?

[fzero]

So will this mean that all the rest of my work is wrong?

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.

I thought about this but surely it can't be that easy. We haven't made use of the fact that \xi=\partial_\phi which seemed to be central to the question!

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.

Also, in the notes, could you give me some advice on how to derive 415?

The stress tensor for a perfect fluid is (175).

And on p128 he says that the universe will expand forever even though the rate of expansion is decreasing (\ddot{a}<0). But surely if \ddot{a}<0 then at some point \dot{a} will change from being positive and become negative and then we will have a contracting universe?

The solutions for a(\eta) are (427). What you claim clearly doesn't happen for \eta>0 in the flat and open universes.

And is there a difference between "the past light cone at p" and the "particle horizon at p"?

Thanks.

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.

 

[latentcorpse]

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?

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?

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?

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?

 

[fzero]

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.

 

[latentcorpse]

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?

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.

 

[fzero]

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.

 

[latentcorpse]

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.