2 added 2092 characters in body edited Apr 19 '11 at 15:40 Lawrence B. Crowell 8,0911717 silver badges2525 bronze badges Addendum: The question was asked about the redshift factor and the cosmological horizon. This requires a bit more than a comment post. On a stationary coordinate region of the de Sitter spacetime $$g_{tt}~=~1~-~\Lambda r^2/3$$. This metric term is zero for $$r~=~\sqrt{3/\Lambda}$$, which is the distance to the cosmological horizon. The red shift factor can be considered as the expansion of a local volume of space, where photons that enter and leave this “box” can be thought of as a standing wave of photons. The expansion factor is then given by the scale factor for the expansion of the box $$z~=~\frac{a(t_0)}{a(t)}~-~1$$ The dynamics for the scale factor is given by the FLRW metric $$\Big(\frac{\dot a}{a}\Big)^2~=~\frac{8\pi G\rho}{3}$$ for $$k~=~0$$. The left hand side is the Hubble factor, which is constant in space but not time. Writing the $$\Lambda g_{ab}~=~8\pi GT_{ab}$$ as a vacuum energy and $$\rho~=~T_{00}$$ we get $$\Big(\frac{\dot a}{a}\Big)^2~=~H^2~=~\frac{\Lambda}{3}$$ the evolution of the scale factor with time is then $$a(t)~=~\sqrt{3/\Lambda}e^{\sqrt{\Lambda /3}t}.$$ Hence the ratio is $$a(t)/a(t_0)~=~ e^{\sqrt{\Lambda /3}(t-t_0)}$$.The expansion is this exponential function, which is Taylor expanded to give to first order the ratio above $$a(t)/a(t_0)~\simeq~1~+~H(t_0)(t_0-t)~=~1~+~H(t_0)(d-d_0)/c$$ which gives the Hubble rule. $$z~=~a(t)/a(t_0)~-~1$$. It is clear that from the general expression that $$a(t)$$ can grow to an arbitrarily large value, and so can $$z$$. On the cosmological horizon for $$d~-~d_0~=~r_h~=~\sqrt{3/\Lambda}$$ we have $$z~=~1$$. Looking beyond the cosmological horizon $$r_h~\simeq~10^{10}$$ly is similar to an observer in a black hole looking outside to the exterior world outside the black hole horizon. People get confused into thinking the cosmological horizon is a black membrane similar to that on a black hole. Anything which we do observe beyond the horizon we can never send a signal to, just as a person in a black hole can see the exterior world and can never send a message out. Addendum: The question was asked about the redshift factor and the cosmological horizon. This requires a bit more than a comment post. On a stationary coordinate region of the de Sitter spacetime $$g_{tt}~=~1~-~\Lambda r^2/3$$. This metric term is zero for $$r~=~\sqrt{3/\Lambda}$$, which is the distance to the cosmological horizon. The red shift factor can be considered as the expansion of a local volume of space, where photons that enter and leave this “box” can be thought of as a standing wave of photons. The expansion factor is then given by the scale factor for the expansion of the box $$z~=~\frac{a(t_0)}{a(t)}~-~1$$ The dynamics for the scale factor is given by the FLRW metric $$\Big(\frac{\dot a}{a}\Big)^2~=~\frac{8\pi G\rho}{3}$$ for $$k~=~0$$. The left hand side is the Hubble factor, which is constant in space but not time. Writing the $$\Lambda g_{ab}~=~8\pi GT_{ab}$$ as a vacuum energy and $$\rho~=~T_{00}$$ we get $$\Big(\frac{\dot a}{a}\Big)^2~=~H^2~=~\frac{\Lambda}{3}$$ the evolution of the scale factor with time is then $$a(t)~=~\sqrt{3/\Lambda}e^{\sqrt{\Lambda /3}t}.$$ Hence the ratio is $$a(t)/a(t_0)~=~ e^{\sqrt{\Lambda /3}(t-t_0)}$$.The expansion is this exponential function, which is Taylor expanded to give to first order the ratio above $$a(t)/a(t_0)~\simeq~1~+~H(t_0)(t_0-t)~=~1~+~H(t_0)(d-d_0)/c$$ which gives the Hubble rule. $$z~=~a(t)/a(t_0)~-~1$$. It is clear that from the general expression that $$a(t)$$ can grow to an arbitrarily large value, and so can $$z$$. On the cosmological horizon for $$d~-~d_0~=~r_h~=~\sqrt{3/\Lambda}$$ we have $$z~=~1$$. Looking beyond the cosmological horizon $$r_h~\simeq~10^{10}$$ly is similar to an observer in a black hole looking outside to the exterior world outside the black hole horizon. People get confused into thinking the cosmological horizon is a black membrane similar to that on a black hole. Anything which we do observe beyond the horizon we can never send a signal to, just as a person in a black hole can see the exterior world and can never send a message out. 1 answered Apr 19 '11 at 13:05 Lawrence B. Crowell 8,0911717 silver badges2525 bronze badges Spacetime can dynamically evolve in a way which apparently violates special relativity. A good example is how galaxies move out with a velocity v = Hd, the Hubble rule, where v = c = Hr_h at the de Sitter horizon (approximately) and the red shift is z = 1. For z > 1 galaxies are frame dragged outwards at a speed greater than light. Similarly an observer entering a black hole passes through the horizon and proceeds inwards at v > c by the frame dragging by radial Killing vectors. The Alcubierre warp drive is a little spacetime gadget which compresses distances between points of space in a region ahead of the direction of motion and correspondingly expands the distance between points in a leeward region. If distances between points in a forwards region are compressed by a factor of 10 this serves as a “warp factor” which as I remember is $$w~=~1~+~ln(c)$$, so a compression of 10 is a warp factor 3.3. The effect of this compression is to reduce the effective distance traveled on a frame which is commoved with the so called warp bubble. This compression of space is given by $$g_{tt}$$ $$=~1~-~vf(r)$$. Of course as it turns out this requires exotic matter with $$T^{00}~<~0$$, which makes it problematic. Universe is also a sort of warp drive, but this is not due to a violation of the weak energy condition $$T^{00}~\ge~0$$. Inflationary pressure is due to positive energy. The gravity field is due to the quantum vacuum, and this defines an effective stress-energy tensor $$T^{ab}$$ with components $$T^{00}~=~const*\rho$$, for $$\rho$$ energy density, and $$T^{ij}~=~const*pu^iu^j$$, for $$i$$ and $$j$$ running over spatial coordinates $$u^i$$ velocity and $$p$$ pressure density. For the de Sitter spacetime the energy density and pressure satisfies a state $$p~=~w*\rho$$ where $$w~=~-1$$. So the pressure in effect is what is stretching out space and frame dragging galaxies with it. There is no need for a negative energy density or exotic matter. Negative energy density or negative mass fields have serious pathologies. Principally since they are due to quantum mechanics the negative eigen-energy states have no lower bound. This then means the vacuum for these fields is unstable and would descend to ever lower energy levels and produce a vast amount of quanta or radiation. I don’t believe this happens. The Alcubierre warp drive then has a serious departure between local laws of physics and global ones, which is not apparent in the universe or de Sitter spacetime. The Alcubierre warp drive is then important as a gadget, along with wormholes as related things, to understand how nature prevents closed timelike curves and related processes.