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PeterDonis said:
Found it:
https://www.physicsforums.com/threa...distance-between-two-spherical-shells.989533/
Q re de Sitter–Schwarzschild metric
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Found it:
https://www.physicsforums.com/threa...distance-between-two-spherical-shells.989533/
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If I haven't made some error, it looks like the geodesic equations for the dss metric should boil down to
$$\dot{t} = \frac{E}{f(r)} \quad \dot{r} = \sqrt{E^2-f(r)}$$
It'd take more work to write ##\dot{t}## and ##\dot{r}## as a function of ##\tau##, however, so it's difficult to directly confirm that these are the correct solutions as-is. I imagine one could use the chain rule, but I haven't done this.
Here E is some constant, representing the energy, and f(r) is, as previously
$$f(r) = 1 - \frac{2a}{r} - br^2$$
It looks like the motion is best understood by the effective potential technique, as used by MTW and on the forurmilab website for the Schwarzschild case, and that the effective potential ##V^2(r)## is just f(r). When b=0, this matches the Schwarzschild effective potential ##1-2a/r## from the fourmilab site / MTW as it should.
The forumilab website is https://www.fourmilab.ch/gravitation/orbits/
a and b are somewhat inconvenient parameter, I wound up normalizing the event horizon to occur at at r=1, and the cosmologcal horizon at r=C, where C is some constant, and then solves for a and b.
Plotting the effective potential for C=10, I get something that looks like this.
If the object can reach the peak of the effective potential at ##r=\sqrt[3]{55} \approx 3.8##, with a coordinate velocity ##dr/d\tau## greater than 0, it will reach the cosmological horizon (and continue on to infinity). I've stop the graph at the cosmological horizon as the coordinates are too confusing beyond it.
$$\dot{t} = \frac{E}{f(r)} \quad \dot{r} = \sqrt{E^2-f(r)}$$
It'd take more work to write ##\dot{t}## and ##\dot{r}## as a function of ##\tau##, however, so it's difficult to directly confirm that these are the correct solutions as-is. I imagine one could use the chain rule, but I haven't done this.
Here E is some constant, representing the energy, and f(r) is, as previously
$$f(r) = 1 - \frac{2a}{r} - br^2$$
It looks like the motion is best understood by the effective potential technique, as used by MTW and on the forurmilab website for the Schwarzschild case, and that the effective potential ##V^2(r)## is just f(r). When b=0, this matches the Schwarzschild effective potential ##1-2a/r## from the fourmilab site / MTW as it should.
The forumilab website is https://www.fourmilab.ch/gravitation/orbits/
a and b are somewhat inconvenient parameter, I wound up normalizing the event horizon to occur at at r=1, and the cosmologcal horizon at r=C, where C is some constant, and then solves for a and b.
Plotting the effective potential for C=10, I get something that looks like this.
If the object can reach the peak of the effective potential at ##r=\sqrt[3]{55} \approx 3.8##, with a coordinate velocity ##dr/d\tau## greater than 0, it will reach the cosmological horizon (and continue on to infinity). I've stop the graph at the cosmological horizon as the coordinates are too confusing beyond it.
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pervect said:
Actually, there is no "infinity" in Schwarzschild-de Sitter spacetime. It is not asymptotically flat. Schwarzschild spacetime is, but Schwarzschild-de Sitter can be thought of, heuristically, as a finite piece of Schwarzschild, centered on the black hole, "glued" to de Sitter spacetime with a finite size spherical "world tube" taken out. So the asymptotic structure of Schwarzschild-de Sitter spacetime is the same as that of de Sitter spacetime. That means there is no "infinity" in the sense of "escape to infinity". The best you can do, relative to the black hole, is to "escape" beyond the cosmological horizon.
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PeterDonis said:
I'll amend my results to omit any statements about what happens after the cosmological horizon, then, since I haven't really analyzed that case.
I'd need some more information to fully appreciate your point, I think. Do you have any references you could share? Aditionally, would you agree that the horizon is a point of no return, that the test particle, once beyond the horizon, does not return to the central mass, nor can signals from the test particle return? You mention an analogy to a sphere. Are there any CTC's in the de-Sitter spacetime? I know there are CTC's in the anti-de-Sitter space time. But I didn't think there were any in the de-Sitter spacetime, though I haven't seen anything definitive.
Additionally, do you have any comment about the solution I found to the geodesic equations, and/or the effective potential diagram?
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pervect said:
Section 2 of the following paper has a good overview of the geometric and conformal properties of de Sitter spacetime:
https://arxiv.org/pdf/hep-th/0110007.pdf
Note in particular the Penrose diagram in Fig. 2. Unlike asymptotically flat spacetimes, which have a spacelike infinity to the side, and separate timelike and null infinites to the past and future (with the timelike being at the bottom and top, and the null being at an angle between the timelike and the spacelike), de Sitter spacetime has only two infinities, one to the past (at the bottom) and one to the future (at the top), and there is no distinction between timelike and null (both kinds of curves reach the same infinities).
The rest of that section has other good diagrams showing how various commonly used coordinate charts cover de Sitter spacetime, and what portions they cover.
The Penrose diagram of Schwarzschild-de Sitter spacetime is even more interesting, as shown in Fig. 1 of this paper:
https://arxiv.org/pdf/1007.3851.pdf
Note how there are still no spacelike or separate null infinities; instead there is an infinite "row" of alternating Schwarzschild and de Sitter regions.
One thing I have not been able to find is a diagram for gravitational collapse of an object to a black hole in a universe that is asymptotically de Sitter instead of asymptotically flat--i.e., a version of the classic Oppenheimer-Snyder model set in de Sitter spacetime.
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It's messy, but my computer algebra package is confirming the solution.
The highlights:
$$g(r) = \frac{df(r)}{dr} \quad \dot{t}(\tau)= \frac{E}{f(r(\tau))} \quad \dot{r}(\tau) = \sqrt{E^2 - f(r(\tau))} $$
$$ \ddot{t}(\tau) = -\frac{E\,g(r)}{f^2(r)} \dot{r}= \frac{-E\, g(r(\tau)) \sqrt{ E^2 - f(r(\tau))} } {f^2(r(\tau))} $$
$$ \ddot{r}(\tau) = -\frac{1}{2} \, \frac{g(r)} { \sqrt{E^2-f(r)}} \dot{r} = -\frac{g(r(\tau)) } {2 }$$
The rest is more algebra, substuting ##\ddot{t}## ##\dot{t}## ##\ddot{r}## and ##\dot{r}## into the geodesic equations.
Note that when g(r)=0, ##\ddot{r}=0##, which is consistent with the idea that f(r) is the effective potential, and the metastable peak of the effective potential occurs when g(r) = df/dr = 0.
From the expression for ##\dot{r}## we could write
$$\frac{dr}{\sqrt{E^2 - f(r)}} = d\tau$$
which could be integrated to solve for ##\tau(r)## which could be inverted to get ##r(\tau)##, but this gets very messy very quickly.
The highlights:
$$g(r) = \frac{df(r)}{dr} \quad \dot{t}(\tau)= \frac{E}{f(r(\tau))} \quad \dot{r}(\tau) = \sqrt{E^2 - f(r(\tau))} $$
$$ \ddot{t}(\tau) = -\frac{E\,g(r)}{f^2(r)} \dot{r}= \frac{-E\, g(r(\tau)) \sqrt{ E^2 - f(r(\tau))} } {f^2(r(\tau))} $$
$$ \ddot{r}(\tau) = -\frac{1}{2} \, \frac{g(r)} { \sqrt{E^2-f(r)}} \dot{r} = -\frac{g(r(\tau)) } {2 }$$
The rest is more algebra, substuting ##\ddot{t}## ##\dot{t}## ##\ddot{r}## and ##\dot{r}## into the geodesic equations.
Note that when g(r)=0, ##\ddot{r}=0##, which is consistent with the idea that f(r) is the effective potential, and the metastable peak of the effective potential occurs when g(r) = df/dr = 0.
From the expression for ##\dot{r}## we could write
$$\frac{dr}{\sqrt{E^2 - f(r)}} = d\tau$$
which could be integrated to solve for ##\tau(r)## which could be inverted to get ##r(\tau)##, but this gets very messy very quickly.