I Does the expansion of the Universe affect orbiting bodies?

[Buzz Bloom]

I don't understand; your MODEL 2 appears to be a matter-dominated universe with no dark energy. It is impossible for such a universe to stay at a constant scale factor "for a while". At most, if it is closed, it will be at a constant scale factor for an instant, at maximum expansion.

Hi Peter:

You are correct, and I am wrong, and I apologize for my mistake. I recall working on calculating the range of Omega values that would allow for

(1) da/dt=0​

and

(2) d²a/dt² = 0.​

If I remember correctly I did this about two years ago, and I had forgotten the need for dark energy. Aside from a diminishing of math skills, my memory is also not as good as when I was younger.

However, I believe this model is sufficient for my purposes with a value of a satisfying (1). I want to calculate what the difference is between the Model 1 and Model 2 orbits.

Regards,
Buzz

 

[PeterDonis]

I interpret the above as saying that you agree that in the McVittie Model 1 there is a distance Dlim such that
(Eq 1) Dlim3 approx = GM/H2at which the orbital velocity of the Earth object relative to the Sun object is zero.

Meaning, in the Schwarzschild-de Sitter metric, since that is what "the McVittie Model 1" is--zero matter, zero radiation, only a cosmological constant. Yes, the statement is true, although I would phrase it somewhat differently, as I did in some thread or other recently (we have several threads now on the same general topic), that there will be some distance from the central mass at which an object can "hover" at a constant altitude with zero proper acceleration, i.e., by simply floating freely in space.

if a particle is at a distance greater than the Dlim of (Eq 1) is is not bound to the central mass M, no matter what it's velocity is

No, this is not correct. The fact that the object can "float" at a constant altitude does not mean it can escape. You would still have to add energy to it for it to escape, where "escape" means "move away from the central mass indefinitely, without ever stopping or falling back".

But there is a more important issue that you are ignoring: in our actual universe, there are multiple isolated gravitating systems in a hierarchical structure, so the idea of "escape" has to be put in context. What does it mean to "escape" from the solar system? If I am halfway between the Sun and Alpha Centauri (supposing for a moment that their masses are equal), have I "escaped"? And even if I am well away from all stars in the immediate neighborhood, I'm still bound inside the Milky Way Galaxy. And even if I "escape" from the galaxy, I'm still bound inside the Local Group. And even if I "escape" from that, I'm still bound inside a larger galaxy cluster.

You are ignoring all of this hierarchical structure, and that's not really valid.

 

[PeterDonis]

for any separation distance
D < Dlim,the escape velocity V (Eq 2) is too large.

I gave a formula for the case where the effects of dark energy are included in some post or other that's been linked to. This discussion is too spread out between multiple threads for me to find it right now.

 

[PeterDonis]

I want to calculate what the difference is between the Model 1 and Model 2 orbits.

First you have to decide how you are going to find "the same" orbit in each model, in order to compare them. How will you do that?

 

[Buzz Bloom]

First you have to decide how you are going to find "the same" orbit in each model, in order to compare them. How will you do that?

Hi Peter:

My thought about this is that the Sun mass M and Earth mass m are the same in the two MODELs. If I calculate the circular orbit velocity for each of the two MODELs, assuming the same distance D between the Earth and Sun, I expect the two orbital velocities to be slightly but non-zero different.

However, I still need to learn how to calculate a circular orbit given the metric. Any suggestions of what I should read would be appreciated.

Regards,
Buzz

 

[PeterDonis]

My thought about this is that the Sun mass M and Earth mass m are the same in the two MODELs. If I calculate the circular orbit velocity for each of the two MODELs, assuming the same distance D between the Earth and Sun, I expect the two orbital velocities to be slightly but non-zero different.

Ok, so to be sure I understand, the models are:

MODEL 1: Sun mass M, nothing but dark energy everywhere else, Earth in circular orbit at proper distance D from the center of the Sun.

MODEL 2: Sun mass M, nothing but ordinary matter (i.e., uniform energy density, zero pressure) everywhere else, Earth in circular orbit at proper distance D from the center of the Sun.

And just for comparison:

MODEL 3: Sun mass M, vacuum everywhere else, Earth in circular orbit at proper distance D from the center of the Sun.

Is the above correct?

 

[PeterDonis]

Is the above correct?

Note, btw, that I carefully said "proper distance D" instead of "radial coordinate D", since they're not the same.

 

[PeterDonis]

MODEL 3: Sun mass M, vacuum everywhere else, Earth in circular orbit at proper distance D from the center of the Sun.

I should also note, btw, that as far as the Earth's orbit about the Sun is concerned, MODEL 3 is exactly equivalent in all its predictions to:

MODEL 3A: Sun mass M, vacuum out to some proper distance A that is larger than the proper distance D of Earth from Sun in its circular orbit, and any spherically symmetric distribution of stress-energy outside proper distance A. (And if we substitute "solar system", and distance A is large enough to be larger than the distance of any object in the solar system from the Sun, the same applies, and similarly for any bound system, up to and including galaxy clusters.)

 

[Buzz Bloom]

Ok, so to be sure I understand, the models are:
...
Is the above correct?
...
Note, btw, that I carefully said "proper distance D" instead of "radial coordinate D", since they're not the same.

Hi Peter:

Those three models, and perhaps a few more I haven't yet decided on yet, are what I currently have in mind to calculate. So I agree that the models you listed "are correct".

I did notice your use of "proper distance", and I agree that may be correct distance I should use. Thank you for mentioning that. However, I am not sure the proper distance from the center of the sun is able to be calculated. It may be necessary to use instead the proper distance from the Sun's Schwartzschild radius (usually denoted as r_s) to the Earth center, or maybe to the Earth's r_s. Or maybe using the coordinate r distance would be OK. After I learn about calculating a circular orbit's velocity, I expect I will be able to chose a reasonable definition for "distance".

Regards,
Buzz

 

[PeterDonis]

However, I am not sure the proper distance from the center of the sun is able to be calculated.

Sure it can. It just can't be expressed in a closed-form formula, unless you make the highly unrealistic assumption that the Sun's density is constant everywhere inside it. But reasonable approximations can be made, especially for the weak field case, which should be sufficient for this discussion.

It may be necessary to use instead the proper distance from the Sun's Schwartzschild radius

The Sun isn't a black hole, so this radius is irrelevant since it doesn't pick out any location of physical significance (same for the Earth).

 

[Buzz Bloom]

The Sun isn't a black hole, so this radius is irrelevant since it doesn't pick out any location of physical significance (same for the Earth).

Hi Peter:

Apparently you missed an assumption regarding the MODELs from my post #34.

Suppose I take the simplest case in which there is a single black hole mass, and a FLRW model as follows:

Regards,
Buzz

 

[PeterDonis]

Apparently you missed an assumption regarding the MODELs from my post #34.

Ah, ok. If the central mass is indeed a black hole, then yes, "proper distance to the center" is not well-defined so you'll need to come up with some other criterion for "same orbit". The Schwarzschild radial coordinate $r$ would be an obvious simple one to use (i.e., "same orbit" means "same $r$"), but $r$ does not directly represent proper distance in the radial direction (for example, the proper distance to the horizon is not the difference in radial coordinates, $r - 2M$).

 

[Buzz Bloom]

the proper distance to the horizon is not the difference in radial coordinates, r−2M

Hi Peter:

I think I understand what M is supposed to represent, but I am not sure. It seems logical that M represents r_s, with some unit constants omitted like G and c. (I feel that I am just not comfortable with equations with missing unit constants.) Is this correct?

I have been working on integrating ds in the Schwartzschild metric from r_s to r assuming that ds and dr are the only non-zero components. I want to compare s with r, for example w/r/t a circular approximation of Earth's orbit. I am making some progress, but I think I have made some math errors, so I am not quite ready to post my results.

Regards,
Buzz

 

[PeterDonis]

It seems logical that M represents r_s,

Actually half of $r_s$. In conventional units what I was calling $M$ would be $GM / c^2$, and the Schwarzschild radius corresponding to $M$ in conventional units is twice that.

I have been working on integrating ds in the Schwartzschild metric from rs to r assuming that ds and dr are the only non-zero components.

Yes, this will give you the proper distance from the horizon to an object "hovering" at $r$. You can then compare it with $r$ itself to see how much error there is in just treating $r$ as "radial distance". For the case of the Earth in orbit about the Sun (or a black hole in place of the Sun), the error will be tiny.

 

[Buzz Bloom]

Ah, ok. If the central mass is indeed a black hole, then yes, "proper distance to the center" is not well-defined so you'll need to come up with some other criterion for "same orbit". The Schwarzschild radial coordinate $r$ would be an obvious simple one to use (i.e., "same orbit" means "same $r$"), but $r$ does not directly represent proper distance in the radial direction (for example, the proper distance to the horizon is not the difference in radial coordinates, $r - 2M$).

Hi Peter:

I have been working on integrating a simplified Schwartzschild metric in which only ds and dr are not zero. This leads to

$ds = \frac {dr} {\sqrt {1-{r_s}/r}}$

When r = r_s, ds = ∞.

This ∞ remains in the value of the integral of ds integrating from r_s to some larger value of r. Therefore it appears to be impossible to integrate ds from r_s.

Is the correct? If so, there is no obvious choice but to use the coordinate r as the variable represening the distance from a gravitating body to a circular orbit. Do you agree? If not, how about using 3r_s/2 as the lower choice for the integration. Wikipedia says this is the orbit with orbital velocity c.

When calculating the proper distance between circular orbits in the same plane, it would be possible to integrate ds.

Regards,
Buzz

 

[PeterDonis]

it appears to be impossible to integrate ds from r_s.

No, it isn't. Look up the integrand in a table of integrals; you will see that it has a well-defined antiderivative even for $r = r_s$, so the antiderivative can just be evaluated at both endpoints and the difference taken to obtain the answer. This is a good example of why you shouldn't use your intuition in calculus, at least not if your intuition is untrained.

 

[nrqed]

It's not a matter of "what length scale". It's a matter of whether a particular object is part of a gravitationally bound system or not.

So if two masses, let's say both of 1 kg, are placed at rest relative to one another say 1 Mpc away from one another and there are no other gravitational forces acting on them. The expansion of the Universe will have no effect on their separation because they are gravitationally bound? Their subsequent motion will be the same as if they were in a static Universe? The proper distance between them will vary with time the same way in both cases?

And if say we give a tiny amount of kinetic energy to one of the masses so that they are no longer gravitationally bound, then in that case the expansion of the universe will affect their separation?

 

[PeterDonis]

if two masses, let's say both of 1 kg, are placed at rest relative to one another say 1 Mpc away from one another and there are no other gravitational forces acting on them.

"No other gravitational forces acting on them" is highly unlikely in the actual universe if they are 1 Mpc apart, because it requires that there are no other gravitating objects between them. (It also requires that the distribution of matter outside of them is spherically symmetric. And that there is no dark energy present.) But for an idealized thought experiment, ok.

The expansion of the Universe will have no effect on their separation because they are gravitationally bound?

No, the expansion of the Universe will have no effect on them because there are no other gravitational forces acting on them. You have eliminated all possible effects of any other matter with that specification. That includes the matter in the rest of the expanding universe.

Their subsequent motion will be the same as if they were in a static Universe?

A static universe containing any matter at all (or radiation or dark energy) is impossible, except for the edge case of the Einstein static universe, which is unstable against small perturbations (like a pencil balanced on its point). But if by "static universe" you mean "a universe entirely empty except for the two objects", then yes, your statement is correct.

if say we give a tiny amount of kinetic energy to one of the masses so that they are no longer gravitationally bound, then in that case the expansion of the universe will affect their separation?

Not as long as the "no other gravitational forces" condition holds.

Say, for example, that the two objects are 1 Mpc apart and are inside a "void" containing no other gravitating objects, and the void is 2 Mpc wide. Then if you give the objects just enough kinetic energy so they move apart at their relative escape velocity (so they're no longer gravitationally bound), their motion will be the same as if they were in an empty universe (as I defined that term above) until they are no longer inside the void and there is other matter exerting gravitational force on them. (There is a technicality here: if the other matter outside the void is expanding with the rest of the universe, the void will be wider than 2 Mpc by the time the objects reach its edge. So they will have to be somewhat further than 2 MPc apart to be no longer inside the void.)

 

[Buzz Bloom]

No, it isn't. Look up the integrand in a table of integrals; you will see that it has a well-defined antiderivative even for , so the antiderivative can just be evaluated at both endpoints and the difference taken to obtain the answer.

Hi Peter:

I have gone over my integration work again, and I found my mistake.

I did lookup integrals - in the 11th edition of CRC Mathematical Tables (1957) section on Integrals (pp 274-307). In particular I used the integral forms #60 and #57 on page 279. In what follows, I substitute r for x.
The group of equations is called:

"FORMS CONTAINING $\sqrt {a+br} = \sqrt {u}$ AND $\sqrt {a'+b'r} = \sqrt {v}$ WITH k=ab'-a'b".​

So with

a = -r_s​

b = 1​

a' = 0​

b'= 1​

k = 1​

the metric becomes

$ds = {\frac {\sqrt v} {\sqrt u} } dr$​

Form #60 is

$\int {\frac {\sqrt v} {\sqrt u} } dr = {\frac {1} {b}} \sqrt {uv} - {\frac k {2b}} {\int {\frac 1 { \sqrt {uv}}} }dr$.​

The integral on the RHS uses form #57, which is

${\int {\frac 1 { \sqrt {uv}}} }dr = \frac 2 {\sqrt {bb'}} {log ({\sqrt {bb'u} + b{\sqrt v})}}$​

Combining these two forms produces
$s = {\frac {1} {b}} \sqrt {uv} - {\frac k {2b}} {\int {\frac 1 { \sqrt {uv}}} }$
$s = {\frac {1} {b}} \sqrt {uv} - {\frac k {2b}} { \frac 2 {\sqrt {bb'}} {log ({\sqrt {bb'u} + b{\sqrt v})}}}$
$s = \sqrt {uv} - log(\sqrt u + \sqrt v)$
$s = \sqrt {r(r-r_s)} - log(\sqrt {(r-r_s)} + \sqrt r)$

Now, if the lower limit of the integral is r=r_s, then
$s(r) = \sqrt {r(r-r_s)} - log(\sqrt {(r-r_s)} + \sqrt r) + log(\sqrt {r_s})$

Thank you for telling me about my error.

Regards,
Buzz

 

[PeterDonis]

You have a sign error and a missing coefficient in the log terms. My go-to online reference for integrals is here:

http://integral-table.com/downloads/single-page-integral-table.pdf

Equation (25) is the relevant one, with $x = r$ and $a = - r_s$. The integral then becomes

$$
s(r) = \sqrt{r \left( r - r_s \right)} + r_s \left( \ln \left[ \sqrt{r} + \sqrt{r - r_s} \right] - \ln \left[ \sqrt{r_s} \right] \right)
$$

Which can be rewritten as

$$
s(r) = r \sqrt{1 - \frac{r_s}{r}} + r_s \left( \ln \left[ \sqrt{\frac{r}{r_s}} + \sqrt{\frac{r}{r_s} - 1} \right] \right)
$$

which makes it clearer what happens for $r_s << r$.

 

[Buzz Bloom]

You have a sign error and a missing coefficient in the log terms.

Hi Peter:

Thank you for the reference to the integral table. I found the #25 relevant integral. It seems much more useful than the forms in CRC.

Where I went astray is not immediately obvious to me, so I will go over my work again until I see clearly where I made my mistake.

Ah ha! I found it. My calcualtion of k was wrong. A quite careless error.

Regards,
Buzz

 

[pervect]

No, the expansion of the Universe will have no effect on them because there are no other gravitational forces acting on them. You have eliminated all possible effects of any other matter with that specification. That includes the matter in the rest of the expanding universe.

I agree with this, but not the following:

A static universe containing any matter at all (or radiation or dark energy) is impossible, except for the edge case of the Einstein static universe, which is unstable against small perturbations

My understanding is that the De-sitter universe (and anti-Desitter, and the De-Sitter Schwarzschild solution, henceforth DSS solution) are all static. The first two have only dark energy, the third has dark energy and matter. This can be seen from the fact that they have a timelike Killing vector, less formally but equivalently one can write the metric in a time-independent manner with the proper choice of coordinates. This choice of coordinates is not the same as the choice made by the FLRW metric, which is time-dependent. But the choice of FLRW metric doesn't remove the static and time independent nature of the universe, it simply obscures it by a different choice of coordinates that are not time-independent.

Basically, I suspect that the OP is struggling with the notion of energy in GR, implied by using the notion of "escape velocity", which can be defined in a static case by asking if an object has enough energy to escape. Unfortunately, this is not an easy topic. A static universe, such as the DSS universe, can be thought of as having a reasonably well defined energy because of the underlying time symmetry, the time-like killing vector, even though it's not asymptotically flat. This is because of Noether's theorem, which relates energy to time-translation symmetry, i.e. time independence.

I'm not sure if the OP is aware of Noether's theorem. I think it might be very useful for them to look at it more, but unfortunately it may be a bit off-topic to go into it in any great length. The formal treatment requires a bit of knowledge of Lagrangian mechnaics, which I'm not sure if the OP has.

Noether's theorem is a very important tool in understanding the notion of energy in GR, and one of the tools I prefer, but it's not the only tool. Asymptotically flat space-times also have a notion of "energy", these give rise to the importance ADM energy, for instance. However, the DSS case isn't asymptotically flat, so we can't apply the ADM notions of energy, though we can apply the notions of energy based on the time translation symmetry. This points out why energy in GR isn't a simple topic- there is sometimes more than one reasonable notion of what "energy" means, and their domain of applicability differs.

I believe that any static universe makes the concept of "escape velocity" reasonably well defined i because of the time indepenence. In the non-static case, one in general needs more information such as the direction of the velocity and at what specific time the velocity was measured it in order to know if the object will "escape" or not, assuming one has a well-defined criterion for "escape". I believe that the OP's criterion for escape is to consider a single massive body, and to ask whether the object escaping continues to increase its distance indefinitely.

In a static case, one can definitely say that an object without enough of the associated energy cannot escape. It's not clear to me if it is _always_ true that an object with enough energy will escape, though. Probably, but perhaps there is some loophole that I am not aware of, currently I can't think of a solid proof that an object with enough energy must escape, though I can't think of any examples in which an object with enough energy does not escape.

 

[Buzz Bloom]

Hi @PeterDonis :

I have done some calculations to determine the ratio of s/r when r is the approximately the radius of a circular Earth orbit. My result is

s/r = 1.00000017862045.​

I will now begin working with the McVittie metric to see what the influence of expansion is on orbital velocity.

Regards,
Buzz

 

[PeterDonis]

My understanding is that the De-sitter universe (and anti-Desitter, and the De-Sitter Schwarzschild solution, henceforth DSS solution) are all static.

They have static regions, but they are not static everywhere. (The Killing vector field that is timelike in the static regions exists everywhere, but is not timelike everywhere.) The Einstein static universe is static everywhere.

It's also worth noting that in dS, AdS, and dSS, the integral curves of the timelike Killing vector field in the static regions are not geodesics. In the Einstein static universe, they are.

The first two have only dark energy, the third has dark energy and matter.

No, the third has dark energy and a black hole. There is no stress-energy other than dark energy (if you consider that to be stress-energy).

A static universe, such as the DSS universe, can be thought of as having a reasonably well defined energy because of the underlying time symmetry, the time-like killing vector, even though it's not asymptotically flat.

For the case of a spacetime that is static everywhere (or more precisely stationary everywhere; "static" actually includes the additional condition that the timelike Killing vector field is hypersurface orthogonal), yes, the Komar energy gives an energy integral for the spacetime (although the integral may not always converge). However, for a spacetime that is only static in a limited region, not everywhere, the Komar energy integral can only be taken over the static region, not the entire spacetime.

I believe that any static universe makes the concept of "escape velocity" reasonably well defined i because of the time indepenence.

No. The concept of "escape" in the first place is only well-defined for an asymptotically flat spacetime, which does not have to be static or even stationary for "escape" to be meaningful. The concept of "escape velocity" implicitly assumes that this velocity is relative to stationary observers, so it requires a spacetime that is both asymptotically flat and stationary. Stationary alone is not enough.

In the case under discussion, the universe as a whole is actually not asymptotically flat, but we are implicitly modeling the "void" region containing the two particles as asymptotically flat, at least to a good enough approximation that "escape" from the two-particle system can be reasonably well defined. But that "escape" does not equate to "escape" from the universe as a whole (which is of course impossible).

 

[pervect]

They have static regions, but they are not static everywhere. (The Killing vector field that is timelike in the static regions exists everywhere, but is not timelike everywhere.) The Einstein static universe is static everywhere.

I see your point, looking over the metric more closely, but the region of of the DSS metric is static is just the region of interest for the escape velocity question.

https://en.wikipedia.org/w/index.php?title=De_Sitter–Schwarzschild_metric&oldid=959698099
The line element is:

$$-f(r) dt^2 + \frac{dr^2}{f(r)} + r^2 \left( d\theta^2 + sin^2 \theta d\phi^2 \right) \quad f(r) = 1 - 2\frac{a}{r} - frac{b}{r^2}$$There is no dependence of the metric on t, so $\partial / \partial t$ is always a Killing vector,i.e. translations of the coordinate t are always a symmetry. However, in spite of the fact that we are using the coordinate name "t" , $\partial / \partial t$ is only timelike in the region where f(r) is positive, i.e. the label "t" is always a number that is a generalized coordinate, but that coordinate number doesn't always represent time.

In the limit of low r, the DSS metric approaches the Schwarzschild metric, and the exterior region of the black hole is static, while the interior is not. But the exterior region is the one relevant for escape, nothing in the interior region below the event horizon escapes so it's not relevant to the escape issue.

In the limit of high r, we have the cosmological horizon of the De-Sitter space, where f(r) becomes zero and then negative at large r due to the 1 - br^2 term of f(r). But something at or beyond the cosmological horizon has already escaped by any reasonable defintion of "escape" that I can imagine. So the relevant region where it's sensible to talk about escape is just the region where f(r) is positive, and we have a timelike Killing vector.

For the case of a spacetime that is static everywhere (or more precisely stationary everywhere; "static" actually includes the additional condition that the timelike Killing vector field is hypersurface orthogonal), yes, the Komar energy gives an energy integral for the spacetime (although the integral may not always converge). However, for a spacetime that is only static in a limited region, not everywhere, the Komar energy integral can only be taken over the static region, not the entire spacetime.

I believe we have a quantity that is conserved everywhere, but it is interpretatble as an energy only when it represents a time-translation symmetry. In the regions where it's a space-translation symmetry, rather than a time-translation symmetry, our conserved quantity represents a conserved momentum rather than a conserved energy.

No. The concept of "escape" in the first place is only well-defined for an asymptotically flat spacetime, which does not have to be static or even stationary for "escape" to be meaningful. The concept of "escape velocity" implicitly assumes that this velocity is relative to stationary observers, so it requires a spacetime that is both asymptotically flat and stationary. Stationary alone is not enough.

I believe it's reasonable to say in the DSS case that "escape" is the ability of a particle to reach the cosmological horizon. This is a semantic issue. I agree it's important to define what we mean by escape, otherwise there may be confusion.

I can't really agree that we need asymptotic flatness to be able to define escape. I would point to the defintion I was using. The basic idea I am using to define "escape" is that if the distance from the central body of a particle in "free fall" aka "natural motion" aka "geodesic motion" increases without bound, the object escapes. We do need a shared notion of "distance" to define escape by this definition, but I don't think that's a real issue. And we don't need asymptotic flatness with this definition of "escape".

I do have concerns about the meaningfulness of "escape" in the actual context of our universe. If we look at an object on Earth, we see that the escape velocity from the Earth is aobut 11 km/sec. The escape velocity from the solar system starting at the positon of Earth is about 42 km/sec. Escape from the galaxy is about 550 km/sec. I would expect that escape from our local super-cluster of galaxies is even larger. As we consider larger and larger distances, the escape velocity keeps going up - I don't think there is any sensible limit. In the end, nothing escapes the universe, by the very definition of universe.

However, the OP's question seems to be about "escape" from a single body. I think we have a sensible defintion of escape in that case.

And I think that considering the static case of "escape" leads to the most insight. So I'd recommend understanding the static case, with a cosmological constant, first. And I think looking at the conserved energy due to the time translation symmetry is the approach that gives the most insight there.

 

[Buzz Bloom]

The line element is:

Hi pervect:

Thanks for your post. I may be mistaken, but your Telex equation for the metric seems to have a typo. The last term on the right has "frac" which I think you intended as "/frac". If that is correct, then the br² is probably "b {r^2}".

I do not have the time now to study the body of your post, but I will get to it as soon as I can.

Regards,
Buzz

 

[PeterDonis]

I believe it's reasonable to say in the DSS case that "escape" is the ability of a particle to reach the cosmological horizon.

I agree that would be a reasonable interpretation for the DSS case. However, DSS is ruled out for this discussion by the OP's requirement of "no gravitational forces" other than those exerted by each particle on the other. The effects of the cosmological constant and the black hole in DSS are "gravitational forces" and so are ruled out. The "void" region occupied by the particles has to have zero cosmological constant as well as zero matter, radiation, or black holes present to meet the OP's specifications.

 

[PeterDonis]

I can't really agree that we need asymptotic flatness to be able to define escape.

The usual definition of "escape" is "escape to infinity", which requires asymptotic flatness, otherwise there is no "infinity" to escape to.

"Escape to the cosmological horizon", as I said in my previous post, would be a reasonable interpretation of "escape" for that particular case, but only if one is willing to accept that "escape" doesn't take you to infinity.

 

[PeterDonis]

I believe we have a quantity that is conserved everywhere, but it is interpretatble as an energy only when it represents a time-translation symmetry. In the regions where it's a space-translation symmetry, rather than a time-translation symmetry, our conserved quantity represents a conserved momentum rather than a conserved energy.

In regions where the Killing vector field is spacelike, the quantity $p_a K^a$ has no obvious physical interpretation at all. The interpretation of $p_a K^a$ as a conserved energy when $K^a$ is timelike requires there to be observers whose worldlines are the integral curves of $K^a$; the quantity $p_a K^a$ is then the kinetic energy plus potential energy relative to those observers. But if $K^a$ is spacelike, there can't be any observers whose worldlines are its integral curves, so the only known basis for the physical interpretation of $p_a K^a$ is lost.

 

[Ibix]

The usual definition of "escape" is "escape to infinity", which requires asymptotic flatness, otherwise there is no "infinity" to escape to.

You could try to define it in terms of return time. Escape velocity would be the limiting case where a test mass thrown radially outwards doesn't return in finite time. You could even duck defining what you mean by "at rest" by considering two outward velocities $v$ and $v-\epsilon$ and finding the minimum $v$ where the two geodesics thus defined don't meet again even in the limit $\epsilon\rightarrow 0$. It might well be a function of time, and all you'll get is a four velocity. Interpretation of that in terms of three velocity does of course depend on a definition of "at rest", but the trajectories should be well defined. Or maybe not in a closed universe?

 

[PeterDonis]

You could try to define it in terms of return time. Escape velocity would be the limiting case where a test mass thrown radially outwards doesn't return in finite time.

Hm, yes, this would work for a spacetime which wasn't asymptotically flat but which was spatially infinite, such as de Sitter (or dSS) or any FRW spacetime that wasn't spatially closed.

You could even duck defining what you mean by "at rest" by considering two outward velocities $v$ and $v - \epsilon$ and finding the minimum where the two geodesics thus defined don't meet again even in the limit $\epsilon \rightarrow 0$.

I don't think this will work since the two should agree in the limit you give even if $v$ is the escape velocity. There is no different "falling back" geodesic that the limit could be for the escape velocity.

Or maybe not in a closed universe?

As above, in a spatially closed universe any radial geodesic will return in finite time (except in edge cases like a spatially closed universe with a cosmological constant that is spatially large enough that a radial geodesic will pass behind the cosmological event horizon before it can return).

 

[docnet]

Back in the old days when probably some older members got their PhDs, the scientific consensus was the expansion of space was slowing down, the expansion was explained by inertia. Observational evidence in the last several decades has shown countless times the expansion is not slowing down, but speeding up. I think most physicists now accept that empty space has energy, and that space is being created and keeping the energy density of vacuum constant. There are books by prominent astronomers and Nobel prize winning theoretical physicists such as Frank Wilzeck and David Gross related to this topic. These are scientists who are currently involved in research and academic professorships. I believe the question of "why" the cosmological constant is what it is is still a mystery to scientists though. Confession: I'm an avid reader of science and not an expert in physics.

 

[PeterDonis]

Observational evidence in the last several decades has shown countless times the expansion is not slowing down, but speeding up.

More precisely, it is speeding up now, and has been doing so for a few billion years (approximately); before that, it was slowing down.