Does the expansion of the Universe affect orbiting bodies?

[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.