I The proper Schwarzschild radial distance between two spherical shells

Buzz Bloom
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I am trying to understand the use of the Schwarzschild metric in measuring the "proper" distance between two non-moving concentric spherical shells which are also both concentric with the event horizon sphere which of course has the Schwarzschild radius.
If what I describe below is correct, I hope someone will confirm this for me. If it is is incorrect, I hope someone with explain my error to me.
For the purpose of this thread the metric is
ds2 = - (1-rs/r) c2 dt2 + dr2 / (1-rs/r)​
where
rs = 2GM/c2.​
(I modified the above from

I assume that the two spherical shells are stationary. Therefore
dt = 0.​
The r coordinate for the radii of the two shells satisfy the relationships:
A1 = 4 π (r1)2
is the surface area of one spherical shell, and
A2 = 4 π (r2)2
is the area of the other spherical shell.

The proper radial distance D between the r1 shell and the r2 shell is:
D = ∫r1r2 (1/(1-r/rs))1/2 dr.​

My purpose in studying this problem is that I am interested in estimating the error if I use Newtonian gravity equations for orbital motion rather than the more accurate (but more difficult to use) Schwarzschild math.
 
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Buzz Bloom said:
I am interested in estimating the error if I use Newtonian gravity equations for orbital motion...
Do remember that the trajectory of an orbiting body is not a curve of constant ##t##
 
Nugatory said:
Do remember that the trajectory of an orbiting body is not a curve of constant ##t##
Hi Nugatory:

Thank you for your post.

I do understand and that time plays a role in orbital behavior. I plan to limit my study (at least for a while) to radial motion only to keep the math simpler. I am assuming that that the particular approximation of error using Newtonian gravity math rather than Schwartschild math will be similar for time as for distance.

Regards,
Buzz
 
Buzz Bloom said:
The proper radial distance D between the r1 shell and the r2 shell is:
D = ∫r1r2 (1/(1-r/rs))1/2 dr.​

My purpose in studying this problem is that I am interested in estimating the error if I use Newtonian gravity equations for orbital motion rather than the more accurate (but more difficult to use) Schwarzschild math.

I'm not sure I understand your motivations, but I don't see anything wrong with what you actually wrote. It would be cleaner if expressed in Latex, looking like this:

$$D = \int_{r_1}^{r_2} \frac{dr}{\sqrt{1-\frac{r}{r_s}}}$$

and is generated by

Code:
$$D = \int_{r_1}^{r_2} \frac{dr}{\sqrt{1-\frac{r}{r_s}}}$$

https://www.physicsforums.com/help/latexhelp/ has more help on Latex on PF.
 
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In Euclidean geometry, given two concentric spheres given areas, there is an implied radial distance between them. In Schwarzschild geometry, the radial distance given two concentric sheres of the same area is different from the Euclidean expectation. What I don't see is what this has to do with calculating orbits, whether you use Newtonian gravity or GR.
 
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Buzz Bloom said:
I plan to limit my study (at least for a while) to radial motion only to keep the math simpler.
If you are limiting yourself to purely radial motion you can actually solve the differential equations for orbital motion in Schwarzschild coordinates and write ##\tau(r)## and ##t(r)## (unfortunately I don't think you can invert them to get ##r(\tau)## or ##t(\tau)##). These coordinate-based expressions aren't directly useful, but you can derive some direct observables (e.g. time of flight for a thrown ball) and make comparisons between Newtonian and relativistic predictions.

I think that's the right way (or a right way, anyway) to go about this.
 
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Hi @pervect, @PAllen, and @Ibix:

Thank you for your posts.

Regarding the use of Latex, I agree that it is a better form of presentation than what I used, but I have had difficulties using Latex. When I try to preview a post with Latex before posting, I experience display issues that make correcting the Latex very difficult.

Regarding the comparing of coordinate radial distance with proper radial distance, I want to be as certain as I can be that with respect to the distances I want to explore the difference is sufficiently small that it can be reasonably ignored. The problem I want to explore involves calculating whether a test object escapes from a gravitational source or is bound to it, involving a combination of gravitational attraction by a Euclidean point, and the outward acceleration from the expansion of the universe. I feel this topic has not been adequately discussed in several threads, for example.
and

I think that I am now ready to start a new thread specifically to explore (limited to radial motion) of what seems to me (possibly incorrectly) to be a correct way to determining whether a test object moving at a specified radial velocity will escape from a single given mass at a given distance. @PeterDonis has made an good argument that (for practical purposes) the Euclidean escape velocity is a sufficient test, even though it ignores the influence of the expanding universe. I want to explore in detail comparing this practical test with a test including the expanding universe. I have not been able to find on the Internet any source which explores this topic.

Regards,
Buzz
 
First, it is only the acceleration of expansion that will have any effect at all on escape velocity (the first derivative of scale factor has exactly zero tidal effect). Second, for any reasonable universe parameters, this effect will be much smaller than the difference between GR and Newtonian escape velocity. Thus, to analyze this effect, you would need to merge a Schwarzschild solution and an FLRW solution, which is a nontrivial undertaking.
 
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Buzz Bloom said:
@PeterDonis has made an good argument that (for practical purposes) the Euclidean escape velocity is a sufficient test, even though it ignores the influence of the expanding universe.

Where did I make this argument?
 
  • #11
PeterDonis said:
Where did I make this argument?
Hi Peter:

I will make a search to find some specific posts. I have forgotten the exact threads where they appear.

Regards,
Buzz
 
  • #12
PAllen said:
First, it is only the acceleration of expansion that will have any effect at all on escape velocity (the first derivative of scale factor has exactly zero tidal effect). Second, for any reasonable universe parameters, this effect will be much smaller than the difference between GR and Newtonian escape velocity. Thus, to analyze this effect, you would need to merge a Schwarzschild solution and an FLRW solution, which is a nontrivial undertaking.
Hi Paul:

I may well have a misunderstanding regarding your post, and I will reply as soon as I can. I am now off to a meeting.

Regards,
Buzz
 
  • #13
Hi @PeterDonis:

I found one post which I quote from below. I still think there were related other posts, but I failed to find any others that were on topic.

Post #57​
... the correct criterion for whether a pair of objects are gravitationally bound is whether their relative recession velocity is greater than escape velocity.​

I understand that the definition you are using for "escape velocity" is the standard Euclidean one:
Ve2 = √(2M/D).​
This ignores the influence of the Hubble acceleration AH:
AH approx = H2D.​
This approximation ignores the changes in H during the time it takes as the test particle moves.I am still working on how to include this change in H, or to determine that the change in H is negligible regarding the issue of boundedness.

Regards,
Buzz
 
  • #14
PAllen said:
Poking around a bit, it seems that the key thing to search for is the McVittie metric, which is an exact solution for a mass point embedded in an FLRW cosmology. Here are two example references:

https://arxiv.org/abs/1508.04763
https://iopscience.iop.org/article/10.1088/1742-6596/484/1/012044/pdf
Hi Paul:

I looked at the first reference and found some strange discussions. First, the date and title:
(2002) Search for a Standard Explanation of the Pioneer Anomaly.​
I did a search on "Pioneer Anomaly" and found (2012)
Here is a relevant quote from this article:
Why was the thermal emission from the spacecraft anisotropic and slowing the spacecraft down? First of all, because the Pioneer spacecraft were spin-stabilized and almost always pointed their big dishes towards Earth. Second of all, because two sources of thermal radiation (heat) were then on the leading side of the spacecraft .​

I confess this seems like it is not relevant to the problem I am exploring, but I will give it and the other article a deeper look over the next few days.

Regards,.
Buzz
 
  • #15
Buzz Bloom said:
I looked at the first reference and found some strange discussions. First, the date and title:
(2002) Search for a Standard Explanation of the Pioneer Anomaly.
This article would appear to be https://arxiv.org/abs/gr-qc/0107022, while the article @PAllen linked is https://arxiv.org/abs/1508.04763, titled "Lensing in the McVittie metric". Perhaps you confused two different browser tabs?

I have to admit I'm not clear what it is you want to do. Are you just trying to determine the escape velocity from some object in an expanding universe? Then the McVittie metric would appear to be what you want to study, from PAllen's description. In particular you'd want to look at radially outgoing timelike geodesics. I suspect that there are quite a few potentially thorny questions on the way to this, though.

Aside: I could murder a chocolate digestive...
 
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  • #16
Buzz Bloom said:
Hi Paul:

I looked at the first reference and found some strange discussions. First, the date and title:
(2002) Search for a Standard Explanation of the Pioneer Anomaly.​
I did a search on "Pioneer Anomaly" and found (2012)
Here is a relevant quote from this article:
Why was the thermal emission from the spacecraft anisotropic and slowing the spacecraft down? First of all, because the Pioneer spacecraft were spin-stabilized and almost always pointed their big dishes towards Earth. Second of all, because two sources of thermal radiation (heat) were then on the leading side of the spacecraft .​

I confess this seems like it is not relevant to the problem I am exploring, but I will give it and the other article a deeper look over the next few days.

Regards,.
Buzz
I don’t understand. My link is to: ”Lensing in the McVittie metric”, from 2015. There is a brief mention of the pioneer anomaly, but that is tangential to the paper’s topic. The paper discusses the background of using the McVittie metric to analyze small scale effects of expansion. Its specific focus on lensing is not relevant, but the purpose of referencing it was to answer my own point that you need a way to combine Scwarzschild metric with FLRW metric to answer the questions you pose. After a little research, I found the McVittie metric is exactly such a proposal. This paper presents the McVittie metric with background and many references, thus perfectly on point from this point of view.
 
  • #17
Buzz Bloom said:
I understand that the definition you are using for "escape velocity" is the standard Euclidean one

"Euclidean" is not the right word for the first formula I gave in that post for ##V_E##. That formula is derived from Schwarzschild spacetime; it is the formula for the escape velocity at radial coordinate ##D## in Schwarzschild spacetime relative to a mass ##M##. For large radial coordinates, the radial coordinate is a good approximation to actual radial distance, so ##D## can be considered the distance to the object.

Buzz Bloom said:
This ignores the influence of the Hubble acceleration

I'm not sure what you mean by "Hubble acceleration", but if you mean the effect of dark energy, go back and read the post you reference again, since the second formula I gave for ##V_E## in that post explicitly takes the effects of dark energy into account. As I noted in the post, this decreases the effective distance ##D## from a given mass at which a comoving object will be moving away from the mass at escape velocity, so if you're just looking for an upper bound to that distance, the first formula is fine.

Buzz Bloom said:
This approximation ignores the changes in H during the time it takes as the test particle moves

##H## is decreasing with time, so if you want an upper bound, just use the value of ##H## at the end of whatever time period you're interested in, since that will be the lowest value (which will therefore give you the largest value of ##D##).
 
  • #18
PAllen said:
a way to combine Scwarzschild metric with FLRW metric

Note that the McVittie metric uses isotropic coordinates, so the form of the Schwarzschild metric it is implicitly using is different from the form I used in the post in another thread that @Buzz Bloom linked to and that I was discussing in my previous post. For large distances from a given mass, the difference in radial coordinates between isotropic and Schwarzschild coordinates is negligible, and both are well approximated by the actual proper distance ##D##, so this doesn't significantly affect what I posted, but it's worth being aware of.
 
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  • #19
PeterDonis said:
the McVittie metric uses isotropic coordinates

Btw, another implication of this is that describing the McVittie metric as a "mass point" embedded in an FRW universe is not quite correct. The metric does not cover the spacetime region at or below the event horizon of the mass at all (it can't since isotropic coordinates do not cover that region). So it is better described as a metric for the exterior of a black hole embedded in an FRW universe.
 
  • #20
PAllen said:
First, it is only the acceleration of expansion that will have any effect at all on escape velocity
Hi Paul:

I do not understand why the above might be true.

Imagine two masses, each of mass M, and each occupying a non-moving point relative to co-moving coordinates. Assume at time t0, the distance between the centers of these two masses is D. The Hubble expansion tells us that at time t0 they are moving apart at velocity HD:
V(t0) = H(t0)D(t0).​
There is also a Hubble acceleration happening. I do not mean the acceleration of the expansion. I mean
A(t) = dV(t)/dt = H dD/dt + D dH(t)/dt.​
If we assume
ΩR = ΩM = Ωk = 0​
and
ΩΛ=1,​
then
dH(t)/dt = 0,​
H(t) = (1/a) d(a(t))/dt,​
and​
a(t) = eHt.​
With these assumptions,
A(t) = HV = H2D.​
I call this the Hubble acceleration, but if that is an ambiguous name, I will have to chose some other name.

Now assume (Newtonian view) the inward gravitational acceleration on a test particle at a distance D from the gravitating point mass M equals the Hubble acceleration.
GM/D2 = H2D​
which leads to
D = (GM/H2)(1/3).​
The implication is that the test particle with no velocity will be stationary with no acceleration. This is, this particle will be in an "orbit" with zero orbital velocity.

It seems natural to assume that the Hubble acceleration applies to a test particle at a distance D from a massless origin point P, when (1) the test particle is moving away from P at velocity HD, and (2) it is accelerating away from P with accleration H2D. Apparently, there is some doubt that (assuming the Newtonian view) if there is a mass M at the origin P, the test particle would not be influenced by H2D, and it would not have a total acceleration of
A = -GM/D2 + H2D.​

Do you know any reference that explains why the Hubble acceleration affects the motion of a test particle at a distance where the gravitational attraction becomes negligible, but this does not apply when the gravitational attraction has a similar magnitude. I have been unable to find any such explanation anywhere on the Internet.

Regards,
Buzz
 
  • #21
Buzz Bloom said:
I do not understand why the above might be true.

It's because expansion (as distinct from accelerated expansion) is just inertia; faraway comoving objects are moving away from us because of inertia. That fact cannot affect the escape velocity for objects from, say, the Milky Way.

A slightly more technical version of the above is that, because of the Shell Theorem, a spherically symmetric distribution of matter surrounding some region cannot affect the motion of anything inside the region. To a very good approximation, the distribution of matter in the rest of the universe with respect to a single gravitationally bound system like the Milky Way is spherically symmetric, so it can be ignored when analyzing the motion of objects within the system. The same applies to analyzing what the escape velocity from the system would be.

Also, please learn how to use the PF LaTeX feature. Most of the rest of your post is basically unreadable because of the equation soup and the lack of proper formatting.
 
  • #22
Buzz Bloom said:
There is also a Hubble acceleration happening. I do not mean the acceleration of the expansion.

Yes, you do. In the formula you derive, you explicitly ignore everything except ##\Omega_{\Lambda}##, which is dark energy, i.e., "acceleration of the expansion".

I already explicitly acknowledged, in the previous thread you linked to, that the presence of dark energy does affect escape velocity, and gave you the correct formula for escape velocity in the presence of dark energy and how that formula is derived.
 
  • #23
@Buzz Bloom , the proof of my statement is simple. Two different arguments are possible. First, only curvature, as an invariant object, can affect physics, and this depends on second derivatives of the metric components, which means only second derivatives of scale factor can contribute to tidal gravity. A more cute argument is to ask what happens if the derivative of the scale factor is constant. The answer is that the curvature vanishes identically, which means you have flat Minkowski space in funny coordinates (Milne coordinates). Choice of funny coordinates cannot possibly affect orbits or escape velocity, this linear expansion rate must have exactly zero local effect.

In fact, it is well known that the following possibly counterintuitive statements are true, that follow from the fact that only the second derivative of scale factor matter for local physics:

- if you have a forever expanding universe, such that the rate of expansion is forever decreasing, then local orbits will be smaller than expected, and escape velocity larger.

-
if you have a forever contracting universe, where the rate of contraction is forever decreasing, then local orbits will be larger than expected, and escape velocity smaller than expected.

Of course, our universe is presumed to have accelerated expansion, so escape velocity will be smaller than expected, and orbits larger than expected.
 
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  • #24
PeterDonis said:
lack of proper formatting
Hi Peter:

I will work on getting familiar with Latex, but I fear that my frustration limit is likely to prevent me from using it regularly.

Aside from no Latex, can you give me some examples of what you mean by my "lack of proper formatting"?

Regards,
Buzz
 
  • #25
PAllen said:
In fact, it is well known that the following possibly counterintuitive statements are true, that follow from the fact that only the second derivative of scale factor matter for local physics:
Hi Paul:

I think it would be very helpful to me to see a lengthy explanation of
"only the second derivative of scale factor matter for local physics". Can you recommend a paper or book that explains the physical and/or mathematical reason for this limitation.

Regards,
Buzz
 
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  • #26
Buzz Bloom said:
Now assume (Newtonian view)

You can't do this the way you are doing it.

The general observation you are struggling towards here is correctly stated as follows: in the Schwarzschild-de Sitter metric (which, if you go back and look at my post #57 from the previous thread, you will see that I referenced), there is some finite radial coordinate ##r## (using Schwarzschild coordinates centered on the gravitating massive body) at which there are geodesic timelike worldlines that maintain constant ##r##. That is because the proper acceleration of a worldline of constant ##r## (and constant angular coordinates, so zero "orbital velocity") in Schwarzschild-de Sitter spacetime is:

$$
a = \left( \frac{M}{r^2} - A r \right) \frac{1}{\sqrt{1 - \frac{2M}{r} - A r^2}}
$$

where ##M## is the mass of the gravitating body and ##A## is a constant related to the dark energy density aka cosmological constant. Obviously ##a = 0## if ##r = \left( M / A \right)^{1/3}##.

Note, however, that this bears no relationship to your formula; this ##r## is independent of the Hubble constant and does not change with time.
 
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  • #27
Buzz Bloom said:
i fear that my frustration limit is likely to prevent me from using it regularly

Then I fear you are likely to find that other people's frustration limit is going to prevent them from reading or responding to your posts.

Buzz Bloom said:
Aside from no Latex, Can you give me some examples of what you mean by my "lack of proper formatting"?

By "lack of proper formatting" I meant "no LaTeX". The whole reason we have the LaTeX feature is that equations are so much more difficult to read without it. Which in turn means people are less likely to read or give useful responses to posts that don't use it.
 
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  • #28
Buzz Bloom said:
Hi Peter:

I think it would be very helpful to me to see a lengthy explanation of
"only the second derivative of scale factor matter for local physics". Can you recommend a paper or book that explains the physical and/or mathematical reason for this limitation.

Regards,
Buzz
I don’t really understand how to simplify it any more than I have. The scale factor as function of time is part of several metric components. It is part of GR foundations that only curvature distinguishes local physics from SR. Curvature is only affected by second derivatives of metric components, because the Levi civita connection can be made to vanish at any point by coordinate transform. But its derivatives cannot be made to vanish, and these are determined by metric second derivatives. Cosmology is an application of GR, so it includes all features of GR, and cannot violate any generic features of GR.

Have you read any systematic treatment of GR? Sean Carroll’s online lecture notes develop everything from pretty basic starting point, yet develop the material using a pretty complete modern mathematical form.
 
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  • #29
PeterDonis said:
Obviously ##a = 0## if ##r = \left( M / A \right)^{1/3}##.

Note, btw, that this is not the same as (though similar to) the formula I gave in post #57 of the previous thread. That formula, in the notation of this thread, is

$$
D = \left( \frac{2 M}{H_0^2 + A} \right)^{\frac{1}{3}}
$$

So that formula has a factor of ##2^{1/3}## extra in the numerator, and the denominator has the ##H_0^2## term as well as the ##A## term. The reason the ##H_0^2## term is there in that formula is that it is comparing the comoving recession velocity with the Schwarzschild-de Sitter escape velocity, and the former does depend on the Hubble constant, while the latter does not.
 
  • #30
  • #31
Buzz Bloom said:
Hi Paul:

I found
https://www.physicsforums.com/threa...-iii-ch-8-resolution-of-vector-states.989615/.

Is this the lecture notes you are recommending?

Regards,
Buzz
No, what does Feynman have to do with Sean Carroll? The latter taught the standard GR course at MIT for a number of years and posted his detailed lecture notes on line. Many link to the arxiv version. However, I prefer the following link to the author's own website, which has all the content of the arxiv reference plus a bunch of useful ancillary material, including a 24 page synopsis, which is still fully mathematical:
https://www.preposterousuniverse.com/grnotes/
 
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  • #32
PAllen said:
No, what does Feynman have to do with Sean Carroll? ... plus a bunch of useful ancillary material, including a 24 synopsis, which is still fully mathematical:
https://www.preposterousuniverse.com/grnotes/
Hi Paul:

I apologize for my carelessness. I did an Internet search of "Sean Carrol lecture notes" and found the right URL at the top of the list, but then I carelessly copied the URL for the next item on the list to include in my post.

When I look at the lecture notes page, I do not see anything about "a 24 synopsis". Can you explain a bit more about what that is and in what section I can find it?

Regards,
Buzz
 
  • #33
Buzz Bloom said:
Hi Paul:

I apologize for my carelessness. I did an Internet search of "Sean Carrol lecture notes" and found the right URL at the top of the list, but then I carelessly copied the URL for the next item on the list to include in my post.

When I look at the lecture notes page, I do not see anything about "a 24 synopsis". Can you explain a bit more about what that is and in what section I can find it?

Regards,
Buzz
It is in the fourth paragraph of the link I gave. A direct link is:
https://preposterousuniverse.com/wp-content/uploads/2015/08/grtinypdf.pdf
 
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  • #34
Buzz Bloom said:
If we assume
ΩR = ΩM = Ωk = 0​
and
ΩΛ=1,​
then

I assume this means you're interested in a case where there is no matter, no radiation, but only a cosmological constant.

This would be represented by the de-Sitter metric. With the addition of matter, so that you have a massive body in a space-time with a cosmological constant and an accelerating expansion, you'd have the De-Sitter Schwarzschild metric, https://en.wikipedia.org/wiki/De_Sitter–Schwarzschild_metric, henceforth the DSS metric.

This would be a simpler alternative (or more precisely, one I happen to be familiar with) to using the McVitte metric, which is interesting, but I'm not familiar with it.

Using this DSS metric, we can get the answer without any questionable (and probably wrong) assumptions that GR is like Newton's theory.

We can find the r coordinate in this metric where an object stays stationary so that it's r coordinate does not change with time.

Specifically, looking at the Wiki page https://en.wikipedia.org/w/index.php?title=De_Sitter–Schwarzschild_metric&oldid=959698099#Metric

we have the geodesic equation, and we want to solve for r=constant, which implies that ##\dot{r} = \ddot{r} = 0## and also ##\dot{\theta} = \dot{\phi} = 0##.

The geodesic equation has a solution of this form when ##f'(r)=0##, which implies
$$r = \left( \frac{a}{b} \right)^{\frac{1}{3}}$$

From the wiki, we have

The two parameters a and b give the black hole mass and the cosmological constant respectively.

So the a is proportional to the black hole mass, and b is proportional to the cosmological constant.

Note that the coordinates used for the DSS metric are not necessarily the same as the cosmological coordinates. There is surely some relationship between the two sets of coordinates, but I havean't seen any detailed exposition of what it is. However, we can make useful qualitative statements without the exact knowledge.

The qualitative statement we can make is that there is some critical radius r. At this critical radius, particles in geodesic motion (free fall) exist which do not get further away or closer to the black hole. Below the critical radius, particles in geodesic motion fall inwards, outside the critical radius they accelerate outwards.

There is an intuitive way of looking at this, involving the quantity ##\rho + 3P##. In the de-Sitter space-time, ##\rho## is positive and P = ##-\rho##. So the quantity ##\rho + 3P## is negative. It's a sort of mass density, called the Komar mass, that applies to any stationary space-times. The De-Sitter and the Schwarzschild De-sitter metric are both examples of stationary space-times, as can be seen by the fact that there is no time dependence of the metric coefficients. The more general FRW space-time is not stationary, unfortunately.

My intuitive way of looking at this is that the balance point where r=constant occurs when the positive mass of the black hole is balanced out by the effect of empty space, which has an effective negative mass due to the pressure term in the cosmological constant.

So the gravitational effect of the negative mass cancels out the positive mass at some specific radius, and you have a net zero effect at this particular radius. Inside this critical radius, particles will fall towards the black hole. Outside this critical radius, particles will accelerate away from the black hole.

Note that for this argument to work, we need to rely on the fact that the gravitational effect inside a hollow sphere is zero, which justifies why we only look at the contributions from the "mass" inside the sphere of radius r.

Like all intuitions, this somewhat vague description is a bit suspect, and I can't quote the exact source that gave me this idea, though it was a paper studying the DSS space-time. Nor can I say that my memory is necessarily good on this point.

While the provenance of this intuitive explanation is suspect, I thought it was useful enough an idea to share, even though it is much less rigorous than the argument I gave using the solution of the geodesic equation.
 
  • #35
pervect said:
This would be a simpler alternative (or more precisely, one I happen to be familiar with) to using the McVitte metric

The McVittie metric is a generalization of the Schwarzschild-de Sitter metric that allows the perfect fluid that is filling the universe (apart from the "mass" in the center) to be something other than pure dark energy. Basically, you can plug in any equation of state that is compatible with an FRW spacetime.

As far as the approach you are taking to the OP question (which is basically the same as what I briefly described in post #26), it will work just fine in a general McVittie metric; you just need to write the metric in Schwarzschild coordinates (instead of the isotropic coordinates it is more often written in).

pervect said:
Note that the coordinates used for the DSS metric are not necessarily the same as the cosmological coordinates. There is surely some relationship between the two sets of coordinates, but I havean't seen any detailed exposition of what it is.

The coordinates used on the DSS Wikipedia page you linked to are Schwarzschild coordinates; these correspond to static coordinates for the de Sitter metric, as described here:

https://en.wikipedia.org/wiki/De_Sitter_space#Static_coordinates

"Cosmological coordinates" are any of the flat, open, or closed slicings described in that article.
 
  • #36
PAllen said:
The scale factor as function of time is part of several metric components. It is part of GR foundations that only curvature distinguishes local physics from SR.
...
Have you read any systematic treatment of GR? Sean Carroll’s online lecture notes develop everything from pretty basic starting point, yet develop the material using a pretty complete modern mathematical form.
Hi Paul:

I have started to read Sean Carroll’s online lecture notes, but I have found it difficult to make progeress in my understanding. i have not yet given up though, but I have given more attention to the McVittie metric

I think I understand that "The scale factor as function of time is part of several metric components," but I have not yet not grasps what are the "several metric components", in particular in the McVittie metric which I found on page 9 of the lecture notes
Black Holes in an expanding universe: the McVittie metric, by Daniel C. Guariento.
www.physics.ntua.gr/cosmo13/Paros2013/Talks/Guariento.pdf
McVitteMetric.png

Are the following two the components you intend as having the sclae factor as part of them?
a2
m/(2a rcap) (rcap represents here the r with the carat on top in the metric)​

Would you please explain the meaning of "local physics" as it is related to GR. I have seen this term used w/r/t quantum mechanics, but not GR.

Regards,
Buzz
 
  • #37
pervect said:
I assume this means you're interested in a case where there is no matter, no radiation, but only a cosmological constant.
...
This would be represented by the de-Sitter metric.
Hi pervect:

Thank you for your post.

One of several cases I am interested in is the one where there is no matter, no radiation, but only a cosmological constant. I have accepted the suggestion of @PeterDonis to work with the McVittie metric.

The thread
posts #43 through #49 is a further discussion of this topic.

Regards,
Buzz
 
  • #38
Buzz Bloom said:
I have accepted the suggestion of @PeterDonis to work with the McVittie metric.

Which, as I have already pointed out in other threads, is just the Schwarzschild-de Sitter metric for the case you describe, with zero matter, zero radiation, and only a cosmological constant.
 
  • #39
Some general comments that might help, from Ned Wright's cosmology FAQ, and some of my own comments:

http://www.astro.ucla.edu/~wright/cosmology_faq.html#SS

Ned Wright said:
Why doesn't the Solar System expand if the whole Universe is expanding?

This question is best answered in the coordinate system where the galaxies change their positions. The galaxies are receding from us because they started out receding from us, and the force of gravity just causes an acceleration that causes them to slow down, or speed up in the case of an accelerating expansion.

To be a bit more specific, I'll add my own comments. Without gravity, the expansion of the universe can be modeled as matter moving away from other matter at a constant rate. As Ned Wright says, cosmological expansion happens because things are moving away from each other. Gravity acts to slow down the expansion (usually), the weird exception being dark energy/ the cosmological constant, which speeds it up. It doesn't cause the expansion in the first place, though. That happens basically because things are moving away from each other. ##\Omega_m##, the contribution of matter and it's gravity, slows the expansion down. ##\Omega_r##, the contribution of radiation, also slows the expansion down. The only thing that causes the expansion to speed up is "dark energy", aka the cosmological constant.

Planets are going around the Sun in fixed size orbits because they are bound to the Sun. Everything is just moving under the influence of Newton's laws (with very slight modifications due to relativity). [Illustration] For the technically minded, Cooperstock et al. computes that the influence of the cosmological expansion on the Earth's orbit around the Sun amounts to a growth by only one part in a septillion over the age of the Solar System. This effect is caused by the cosmological background density within the Solar System going down as the Universe expands, which may or may not happen depending on the nature of the dark matter. The mass loss of the Sun due to its luminosity and the Solar wind leads to a much larger [but still tiny] growth of the Earth's orbit which has nothing to do with the expansion of the Universe. Even on the much larger (million light year) scale of clusters of galaxies, the effect of the expansion of the Universe is 10 million times smaller than the gravitational binding of the cluster.Also see the Relativity FAQ answer to this question.

So let me amplify on this, as well. If we look at the effect of dark energy specifically, the effect is static, it doesn't vary with time. So the effect of dark energy by itself is that it theoretically changes the size and orbital velocity of orbits by a tiny amount. WIth a large enough distance, the angular/tangential orbital velocity can even drop to zero, which was one of the questions I addressed in my last post.

Time variation of size of orbits doesn't come from an unchanging cosmological constant. We currently assume the cosmological constant is indeed constant, I'm not sure what if any experimental limits exist on this constant potentially varying with time. I don't think there's any theoretical reason why it should vary with time.

Size changes in orbits potentially occur from the other factors. These include matter leaving the solar system (as the sun radiates some of it's mass away), dark matter potentially leaving the solar system (this is something we have to theoretically model, we as of yet don't have any way to directly observe this). Thee effect that Ned Wright talks about, referencing Cooperstock, http://xxx.lanl.gov/abs/astro-ph/9803097, which is I believe now the paper I was trying to recall. Currently the site seems to be down, though :(.You were also interested in the magnitude of the effect. Ned Wrigh provides some numbers about this . I believe some of the justification for these size estimates were in Cooperstock's paper, the one that I couldn't access.

The effect of matter to slow the expansion down can be more-or-less understood in Newtonian terms. If you don't mind a factor of a 2:1 error, radiation can be regarded as being pretty much like matter. Explaning the 2:1 factor requires going beyond Newton to GR.

The "Newton-friendly" approach to the effect of the cosmological constant is to imagine that empty space is pervaded with somethign akin to a negative Newtonian mass density, a sort of anti-gravity. This isn't really quite right, but it's tough to do better with a Newtonian mindset. If you want a better and more accurate explanation, you'll need to move beyond Newtonian thinking, which involves learning General Relativity. Then you can potentially see why pressure, and not just mass and energy, causes "gravity". This will explain the 2:1 discrepancy in the radiation terms as opposed to the matter terms, and it will also explain why the anti-gravity filling empty space has a positive energy density (but a negative pressure), and why that combination acts as "anti-gravity".
 
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  • #40
pervect said:
I don't think there's any theoretical reason why it should vary with time.
Hi pervect:

In what follows, ΩΛ represents the density of dark energy divided by the critical mass density. Using the equation below, which determines how a and H change with time, also implies that the Ω terms can also change with time.
FriedmannEqWithOmegas.png

What is given is the the sum of the four Ωs initial values, corresponding to a=1, must always equal 1. This implies that as a approaches ∞, ΩΛ approaches 1.

If for some value of a, da/dt=0 and da2/dt2=0, then for that value of a H is a constant, and the four Ωs are constants.

Regards,
Buzz
 
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  • #41
Buzz Bloom said:
What is given is the the sum of the four Ωs initial values, corresponding to a=1, must always equal 1. This implies that as a approaches ∞, ΩΛ approaches 1.

No, it doesn't. The four ##\Omega## values do not change at all; they are constants. All of the time dependence is contained in the powers of ##a##. So what this equation is telling you is that, over time, all of the contributions except the dark energy one become negligible. Or, in terms of the Hubble Constant, ##H## approaches a constant value as ##t \rightarrow \infty##, often called ##H_{\infty}##, which is given by

$$
\frac{H_{\infty}^2}{H_0^2} = \Omega_\Lambda
$$

So if ##\Omega_\Lambda = 0.7## (which is, AFAIK, a reasonable value based on our best current data), then as ##t \rightarrow \infty##, we will have ##H^2 \rightarrow 0.7 H_0^2 = H_\infty^2##.

What might be confusing you is the fact that, if we pick a different time from today as the "now" to determine the values for the ##\Omega## constants, we will get different values for them. But if we pick a different time from today as our "now" to determine those values, we will also get a different value for ##H_0##.

For example, suppose we pick a time so far in the future that the only non-negligible contribution is dark energy, i.e., all of the ##\Omega## constants are practically zero except for ##\Omega_\Lambda##. This would mean ##\Omega_\Lambda = 1## if we picked that time as our "now" time, but it would also mean that our ##H_0## would be smaller, because it would be the value of the Hubble constant ##H## at that far future time, which would be ##\sqrt{0.7}## of the value it has today. So our equation would still end up telling us the same value for ##H## for that far future time, i.e., ##H_\infty##.

This also means that the actual dark energy density in that far future time is the same as it is now. The value of ##\Omega_\Lambda## will have changed if we pick that far future time as our "initial" time to determine the ##\Omega## constants, but the dark energy density itself will not.
 
  • #42
PeterDonis said:
No, it doesn't. The four ##\Omega## values do not change at all; they are constants.

Hi Peter:

I already have edited my post while you were writing your response. The following is the change I made.

Buzz Bloom said:
Using the equation below, which determines how a and H change with time, also implies that the Ω terms can also change with time.
Note that I changed "values" to "terms". In what you quoted, i underlined "initial values" to make this clear, but then I decided to also make the other change to make it even clearer.

Is this revision satisfactory?

Regards,
Buzz
 
  • #43
Buzz Bloom said:
Is this revision satisfactory?

No, it's wrong. Did you read my post?
 
  • #44
PeterDonis said:
This also means that the actual dark energy density in that far future time is the same as it is now. The value of ΩΛ\Omega_\Lambda will have changed if we pick that far future time as our "initial" time to determine the Ω\Omega constants, but the dark energy density itself will not.
Hi Peter:

Apparently I had another misunderstanding. To explain it I need I need to create a symbol for dark energy density, say ρDE.

First, perhaps my use of "critical mass density" is bad usage. How about ρc to represent the cosmological critical density.
ρc/ = 3H2/8πG

ΩΛ = ρDEc
While ρDE does not change with a and H, ΩΛ does change since ρc changes.

Note pervect's quote below.
pervect said:
Time variation of size of orbits doesn't come from an unchanging cosmological constant. We currently assume the cosmological constant is indeed constant, I'm not sure what if any experimental limits exist on this constant potentially varying with time. I don't think there's any theoretical reason why it should vary with time.
I suppose the misunderstand is based on the meaning of the pronoun "it" in the quote. (I bolded "it" for clarity.) I misunderstood understood "it". I mistakenly assumed it referred to ΩΛ rather than the cosmological constant ρDE.

I apologize for my misunderstanding.

Regards,
Buzz
 
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  • #45
Hi Peter:

Thank you for your post. I may have made another mistake in my thought process, but I am definitely a bit confused about the math that would justify that the form of the Friedmann equation (see my post #40) does not remain the same if we choose a different time at which a = 1.

I said:
While ρDE does not change with a and H, ΩΛ does change since ρc changes.​

I think I have missed understood (and continue to misunderstand why) that the adjusted sum of the four Ωs does not remain equal to 1 if I chhose a different time at which a = 1. I believe (although I may be mistaken) that it is possible to create a new set of four Ωs at a later time and redefine a later value of a, say anew, to become anew=1. Why is this not possible for all situations except when H=da/dt=0.
If this is done, the calculated value of H corresponding to the new value of a, say anew, will become the new value of H0, say H0new. The new values for ΩR, ΩM, and Ωk, say ΩRnew, ΩMnew, and Ωknew, respectively become
ΩRnew = ΩR/a4,​
ΩMnew = ΩR/a3, and​
Ωknew = Ωk/a2.​
(I think I may be omitting another needed multiplier for each RHS, but I don't have the time now to explore that possibility.)
Then the new value of ΩΛ, say ΩΛnew becomes
ΩΛnew = 1 - ΩRnew - ΩMnew - Ωknew.​

If this is wrong, I hope you can help me understand the math that explains why.

Regards,
Buzz
 
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  • #46
Buzz Bloom said:
I said:
While ρDE does not change with a and H, ΩΛ does change since ρc changes.

Ah, I misread your post before. Yes, this is correct.
 
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