The proper Schwarzschild radial distance between two spherical shells

[Buzz Bloom]

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​

Are the following two the components you intend as having the sclae factor as part of them?

a²​

m/(2a r_cap) (r_cap 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

 

[Buzz Bloom]

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

https://www.physicsforums.com/threa...f-the-universe-affect-orbiting-bodies.988864/​

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

Regards,
Buzz

 

[PeterDonis]

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.

 

[pervect]

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

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".

 

[Buzz Bloom]

I don't think there's any theoretical reason why it should vary with time.

Hi pervect:

In what follows, Ω_Λ 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.

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 ∞, Ω_Λ 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

 

[PeterDonis]

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.

 

[Buzz Bloom]

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.

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

 

[PeterDonis]

Is this revision satisfactory?

No, it's wrong. Did you read my post?

 

[Buzz Bloom]

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 ρ_DE.

First, perhaps my use of "critical mass density" is bad usage. How about ρ_c to represent the cosmological critical density.

https://www.astronomy.swin.edu.au/cosmos/C/Critical+Density​

ρ_{c/ = 3H²/8πG}​

Ω_Λ = ρ_DE/ρ_c​

While ρ_DE does not change with a and H, Ω_Λ does change since ρ_c changes.

Note pervect's quote below.

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 Ω_Λ rather than the cosmological constant ρ_DE.

I apologize for my misunderstanding.

Regards,
Buzz

 

[Buzz Bloom]

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 ρ_DE does not change with a and H, Ω_Λ does change since ρ_c 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 a_new, to become a_new=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 a_new, will become the new value of H₀, say H_0new. The new values for Ω_R, Ω_M, and Ω_k, say Ω_Rnew, Ω_Mnew, and Ω_knew, respectively become

Ω_Rnew = Ω_R/a⁴,​

Ω_Mnew = Ω_R/a³, and​

Ω_knew = Ω_k/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 Ω_Λ, say Ω_Λnew becomes

Ω_Λnew = 1 - Ω_Rnew - Ω_Mnew - Ω_knew.​

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

Regards,
Buzz

 

[PeterDonis]

I said:
While ρ_DE does not change with a and H, Ω_Λ does change since ρ_c changes.

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