What is the distance at which gravity and the cosmological force are equal?

[J O Linton]

TL;DR Summary

A formula is derived for the force associated with cosmological expansion and for the size of the sphere inside which gravity prevails

In considering the excellent answers given to my previous query about a photon in a box, I was led to consider what force was needed to hold the walls of the box stationary (w.r.t. an inertial observer inside the box.).

If you place a mass m on the end of a very long string of inextensible length D, I believe the tension in the string will be $2mD/T_0^2$ where $T_0$ is the age of the universe (this is assuming that the expansion of the universe has been linear throughout). If we compare this force to the force of gravity exerted on the mass by a galaxy of mass M then we find that the distance at which the two are equal (and opposite) is given by the expression $D = \sqrt[3]{GMT_0^2/2}$. Putting in the appropriate figure for the Milky Way, this gives a distance of 2.8 million light years - a fugure which is consistent with the fact that the Andromeda galaxy is 2.5 million light years distant and is part of our Local Group.

Can someone confirm my analysis - and is there an official name for the distance at which gravity and the cosmological force are equal?

 

[PeroK]

Summary:: A formula is derived for the force associated with cosmological expansion and for the size of the sphere inside which gravity prevails

In considering the excellent answers given to my previous query about a photon in a box, I was led to consider what force was needed to hold the walls of the box stationary (w.r.t. an inertial observer inside the box.).

If you place a mass m on the end of a very long string of inextensible length D, I believe the tension in the string will be $2mD/T_0^2$ where $T_0$ is the age of the universe (this is assuming that the expansion of the universe has been linear throughout). If we compare this force to the force of gravity exerted on the mass by a galaxy of mass M then we find that the distance at which the two are equal (and opposite) is given by the expression $D = \sqrt[3]{GMT_0^2/2}$. Putting in the appropriate figure for the Milky Way, this gives a distance of 2.8 million light years - a fugure which is consistent with the fact that the Andromeda galaxy is 2.5 million light years distant and is part of our Local Group.

Can someone confirm my analysis - and is there an official name for the distance at which gravity and the cosmological force are equal?

The gravitational attraction between two objects depends on their mass. Two galaxies a certain distance apart may be gravitational bound; whereas, two rocks the same distance apart would not. You cannot, therefore, equate gravitational force with cosmological expansion in this way.

For a fixed distance, this cosmological "force" does not change over time - this is again a common misconception. The dynamics of the solar system have been the same since its formation and will remain the same.

Note that if two objects are not gravitational bound and are comoving, then as they get further apart, the universal expansion leads to increasing recessional velocity. But, any system that does not comprise comoving galaxies is subject to local dynamics that do not change over time according to the universal expansion rate.

 

[vanhees71]

I would abandon this early idea by Hubble that the gravitational redshift is a Doppler effect. It's correct for not too far distances/red shifts as a linear approximation of the cosmological redshift. There's a concise analysis about this question in the good old usenet FAQ:

https://math.ucr.edu/home/baez/physics/Relativity/GR/hubble.html

 

[J O Linton]

The gravitational attraction between two objects depends on their mass. Two galaxies a certain distance apart may be gravitational bound; whereas, two rocks the same distance apart would not. You cannot, therefore, equate gravitational force with cosmological expansion in this way.

If you prefer, the cosmological acceleration of a point at a fixed (proper) distance from the origin in a linearly expanding universe is $2D/T_0^2$. This can usefully be compared with the gravitational acceleration of any mass at a distance D from a mass M of $GM/D^2$

For a fixed distance, this cosmological "force" does not change over time - this is again a common misconception. The dynamics of the solar system have been the same since its formation and will remain the same.

The cosmological "force" as I have defined it does indeed change over time, being inversely proportional to the square of the age of the universe. Different models of the universe generate different relations but in general, the cosmological force decreases with time and in consequence, the sphere of gravitational influence imposed by a particular galaxy will gradually increase over time.
The solar system is a bound system (the Sun's sphere of influence extends to about 250 light years) and is therefore immune to the tiny effect due to cosmological expansion.

Note that if two objects are not gravitational bound and are comoving, then as they get further apart, the universal expansion leads to increasing recessional velocity. But, any system that does not comprise comoving galaxies is subject to local dynamics that do not change over time according to the universal expansion rate.

This is incorrect. In a linearly expanding universe the recession velocity of a galaxy is constant. (It does increase in an exponential universe and it may be that our universe is, indeed, expanding like this, but in the OP I stated that I was considering a linear universe.)

I would abandon this early idea by Hubble that the gravitational redshift is a Doppler effect.

This post is not about gravitational red shift.

 

[PeroK]

In a linearly expanding universe the recession velocity of a galaxy is constant.

Even so, those linear dynamics only apply to comoving objects.

 

[PeroK]

If you prefer, the cosmological acceleration of a point at a fixed (proper) distance from the origin in a linearly expanding universe is $2D/T_0^2$. This can usefully be compared with the gravitational acceleration of any mass at a distance D from a mass M of $GM/D^2$

Okay, here's the point. Suppose we take two comoving galaxies, a distance $D$ apart, and measure the recession velocity at some time $t$. What if we take two galaxies and put them relatively at rest the same distance $D$ apart at the same time $t$? What happens then?

The gravitational force in these two cases is the same: $\frac{GM_1M_2}{D^2}$. Does the universal expansion resolve itself into the equivalent of a common force for the comoving and relatively at rest cases? And, indeed, for all objects a distance $D$ apart at time $t$, regardless of their relative state of motion?

Justify your answer!

 

[PeroK]

PS My answer is no, because the universal expansion is determined by the Friedmann equation and if two objects are not comoving at time $t$, then the solution of the Friedmann equation cannot simply be applied to them at that time.

 

[J O Linton]

My argument goes as follows. In a linearly expanding universe the proper distance D between two comoving galaxies separated by a fixed comsological distance $\theta$ is $$D = \theta \frac t T_0$$
If we keep D fixed (by using an inextensible string) then $$ \theta = D \frac {T_0} t $$
Differentiating twice $$ \frac {d^2 \theta} {dt^2} = 2D \frac {T_0} {t^3} $$
At the time $t = T_0$ the acceleration of the 'fixed' galaxies towards each other will be $2D/T_0^2$
In my original scenario, I was not talking about the mutual attraction of two galaxies - rather the gravitational effect of a large galaxy on a distant test mass. To be honest I am not sure what the 'tension' in the string would be if it connected two galaxies of different mass. Also I have no doubt that my simple derivation is not strictly correct but I am quite sure that there will be a real tension in the string (otherwise what is to stop the test mass moving away from the galaxy) and even if my simple argument is incorrect, I am convinced that the expression I have derived will be of the correct form and of the correct order of magnitude.

 

[PeroK]

Also I have no doubt that my simple derivation is not strictly correct but I am quite sure that there will be a real tension in the string (otherwise what is to stop the test mass moving away from the galaxy) and even if my simple argument is incorrect, I am convinced that the expression I have derived will be of the correct form and of the correct order of magnitude.

It certainly breaks down where $D$ is large enough that the recessional velocity is greater than $c$.

Also, as the recessional velocity approaches $c$, the local force required to overcome this would follow the relativistic equation (not the Newtonian one).

I still suspect that if two objects are held a fixed proper distance apart, then however that is achieved is equivalent to being gravitationally bound and wouldn't change over time. Perhaps including the stress-energy of the interaction into the equations is what would inevitably override the Friedmann equation?

 

[Bandersnatch]

This is the tethered galaxy problem. Under linear expansion there is no tension in the string. It takes accelerated expansion to induce tension. Analysis here:
Solutions to the tethered galaxy problem in an expanding universe and the observation of receding blueshifted objects; Davis, Lineweaver, Webb

My argument goes as follows. In a linearly expanding universe the proper distance D between two comoving galaxies separated by a fixed comsological distance $\theta$ is $$D = \theta \frac t T_0$$
If we keep D fixed (by using an inextensible string) then $$ \theta = D \frac {T_0} t $$
Differentiating twice $$ \frac {d^2 \theta} {dt^2} = 2D \frac {T_0} {t^3} $$
At the time $t = T_0$ the acceleration of the 'fixed' galaxies towards each other will be $2D/T_0^2$
In my original scenario, I was not talking about the mutual attraction of two galaxies - rather the gravitational effect of a large galaxy on a distant test mass. To be honest I am not sure what the 'tension' in the string would be if it connected two galaxies of different mass. Also I have no doubt that my simple derivation is not strictly correct but I am quite sure that there will be a real tension in the string (otherwise what is to stop the test mass moving away from the galaxy) and even if my simple argument is incorrect, I am convinced that the expression I have derived will be of the correct form and of the correct order of magnitude.

If I understand correctly what you're doing here is, you begin by taking the Hubble's law, express the recession velocity as D/t, and the Hubble constant as the inverse of Hubble time.
Which has two three four errors right off the bat: 1) in Hubble's law the D and theta distances are the same proper distance, so it makes little sense to keep one constant and expect the other to vary; 2) under linear expansion the Hubble constant (and by extension, Hubble time) are not constant, so it makes little sense to vary time and expect those to stay the same; 3) if you keep either D or theta constant, then the object at that distance is not moving with the Hubble flow, so it makes no sense to apply Hubble's law; 4) coasting expansion doesn't involve any forces, it's just inertial motion with velocities set by the initial conditions, so it makes little sense to start with it and expect any forces to be found.

 

[PeroK]

The tethered galaxy problem. I love it!

 

[PeterDonis]

a linearly expanding universe

Is empty (zero density), so there is no "gravitational force" anywhere.

 

[J O Linton]

This is the tethered galaxy problem. Under linear expansion there is no tension in the string. It takes accelerated expansion to induce tension. Analysis here:
Solutions to the tethered galaxy problem in an expanding universe and the observation of receding blueshifted objects; Davis, Lineweaver, Webb

This looks very interesting. I shall study it.

If I understand correctly what you're doing here is, you begin by taking the Hubble's law, express the recession velocity as D/t, and the Hubble constant as the inverse of Hubble time.

The equation $D = \theta \frac t T_0$ is NOT Hubble's law. It is the standard representation of a linearly expanding universe in which the proper distance from us to a distant galaxy increases linearly with time, it being zero at t = 0 and with the scale factor such that the cosmological distance to the galaxy $\theta$ equal to D at the current epoch.. Hibble's law is a relation between the observed recession velocities of nearby galaxies with distance at an instant in time and has nothing to do with the issue here.

Is empty (zero density), so there is no "gravitational force" anywhere.

I am well aware of this. Do not try to score points by being obtuse. We are talking about an idealised universe which is expanding linearly with, perhaps, just one galaxy in it and a small mass tethered to it. Our own universe has done a pretty good job of expanding linearly since the big bang anyway.

 

[PeterDonis]

I am well aware of this.

Sorry, I should have said "cosmological force", not "gravitational force". Are you aware that there is no such thing as "cosmological force" in an empty universe? (Arguably there is no such thing in any universe, but for models with nonzero density one can at least attribute some kind of interaction to the nonzero density. But if the density is zero, one can't even do that.)

Our own universe has done a pretty good job of expanding linearly since the big bang anyway.

No, it hasn't. The expansion was decelerating until a few billion years ago; now it is accelerating.

 

[PAllen]

I am well aware of this. Do not try to score points by being obtuse. We are talking about an idealised universe which is expanding linearly with, perhaps, just one galaxy in it and a small mass tethered to it. Our own universe has done a pretty good job of expanding linearly since the big bang anyway.

No, this is wrong. Our universe has had nearly spatially flat surfaces of constant cosmological time. However, the spacetime is not at all nearly flat, and the expansion history is not at all linear. Linear expansion is an extremely nice conceptual model to distinguish what features of cosmology are really related to curvature and which ones are not, because linear expansion is consistent only with absence of matter or energy, i.e. a Minkowski spacetime with no curvature. It is useful to demonstrate that superluminal recession rates have nothing to do with curvature (and are primarily a coordinate feature readily producible in Minkowski spacetime) and that to first order, cosmological redshift is nothing but SR Doppler. But there is no gravity whatsoever in a linearly expanding cosmology.

 

[J O Linton]

Thank you all for your comments on this post. The article referred to by Bandersnatch (#10) is particularly enlightening and I am now persuaded that in a linearly expanding universe there will actually be no force on a tethered galaxy. An analogy which I like to use is that of a long rubber belt, fixed at one end, being strectched by a lorry traveling at constant speed at the other. Comoving galaxies are like weights placed on the belt. The proper distance D from the origin increases but the comoving coordinate (what I call the cosmological distance $\theta$) stays the same. A tethered galaxy is like a lawn roller fastened to a fixed string. As the belt moves under it the roller rolls along the belt. Its proper distance is fixed but the cosmological distance decreases and the rate of change of cosmological velocity is as I have derived it. My error was to suppose that this 'cosmological acceleration' would necessitate the existence of a force. This is not the case. There is no tension in the string and if the string is cut, the (assumed frictionless) roller will continue to roll along the belt maintaining a constant proper distance from the origin.

However, this still leaves me with a quandry. As I stated in the OP the Andromenda galaxy is part of the Local Group but galaxies beyond about 6 million light years are not considered part of the Local Group. It seems too much of a coincidence that the expression which I derived for the effective 'range' of gravitational influence $$ \sqrt[3] {GMT_0^2/2}$$ gives an result which is of exactly the right order of magnitude.

Now it would not be the first time that an totally false derivation generates the right result - witness John Michell's derivation of the radius of a black hole in 1783 - but can anyone justify this expression some other way? Is it, perhaps, just the simple observation that the time it would take for our galaxy to 'capture' a distant galaxy increases exponentially with distance and that galaxies beyond this limit are effectively never going to be captured?

 

[Bandersnatch]

However, this still leaves me with a quandry. As I stated in the OP the Andromenda galaxy is part of the Local Group but galaxies beyond about 6 million light years are not considered part of the Local Group. It seems too much of a coincidence that the expression which I derived for the effective 'range' of gravitational influence $$ \sqrt[3] {GMT_0^2/2}$$ gives an result which is of exactly the right order of magnitude.

When you compare the escape velocity at a distance from mass M, with the recession velocity from Hubble's law:
$$V_E = \sqrt{2 GM / D}$$
and
$$V_R=H_0D$$
when $V_E=V_R$ we get
$$D = \left( \frac{2 GM}{H_0^2} \right)^{\frac{1}{3}}$$
and since $H_0=1/T_0$, we get the same expression as yours (albeit with the factor of $\sqrt[3]{2}$ in the numerator instead of denominator - I did not check your maths, so maybe it's just misplaced).

Intuitively, it is what it says on the tin - a distance at which an object moving with the Hubble flow would have higher recession velocity than the maximum velocity that the same object can have and still be bound to the mass M. It's not that far off what you write here:

Is it, perhaps, just the simple observation that the time it would take for our galaxy to 'capture' a distant galaxy increases exponentially with distance and that galaxies beyond this limit are effectively never going to be captured?

It's a nice and simple way of gauging the scale at which the motion of galaxies becomes dominated by expansion instead of peculiar motions. For a bit of an improvement over what's in the OP, one would have to consider all the mass contained within the sphere of radius D, rather than just the mass of the MW (or, just use the mass of the entire cluster). I don't remember off the top of my head what it comes out to, but certainly much more than a few million lyrs. I want to say ~500 Mlyrs, but I really should run the numbers first.

 

[PeterDonis]

As I stated in the OP the Andromenda galaxy is part of the Local Group but galaxies beyond about 6 million light years are not considered part of the Local Group.

But they are still part of the larger cluster that contains the Local Group, and that cluster is still gravitationally bound. You have to get up to distance scales of hundreds of millions of light years before you start to reach the boundaries of gravitationally bound systems in our universe.

 

[J O Linton]

When you compare the escape velocity at a distance from mass M, with the recession velocity from Hubble's law:
$$V_E = \sqrt{2 GM / D}$$
and
$$V_R=H_0D$$
when $V_E=V_R$ we get
$$D = \left( \frac{2 GM}{H_0^2} \right)^{\frac{1}{3}}$$
and since $H_0=1/T_0$, we get the same expression as yours (albeit with the factor of $\sqrt[3]{2}$ in the numerator instead of denominator - I did not check your maths, so maybe it's just misplaced).

Intuitively, it is what it says on the tin - a distance at which an object moving with the Hubble flow would have higher recession velocity than the maximum velocity that the same object can have and still be bound to the mass M. It's not that far off what you write here:

It's a nice and simple way of gauging the scale at which the motion of galaxies becomes dominated by expansion instead of peculiar motions. For a bit of an improvement over what's in the OP, one would have to consider all the mass contained within the sphere of radius D, rather than just the mass of the MW (or, just use the mass of the entire cluster). I don't remember off the top of my head what it comes out to, but certainly much more than a few million lyrs. I want to say ~500 Mlyrs, but I really should run the numbers first.

That's brilliant! I knew there would be some way of deriving something similar. Of course, I understand that the expression you have derived does not determine an absolute boundary of the sphere of gravitational influence but it does give one an idea of the scales of distance we are talking about. The numerical constant is pretty irrelevant. Your expression gives a figure of 4.5 million light years for our galaxy. Because of the cube root, the figure will not be much different if we include the whole mass of the Local Group. I am quite prepared to accept PeterDonis' comment that you have to go to distances of up to 100 million light years before the gravitational influence of our galaxt is totally negligible.

Many thanks for this answer.

 

[PeterDonis]

I am quite prepared to accept PeterDonis' comment that you have to go to distances of up to 100 million light years before the gravitational influence of our galaxt is totally negligible.

It's not just our galaxy; as I said, the largest gravitationally bound system of which our galaxy (and Local Group) is a part is a larger galaxy cluster, whose distance scale is roughly 100 million light years IIRC. If our galaxy were by itself, not gravitationally bound to anything else, its gravitational influence would indeed become negligible at a much shorter distance; and if our Local Group were by itself, not gravitationally bound to anything else, its gravitational influence would become negligible at a distance somewhat larger, but still smaller than the 100 million light years or so that our galaxy cluster spans.

 

[pervect]

If what I did 10 or so years ago in https://www.physicsforums.com/threads/weyl-curvature-and-tidal-forces.433916/post-2910883 is right, for the flat FLRW metric

$$ds^2 = -c^2 dt^2 + a(t)^2 \left(dx^2 + dy^2 + dz^2 \right)$$

The "tidal force" stretching the string will be proportional to $\frac{a''}{a}$ where a'' = $\frac {d^2 a(t)} { dt^2}$

The justification for this is the geodesic deviation equation, which says that the relative acceleration between geodesics is $R^{a}{}_{bcd} u^b u^d D$, where D is the separation between geodesics. It's convenient to let D be the distance between the middle of the long string and the end. Then the proper accleration of the middle of the string is zero, and the proper acceleration of the end of the string must be of the above magnitude to keep the length of the string constant.

However, to get a'' we need to get into more details of the Friedmann equations, https://en.wikipedia.org/wiki/Friedmann_equations

Skimming through the appropriate wiki links will give the relevant equations, most particularly
https://en.wikipedia.org/wiki/Friedmann_equations
https://en.wikipedia.org/wiki/Hubble's_law
https://en.wikipedia.org/wiki/Friedmann–Lemaître–Robertson–Walker_metric

The first link, to Friedmann's equations, is the most directly useful, it gives:

and

$$ {\displaystyle {\dot {H}}+H^{2}={\frac {\ddot {a}}{a}}=-{\frac {4\pi G}{3}}\left(\rho +{\frac {3p}{c^{2}}}\right).} $$

As I've already assumed we have a flat spatial slice, we set k=0 in the above.

You'll have to wade through the math to get more details, but basically Hubble's constant changes with time and gives a' / a, while what we really want is a'' / a.

 

[PAllen]

If what I did 10 or so years ago in https://www.physicsforums.com/threads/weyl-curvature-and-tidal-forces.433916/post-2910883 is right, for the flat FLRW metric

$$ds^2 = -c^2 dt^2 + a(t)^2 \left(dx^2 + dy^2 + dz^2 \right)$$

The "tidal force" stretching the string will be proportional to $\frac{a''}{a}$ where a'' = $\frac {d^2 a(t)} { dt^2}$

You may have assumed flat spatial slices in your derivation, but I think your conclusion is true irrespective of this fact. I certainly know it is true for solutions with hyperbolic slices.

 

[J O Linton]

No, it hasn't. The expansion was decelerating until a few billion years ago; now it is accelerating.

My claim that the universe 'has made a pretty good job of expanding linearly' was based on the following diagram. It shows the currently popular (0.3, 0.7) model in red and the linear (empty) universe in brown. As Peter Donis said, the current model has departed a bit from the linear model but the most remarkable thing about the two models is that they both appear to agree almost exactly on the age of the universe.

Now the linear model predicts the following relationship between proper distance D and red shift z of $$D = cT_0 log(z+1)$$. It also predicts that (in principle) the universe has no observable horizon. I believe that both these statements are very nearly true of our actual universe and that therefore the linear model is a pretty good approximation to the real situation. Correct me if I am wrong.

 

[PAllen]

My claim that the universe 'has made a pretty good job of expanding linearly' was based on the following diagram. It shows the currently popular (0.3, 0.7) model in red and the linear (empty) universe in brown. As Peter Donis said, the current model has departed a bit from the linear model but the most remarkable thing about the two models is that they both appear to agree almost exactly on the age of the universe.

Now the linear model predicts the following relationship between proper distance D and red shift z of $$D = cT_0 log(z+1)$$. It also predicts that (in principle) the universe has no observable horizon. I believe that both these statements are very nearly true of our actual universe and that therefore the linear model is a pretty good approximation to the real situation. Correct me if I am wrong.

The very large difference is that the geometry of the space (not spacetime) of constant cosmological time is exceedingly close to flat in our universe, while for linear expansion, space (as just defined) must be hyperbolic, to a substantial degree. This is not at all consistent with observation.

 

[PeterDonis]

My claim that the universe 'has made a pretty good job of expanding linearly' was based on the following diagram.

And that diagram does not support your claim. It only supports the much weaker claim that the age of the universe now happens to be about the same as it would have been under a linear expansion model that has the same Hubble constant (slope in the diagram) now. Your claim requires that the slope of the curve for our actual universe would have to be close to the slope of the linear curve everywhere from the start of the universe to now, which, as the diagram shows, is not the case.

 

[PeterDonis]

the current model has departed a bit from the linear model

It's much more than "a bit". The slope of the $\Lambda CDM$ curve is much steeper than the linear curve in the early universe, becomes shallower up until a few billion years ago, then increases to be the same now (but will continue to increase as the expansion continues to accelerate in the future, diverging much more from the linear curve once again).

You are putting way too much weight on the ages being close. That is a very weak correspondence.

It also predicts that (in principle) the universe has no observable horizon.

I'm not sure what you mean by this. If you mean it predicts that the universe has no event horizon, that is obviously different from the prediction of the $\Lambda CDM$ model.

 

[J O Linton]

Thanks for these comments - but I must point out that none of you have produced any observational evidence to back your claims. A graph of D against log(1+z) for very large z is consistent with the assumption that they are proportional (although the error bars are huge) and the observable universe is at least as big as the CMBR.

 

[pervect]

Thanks for these comments - but I must point out that none of you have produced any observational evidence to back your claims. A graph of D against log(1+z) for very large z is consistent with the assumption that they are proportional (although the error bars are huge) and the observable universe is at least as big as the CMBR.

There is a fair amount of observational evidence for what is called the $\Lambda CDM$ model, $\Lambda$ being the cosmological constant, and CDM standing for cold dark matter.

I see others have already made this point.

Wiki has a brief discussion of this model, and some of the evidence that supports it, in https://en.wikipedia.org/wiki/Lambda-CDM_model.

That said, I don't think there's any direct observational evidence for the particular scenario you are interested in. I haven't run the numbers, but I doubt it's feasible to create a long enough string to measure the tension in it, even if we could make the mass of the string negligible. Making the mass of the string negligible would require exotic matter (for non exotic matter, the density is always larger than the tension in geometric units). Since said exotic matter may or may not even exist, it's rather likely that the self-gravitation of any such string composed of non-exotic matter would swamp the effects we're talking about. There may be ways around this, but basically I think this whole discussion is more of a thought experiment than a practical one. However, it is a thought experiment that current cosmological models do make predictions for.

Let's go back to the question of evidence, though.

There is evidence supports the $\Lambda$ CDM model, which has been successful in making some predictions as discussed in the wiki, among which are measurments of the CMB spectrum by WMAP which are related to acoustic oscillations in the cosmic medium due to the big bang.

And this model makes predictions about the scenario you are interested in.

One other comment I wanted to make is that some of the predictions made by this model become simpler if one assumes that one is in an era where the radiation pressure p is "small". In this case $\dot{H}$ <<$H^2$, so for the purposes of calculating the tension in a long string, we can neglect the former term.

This results in the tidal force for a unit mass and a unit separation being proportional to $H^2$, modulo various constant factors that I haven't worked out in detail. This makes sense from a dimensional point of view, H gives recession velocity as a function of distance, so it's SI units would be 1/second. H^2 has units of 1 / second^2. Tidal force would have units of kg m / s^2, tidal force / unit mass units of m / s^2, and tidal force / unit mass / unit separation would have units of 1 / s^2, which checks.

 

[PAllen]

The very large difference is that the geometry of the space (not spacetime) of constant cosmological time is exceedingly close to flat in our universe, while for linear expansion, space (as just defined) must be hyperbolic, to a substantial degree. This is not at all consistent with observation.

I misspoke here. Linear expansion requires hyperbolic spatial slices if the the spacetime curvature is to end up 0 (Minkowski space). However, you can posit linear expansion with flat or positive curvature spatial slices; such spacetimes have nonzero Ricci curvature. However, they still don't agree with observation of the history of expansion rate.

 

[pervect]

I had one more comment I wanted to make, and that's about sign issues, in this case whether the string is under tension or under compression. I have mostly glossed over the sign issues in my previous posts, not really paying attention to them.

If it were not for the cosmological constant $\Lambda$, a very long string would be under compresssion, not tension, even in an expanding universe.

The universe might be expanding, but the cold dark matter part of the model would make the rate of expansion slow down, and this would cause two points following geodesics to accelerate twoards each other, even though they might be moving apart due to cosmological expansion. It's not the cosmological expansion itself that's important to the question of the tension in the string, but the rate of change.

The cosmological constant $\Lambda$ is which fuels the acceleration of the expansion of the universe, and which also makes the prediction of the model change signs to be one where the string is under tension, rather than compression.

 

[PeterDonis]

I must point out that none of you have produced any observational evidence to back your claims.

This comment is out of line. The $\Lambda CDM$ model has a voluminous literature dealing with its predictions and comparison of those with observational evidence. Even in a "B" level thread, it is quite fair for other posters to expect you to have at least a basic knowledge of that. Asking for some references so you can learn more about the $\Lambda CDM$ model is fine (the Wikipedia article on the model, which @pervect linked to, has plenty of references to the literature, so that's a good place to start). Refusing to accept basic descriptions of what the $\Lambda CDM$ model says unless people produce observational evidence directly in this thread is not.

A graph of D against log(1+z) for very large z is consistent with the assumption that they are proportional

I have no idea what graph you are referring to, since the only graph that has been given so far in this thread directly contradicts your claim here, as I have already pointed out in post #26 of this thread.

 

[J O Linton]

I am well aware of the arguments in favour of the $\Lambda CDM$ model and am quite willing to accept that it is the best model we have at present. Please do not deliberately misunderstand me. I am not arguing that the expansion of the universe has been linear - only that in order to understand the basic properties of the universe we live in, a linear model is perfectly adequate, at least to a first approximation.
I attach a graph of log(Z) (Z = 1+z) against D. the data is taken from the NED-4D files available at https://ned.ipac.caltech.edu/level5/NED1D/intro.html.

In resepct of your other comment in post #26, by 'observable universe' I meant that part of the universe within the particle horizon. A linear universe has an infinite particle and an infinite event horizon. I believe I am right in saying that the $\Lambda CDM$ universe has an infinite particle horizon but a finite event horizon - i.e we can see all of the past universe but may not be able to communicate with all of the future universe. As I have already pointed out, the particle horizon of our universe extends at least as far as the CMBR.

 

[vanhees71]

The linear model is not adequate to describe the observations. As you say yourself to fit all (!) data, i.e., at least both the redshift-distance measurements (e.g., from Hubble space telescope) and the CMBR-fluctuation data (e.g., from the Planck satellite measurements) the cold-dark-matter model with a cosmological constant (including acceleration!) is the far best model, and the linear model fails. Whether or not this really is the final answer, nobody can say. There's at least the issue about the value of the Hubble constant when measured with different methods that has to be understood and be resolved.

 

[J O Linton]

I doubt if anyone has seriously tried to reconcile the observations with the linear model. The reason for this is, of course, that there are good theoretical reasons why the expansion cannot be linear because a linear universe is theoretically empty - which our universe patently is not! I am not trying to argue that the $\Lambda CDM$ model is wrong - only that, to a first approximation, the linear model makes a pretty good job of explaining the erssential features of our universe.

 

[Bandersnatch]

I doubt if anyone has seriously tried to reconcile the observations with the linear model. The reason for this is, of course, that there are good theoretical reasons why the expansion cannot be linear because a linear universe is theoretically empty - which our universe patently is not!

It's not just theoretical reasons. There are direct observables that can't be reconciled with the linear model. Most obviously the shape of the Hubble diagram (e.g. see Reiss' work, past and current). As one looks towards higher z's the shape of the curve will be different, depending on the value of the second derivative of $a$ (in particular, it will curve in opposite directions during acceleration/deceleration phases).
That's what first prompted the inclusion of Lambda into the model. And if you think about it, it couldn't have come from what you call theoretical reasons, as with matter - we can look around us and see that the universe is not empty, and decide that only models with matter in it should be considered; but the cosmological constant is not conspicuously advertising its existence, so there's no a priori reason to favour models with it over those without. Its inclusion in the model must come from observations of how the expansion unfolds over time.

Now, it's true that you can approximate a curve that slopes one way for about half it's length and the other way for the other half, as a line. This is contingent on not looking too closely and being fortunate enough to find oneself at the cosmic time when the two opposing contributions are about equal.
But you're doing more than that here - you're assuming the approximation will hold just as well in the future, which is what you're doing when you're making deductions about actual horizons.

If you forgive me a hyperbolic analogy, it's like acknowledging that the Earth being flat is a good local approximation, and - even though one is aware that the approximation diverges the farther (or closer) one looks, use the flat Earth model to make deductions about the entire planet.

A linear universe has an infinite particle and an infinite event horizon. I believe I am right in saying that the $\Lambda CDM$ universe has an infinite particle horizon but a finite event horizon - i.e we can see all of the past universe but may not be able to communicate with all of the future universe.

It is not correct. Yes, the particle horizon in the LCDM universe extends to infinity in terms of proper distance, but it's finite in terms of comoving distance. Which means there is a finite number of galaxies whose past states we can ever observe. Which is another way of saying that there exists an event horizon - the EH limits not only what galaxies you can send a signal to, but also what galaxies can send a signal to you.
This is different in a linear universe, where you can get to see all the galaxies, even at infinite distance, if you just wait long enough.