Lambda-CDM theory - Woo or not?

[W.D.Clinger]

Although that single exception expands forever, I believe its spatial size is bounded.

Nope!

(It's a bit of a trick exception, in a way.)

With that hint, I'll return to the yellow book. Thanks!

 

[Vorpal]

Perhaps de Sitter was a sorcerer, channeling dark energies to bend the Einstein static universe by a spatially-independent conformal factor.?

 

[ben m]

Does anyone else have the impression that Farsight is scurrying over to Wikipedia before each post, looking up whatever keywords Sol and WD just used, and hoping to learn enough relevant factoids to make his replies sound well-informed?

 

[W.D.Clinger]

Does anyone else have the impression that Farsight is scurrying over to Wikipedia before each post, looking up whatever keywords Sol and WD just used, and hoping to learn enough relevant factoids to make his replies sound well-informed?

That would not surprise me, but it goes deeper than that:

Thanks. I think the issue with boards like this tends to be intellectual arrogance and dishonesty. Some people who think they understand relativity and cosmology cannot bear a hard evidential argument backed by cool clear logic. When their belief and expertise are challenged, they sometimes start getting nasty. I don't. My name (John Duffield) is public knowledge, and I have my reputation to think about.

Farsight's identity is irrelevant, and I don't mean to suggest we have any more reason to believe his name is John Duffield than to believe anything else he writes, but Farsight does appear to be reading from a script written by this particular John Duffield:

John Duffield. A qualitative 3+1 dimensional geometrical model.​

See section 6, "Gravity". To give you some idea of what that paper's like, here's an extract from the two paragraphs immediately preceding section 6 (with italics as in the original):

Charge is curl, charge is twist. The electromagnetic field is a region of twisted space, and if we move through it we perceive a turning action which we then identify as a magnetic field.

Since charge can be created and destroyed in pair production and annihilation along with mass, charge cannot be viewed as fundamental. It is instead, along with the electron, its mass, and its associated electromagnetic field, the product of a particular geometric configuration of the photon....

You may also enjoy this television interview, in which John Duffield promotes his book:
http://www.richplanet.net/starship_main.php?ref=7&part=1
(Skip the first 3:25, which is devoted to UK news.)

Here's the part of the interview in which Duffield explains how gravity makes light curve:
http://www.richplanet.net/starship_main.php?ref=8&part=2

 

[Vorpal]

You need to read http://arxiv.org/abs/physics/0204044.

I wondered what this was about, and based on the abstract, yes, that's basically right (no idea about the content, though). I've posted a quote and translation about Einstein's interpretations of what the 'gravitational field' is in the past.
But it doesn't mean what you think it means. The Riemann curvature tensor has a very clear meaning: it's the gravitational tidal forces. That's what geodesic deviation means, and it is a completely objective, coordinate-independent effect. So if we're talking about whether something is curved, the above considerations change nothing.

I know what it really is: inhomogeneous space. That’s what Einstein said a gravitational field is.

Einstein said that the gravitational field is the connection, since he explicitly treated the connection coefficients as the components of the gravitational field. There's a good conceptual reason for this, since it explicitly dissolves any difference between gravity and inertia. But it doesn't affect Clinger's point here, since we're talking about curvature.

I'm really not seeing the difficulty here. Spacetime being curved is by definition having non-zero curvature. Since curvature is given by a tensor, you can't make it go away. If it's nonzero in some frame, it's nonzero in all frames. (Unlike the connection coefficients, which do not form a tensor, and can be transformed away [edit: locally]... which goes right to Einstein's point about things in freefall locally acting just like STR.)

This is wrong I’m afraid. Start with a high-energy-density universe. It expands, reducing the energy density if you adhere to conservation of energy. What happens next? It expands, reducing the energy density. What happens next? It expands, reducing the density. At all times the universe is homogeneous, meaning there’s no gravity in it. The universe didn’t collapse when it was small and dense. It's going to sail right on past the critical value without a care in the world.

If one were to take Einstein's conceptualization seriously, then it follows trivially that every spacetime has gravity. Even the completely flat Minkowski spacetime, because sure enough, there's inertia there and a connection defining the geodesics. So it makes little sense to invoke Einstein to say that some spacetime has no gravity because according to him, every spacetime does.

And it doesn't really affect the disagreement here, because what was discussed was curvature anyway.

 

[sol invictus]

[ QUOTE=W.D.Clinger;7632636]With that hint, I'll return to the yellow book. Thanks![/QUOTE]

Just to extend the hint a little, we're looking for metrics of the form

[latex]$ds^2=-dt^2+a(t)ds_3^2$[/latex]

where ds_3^2 is the metric of either Euclidean 3-space, a unit sphere, a unit RP3, or a unit hyperboloid (those are all the homogeneous and isotropic possibilities). We want to restrict to cases where the time derivative of a is strictly positive (expanding).

 

[edd]

Ya, actually it does IMO. The particle kinetic energy that is contained in a QM definition of "space" would tend to stay in motion. It would work much like ordinary particle pressure. If you pop a pressurized balloon, the more dense air from inside the balloon will seek to expand outward until equilibrium is achieved.

If ABSOLUTE zero space (containing no kinetic energy) is the limit, then any space with a particle kinetic energy that is greater than absolute zero would in fact seek to expand toward zero like ordinary pressure IMO. When you folks discuss your mythical "negative pressure", you're ultimately talking about a force that PULLS on the universe from outside, or a particle kinetic energy inside of a given space that is less than absolute zero (physically impossible).

The only problem I see is that you'd have a hard time exceeding the absolute speed limit of the fastest particle in the QM definition of space.

We can leave negative pressure out of it (especially since we know you don't like it). I simply mean that the effects of pressure in GR are pretty clear and don't work the way Farsight seemed to be saying.

 

[W.D.Clinger]

Just to extend the hint a little, we're looking for metrics of the form

[latex]$ds^2=-dt^2+a(t)ds_3^2$[/latex]

where ds_3^2 is the metric of either Euclidean 3-space, a unit sphere, a unit RP3, or a unit hyperboloid (those are all the homogeneous and isotropic possibilities). We want to restrict to cases where the time derivative of a is strictly positive (expanding).

Yes. In my earlier guess, I said "bounded" when I meant finite. I was thinking of a 3-sphere with a(t) converging asymptotically from below toward the steady state solution, but (as I actually explained to Farsight, d'oh!) that isn't flat. I hadn't thought of RP3, but I doubt whether that's flat either, which leaves Euclidean 3-space and the obvious solution to the field equations with Minkowski metric, positive Λ, and the consequent values for T_μν. (That doesn't seem tricky enough, but negative matter density might qualify.) I haven't worked out the consequences of that solution for a(t), so I don't know whether it satisfies the "expand forever" part. (ETA: but it seems likely just on an intuitive basis.)

I'm not going to think about it any more this weekend. I wasted too much time yesterday watching the Duffield interview and reading part of his paper. I have a conference presentation to prepare and can no longer risk this avenue of procrastination.

 

[Michael Mozina]

Does anyone else have the impression that Farsight is scurrying over to Wikipedia before each post, looking up whatever keywords Sol and WD just used, and hoping to learn enough relevant factoids to make his replies sound well-informed?

Why is it that you folks have the emotional need to immediately attack the individual instead of focusing on the material at hand? That emotional need to attack the dissenting individual is exactly the same emotional need every cult/region uses to keep the flock in line. How pathetic. This is supposed to be "science", but you folks collectively and empirically act just like any other "dark religion".

 

[Michael Mozina]

Ya know.....

This is EXACTLY why I have never tried to "bark math" or "argue math" with you folks!

Even when someone comes here that's willing to "speak your language" and look into the mathematics, you belittle and berate them every single step of the way. You automatically ignore the beliefs of the folks that wrote the formulas you're talking about too. You really don't give one hoot about the math. You simply use it as a "weapon", as a way to belittle and put down individuals. That's exactly what I expected. Thanks for demonstrating exactly how pointless it is to attempt to publicly discuss mathematics with you folks. Wow!

 

[ben m]

This is EXACTLY why I have never tried to "bark math" or "argue math" with you folks!

Even when someone comes here that's willing to "speak your language" and look into the mathematics, you belittle and berate them every single step of the way.

That's a good way of putting it. Farsight, like you, is merely barking math; he thinks that throwing out a few equations will make him sound like he's "speaking our language". Like you, he is not using math to find out whether his hypotheses are right or wrong.

It's like a stranger walking into a big band rehearsal.

S: "You guys are terrible. I'm your new first chair."
B: "Is that so? Let's hear it. Take the first solo on 'Beguine'. One, two ..."
S: "Wait, wait, I don't have this 'sax solo', I have an interpretive dance." <dances>
B: "That dance is terrible, but if it inspires a good solo then you're in. Let's hear the solo."
S: "But my dance is already the best possible sax solo." <begins dancing. Appears incompetent.>
B: "Baloney. Let's hear something."
S: "I don't bark on command." <dances> "But I'm clearly the new first chair."
B: "This is a band. The first chair needs to play sax solos."
S: "OK, blowing air into metal tubes is your membership totem? Fine. I'll play along. hissssss HONK SPLAT SQUEEEEEEONK BLAAAT. There, see? I can play sax." <dances>
B: "Ha ha ha ha! Go away."
S: "First you tell me to play music, and when I do you're all critical! This is why I don't bother with the music."

 

[W.D.Clinger]

weirdness in Hawking&Ellis equation (5.12)

Okay, I'm procrastinating. (I hit a technical snag I can't fix from home.)

I used the equations in Hawking&Ellis §5.3 to check out the potential FLRW solution I described as follows:

...Euclidean 3-space and the obvious solution to the field equations with Minkowski metric, positive Λ, and the consequent values for T_μν.

Equation (5.10) says the density is constant.
Equation (5.11) says the second derivative of S is zero.
Equation (5.12) says the first derivative of S is zero.

That tells me S is constant, which means I've found a steady-state FLRW solution I didn't know about. That isn't too surprising, because the negative density contradicts observation, but repeating the calculation with a negative Λ yields the same results, which suggests rather strongly (!) that I've made a mistake.

Equation (5.12) looks weird, so I suspect I'm misreading it:

[latex]
\[ 3 {\dot S}^2 = 8 \pi (\mu S^3)/S + \Lambda S^2 - 3 K \]
[/latex]​

Why didn't Hawking&Ellis simplify the (μS³)/S to μS² ?

They say equation (5.12) is essentially the same as equation (2.35), but that isn't obvious and I know my limitations: Even if I were able to reproduce that derivation, it might take me several days, so I'm asking for help here.

 

[sol invictus]

[latex]
\[ 3 {\dot S}^2 = 8 \pi (\mu S^3)/S + \Lambda S^2 - 3 K \]
[/latex]​

That's looking fine. Let's clean it up a little:

[latex]
\left({\dot S}/S \right)^2 = 8 \pi \mu/3 + \Lambda/3 - K/S^2
[/latex]

That's the standard form for the Friedmann equation. Now, we want a solution with \dot S > 0, so the right hand side needs to be non-zero. But we also want a solution with zero spacetime curvature. This equation alone doesn't tell us what the curvature is, but Einstein's equations were posted earlier, and they say that if either \mu or \Lambda is non-zero, then at least some components of the Einstein tensor are non-zero, which means the curvature is non-zero. So.... ?

 

[W.D.Clinger]

[latex]
\[ 3 {\dot S}^2 = 8 \pi (\mu S^3)/S + \Lambda S^2 - 3 K \]
[/latex]​

That's looking fine.

As it should be: It's Hawking&Ellis equation (5.12), just as they write it. I have found only two errors in Hawking&Ellis, both quite minor, but I thought I'd better ask about that equation before spending too much more time on this. (In particular, I thank you for saving me from deriving equation (5.12) from (2.35)!)

Let's clean it up a little:

[latex]
\left({\dot S}/S \right)^2 = 8 \pi \mu/3 + \Lambda/3 - K/S^2
[/latex]

That's the standard form for the Friedmann equation.

Yes. For example, it matches Wald's equation (5.2.14), except he leaves off the Λ term. (I'm just thinking out loud because that might make it easier for you to spot my mistakes.)

Now, we want a solution with \dot S > 0, so the right hand side needs to be non-zero. But we also want a solution with zero spacetime curvature. This equation alone doesn't tell us what the curvature is, but Einstein's equations were posted earlier, and they say that if either \mu or \Lambda is non-zero, then at least some components of the Einstein tensor are non-zero, which means the curvature is non-zero. So.... ?

So I have Λ = -8πμ = 8πp (using Hawking&Ellis's notational conventions, with c=G=1, except I prefer to use 0 for the timelike coordinate so μ=T₀₀, and p=T_ii for i > 0). I started with those values, and plugged them into Hawking&Ellis equations (5.10) through (5.12).

Everything worked out fine, except I get \dot S = 0 when I want \dot S > 0. My conclusion is that the space isn't expanding, so this isn't the example you're looking for.

At this point, my best guess is that your example consists of an empty universe with Minkowski metric and no cosmological constant. The trick part of your question is that we usually think of that solution as steady state, but it's still an FLRW solution even if the empty space is expanding.

Did I get it, finally?

 

[sol invictus]

At this point, my best guess is that your example consists of an empty universe with Minkowski metric and no cosmological constant. The trick part of your question is that we usually think of that solution as steady state, but it's still an FLRW solution even if the empty space is expanding.

Did I get it, finally?

That's correct, but you skipped a step! First, you were supposed to note that if \mu=\Lambda=0, there is still a solution with \dot S>0 - namely when K=-1, in which case one simply obtains S=t, so that \dot S=1.

Therefore, there is a solution with negative SPATIAL curvature (since K=-1), infinite volume, and a scale factor that's linear in time. You're completely correct that this is nothing other than Minkowski space, but written in a coordinate system that makes it look like an expanding, negatively curved FRW cosmology.

Incidentally, you can make it an honest-to-goodness expanding universe by making the negatively curved space compact, or in any case by choosing a hyperbolic 3-manifold other than the simplest one. But those spacetimes aren't globally isotropic.

 

[W.D.Clinger]

At this point, my best guess is that your example consists of an empty universe with Minkowski metric and no cosmological constant. The trick part of your question is that we usually think of that solution as steady state, but it's still an FLRW solution even if the empty space is expanding.

Did I get it, finally?

That's correct, but you skipped a step! First, you were supposed to note that if \mu=\Lambda=0, there is still a solution with \dot S>0 - namely when K=-1, in which case one simply obtains S=t, so that \dot S=1.

Gotcha. Thanks.

Therefore, there is a solution with negative SPATIAL curvature (since K=-1), infinite volume, and a scale factor that's linear in time. You're completely correct that this is nothing other than Minkowski space, but written in a coordinate system that makes it look like an expanding, negatively curved FRW cosmology.

Before Michael Mozina and any other lurkers who've been following along can jump on this, let me point out that the spacetime curvature is zero even though the "SPATIAL curvature" is negative. Because the space is empty, the negative spatial curvature is nothing little more than a mathematical artifact of the Friedmann equations and their conventions for classifying the spatial geometry.

Incidentally, you can make it an honest-to-goodness expanding universe by making the negatively curved space compact, or in any case by choosing a hyperbolic 3-manifold other than the simplest one. But those spacetimes aren't globally isotropic.

I'll think about that later. Thanks again!

 

[Farsight]

Still with me, at least kind of?

Yes, but you've totally ignored my salient comments. You've totally dodged them. Now respond properly to my post or I'm writing you off as a naysayer.

 

[Farsight]

I don't think this works the way you believe.

It may not. But I'm confident that in a homogeneous universe there is no gravity to slow down the expansion, or cause a big crunch.

Although that single exception expands forever, I believe its spatial size is bounded.

I've rather avoided discussing the boundaries of our flat universe. IMHO the situation is something like a fish in a globe of water, only the fish is made out of waves in the water. Water is space. Beyond it, is no space. Which means there is no beyond it.

By the way, you might just want to just consider our conversation a "friendly side conversation" and put me on the back burner. FYI, that assumption about the nature of space seems to be where you and I have a difference of opinion.

Noted, MM. Some of the other comments I can see aren't quite so friendly. Rather living down to my expectations I'm afraid. Such is life.

"Space", as in the the space in the sky, is filled with particle kinetic energy, including photons of all wavelengths, neutrinos, ect. It's not technically empty, nor devoid of all kinetic energy and all matter.

It gets interesting when you consider the infinite time dilation at a black hole event horizon.

While I buy your basic argument that the quantum mechanical kinetic energy will tend to "expand", I see no reason why anything should expand faster than light.

Don't worry about whether it expands faster than light or not. It isn't material to the central issue, which is the issue of dark matter. Just think on this: if a gravitational field is inhomogeneous space where the energy density varies, and if galaxies are gravitationally bound and do not expand whilst the intervening space does, and if you adhere to conservation of energy... then what will every galaxy be surrounded by?

 

[edd]

I rest my case.

 

[sol invictus]

Yes, but you've totally ignored my salient comments. You've totally dodged them. Now respond properly to my post or I'm writing you off as a naysayer.

If you insist....

That’s Gauss’s law for gravity in differential form, see wiki. Newtonian gravity is usually described in a different fashion, such as... (is there a problem with frac and over on this latex? F = G \frac{m_1 m_2}{r^2} and another expression doesn't seem to parse, maybe it's me).

Already dealt with above. The expression you gave is a special case; mine is the general form.

I’m not fond of that mapping. Yes [latex]\nabla \cdot \mathbf{E} = \frac{\rho}{\varepsilon_0}[/latex] but remember that it’s the electromagnetic field, which has curl, whilst a gravitational field hasn’t. This could take into the issue of curved space versus curved spacetime, but let's come back to that.

Sorry, I’m not familiar with that expression. But given that [latex] u_{em} = \frac{\epsilon_0}{2}E^2 + \frac{1}{2\mu_0}B^2 \,[/latex], no matter, an electromagnetic field has an energy density, the energy is there in the space, and it’s related to electromagnetic four-potential A^α in which Φ plays a part.

Yes.

Pause there. Don't guess. Think about a single electron. There’s electromagnetic energy in the space surrounding the place where we say the electron is located. Now add a proton. It also comes with electromagnetic energy in the surrounding space. Now you’ve got a hydrogen atom, still with electromagnetic energy in the surrounding space. The two opposite fields mask one another, the energy doesn’t vanish from the space just because the electron and the proton are close to one another. So recast your expression saying [latex]$\rho_g = (2/2)\vec E^2[/latex]

The gravitational version of that is my expression, apart from the (2/2). You were interested in how gravity acts on gravity, and that's what I'm trying to explain to you. As I pointed out, it's true that E^2 is an energy density that gravity acts on, and it should also be included in Newton's equation. I'm happy to add it in if you like, but it's not needed for what I'm attempting to explain to you. Just think of it as part of rho_m.

Sorry, you lost me there. How do you go from a constant alpha to c²? Because it relates energy E to mass m and you’re talking in terms of gravitational charge? Like I said, I’m not fond of that, and the point I was making earlier still holds: Einstein asserted that c varies with gravitational potential, which means it isn’t a constant.

The expresion I gave is what one obtains from the weak field limit of Einstein's GR. The non-linear term does indeed vary with gravitational potential; that's the whole point.

That’s sounds like circular reasoning. If you look at a plot of gravitational potential there’s a flat bit at the bottom of the upturned hat. That’s where gravitational potential is lowest and energy density is at a maximum. When you’re in a void at the centre of the earth there’s no discernible gravity because you’re in region where there’s no gradient in energy density. And you can’t tell locally what the energy density is.

In a void at the centre of the earth the field g is so weak it’s undetectable. It’s essentially zero. But the energy density in the space at that region is higher than anywhere else.

You seem to be referring to the potential of a spherically symmetric mass distribution. What you say about it is mostly correct. I'm not sure what you want me to respond to.