[Ma-Dai]

Are we speaking or not about a drive that can go faster than light or not?

Are these observable galaxies moving faster than light or not?

The answer to your above question is both "Yes and no."  Again, and with elaboration having been already provided, faster than light motion does not violate general relativity unless it is faster than light locally.  Such motion does not occur anywhere in the Alcubierre solution as far as anyone has yet been able to demonstrate in a convincing manner.

There's one point in your article I don't quite get:

Superluminal recession is a feature of all expanding cosmological models that are homogeneous and isotropic and therefore obay Hubble's law. This does not contradict special relativity because the superluminal motion does not occur in any observer's inertial frame. All observers measure light locally to be travelling at c and nothing ever overtakes a photon.

Isn't that exactly what would happen with an alcubierre drive? Wouldn't our alcubierre-ship overtake photons travelling in the same direction? Or is there some curvature of space-thingy going on?

To clarify: In my imagination of an inflationary process, everything is moving away from everything else. So even if space is inflating superluminally, nothing could ever possibly overtake something else at a relative velocity higher than c.

[Phantom Hoover]

maybe the alcubierre drive is like a photon snowplough

[Dragon]

Is that possible you're right a second time in a single thread? Well, I dunno (I don't have a PhD in relativistic physics, or at least not yet), but it is predicted that an FTL Alcubierre drive would produce a massive burst of gamma radiation in front of the ship once it stops (something to keep in mind if we ever end up at war with aliens. Or if we ever encounter aliens, for that matter...). It seems to me that it might actually stem from photons "accumulated" in front of the warp bubble. It's probably a huge oversimplification, if it's even remotely on target, but it might be that sort of effect. It can't "fly past" photons it encounters (it'd violate relativity), and they will be "scooped up" by the bubble, that's for sure. Something has to happen to them, it makes sense that they'll travel with the bubble and accumulate in front. And once it's turned off, the space stops being curved, but the photons are still going at c... I can't see how there could not be a huge, multispectral photon flash every time the thing is turned off. It shouldn't damage the ship (dunno what the exact geometry would be, but the photons would probably not be moving towards the ship), but woe to anything that happens to be in front of it.

Isn't that exactly what would happen with an alcubierre drive? Wouldn't our alcubierre-ship overtake photons travelling in the same direction? Or is there some curvature of space-thingy going on?

Your picture of inflationary process is correct. Also, an Alcubierre ship certainly can't "overtake" photons in a traditional sense. Nothing can, in fact, it's inherent in the geometry of the universe. As such, it must warp space in such a way that it doesn't overtake photons, yet it arrives at destination in shorter time than it'd take to traverse that distance at c was it a flat region of spacetime. See the above - it seems consistent with the gamma burst that is predicted to be emitted in front of such a ship once it stops. Remember, it warps space, so everything within the bubble is affected, including light. There's also nothing stopping the light from entering the bubble.

[Meneldil]

as an aside that I hope doesn't derail the thread too much, I've wondered for a while now, what the **** is the difference between a tensor and a matrix?

A matrix is simply a grid of quantities (which can themselves be almost anything -- scalars, vectors, derivatives, etc), and they may represent something physical (like pixel values in an image, or temperature on a map) or they can be totally random entries.

A tensor on the other hand (which can be expressed with a matrix!), is a geometric object which describes quantities that transform under a particular set of rules.  Another popular way of thinking of tensors is that they are 'machines', which accept certain inputs and spew out outputs, again according to a particular set of rules.  This is pretty vague (e.g. what rules?), but basically what they do is allow us to describe things irrespective of choice or change in coordinate system.  Thus their importance in relativity and other branches of physics.

So to be brief, a matrix is a grid of quantities, while a tensor is a grid of quantities which follow a particular structure.

One could rather say tensors don't transform at all, that they are fixed properties of, or extra structure on space-time. When you choose a particular way of looking at a space-time (ie. when you choose a particular set of coordinates) then you can describe them by (in general n-dimensional) arrays of numbers, and when you change the coordinates, those arrays change in a certain way.
It's exactly analogous to how the boiling point of water is either 100°C or 373 K or whatever the hell °F. But you don't say that temperatures are numbers that change in a particular way (depending on the temperature scale you're using), even though technically you could. You think of temperature as being what it is, and the numbers that change in certain way depending on the change of scale are just our way of representing it.
In other words, a representation of a tensor in GR changes when you change the space-time coordinates, and a representation of a temperature changes when you change the "temperature coordinates", but it's probably best to view both as things that just are.


As for the Alcubierre's drive, it may be a solution to the Einstein's field equations, but isn't that kind of a non-brainer? I mean, isn't every metric a solution, depending on what we are willing to take the stress-energy tensor to be? I'm sure the Alcubierre's solution makes infinitely more sense than what you'd get by just plugging in numbers randomly, but just saying it's a solution doesn't seem to mean that much.
In any case, I also don't understand how an Alcubierre's drive you can turn on and off doesn't lead to exactly the same causal consequences the superluminal travel does in SR, but I'm not a physicist and I've never studied GR, so I'll admit I'm talking out of my ass on this one.

[watsisname]

Isn't that exactly what would happen with an alcubierre drive? Wouldn't our alcubierre-ship overtake photons travelling in the same direction?

It depends on if we qualify this question with "traveling in the same direction (parallel) to the ship's trajectory inside the Alcubierre field or outside.  If it is outside the field (flat space-time), then yes, the ship overtakes the photon.  If it is inside the field, in same local space-time geometry as the ship itself, then no, the photon overtakes the ship.

Basically this comes down to an analysis of null geodesics (light-like paths) throughout the Alcubierre solution.  Clark et. al (1999) were the first to do this IIRC, though this paper is not free to view.

Added:
@Dragon:  Your descriptions of radiation effects when the ship stops are qualitatively correct.   There is a horizon-like structure in front of the ship, upon which photons leaving the ship's bow become trapped, and photons moving parallel with the ship but initially outside the field also become trapped.  (But photons moving anti-parellel pass through this horizon and meet the ship, with curvature effect making the destination seem closer than it "really" is.)  At the terminus of its trajectory, these photons would be released as a powerful pulse of EM radiation.

There is also a horizon-like structure behind the ship, on which photons moving parallel to the ship but initially behind and outside the field will never reach.

So, @Phantom Hoover, yes, the Alcubierre ship is sort of like a photon snow-plow, albeit a very strange one.

[watsisname]

As for the Alcubierre's drive, it may be a solution to the Einstein's field equations, but isn't that kind of a non-brainer? I mean, isn't every metric a solution, depending on what we are willing to take the stress-energy tensor to be?

Great question.  Basically you are asking: "For every possible metric of space-time, does there exist a unique curvature and stress-energy tensor which satisfy the set of partial differential equations that are the Einstein Field Equations?" 

I'm fairly sure the answer to this is going to be "no", but I am not certain and don't have a source to check/cite at the moment.  I am also fairly sure that it will not be guaranteed that for all choice of metrics you will preserve the conditions of energy/momentum conservation, Newtonian approximation, etc.  (If anyone happens to know please feel free to add to or correct me.)

To give an analogue of this situation, consider the wave equation.  There are many (an infinite number) of functions psi that you could try to plug into it that will not be solutions.  To check if it is a valid solution, you check if the second derivatives with respect to time equal the second derivatives with respect to position.  If it holds true, your result will look like some sort of wave.  Naturally, solutions to the wave equation are sinusoidal in form because the sine/cosine functions interchange through their own derivatives.

[Phantom Hoover]

As far as metrics go it's not hard to come up with something way too weird to qualify, but that's really just pedantry.

[Meneldil]

To give an analogue of this situation, consider the wave equation.  There are many (an infinite number) of functions psi that you could try to plug into it that will not be solutions.  To check if it is a valid solution, you check if the second derivatives with respect to time equal the second derivatives with respect to position.  If it holds true, your result will look like some sort of wave.  Naturally, solutions to the wave equation are sinusoidal in form because the sine/cosine functions interchange through their own derivatives.

As you describe it, the wave equation is of the form Aψ = Bψ, where A and B are two differential operators, so it's obvious that plugging in random ψ's won't do you any good.
On the other hand, if you take an inhomogeneous equation, like eg. Poissons Δψ = f, you can say that any (sufficiently smooth) ψ solves a Poisson's equation. You just calculate Δψ, and set f = Δψ. Of course, that's exactly the opposite of what you usually need to do: usually you're given f, and then need to reconstruct what ψ is from knowing Δψ.
I thought the Einstein's equation is analogous; disregarding the cosmological constant and with the right choice of units, it's just G = T, where G is the Einstein tensor, and can be computed directly from the metric tensor g, and T is the stress-energy tensor.
So if a metric g_0 is given, you can just calculate what G is and then say g_0 solves the field equations for T = G. This would work if T doesn't itself depend on the metric, which in retrospect sounds unreasonable; the distribution of matter changes in time due tue the effects of among other things gravity, which in GR is nothing but the metric, so... yeah. In my defense, I've seen someone claiming to be a physicist describe it this way (ie. that anything is a solution to GR with the appropriately chosen stress-energy tensor.)

It would be interesting to know then what is known about the space of solutions, at least for a fixed space-time topology, but the answer's probably "hardly anything" besides the fact that it includes some very weird things (eg. Gödel's metric).

[watsisname]

I thought the Einstein's equation is analogous; disregarding the cosmological constant and with the right choice of units, it's just G = T, where G is the Einstein tensor, and can be computed directly from the metric tensor g, and T is the stress-energy tensor.
So if a metric g_0 is given, you can just calculate what G is and then say g_0 solves the field equations for T = G.

Certainly.  I usually write the field equations in the form G + Λg = kT, but since G is technically determined uniquely by g, you could say G = kT if you like.  The problem is there is a lot of mathematical machinery hidden within this relation.  You could make up any kind of metric g, but not be able to compute G from it, or find that a corresponding G may not exist at all.  For this reason you will almost never find this method described or used in the literature.  Generally, people will first compute the Ricci curvature tensor R, or a more simple route if the situation makes it possible.

Exact solutions to the field equations turn out to be a pretty big deal because they are so difficult to find.

[Meneldil]

You could make up any kind of metric g, but not be able to compute G from it, or find that a corresponding G may not exist at all.

I'm not sure I understand how could this be... Riemann tensor certainly exists and contracting from that you get the Ricci tensor, and contract again to get the Ricci scalar, which is what you need to get G. In coordinates all you need to do is calculate the christoffels and then multiply the whole lot of them, which I wouldn't recommend doing by hand, but in the end is just a routine calculation. The reverse (which is what you need to do when actually trying to solve for metric) is of course impossible except in the simplest of cases.
But this is starting to get awfully off topic, sorry for that.

[Mongoose]

If nothing else, I want to thank you guys for reminding me why I didn't continue my physics education into graduate work.

[watsisname]

I'm not sure I understand how could this be... Riemann tensor certainly exists and contracting from that you get the Ricci tensor, and contract again to get the Ricci scalar, which is what you need to get G.

Ah, this is a nuance of G that often gets missed.  I don't know if this will make it more clear or less, but I'll try my best here.  Let's start with what we already know.

We have a pretty good understanding of what the field equations are saying:  mass-energy produces space-time curvature, AKA gravitation.  We describe the source of gravitation with the frame-independent stress-energy tensor T.  We know this cannot be any arbitrary tensor.  For one, it must have zero divergence,
∇·T = 0
which is another way of saying that momentum-energy are conserved.  So far so good.

We place this on the right hand side of the general form of the field equations (with some proportionality constants k).  On the left hand side we have the tensor description of the curvature, G, which we may call the "Einstein tensor".  This gives us
G = kT.
"Curvature = mass/energy"

But is this G the same Einstein tensor that we form by contraction of the Riemann tensor as in the usual analysis you're describing above?  The answer is no, when the field equations are written in this form they are not the same.

We know of course that G must also be a divergence-free tensor, but this is not set by T.  Rather, it is a consequence of the geometry -- it is true for any smooth and Riemannian space-time.  So for a valid solution, we do not have
∇·G = 0,
but rather
∇·G ≡ 0.

Thus G must have the following properties:
-reduces to 0 in special-relativistic limit
-constructed from curvature tensor and metric, and nothing else
-unique from other tensors which could be built from R and g in that it must
(i) be linear in R
(ii) Like T, is 2nd rank, symmetric, and divergence free.

Given a metric, there is, (I'm pretty sure), a unique curvature tensor for it.  The curvature tensor is simply a way of describing how non-flat the metric is.  If it is flat then there is no curvature, R vanishes, and we're back to special relativity.  But I am also pretty sure that for any arbitrary metric, you will not find a G which satisfies the field equations.

Added:  I think anything pertaining to relativity and in particular the field equations and their solutions is both sufficiently on-topic and interesting to be posted here, though granted I am not the OP.

[Meneldil]

Given a metric, there is, (I'm pretty sure), a unique curvature tensor for it.  The curvature tensor is simply a way of describing how non-flat the metric is.  If it is flat then there is no curvature, R vanishes, and we're back to special relativity.  But I am also pretty sure that for any arbitrary metric, you will not find a G which satisfies the field equations.

Oh no, I completely agree, I gave up on that a few posts back
And yes, given a metric, curvature not only exists but is easily if tediously calculated. The Riemann and Ricci tensors are actually defined for a structure on manifolds a lot more general than (pseudo-)Riemann metrics, it's only the Ricci scalar that requires rising and lovering of the indices (to contract the rank (0, 2) Ricci tensor). So yeah, you can always calculate what R - 1/2gR is, but I agree there's no telling if it satisfies the field equations.

As for the graduate work in physics, seeing General Relativity actually sometimes makes me wish I went for that instead of just pure math but i understand that going straight from "vector calculus" to full-blown differential geometry must be daunting. I'm honestly amazed by how much stuff are physicists expected to just pick up along the way.

[watsisname]

No worries, I was just wording things poorly earlier.  Basically I was trying to explain that while you can always compute "G", it is not necessarily the right "G" for G=kT.  But I had simply asserted "G might not exist at all" which, without understanding the difference, sounds really odd.

But anyway, I think that at least in vector calculus things remain fairly easily graspable in a visual sense.  In differential geometry, there may still be a visual interpretation, but it becomes much harder to see, so everything seems quite abstract.  On the physics side it never fails to amaze me how even really high level concepts, like general relativity, can be distilled down to a really simple set of statements.  But of course, to fully grasp those statements so as to make use of them for calculations and predictions, the mathematics and rigor can be daunting.

On a slight tangent to the viability of the Alcubierre solution, I sometimes ponder about how it works in the sense of a toggle-able drive on a spaceship, versus having an "always on" state such as a particle. 

When the ship activates its drive, it is changing the curvature around it, and so GR says the mass/energy distribution must be changing.  But how is it changing?  The energy being output or converted by the drive must come from some kind of reservoir within the ship, so the total mass/energy must be constant.  This makes me think the solution makes the most sense in describing what is effectively a tachyon particle, rather than a spaceship drive.  That is, if this type of localized anti-symmetric curvature was an intrinsic property to a particle, then it would naturally exhibit faster-than-light motion relative to non-local observers.  I see absolutely no reason whatsoever to suppose such a particle actually exists, but it is an interesting way of thinking of it.

The explanation for how it works as a drive given by Alcubierre and others is that the ship's mass/energy reserves are being transformed into (rather exotic) forms which produce this style of curvature, in way consistent with GR's mass --> curvature description and without violating mass conservation.  So I suppose the drive is not so much "producing" this curvature from scratch, but "redistributing" the curvature which was already present all along.  It might make sense then that the weak but infinite-range 1/r² standard gravitational field might become a very strong Alcubierre field if it is condensed to a small volume around the ship, similarly to how the gravitational field of the Earth would become equivalent to a black hole if it were compressed down to the size of a marble.

Granted, I have no idea if these thoughts actually make much sense or if someone who has studied the Alcubierre solution in detail would have words to say.

[Bobboau]

there are only a handful of ways I am aware of that come close to making anything that could in any way possibly be described as negative energy. one of those ways would be the casimir effect, if this could count as the required negative energy then the way you would "make" it would probably be via a meta-material that when energized alters it's molecular structure to have a vast number of nano-scale cavities. Another way to produce negative energies is via a rapidly spinning singularity

[watsisname]

I would say that as far as we can tell, the cosmological constant (dark energy) is a very good candidate for 'negative energy', in the sense that when described as a fluid it exerts negative pressure.  But it has a really tiny value which only becomes relevant on large scales.  How tiny is it?  Let's imagine a sphere of space around the Earth whose boundary lies at the Moon's orbit.  Then let's calculate "how much" dark energy lies within this sphere.  Call this value Λ_moon:



Taking H = Hubble's constant ≈ 68km/s/Mpc
Ω_Λ = ratio of dark energy density to critical density ≈ 0.69
r_moon = moon's semi-major axis (really should have called this a rather than r, but whatever) ≈ 385000km
G = Gravitational constant ≈ 6.67x10^-11m³*kg^-1*s^-2,
ρ_c is the current critical density of the universe. (Kind of fudging this, but not by much).

WolframAlpha tells us this is about one kilogram.  One measly kilogram of dark energy spread through all of the Moon's orbital space!  (By the cubic meter, it's on the order of 10^-26 kilograms, or a few hydrogen atoms equivalence).

If it is possible to actually harvest dark energy (highly dubious), it wouldn't be easy.

[Herra Tohtori]

WolframAlpha tells us this is about one kilogram.  One measly kilogram of dark energy spread through all of the Moon's orbital space!  (By the cubic meter, it's on the order of 10^-26 kilograms, or a few hydrogen atoms equivalence).

Well, if we consider that intergalactic space is commonly cited to have baryonic matter density of less than one hydrogen atom per cubic metre, that's actually quite substantial amount of dark energy. In a cubic metre of intergalactic space, there's more dark energy than baryonic matter energy!

Which, actually, explains why it currently appears as though majority of energy in the universe is in the form of dark energy.


Ah, cosmology and quantum physics, the two branches of science that regularly make me think "go home universe, you're drunk".

there are only a handful of ways I am aware of that come close to making anything that could in any way possibly be described as negative energy.

Gravitational or electric potentials? 

[watsisname]

Which, actually, explains why it currently appears as though majority of energy in the universe is in the form of dark energy.

Precisely.   Normal baryonic mass density is only ~5% of the total density, while dark energy is a whopping ~70%, or 14 times as much.  The other substantial difference between the two is that dark energy seems to be uniformly distributed, while baryonic matter is certainly not.  So on scales like planets, solar systems, and even galaxies, dark energy doesn't amount to much. Over cosmological distances, though, it dominates, and makes the universe expand faster.

Strange universe is strange; nobody expected this result.

That citation for the matter density by the way is actually made through the same kind of method as above: we calculate the critical density (that which would make the universe flat) via some equations, and measure what the total density is relative to this value (which turns out to be very close to 1), then chop it up into the components (baryonic matter, dark matter, dark energy, radiation).

[Dragon]

Yeah, a bit too strange for my liking. One alternate theory I once took a liking to (dunno what become of it, I've heard about it quite some time ago) rejected the Copernican principle and made universe non-uniform. It makes sense to me - afterall, cosmic background radiation isn't uniform, either. Neither was the Big Bang. Perhaps the structure of space might be as well, we know it's quite more malleable than it seems. IIRC, at least a part of this theory was disproved at some point, but dunno which. That said, I've always found it very "untidy" that there's so much stuff that is undetectable, doesn't interact and might exerts negative pressure to boot (under tension without anything to keep it that way?). Remember, a whole lot of this is very theoretical, and subtle changes to certain assumptions can produce different results. I have a feeling that somewhere, we missed something very subtle, but essential.

That said, I haven't heard anything about this theory since then, and I'm not sitting deep enough in cosmology to work on it myself.

Ah, cosmology and quantum physics, the two branches of science that regularly make me think "go home universe, you're drunk".

My thoughts exactly. One lets you do the weirdest things with matter, the other with spacetime itself. That said, it's fun once you get used to it, and is logical in strictly mathematical sense. As I often joke, I prefer working with quanta rather than with people. In both cases you never know what they'll do, but in the former case, you can at least calculate how much you don't know it.

[watsisname]

Ah yes, there have been alternative non-uniform or 'fractal' cosmological models, but they have more or less died out over the last couple decades with observations of the cosmic microwave background (CMB).  The background is not perfectly isotropic, true, but it is isotropic to about one part in 100,000.  Surveys of the distribution of galactic clusters also indicate homogeneity on scales larger than about a hundred million light years.

In other words, the slight anisotropy of the CMB is crucial to understanding the origin of structure and its evolution, but it does not mean that the Copernican principle (or more generally, the cosmological principle, which states the universe is homogenous and isotropic on large scales and naturally leads to Copernican principle) is wrong.  It is rather a nice proof that it is correct.  The universe began in a hot, dense, and extremely uniform state, and structure arises because of the tiny variations from perfect homogeneity collapse gravitationally over time.  Expansion limits the size of collapsing regions to ~100 million light years at present, and so this is the scale in which we see the largest structures.

I also must disagree on the notion of dark energy not being detectable or interacting.  If it didn't interact, we would not know that it is there, and we wouldn't be talking about it (at least not as a scientific discussion).

First, some historical context:
Dark energy is another name for the cosmological constant which appears very naturally in the field equations.  Einstein first used it as a parameter to try to force a static solution out of the equations, because he (and pretty much everyone else at the time) thought the universe was static and eternal.  But the equations show that the total mass content of the universe would cause it to collapse via gravity.  Since evidently it is not collapsing, Einstein used the cosmological constant as a repulsive field which would exactly counter the gravitational collapse and preserve a static universe.  Nice try, but unfortunately, such a solution is precariously unstable, like a pencil balanced on its point. The tiniest fluctuation would cause it to fail.  People pointed this out, and before long Einstein abandoned the idea, especially after Hubble's galactic redshift surveys indicated the universe was indeed not static, but expanding. 

If Einstein had been a bit more bold and trusted the equations he had derived, he might have predicted the non-static nature of the universe, though not necessarily in which direction (expansion or contraction).

The cosmological constant remained in the equations, but as a term most people assumed to be zero, because with available data it wasn't necessary to consider otherwise.  We knew with increasing confidence that the universe was expanding, a model developed which described this as being due to a 'Big Bang', causing the prediction of the CMB, which was later discovered. The equations describe this all very well, with an expanding universe whose expansion rate slowly diminishes due to the mass it contains.  Concentration lay on modeling the future behavior of expansion (is universe open, closed, or flat?), depending on how much mass there is.  Out of curiosity/thoroughness, models were also worked out under the premise of a non-zero value for the cosmological constant, but nobody thought these would actually be valid.

Imagine our surprise then when surveys of distant Type-Ia supernovae indicate that the distant ones are much fainter than expected.  There are several plausible explanations for this, ranging from mundane to mind-blowing.  Maybe intergalactic dust is simply blocking out more of the light than we thought.  Or perhaps some physics is going on which causes supernovae to be more luminous today than in the distant past.  Or, maybe our assumption of an expansion rate governed by mass and a zero-valued cosmological constant is wrong.

We've tested all of these ideas, and the simple, most easily believable ones do not seem to work.  The non-zero cosmological constant best fits the data.  You would be right to think that this in itself would not be quite compelling enough.  But now with the latest generation of CMB mapping satellites (esp. WMAP and Planck), we can very precisely determine the value of the cosmological constant, or 'how much' "dark energy" there is.  This works because, thanks to the modeling work done long before, we know that the relative abundance of the various types of mass/energy (matter + dark matter, dark energy, and radiation) in the universe has observable consequences on the appearance of this background radiation.  We now know that we live in a universe currently dominated by dark energy, and increasingly so as it ages.

That's the history, here's the science:
Dark energy is weird.  Yep, that's the technical term.  When it is described as a fluid, it is one which exerts negative pressure.  That's really weird.  How does a fluid exert negative pressure?  What does that even mean?

If you think in context of particles of matter or radiation zipping around, bombarding the sides of a container, they exert forces due to the change in momentum upon rebounding off the surface.  Force over area is pressure.  There's something else going on, too.  From the field equations, we find that a medium of particles zipping around has a "mass" to it which is more than what you get from simply summing the individual particle rest-masses.  Why?  Because each particle has a momentum, and the momentum factors in to the stress-energy-momentum tensor of the field equations.  You can say it's a consequence of E=mc².  Thus, momentum acts as a source for gravitational field or space-time curvature.  That's interesting.  That means even a photon produces a (very) weak gravitational field.  It also yields a very counter-intuitive result: in cosmology, a uniform field of radiation or moving particles, which exerts pressure, actually decreases the expansion rate, rather than increasing it.     Weeeird.

The established value for the cosmological constant produces an accelerating expansion.  This again follows straight from the field equations.  We can, if we like, choose to describe the cosmological constant as a uniform matter/energy distribution, AKA "dark energy".  In this case it enters into the stress-energy tensor, with a negative contribution to the space-time curvature, a negative pressure/momentum, and, if you want to treat it as particles, they have negative mass. 

Sorry, that was a long post.  Cosmology is weird. 


edit:  Fixed where I said non-zero when I meant to say zero.