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View Full Version : Dark energy, General Relativity, etc - a fun paper for many readers (like publius)!



Nereid
2007-Oct-28, 04:11 PM
"A New Independent Limit on the Cosmological Constant/Dark Energy from the Relativistic Bending of Light by Galaxies and Clusters of Galaxies", by Mustapha Ishak, Wolfgang Rindler, Jason Dossett, Jacob Moldenhauer, Chris Allison (The University of Texas at Dallas); arXiv preprint here (http://arxiv.org/abs/0710.4726).

Yes, that's Wolfgang Rindler, he of the Rindler metric!

The abstract is somewhat dry, but raises a very interesting possibility:
We derive new limits on the value of the cosmological constant, $\Lambda$, based on the Einstein bending of light by systems where the lens is a distant galaxy or a cluster of galaxies. We use an amended lens equation in which the contribution of $\Lambda$ to the Einstein deflection angle is taken into account and use observations of Einstein radii around several lens systems. Interestingly, we find that the contribution of the cosmological constant to the bending angle can be as big as 27% of the magnitude of the first-order term in the deflection angle and a few orders of magnitude larger than the second-order term. We use these observations of bending-angles to derive limits on the value of $\Lambda$ and find them to be competitive with the value determined from cosmology.Gravitational lensing observations as a possible way to independently estimate Λ? Very interesting indeed.

Fortunate
2007-Oct-28, 07:43 PM
Interesting. The paper derives an upper limit on the value of lambda that is competitive with the one inferred from cosmological considerations. Do we have a good lower limit?

publius
2007-Oct-28, 11:48 PM
Nereid,

Wonderful. Yes, indeedy do, I'm enjoying that. Here is the earlier paper by Rindler and Ishak showing that Lamda does indeed contribute to the observered gravitational bending of light:

http://arxiv.org/abs/0709.2948

This is correcting a previous notion that lamba would have no effect on the bending angle. I'm really enjoying this because it is a perfect example of how coordinates can trip you up in our favorite metric theory of gravity. That still gets even the high priests.... This is a classic example of that. *Classic*.

Consider Schwarzschild. What is that? That is the space-time of a spherically symmetric mass, and the limit of that as that mass goes to a "point". It comes from plopping down a mass into flat space-time. What do you get with zero source terms in the EFE? You get Mr. Minkowski's metric, the playground of SR.

So Schwarzschild is basically dropping a "point mass" (massenpunkt as Schwarzschild himself called it) against a flat Minkowski background.

What if the background weren't Minkowski? Well, if you add Lamba to the EFE, and still set the mass source to zero, you get something different from Minkowski. That's the deSitter space-time, an expanding space-time, although it has a *static* form, which blows your mind. That's the simplest "lamba vacuum" you can have. How can expanding space have a static form. That is coordinates for you! That metric simply says that if you drop something, it flies away from you, and the farther it gets, the faster it goes, until it gets to a cosmological horizon thingy. However, it doesn't see anything funny, it just sees you flying away and thinks you hit a cosmological horizon. Your own ruler and clock's notion of space-time is completely static. Stuff just tends to fly away from you in all directions if you drop it. :lol:

Now, what happens when you drop a massenpunkt down in that thing? That's the Schwarzschild-deSitter space-time. Pondering that thing, you'll notice it looks very similiar to the original. The magic metric factor there, 1 - R/r, gets another term on it. -Lr^2/3, where L is lamba.

Now, note that gives a local maximum to clock rates, which then drop off to zero again at some high r. But it still goes to zero at R, just like the original. That means a black hole still has the same event horizon, even in a Lamba vacuum! A black hole remains unperturbed by Lamba (but that doesn't hold for "dark energy" type models that expand "harder than lamba", mind you).

However, its influence is opposed. Get far enough away, right where the magic metric factor is maximum, and you are at a balance point. Stay there, and you stay a constant notion of distance away from the black hole.

Note this is a very different space-time from Schwarzschild alone. That sucker is asymptotically flat. This thing is NOT. Get far enough away, and you hit another horizon, that cosmological one. Schwarzschild coordinates correspond to the ruler and clock of an observer, stationary to the massenpunkt, but infinitely far away.

There ain't no such observer here! So just whose coordinates are these, anyway? :) Well, you can use coordinates that don't correspond to any observer, actually. These coordiantes, this r, theta, phi and t, are sort of imaging an inertial observer in deSitter space-time, giving him some spherical coordinates, then plopping the mass down at his r = 0. Don't hold me to this, because I just know enough to be dangerous, but I think those coordinates will correspond to the local ruler and clock of an observer sitting right at the balance point.

Now, the error everyone made apparently in thinking Lamba didn't matter for gravitational lensing was assuming these coordinates were just like Schwarschild, and applied to someone "very far away". They don't at all.

It turns out that lamda cancelled for null geodesics in a cute way, so the path of light was exactly the same as it is in Schwarszchild. So, the far away observer should see exactly the same thing.

That was wrong, because those far away observers have very different ruler and clocks.

We can thank Rindler for straightening that out, among his many other accomplishments.

-Richard

publius
2007-Oct-29, 01:45 AM
As a sidenote from that (earlier) paper, Rindler shows the effect of lamba on the bending of light by the sun is only 1 part in 10^28, using the currently accepted range for lamba. However, the effect on the advance of the perihelion of Mercury should be "much" better, 1 part in 10^15.

-Richard

publius
2007-Oct-29, 02:10 AM
Sorry folks, that's lambda. Lambda. 'b' followed by 'd'. I know that. It's just apparently very hard for me to type. Let me practice. lambda, lambda, lambda, lambda, lambda.

-Richard

Fortunate
2007-Oct-29, 05:55 PM
Thanks, Nereid and Publius. I "read" the paper cited in post #1 and its prequel, cited by Publius in post #3. Partly because of Publius' edifying remarks, and partly because the papers are rather well-written, I obtained a decent (relative to the time I invested) level of understanding.

dhd40
2007-Oct-29, 09:24 PM
Sorry folks, that's lambda. Lambda. 'b' followed by 'd'. I know that. It's just apparently very hard for me to type. Let me practice. lambda, lambda, lambda, lambda, lambda.

-Richard

I hate this post, because I was just about to teach you the correct spelling. :sad:
On the other hand, I like your posts, although Iīm understanding only 50% of it (maybe 51%) . :)
But thatīs good enough for me to learn, learn, ...

Fortunate
2007-Oct-29, 10:03 PM
Mary had a little lambda,
Fleece was white as snow.

Kaptain K
2007-Oct-29, 11:07 PM
Mary had a bicycle,
Its color red as fire,
Everywhere that Mary went,
The bicycle had a flat tire.

Fortunate
2007-Oct-30, 03:35 AM
Gravitational lensing observations as a possible way to independently estimate Λ?

Nereid knows how to spell "lambda".

publius
2007-Oct-30, 05:11 AM
All right, enough of this horse -- uh, lamb*d*a play, here. Yeah, quit lambda-ing around here. :)


Well, let's lambda around in a way related to the OP. DeSitter space, the empty (devoid of mass-energy) "lambdavacuum", is very much related to the famous metric that bears Rindler's name. Well, it is related to what Rindler's metric is based on/related to itself, I should say.

100 points to who can tell us that relation. Hint: Balloon, embedding space. As you know, expanding space-time is often described as like the surface of a balloon. Well, that can be made rigorous, actually, for a lambavacuum. :)

-Richard

Fortunate
2007-Oct-30, 03:00 PM
I've decided to maintain a judicious silence on some of these matters, a silence born of my total ignorance of the Rindler metric.

Well OK, the "embedding space" part of the hint is intriguing, because in the typical balloon analogy, the embedding space is there just to help create the picture of an expanding balloon. GR views spacetime intrinsically, with no embedding space.

Fortunate
2007-Oct-30, 03:13 PM
Now that I think about it some more, I don't think that spacetime expands, because spacetime includes all past and future times within it. So the balloon at a particular point in its expansion represents space, a cross section of spacetime. Then the balloon analogy is flawed in that it suggests that such a splitting is legitimate. But each point of the embedding space actually represents a point of spacetime, since the balloon will at some point in time contain that point of the embedding space.

Fortunate
2007-Oct-30, 03:31 PM
Well, I Googled "De Sitter Space" and was informed that De Sitter Space can be considered as a submanifold of Minkowski Space of one higher dimension, so I guess we can consider Minkowski space as an embedding space for De Sitter Space.

Fortunate
2007-Oct-30, 04:27 PM
OK, back from a lunch run.

As someone on the balloon travels along a geodesic to another point on the expanding balloon, his or her path in the embedding space traces out a segment of a spiral, always in the same plane, but of increasing radius. But remember that the embedding space corresponds spacetime. I think that the angle between two vectors tangent to the balloon at the same point would remain constant as the ballon expanded.

publius
2007-Oct-30, 05:05 PM
I'll give you 50 points so far. :)

What is the following expression?

(ct)^2 - (x^2 + y^2) = -(c^2/a)^2 = constant. {ETA: squared the right hadn side of that, to make it correct}


This is in 1T, 2D Minkowski. Space is a plane, plus time (for us). What is a "point" (x/y = constant) on that doing?

Hint: hyperbolic motion...proper acceleration

-Richard

publius
2007-Oct-31, 12:28 AM
Ok, what is hyperbolic motion. That's constant proper acceleration motion. Speed only asympotically approaches 'c', but the accelerating observer feels a constant proper acceleration the whole time. It's just the coordinate acceleration of the Minkowski observer goes to zero asymptotically.

(ct)^2 - r^2 = -(c^2/a)^2 describes that curve in the r-t plane, where r is the spatial coordiante.

You will note the points on that world line are a constant, *space-like* norm from the origin, c^2/a. That family of curves, of varying a, which is the proper acceleration, forms the "Rindler family" of observers. The Minkowksi observer there, at r = 0, is at the event horizon, called the Rindler horizon here. All those observers maintain a constant notion of distance between themselves, all feeling a gradient in proper acceleration. See the rocket and rope "paradox". :lol:

So if x^2 + y^2 is r^2, then what does this mean the radius of a circle in the plane is doing?

-Richard

publius
2007-Oct-31, 12:48 AM
Fortunate,

You are quite correct. GR doesn't need an embedding space to describe curvature, and in general, with machinery of differential geometry, one doesn't need embedding spaces. However, they can be useful and even insightful to picture something. That doesn't mean the embedding space is real in any physical sense, but it's just a way of seeing things.

And I hope you'll agree that it gives us a pretty slick way to visualize deSitter space-time. :)

And finally, you're also correct that one shouldn't say space-time is expanding. Give me a big red "-1" there. I get those from to time (but everybody does -- beware those who don't realize how easy it is to get those -1s or worse with this stuff!) :) Change that to "space-time of an expanding universe".

-Richard

Fortunate
2007-Oct-31, 01:52 AM
Thank you, Richard. I enjoy both your enthusiasm and your insight.

Brian

publius
2007-Oct-31, 09:45 PM
Since no one has taken a stab yet, here's what's going on. Let's do this in a Minkowski plane, 1 time dimension, two space dimensions, which we'll call a (1, 3) Minkowksi space. The equation

(x^2 +y^2) - (ct)^2 = (c^2/a)^2

describes a circle in the spatial plane whose radius, r^2 = x^2 + y^2 is accelerating with constant proper acceleration. Each observer on that radius is simply a Rindler observer.

Note we can do that with as many spatial dimensions as we want. In
(1, 3) space we could specify a sphere of accelerating radius. An expanding balloon. But note the rate of expansion is not constant, but is accelerating in hyperbolic, Rindler fashion.

Here's the neat thing. The above equation is a constraint that reduces the total dimensions by 1. It turns out that the subspace formed by this is a (1, N - 1) space that *is a solution to the EFE with positive lambda in (1, N - 1) space*. Lambda is simply proportional to a.

So, taking the expanding balloon surface in (1, 3) space, that would be a lambda expanding universe space for 2D spatial flatlanders living on the surface of the balloon.

The proper time of each co-mover on that surface is exactly the Rindler proper time. That would simply be the clock of a co-mover in the co-moving coordinate system, relative to an observer in the higher dimensional Minkowski space.

That observers see the N - dimensional space as an expanding N sphere. However, just as the Rindler observer's notion of space is something different, so the flatlander's own notion of that "surface" space will be different.

To get a (1, 3) lambavacuum, we simply do that from a (1, 4) Minkowski space and have a hypersphere expand in Rindler fashion. Our notion of space is the expanding "surface volume" of that hypersphere, as we see it using our own coordinates.

That just strikes me as cute and pretty slick. There is a rigorous notion of the "expanding balloon", but it's a bit more complex than you think, and maybe doesn't really mean what you originally hoped it would mean. And that's the story of GR. :)

-Richard

Fortunate
2007-Nov-01, 07:54 PM
Cute and slick. Thank you for the perspicuous explanation.:)

The hints were a lot of fun, too. I didn't devote much time to thinking about
(ct)2 - (x2 + y2) = constant. I thought it would be simpler to just look at (ct)2 - x2 = constant first and then add an extra spatial dimension afterwards, so I wound up staring at a family of hyperbolas. I saw a point accelerate along the right-hand branch of each hyperbola but didn't realize the significance of both branches together (expanding boundary of 1-dimensional disc). I put the y back in, but just saw a frozen figure, not realizing that I was looking at an expanding circle.

I know I haven't grasped the significance of the proper acceleration yet. Google tells me the accelerating observer will "feel" a constant acceleration (I think that's what it told me). Maybe that's related to the constant energy density.

Anyway, thanks again.

publius
2007-Nov-01, 08:24 PM
The proper acceleration turns out to be proportional to Lambda. Higher lambda means a faster accelerating boundary.

And note this is just a cute mathematical picture. In "real" space-time, observers would have to feel a force to accelerate, of course. But co-movers in a lambdavacuum are inertial and don't feel any force. So, from our embedding space perspective, we have to say that somehow they are flatlanders bound to ride the accelerating surface. They are stuck on it and are dragged along.

It is a neat embedding space picture, but we're certainly going outside of the physics of the subspace we've embedded in it. :)



-Richard

Nereid
2007-Nov-08, 09:03 AM
Kayll Lake seems to see things somewhat differently: More on the bending of light ! (http://arxiv.org/abs/0711.0673)
Recently, Rindler and Ishak have argued that the bending of light is, in principle, changed by the presence of a cosmological constant since one must consider not only the null geodesic equation, but also the process of measurement. I agree with the fact that both must be considered. Here, on the basis of the mathematically exact solution to the classical bending problem, and independent of the cosmological constant, I show that the approximate argument found in the vast majority of texts (new and old) for the measured value of the bending of light for a single source is, despite getting a good answer, bogus. In fact, the measured value for a single source is in part the result of the almost perfect cancelation of two terms, one of which is seldom considered. When one considers two sources, this cancelation is of no consequence, and if the sources are opposite with the same associated apsidal distance, the approximate argument gives the rigorously correct answer (up to numerical evaluation), an answer which is unaffected by the presence of a cosmological constant.

Fortunate
2007-Nov-08, 02:56 PM
Interesting that there can be so much discussion even among "experts" over how to obtain the correct value.


Recently, Rindler and Ishak have corrected a long-standing error in the literature concerning the cosmological constant and the bending of light for a single source, an error perpetuated by the author. Here I reexamine the classic subject of the bending of light...and arrive at some further results which should be of wide-spread interest.

I wonder if we have heard the last of this.