Two World-theories (neither one especially stringy)

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  • #26
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Marcus, I just found http://yolanda3.dynalias.org/wb/whoiswb.html [Broken]; check him out.
 
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  • #27
marcus
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Marcus, I just found
http://yolanda3.dynalias.org/wb/whoiswb.html [Broken]
check him out.
-----------
thanks, I will do so at once
strange URL
-----------

with Fahrenheit near zero by Lake Winnebago
you are well situated to appreciate the photo album
of life in the Bahamas. Wolfgang Beirl has a nice
offshore lifestyle.

his real business appears to be modeling financial markets
like the US stock market
his most recent lattice gravity posting (that he gives a link to)
is mid 2003
 
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  • #28
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http://arxiv.org/find/hep-lat/1/au:+Beirl_W/0/1/0/all/0/1

19 simplicial/lattice quantum gravity papers

3 in 1996 but only two since then

Here is Beirl's record of some communications with Lubos Motl
http://yolanda3.dynalias.org/tsm/tsm.html [Broken]
that began with Lubos comment on the recent AJL paper
 
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  • #29
nightcleaner
Marcus, in regard to LQG and DT, you said:
"they are kindred approaches, even though they do not agree about the
discrete spectrum of area and volume (which I think makes it likely that only one can be successful in the long run)"

Marcus could you take a moment to expand on this? I would like to understand more about what is meant by discrete spectrum of area and volume. I think I know what area and volume are. I have a notion of discrete spectrum but need to verify. And then, in what respects do not agree?

Thanks

nc
 
  • #30
nightcleaner
marcus said:
Instead of saying 4D simplex I will say "chunk"
A chunk has 5 points and 5 tetrahedron for its sides.
the easiest way to picture a chunk is to put down a tetrahedron as a base
(like the base of a pyramid, sort of) and then put a new apex point "up in the air" and imagine drawing lines from each point of the tetrahedron base up to the the new point.

here is something you can visualize because it is in ordinary 3D:
take a tetrahedron and put a point in the middle of it and connect the 4 orig points to the centerpoint. Presto you have divided the orig tet
into FOUR tets.

each of the four orig. faces becomes like the base of one of the four new tets and the centerpoint becomes like the common apex
Now to business:

(2,8)
now we are in 4D and we have a spatial tetrahedron in the present and two apexes one up in the future and one down in the past. So we have a CLUSTER OF TWO CHUNKS meeting at a shared tet.

put a centerpoint into that shared tet, dividing it into FOUR tets, and connect each of them to the two (future and past) vertices.
pretty clearly we now have a CLUSTER OF EIGHT CHUNKS.
so that is what is called the move (2,8)
and there is an obvious reverse move (8,2)

(4,6)

there is a 3D thing that is easy to visulize where you start with TWO tets butting together at a shared triangle, like two pyramids base-to-base with one's apex out East and the other's apex out West. And you erase that triangle and draw a line between the east and west apex points and suddenly you see that you have THREE tetrahedrons.

well now in 4D suppose you have two tets in the present, butted together like that, and each connected to an apex up in the future and to an apex down in the past-----so you have a cluster of FOUR chunks

if you do that 3D redivision of the spatial pair of tetrahedrons so you now have 3 tetrahedrons in the present----and then connect each of them up and down to the future and past apexes as before, then you have a
cluster of SIX chunks. that is the (4,6) move and the reverse move is obvious.

The (3,3) and the (2,4) moves involve cluster of fewer chunks. I guess I will take a break here and describe them later.
These are spelled out and depicted around page 23 of hep-th/0105267
Marcus,
You said
"here is something you can visualize because it is in ordinary 3D:
take a tetrahedron and put a point in the middle of it and connect the 4 orig points to the centerpoint. Presto you have divided the orig tet
into FOUR tets."

Are you accounting here for the condition that the new point has to be non-co-spatial with the original 4 of the tetrahedron?

You said
"each of the four orig. faces becomes like the base of one of the four new tets and the centerpoint becomes like the common apex"

The four new tets each have a 2-simpex base, and these four bases share a common 3-space. The apex however is in 4d, and is not really in the common 3-space of the bases. The apex is best thought of as being infinitely removed from the 3-space of the bases, which is to say it is not in the common 3-space of the bases at all. So the lines that join the points of the bases to the apex are infinitely long parallel time-like lines, and the apex itself is not a point, but another 3-space tetrahedron offset an infinite distance, so that it could be represented as a point anywhere in the 3-space of the bases. You chose to represent it as if it were in the center of the original tetrahedron, which is a good choice since it represents each of the infinitly long timelike lines as equal in length in the 3-space, but of course infinity does not equal infinity, so the apparent equality does not hold, but is best remembered as an artifact of the visualization process, much as in a two dimensional drawing of a three dimensional object two points that seem to be close together in the drawing can actually represent points that are far apart in the 3-space. Think of a drawing of a cube. The near corner and the far corner can seem to be in the same place on the drawing, but we mentally recall that they are really in separate planes.

I am trying to follow your analysis of the chunks and moves, but have worked myself into some kind of a cross-eyed cross-legged 4-d stance and need to take time out for sugar and protien to balance the caffeine fix. As the Arnold says, with a steely gleam in his android eye, "I'll be back."

nc

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  • #31
marcus
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in ordinary 2D, on a sheet of paper, draw an equilateral triangle.


then put a point in the middle and connect it to each of the orig. vertices.
this divides the triangle into 3 triangles (still on the orig. 2D piece of paper)
everything is "co-spatial"

do the same with a tetrahedron,
 
  • #32
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nightcleaner said:
Are you accounting here for the condition that the new point has to be non-co-spatial with the original 4 of the tetrahedron?
...
no I am counting on the new point being actually inside the tetrahedron
and so part of the same 3D space that the tetrahedron occupies

[clarification: in this case we are not making a 4simplex, we are
DIVIDING UP an existing tet to make 4 smaller tets, by placing a point in the center]
 
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  • #33
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nightcleaner said:
Marcus, in regard to LQG and DT, you said:
"they are kindred approaches, even though they do not agree about the
discrete spectrum of area and volume (which I think makes it likely that only one can be successful in the long run)"

Marcus could you take a moment to expand on this? I would like to understand more about what is meant by discrete spectrum of area and volume. ..., in what respects do not agree?
...
Here is a brief response to this question. I would be glad if anyone wants to explain it in more detail.
this issue points to a crucial difference between LQG and DT.

DT is fairly new and i do not know of the area and volume operators being studied yet. I suspect that when they are defined they could turn out NOT to have discrete spectrum.

But in LQG associated with any material surface, like a desktop, there is an operator on the hilbert space (the quantum states of geometry) which corresponds to measuring the area of that surface----it is an operator called an "observable" and it has a discrete set of possible values, including a smallest positive value.

the language is a bit technical, even a bit awkward, and gives the impression of more difficulty than there really is.

the essential is that from LQG one has learned to expect that measuring the area will give discrete levels of area, like the energy levels of an atom.

as an atom can be excited and made to go up into a higher energy level (or an electron of the atom can, however you think about it)
so also the gravitational field can be excited and go up into a stage where things have more area, and also more volume.

but the areas and volumes only go up in little "jumps"
(too small to measure with today instruments)

and this is a characteristic of LQG-------indeed in LQG COSMOLOGY even the size of the universe goes up in miniscule "jumps" and at the time of the bigbang this turns out to be significant (even tho the steps are very very small it still matters)

But I have not seen any hints that when they get DT more developed and get area and volume operators, that they will have a discrete set of possible values---a discrete spectrum---and that areas and volumes will increase and decrease in jumps (as the gravitational field changes).
So this could be a serious disagreement between DT and LQG
 
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  • #34
nightcleaner
Hi all

Some more on 4d visuals.

We have seen how the 3d simplex is a tetrahedron. Now we wish to consider the shape of things in the 4th dimension. We have discussed how logical development by geometric principles lead us to think that the 4d simplex will have five points. Moreover, these five points consist of a 3d simplex tetrahedron, and one point which is not in the same 3d space as the tetrahedron. This follows from the condition placed on upper dimensional points that they not be part of the lower dimensional simplex.

One way to model the 4space simplex, which will have five points, is to place the fifth point somewhere in the same 3space as the tetrahedron and then try to remember that it is not really a part of the same 3space. We then draw lines from each point of the tetrahedron through 3space to the new point. One convenient way to do this is to place the point at the center of the tetrahedron, thus dividing the tetrahedron into four 3spaces which are interior to the tetrahedron.

This model may prove useful, but we need to remember that the central point is not allowed to be in the same space as the original tetrahedron. I have suggested that we think of the fifth point as being offset in time, or alternately, offset to another 3space which is at a sufficient distance so that there can be no contact with the original 3 space. This offset distance is, for any practical purpose, infinite. This means that the interior lines to the central point in the tetrahedron are infinite. We could represent this infinity either by a gauge variance or by a space-like curvature.

Infinities are not welcome in physics problems because they lead to divergent conclusions. In the math sense, infinity is not equal to infinity so calculations are nearly impossible, making any theory which relies on infinities non-physical. Most serious physics reasearchers rule out any such theory on the grounds that it cannot describe the physical processes we find around us in this universe. I would like to suggest that we hold on to the tetrahedron with its infinitely removed center for a moment, if for no other reason than it gives us the most symmetric possible 3space model of this 4space system.

Meanwhile, let us return to the idea that the offset is not in space, but is in time. This has the advantage of allowing us to place the fifth point in the same space as the tetrahedron but offset one unit of time, thereby removing the infinities. We can still use our model of the tetrahedron with a central point, with a few modifications.

First, we must keep in mind that this central point is a representative point, and our choice of placement at the center is merely a convenience. The point could equally well be placed anywhere in the 3space of the model, because it is not really in the 3space of the model at all, but is offset by one unit of time. The only limitation to the placement comes from the speed of light, which causes the set of possible placements in the next unit of time to be limited to a three dimensional sphere. Any placement outside the limits of the sphere results in a discontinuity between time units, and perhaps it would be best for now to regard time as continuous.

Now we can take our tetrahedron simplex in 3space, and displace it one unit of time, and regard them side by side, as if they were two tetrahedrons side by side in 4space. We can say that the tetrahedron has not actually moved at all in 3space, so all of its points in the offset space are in the same relationship to each other as they were in the original 3space tetrahedron.

But what relationship do the two tetrahedrons have, in the 3space model, to each other? That is, if we represent the original tetrahedron, call it tet1, and the offset tetrahedron, call it tet2, in the same 3space, do we have any justification for saying that the lines in tet1 are parallel to the lines in tet2? It would be convenient for visualization purposes if we could say that tet2 is not rotated compared to tet1 in the 3space model, but is it justified?

We have to remember that tet2 can be placed anywhere in the 3space model. It could be placed offset to the right of the viewer and then up, and then forward or back, with no preferred position. A sequence of these moves can result in any desired rotation, at any desired location, so we cannot justify the convenient proposition that the two tetrahedrons should have parallel lines in the 3space model

Moreover, tet2 can have any size compared to tet1, within the 3space limits set by the continuity provision as determined by the spacetime ratio, c. Tet2 could be represented as entirely within or entirely outside of tet1, and any size from a single point to the full extension of 3space surrounding tet1. Within the limits of the tet1 3space set by the discrete unit of time, every point has to be considered as equal in terms of representation for the offset.

The model now seems rather blurry, and of little use in visualization. However, we can make some improvements. We can justify some preferred conditions. For example, no matter where we place tet2, and no matter what rotation, there is always a one to one correlation between the four points in tet1 and the four points in tet2. From any apex in tet1, there are four lines leading to tet2. Likewise from any apex in tet2, there are four lines leading to tet1. We can count these lines. There are sixteen lines leading from tet1 to tet2, and sixteen lines leading from tet2 to tet1. Can we say that the sixteen lines 2=>1 are the same sixteen lines 1=>2? No.

We have to remember that there is no preferred orientation. All rotations must be considered. The lines from 2 to 1 are therefore not simple one dimensional lines. They are cones. They start as a single point in the tet of origin, but by the time they arrive at the tet of destination, they are no longer zero dimensional in cross section. Because of this fact, we cannot say that a line from tet1 to tet2 is matched by any line from tet2 to tet1. We are left with the unavoidable conclusion that there are thirty two cones.

These cones are not without structure. They expand from the origin to the offset, and moreover, they are not of consistant internal density. Rather, there is a spectrum of preferred densities within the cone. This results from the fact that the offset tetrahedron can take any rotated position. To draw the cone, we must consider all possible rotations. A moment of consideration will convince you that not all points in three space will be equally represented in these rotations.

The tet2 can be represented, in its own 3space, as unchanged in any way from tet1, except for the offset in time. This identity can be asserted as long as we keep the two 3spaces apart in our mind. The blurring of the model only comes about when we try to represent the two tets as if they were in one 3space. But it is allowed, for example, to indicate the two tets on one sheet of paper, so long as we keep in mind that they are seperated by one unit of time. We can do this by drawing a circle around one of the tets to remind is that it is not in the same space as the other tet. In fact it might be a good idea to draw circles representing their spaces around each of the two tets. Then we could label one circle 3space1 and it contains tet1, and the other circle is 3space2 containing tet2. Now when we draw the lines, we have to draw not from point one tet1 to point one tet2, but from point one tet1 to the entire circle containing tet2. Each resulting cone then represents all four lines from point one to the four apexes of tet2.

Now lets consider all the possible rotations of one tet in 3space.

First we have to choose a point of origin for the rotation. The simplest choice would be the center point of the tetrahedron. This choice gives us a sphere when the tet is rotated in every possible way around it. The sphere is equal density everywhere on its surface, but the interior of the sphere has a density structure.

This density structure is definable from the existance of edge lines and surfaces of the tetrahedron. Consider for example the density of the region close to the center. This density is defined only by lines radial from the center point to the apexes of the tet. Then consider a region near to the interior surface of the sphere formed by the rotated apex points. The density of this region is also defined by the rotation of the radial lines, but in addition has definition provided by the existance of lines between the apexes. These lines are the edges of the tet. When rotated, these lines define a region limited to the difference between the radial distance to the center of the cord defined by two apexes and the radial distance to the apex itself. The center of the cord is closer to the center of the tet (origin of rotation) than is the apex.

This results in a density of definition which is not merely a spherical surface of points equidistant from the center, but is a sphere with a surface that has thickness, an inner and an outer surface with an incremental space defined between them.

But we also have to consider rotating the tet around the apex points. This gives another equal surface density sphere, but one which is larger than the rotation from the center point. It is larger because the distance from the center of the tet in the first rotation is less than the distance of each apex from the other three apexes. The sphere also has its own internal density structure formed in a similar manner to that described above. Then we have to rotate the tet around each of the other apex points.

Now we begin to see the 4d stucture in some detail. There are spheres within spheres, and spheres intersecting spheres. There are unique definable points within this 4d structure, and it has a spectrum of densities in different regions.

In this discussion, I have tried to show the limitations and conditions which can be placed on a three dimensional model of a four dimensional structure. The simplest model in three dimensions, that of a tetrahedron with a central point, is certainly useful, as long as we keep the conditions in mind.

I have also shown that a four dimensional structure is not merely an undifferentiated space, a blur of 3space, but can be shown to have definable geometric points, lines, planes, surfaces, densities, and spectra.

I have shown that mappings from points in one 3space to their corresponding points in another 3space, even when there is a one to one correspondance between the points, can not be taken as one dimensional lines but have to be considered as three dimensional objects (cones) with internal structure.

In conclusion, this discussion has application to the Regge calculus in that it shows that it is not sufficient to model 4space objects in 3space by a dynamic triangulation with variable but discrete side lengths. I suggest that a better fit to physical measurements can be achieved by spectral analysis of rotated 3space objects.

Richard

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  • #35
marcus
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I am still working on the small project of reviewing the 4D Monte Carlo moves. BTW other sets of moves would probably be OK, the moves just have to be simple and easy to program in a computer and ergodic in the sense that if you do them enough it thoroughly explores the space of geometries Geom(M).

Part of this is a re-edit of post #25 to make it clearer. In that post I got halfway thru the list of moves and then got distracted, so i will review that and then try to proceed further with it.

To have fewer syllables to say: Instead of saying 4D simplex I will say "chunk" and instead of tetrahedron I will sometimes say "tet"
A chunk has 5 vertices and 5 sides, which are tets.
the easiest way to picture a chunk is to put down a tetrahedron as a base
(like the base of a pyramid, sort of) and then put a new apex point "up in the air" and imagine drawing lines from each point of the tetrahedron base up to the the new point.

As preliminaries, here is some things you can visualize in ordinary 3D and THEY DONT REQUIRE 4D:
take a tetrahedron and put a point in the middle of it and connect the 4 orig points to the centerpoint. Presto you have divided the orig tet
into FOUR smaller tets.
(each of the four orig. faces becomes like the base of one of the four new tets and the centerpoint becomes like the common apex)

here is another 3D thing where 4D IS NOT REQUIRED. You start with TWO tets butting together at a shared triangle, like two pyramids base-to-base with one's apex out East and the other's apex out West. And you erase that triangle and draw a line between the east and west apex points and suddenly you see that you have THREE tetrahedrons. they all share that eastwest line.

we will use these two maneuvers----one is a purely 3D way to divide ONE tet into FOUR (by adding a point) and the other is a purely 3D way to divide TWO tets into THREE (by adding a line)


Now let's consider some 4D moves:

(2,8)
Suppose we have a CLUSTER OF TWO CHUNKS meeting at a shared tet. Imagine it is a spatial tet in the present plus two apexes one up in the future and one down in the past.

Put a centerpoint into that shared tet, dividing it into FOUR tets.
(as described earlier)
Then connect each of them to the two (future and past) vertices.
We now have a CLUSTER OF EIGHT CHUNKS.
That is what is called the move (2,8)
and there is an obvious reverse move (8,2)

(4,6)

Now suppose you have two spatial tets in the present, butted together at a common triangle. Suppose each is connected to an apex up in the future and to an apex down in the past-----so you have a cluster of FOUR chunks

Purely in 3D, we can redivide the spatial pair of tets to make 3 tets (as described earlier). So we now have 3 tetrahedrons in the present, and we connect each of them up and down to the future and past apexes as before, so that we have have a cluster of SIX chunks. That is the (4,6) move and the reverse move is obvious.

-----------that finishes that, the rest is just a comment------

I still have to review the (3,3) and the (2,4) moves.
All these are spelled out and depicted around page 23 of hep-th/0105267

Maybe it is worth mentioning that the moves discussed already both require 7 points. In the case of the move (2,8) the initial cluster needed only 6 points to define but one had to add a point, and in the other move (4,6) the initial cluster already needed 7 points to define-----5 for the pair of abutting tets in the present, plus two apexes in future and past.

But the moves I still have to describe, namely (2,4) and (3,3),
are simpler in the sense that they take place in the context of just 6 points.
Also these moves are interesting because THEY DONT CHANGE THE PRESENT AT ALL. More is true: they dont change spatial layer triangulation either in the present or in the future. They only change HOW YOU GET FROM ONE LAYER TO THE NEXT.

they just change the timework links that are sandwiched between spacework layers.

so these two remaining types of moves are in a way more simple, but actually I had more struggle visualizing them from the pictures in
http://arxiv.org/hep-th/0105267 [Broken]
maybe this is their fault (the moves really arent simpler) or else my fault, or maybe the picture's fault (maybe better pictures could be made, say by coloring)
 
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  • #36
marcus
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Imagine 6 points, 3 in the present and 3 in the future------from here on we only are dealing with 2 layers at a time so I will choose them to be present and future and not refer to past.

So there are two triangles, call the downstairs triangle 134 and the upstairs triangle 256.

Now in this little venue marked by these 6 points, we can imagine a tet named 3456------this is not a spatial tet, it is timelike! It goes slanting up between layers,

and by joining this tet to apexes 1 and 2, we can make two chunks!

HERE IS WHAT MOVE (2,4) DOES.

It erases tet 3456, and shoots a line from 1 to 2, and the same bulk (which used to be TWO chunks butted together at a shared tet) is now divided up into FOUR chunks all meeting at the common line 12.

The authors write this move like this, where the underline shows what the chunks have in common

13456 + 23456 goes to 12345 +12346 + 12356 + 12456

-----analogy in 3D----
this is analogous to a move we described earlier that you can do entirely in 3D where you have two tets meeting at a triangle and you erase the triangle and shoot a line from the east vertex to the west vertex and then you have three tets meeting at that line.
---------

Now let's see what is left to do. We have dealt with at least one instance of the move (2,4)

-----comment-----
keep remembering the analogy of shuffling a deck of cards.
if you do enough shuffles you will explore all the possible orderings of the deck.
these moves are very simple modifications of a 4D triangulated geometry ("all the geometry is in the gluing") and they are only dealing with one small local cluster of chunks chosen randomly from perhaps zillions.
these moves are like a shuffle so simple that it only swaps two cards or permutes 3 or 4 cards.
but if you do enough of these very simple shuffles then in the end you make a kind of random walk thru the whole space of possibilities.

this is at the core of the MonteCarlo method which the authors have programmed, which explores the 4D geometries of their small universes
(so far at most a third of a million chunks) evolving under the Einstein rules of dynamic geometry------which rules Tullio Regge translated into rules about simplexes.

you might like to check out the animations at Jan Ambjorn website.
they give some of the flavor.
 
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  • #37
marcus
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there is a minor variation on the move (2,4) just described
I mentioned that the set up was 3 points in the present and 3 in the future and it can also be set up with 4 and 2-----for example 4 points in the present and 2 in the future.

the move is written exactly the same way

13456 + 23456 goes to 12345 +12346 + 12356 + 12456

and the same thing happens as described in the previous post.
 
  • #38
godzilla7
hi guys not sure if this is appropriate to this discussion, but seeing as you mentioned the big bang; read an article about the IMAP data being flawed because of polorizations in our solar system, if this is true does it have implications on the amount of matter we have calculated, and if it does would dark matter be irrelevant, kind of a naive question but the article was a little vague can someone clarify the errors and make a guess at the implications :confused:
 
  • #39
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godzilla7 said:
... read an article about the IMAP data being flawed because of polorizations in our solar system, if this is true does it have implications ...
I read an article about that too. I think you mean WMAP.
the first few bumps in the CMB seem "too" alligned with the solar system for that to have been an accident. It could be coincidence, or it could indicate that something we dont know about in the the solar system is acting as a source or sink of microwave and affecting the low-order poles.

My take on it (just my personal reaction) is that it is overblown----the estimates of dark energy and other good cosmological stuff depend on the higher order poles----the smaller bumps that make the CMB skymap all speckly. Cosmologists estimates about the universe dont depend on those few low-order huge bumps. they might be affected by some dust or crud or other unexpected local effect and one would then just factor that out and throw that away and one would still have the essential WMAP picture of the microwave background and all that can be inferred from it.

so if there is something in that coincidence then it just tells us some minor new detail about the solar system that we didnt know, and doesnt affect the picture of the universe at large (guess).
 
  • #40
marcus
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we should try to maintain our focus on quantum gravity in this thread.

quantum gravity means quantum models of the universe
specifically of the GEOMETRY of the universe

the 1915 Gen Rel insight was that gravity equals geometry
ultimately you cannot have a quantum theory of gravity unless you
have a model of the evolving geometry of the universe

oddly rather few approaches to quantum gravity
actually model the universe
or even offer a quantum version of Einstein's 1915 Gen Rel equation
(which describes how the geometry of space evolves, along various possible paths called spacetimes)

some of the most visible lines of theory do not bother to quantize the main equation and do not come up with a model of evolving geometry.

so given this anomalous situation and I want to focus on an approach that DOES do the requisite stuff, namely Lorentzian DT.

In DT ("dynamical triangulations") you have a spacetime manifold M and you have a bigspace of all the possible geometries on M, called Geom(M).
And for these researchers Geom(M) is not just an abstract idea but they are able to get a handle on the possible geometries and express them
and code them as data structures into a computer
and RUN the little mothers
and do various kinds of counts and extract statistics on them and get specs.

so this is a hard-edge hands-on approach to quantizing the evolving geometry of the universe

and the bizarre thing is even tho it looks like an obvious thing to do the main bulk of theorists work on stuff with no connection or relevance to it
 
  • #41
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I want to recall a quote from the most recent DT paper

"The idea to construct a quantum theory of gravity by using Causal Dynamical Triangulations was motivated by the desire to formulate a quantum gravity theory with the correct Lorentzian signature and causal properties [14], and to have a path integral formulation which may be closely related to attempts to quantize the theory canonically...."

this is page 3 of http://arxiv.org/hep-th/0411152 [Broken]
Semiclassical Universe from First Principles
by Ambjorn, Jurkiewicz and Loll

it begins a section called "Observing the bounce"

this points up some unresolved issues for me.
one of the two main aim of developing Lorentzian DT, the authors say, is to get close to Loop Quantum Gravity (the canonical approach to quantizing Gen Rel)

and yet there are conspicuous differences
OF COURSE THE BIG BANG SINGULARITY GOES AWAY IN EITHER CASE
but with Loop it is replaced by a real bounce, where a (possibly very small) contracting phase turns inside out to become an expanding universe.

In DT there is also no singularity but in this case the "bounce" seems to get started from nothing-----you dont see a prior contraction.

the cosmological constant appears naturally in Lorentzian DT and must be positive every computer run they do they choose a value for Lambda . I do not understand this. for them it seems related to volume of their model universe.

another conspicuous difference is that in DT I do not see area and volume observables, and I dont see any indication that they will turn out to have discrete values (when and if constructed)

with DT I see a real direct relation to matter and field theory
because DT already has a kind of lattice that QFT is often defined on.
the only real novelty is that it is not a fixed, pre-arranged, lattice, but instead an evolving one.

Anyway, what is puzzling me right now is that there are important differences from LQG, even though the aim is to get a workable path integral approach that connects with LQG
 
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  • #42
nightcleaner
marcus said:
this is page 3 of http://arxiv.org/hep-th/0411152 [Broken]
Semiclassical Universe from First Principles
by Ambjorn, Jurkiewicz and Loll

it begins a section called "Observing the bounce"

Hi Marcus

I have been thinking about the bounce and was wondering if my insight could be correct, or if it is contradicted by the math. My idea is that the bounce is not really a bounce at all, in the same sense that the event horizon is not really a membrane. As I recall an observer falling through an event horizon doesn't really see any "there" there. No observable thermal barrior or gravitational tidal effect is measurable by the observer because of distortions in spacetime. Basically, the horizon effect is not observed by a free falling observer because all the gauges are distorted along with the spacetime distortion.

Anyway the horizon which appears to be present in embedding diagrams and in the universe as seen by an outside observer is never actually reached by the free-falling observer, just as you can see the earth's horizon quite easily but if you set out to find the horizon you discover that you never acturally get there....because there isn't any "there" there, or maybe more precisely, the there that is there is everywhere, so that even when you go there, you find the there that is there is the same as the there that was there before you went there. Ha!

So if the changeover from the approach to infinity to the approach to unity is a horizon, then there will be no locally observable effect on crossover, no bounce. Bounce, after all, implies a change in acceleration, or delta L over T^3. The observer doesn't feel the third factor in the inverse of time because that dimension of time goes to zero as the observer passes through the limit. I am not sure what to call the limit in LQG. It seems to me that it is quite symmetrical, so we could speak of a concave limit and a convex limit, we could speak of a limit approaching infinity (never gets there because of restrictions on the maxumum value of c, which is L/T) and a limit approaching unity (never gets there because of discrete quantum intervals). These limits are different only to the observer watching the fall, and cannot be determined or felt by the falling observer.

Correct me if I am missing something.

Meanwhile, you seem to have a better grasp of the math than I have. Would it be possible for you to translate into words the terms in equation one from your reference, the partition function for Quantum gravity? I mean a literal translation of the formula into English. If it would not be too much trouble for you to do so, I am sure it would help me understand, and perhaps help others also.

I sat in one of Brian Greene's topology classes and was very excited to find that I could understand the formulas he was writing on the board, because as he wrote them he also spoke the meaning of the symbols. For some reason this made an incredible difference to me, and gave me renewed confidence that i could understand the math if only someone would walk me through it a few times. I have been longing desperately for a return to that fluidity of understanding.

When you speak of dynamics in the spacetime structure, as if spacetime itself changes over time in a manner of a history, or of a path integral, how many dimensions are you counting? I guess i could ask, what order equation? For example, velocity is length divided by time, L/T, a first order equation, but acceleration is lenght over time squared, L/T^2, a second order equation. Spacetime in the Minkowski-Einstein sense is four dimensions of spacetime equivalence, in which case L^3/T, a third order equation? So if we take that as changing over time, are we talking of a mathematical expression involving something like L^3/T^2?

I am feeling confused by this and hope you can help me get clear.

Thanks,

nc

484
 
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  • #43
Kea
859
0
QG Partition Function

nightcleaner said:
Would it be possible for you to translate into words the terms in equation one from your reference, the partition function for Quantum gravity? I mean a literal translation of the formula into English. If it would not be too much trouble for you to do so, I am sure it would help me understand, and perhaps help others also.

When you speak of dynamics in the spacetime structure, as if spacetime itself changes over time in a manner of a history, or of a path integral, how many dimensions are you counting?
Richard

Your physical intuition is much better than many people I know with
the mathematical expertise.

The first equation in

http://arxiv.org/abs/hep-th/0411152

is [tex] Z ( \Lambda , G ) = \int \mathcal{D} [g] e^{i S [g]}
\hspace{4mm} S[g] = \frac{1}{G} \int \textrm{dx}^{4}
\sqrt{| \textrm{det} g |} (R - 2 \Lambda) [/tex]

[itex] \Lambda [/itex] is the cosmological constant. [itex] G [/itex]
is Newton's constant. The fancy [itex] \mathcal{D} [g] [/itex] means
that the integral is over some ridiculously huge space of fields; in
this case the metrics [itex] g [/itex].

The [itex] e^{i S [g]} [/itex] is the 'weight' for the contribution
from any given field. [itex] S [/itex] is the action functional, an
integral over a four dimensional manifold. The dimension could be
altered.

Under the square root
is the determinant of [itex] g [/itex], which recall at a given point
is like a matrix. [itex] R [/itex] is the Ricci scalar, defined by
[tex] R = g^{\mu \nu} R_{\mu \nu} [/tex] where one uses the Einstein
convention that one sums over matching indices, which range from
1 to 4. Here [itex] R_{\mu \nu} [/itex] is the Ricci tensor, defined by
[tex] R_{\mu \nu} = R^{\lambda}_{\mu \lambda \nu} [/tex] in terms
of the Riemann tensor
[tex] R^{\rho}_{\sigma \mu \nu} = \partial_{\mu}
\Gamma^{\rho}_{\nu \sigma} - \partial_{\nu}
\Gamma^{\rho}_{\mu \sigma} + \Gamma^{\rho}_{\mu \lambda}
\Gamma^{\lambda}_{\nu \sigma} - \Gamma^{\rho}_{\nu \lambda}
\Gamma^{\lambda}_{\mu \sigma}[/tex]
using the Christoffel symbol
[tex] \Gamma^{\lambda}_{\mu \nu} = \frac{1}{2}
g^{\lambda \sigma} ( \partial_{\mu} g_{\nu \sigma} +
\partial_{\nu} g_{\sigma \mu} - \partial_{\sigma} g_{\mu \nu}) [/tex]

Hope this is a little helpful. As you will see, this mathematics falls far short
from being a satisfactory description of 'cosmic bounces', which is
a term combining the 'bounce' of CT (not a great idea as you say)
and the 'cosmic duality' associated to Strings or Cosmic Galois groups
and other fun things.

Regards
Kea

:smile:
 
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  • #44
nightcleaner
Hi Kea. Thank you for the kind words.

So we might say something like "The partition function for quantum gravity in a universe with a given cosmological constant and gravitational constant [tex]\zeta_(\lamda G)[/tex] is the integral over the metrics (g) multiplied by the weight factor expressed as the natural log of the action functional for any given field, where the action functional is the inverse of the gravitational constant times the integral in four dimensions of the square root of the absolute value of the determinants of (g) times the ......ok I'm lost.

Is that gamma the inverse of the hyperbolic tangent function? I ran into it once before and came out with the right table of numbers after some coaching.

Anyway I am grateful for your exposition and am still working on it. Just to let you know I am here paying attention.

thanks,

nc

573
 
  • #45
Kea
859
0
nightcleaner said:
nc

573
What are the numbers?

[itex] \Gamma [/itex] hasn't anything to do with such functions. It
is the Christoffel symbol associated to a connection, which we need
to understand how to move about the manifold, and to see the
'curvature'.

Try this webite:
Carroll's lecture notes on General Relativity

http://pancake.uchicago.edu/~carroll/notes/

Kea
 
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  • #46
nightcleaner
The numbers are pure vanity. Silly, really, but that is how many views there were when i posted. So now there are 608, and i know that there are maybe as many as thirty people checking in.

I will check out the Christoffel symbol. Thanks.

Later..... I did check out the link, but my browser didn't want to read that file type. Anyway I looked it up in Wikipedia and amazing coincidence, it is the same topic Brian Greene lectured on in his topology class the day I was there. And it might even be the same gamma I remembered from my work with DW on the relitivity board. I noticed on Wiki, although I have not had time to study it, that gamma deals with tangential vectors. Tanh would just be the same vector under relitivity, I guess.

nc
 
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  • #47
Kea
859
0
You're quite right about tanh, of course. Are you sure it wasn't a
course in Differential Geometry?
 
  • #48
223
1
Kea said:
Are you sure it wasn't a
course in Differential Geometry?
No it was an undergrad topology class.
 
  • #49
Kea
859
0
Shoshana

Shoshana said:
No it was an undergrad topology class.
You're lucky. I never had a proper topology course as an
undergrad.

So - do you know Nightcleaner?
 
  • #50
223
1
Kea said:
You're lucky. I never had a proper topology course as an
undergrad.

So - do you know Nightcleaner?
Correct. It is difficult to find a topology course at many Universities, but Columbia University offers undergrad topology. My first topology professor at Columbia University was Michael Thaddeus. Amazing speaker!

Yes I am of good fortune to know Richard. He is very fine and uniquely talented.
 

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