Redshift z

quantumtopology · June 1, 2010

The latter is what Hubble did well into the 1950's.

I find highly ironic that the one person that the scientific stablishment has sanctified and turned into the bastion of expansion was so doubtful to this interpretation til the end of his days. Certainly a clever and most peculiar man.

I don't think we need to go that far. :rolleyes:

I don't know, maybe not yet, and that would deserve a new thread :rotfl:

modest · June 1, 2010

The diagram you posted is not a good example of what is happening in the real world. The idea here is not to make a projection onto a flat surface, which would end up looking like an illustration by M. C. Escher of a hyperbolic sphere, where distances appear to become smaller with increasing distance from the origin (at the center).

This,

is just a spacetime diagram of a low-density universe. I pulled it off of Ned Wright's cosmology tutorial. Nothing is really being projected. The green dots are where the world lines of galaxies intersect our past light cone.

The physical interpretation is very straightforward, so we should focus on that.

You say that you're talking about look-back time. That is the distance measure you're talking about. How do you measure it? A simple question. You look at a galaxy through a telescope. You measure its brightness and redshift and angular diameter and whatever else. How exactly does looking at that galaxy tell you how long light took to get from it to you?

Modest , I've seen the thread you link; interesting , if a little confusing for me ,though.
The way I see it the Kashlinsky work is a new problem for L-CDM model, and as that is being fought by many cosmologists like ned wright and others that try to dismiss it quickly.

I've answered you here, if that's ok.

~modest

coldcreation · June 1, 2010

This, [snip] is just a spacetime diagram of a low-density universe. I pulled it off of Ned Wright's cosmology tutorial. Nothing is really being projected. The green dots are where the world lines of galaxies intersect our past light cone.

The reason why there is still a confusion regarding hyperbolicity was explained in this post.

In brief, once again (and I know it's not easy), when you project orthographically a hyperbolic surface on to a flat plane, of the polar coordinate type, you end up with a projection that resembles Escher's famous hyperbolic sphere drawings, where the outer boundary of the sphere is at infinity, and all the figures are actually the same size (though they are drawn smaller towards the edges). That is because of a process called 'parallel projection' (the view direction is orthogonal to the projection plane).

The illustrations I am proposing are not orthographic projections, in that they are not parallel projections, i.e., the view direction is not orthogonal to the projection plane, and infinity is where it's supposed to be: out of sight, far, far, away. :shrug:

The view direction is from the origin, from the center looking outwards onto a polar coordinate system that spans 360 degrees and extends to the visual horizon (again, not to infinity). These are reduced dimension coordinate systems in which each point on a plane is determined by a distance from a fixed point (the rest-frame of any observer, at the origin). Note, the origin is the only rest frame present in the manifold of each diagram. All other points are in the past relative to the observer. See the illustration below:

Figure 4A

Note: This diagram is virtually the same (with the same characteristics) as Figure A above, though I've added a little perspective, an oblique angle, and some effects, just for fun. :)

Figure 4A represents the visible universe, i.e., it spans from the observer to the horizon.

The observer is at the origin of the polar grid/coordinate system. She looks outwards in the past light-cone, in any direction. What she sees are photons, and with a variety of methods, mentioned above (using the rest-frame UV spectral energy distribution, flux-averaging, brightness, lightcurve shape parameter(s), colors in the supernova rest-frame, angular diameter, comparing apparent and abolute magnitudes of SNe Ia, and for convenience distances can measured in km s-1 according to redshift and absolute magnitudes, even in a non-expanding models. Too, the expanding photosphere method of core-collapse SNe can be used with the calibration of the plateau luminosity to determine relative distances. Not to mention, again, the use of Cepheids for distance determinations.

The physical interpretation of Figure A4 is straightforward, as it directly relates to empirical observations. This is virtually what you would have if you flattened out a hyperbolic paraboloid. The understanding of such is by far more intuitive physically than Escher's or Wright's projection (even though they are practically identical when interpreted correctly). So we should focus on this type of manifold.

You say that you're talking about look-back time. That is the distance measure you're talking about. How do you measure it? A simple question. You look at a galaxy through a telescope. You measure its brightness and redshift and angular diameter and whatever else. How exactly does looking at that galaxy tell you how long light took to get from it to you?

Fair question: We know from relativity that the speed of light is independent of the velocity of the source. (the speed of light is the same for all observers), i.e., the speed of light is constant in any inertial reference frame. The invariance of the speed of light, is enough to assure that the coordinate transformations are the Lorentz transformations. The speed of light must be independent of the reference frame (of the velocity of the light source).

Because of the velocity of light constancy, we can measure light-travel distance: the speed of light times the cosmological time interval, i.e., integral of c dt, while the comoving distance is the integral of c dt /a( t ), or, in another way, S = D/T (Speed = Distance/Time).

So by measuring light travel distance (the light travel time times the speed of light) the 'comoving distance' (the distance between two points measured along a path defined at the present cosmological time) or the angular diameter distance should be equal.

Another way to measure distance is to use 'clocks' with known rates (such as pulsars, rotating neutron stars that periodically emit radio pulses), see what the speed of light predicts for observations of these rates, then compare predictions with the data.

This is why I say that observations could be used to determine the degree of curvature (independent of expansion that is). The deviations from linearity would emerge as comparisons are made between the expected light travel time, or light travel distance, measured against time/distance predictions in a flat, Euclidean universe.

CC

modest · June 2, 2010

In brief, once again (and I know it's not easy), when you project orthographically a hyperbolic surface on to a flat plane, of the polar coordinate type, you end up with a projection that resembles Escher's famous hyperbolic sphere drawings

No, I think Escher's drawing would not be orthogonal, it'd be stereographic, like this. An orthogonal projection of a hyperbola would be your diagram A.

If you're talking about lookback time then:

the green dots would be projected onto the y axis (the time axis) orthogonally—which is just another way of saying that the y axis measures time. This would correspond to diagram C. I think we're over-complicating this tangent (get it... "tangent" :shrug:). Sorry, I'm cracking myself up ;)

Note: This diagram is virtually the same (with the same characteristics as Figure A above), though I've added a little perspective, an oblique angle, and some effects, just for fun. :)

Beautiful image. I've gotta give ya props.

Perhaps we could clarify things if I ask: the number of galaxies along a radial line between two outer concentric circles is more than, less than, or the same as the number of galaxies along a radial line between two inner concentric rings (assuming the physical size of a galaxy as measured by someone in the galaxy is the same as all others).

We know from relativity that the speed of light is independent of the velocity of the source. (the speed of light is the same for all observers),

But, that is not true in general relativity.

In the second place our result shows that, according to the general theory of relativity, the law of the constancy of the velocity of light in vacum, which constitutes one of the two fundamental assumptions in the special theory of relativity and to which we have already frequently referred, cannot claim any unlimited validity. A curvature of rays of light can only take place when the velocity of propagation of light varies with position. Now we might think that as a consequence of this, the special theory of relativity and with it the whole theory of relativity would be laid in the dust. But in reality this is not the case. We can only conclude that the special theory of relativity cannot claim an unlimited domain of validity ; its results hold only so long as we are able to disregard the influences of gravitational fields on the phenomena (e.g. of light).

Albert Einstein—The General Theory of Relativity—Chapter 22

Where gravitational fields are present the speed of light is not independent of the source.

Because of the velocity of light constancy, we can measure light-travel distance: the speed of light times the cosmological time interval, i.e., integral of c dt, while the comoving distance is the integral of c dt /a( t ), or, in another way, S = D/T (Speed = Distance/Time).

The comoving distance is NOT the light-travel distance. If you want the light-travel distance then the question remains: how do you know by looking at a galaxy for how long the light traveled? If you want the comoving distance then how do you know the cosmological time that the light was emitted and the scale factor at that time?

In essence: by what exact process can you look at a galaxy through a telescope, make measurements, and decide where to put it on your diagram? How do you know the look-back time or the light-travel time distance.

So by measuring light travel distance (the light travel time times the speed of light) the 'comoving distance' (the distance between two points measured along a path defined at the present cosmological time) or the angular diameter distance should be equal.

The only one of the things you just mentioned which can actually be directly measured is the angular diameter. The relationship between angular diameter and angular diameter distance depends on the shape, or the curvature, of the manifold (spherical, flat, or hyperbolic). So, you cannot look through a telescope and decide how far a galaxy is by measuring its diameter without assuming the curvature of space.

The same is true of the brightness and redshift.

I think this gets at the heart of things. You presented diagrams A, B, and C saying that they represent observation alone (model-independent so to speak). But, I really think we need to decide what exact observation, or method of observation, is meant to end up looking like one of those diagrams when done.

Another way to measure distance is to use 'clocks' with known rates (such as pulsars, rotating neutron stars that periodically emit radio pulses), see what the speed of light predicts for observations of these rates, then compare predictions with the data.

But, look-back time is not proportional to time dilation (or, at least, there is no reason to assume it is). Time dilation is proportional to redshift, but redshift is already directly observable and doesn't give look-back time.

This is why I say that observations could be used to determine the degree of curvature (independent of expansion that is). The deviations from linearity would emerge as comparisons are made between the expected light travel time, or light travel distance, measured against time/distance predictions in a flat, Euclidean universe.

But, that is not model independent. You can measure the brightness and redshift (for example) of a galaxy and that does not directly tell you the curvature. There are multiple models with different curvatures which would give the same redshift / brightness ratio for a single galaxy.

If the distance between concentric circles in your diagram is the difference in look-back time then it remains unclear how exactly you want to measure that value in order to plot it without first assuming a particular model or assuming curvature (which is what you're trying to determine and would therefore be circular).

I really think, before we discuss the implications of these deviations from linearity we need to answer this first. How does one use a telescope or other instruments to plot data for your diagram—described well enough that we can do it with some real data?

~modest

coldcreation · June 2, 2010

No, I think Escher's drawing would not be orthogonal, it'd be stereographic, like this. An orthogonal projection of a hyperbola would be your diagram A.

If you're talking about lookback time then: [snip] the green dots would be projected onto the y axis (the time axis) orthogonally—which is just another way of saying that the y axis measures time. This would correspond to diagram A. I think we're over-complicating this tangent (get it... "tangent" :shrug:). Sorry, I'm cracking myself up ;)

I just edited my last post then saw this one. So I'll post my edit here instead.

The hyperbolic projection from Wright's tutorial is most certainly a projection. The physical interpretation is not very straightforward, since it gives those unfamiliar with non-Euclidean geometry the false impression that things become (or appear) closer with increasing distance away from an observer, when in actual fact it would be the exact opposite.

That drawing is a projective model for two-dimensional hyperbolic space. The image plane is at the pole of one sheet of the surface and the point of view is where the asymptotes meet. Note, the projection near the pole is practically undistorted, but apparent 'shrinkage' increases as the line of sight tends further up the hyperbola. The entire hyperbola is projected onto finite Euclidean open line segments, tangent to the pole. The asymptotes of the hyperbola are the boundaries of the line segment and cross at the eye point.

Objects drawn on the hyperbola itself (in this case, green) are projected to the line segments. The projections of green objects exactly at the pole are undistorted (in the Euclidean sense), but objects far from the pole project to a vanishingly small section of the line segment (at infinity). Objects in the projection that translate along the hyperbola will appear to grow to a maximum when at the pole (where the observer would be) and then shrink after translating infinitely up the hyperbola. At those points their projection will be infinitely close to the line segment border (again at infinity).

The Escher drawings (though not the same type of projection) are similar in that they have infinity on the edge, right there in full view, kind of...:)

I'll be back for the rest... :wine break:

CC

coldcreation · June 2, 2010

Hey modest, I just caught this.

Earlier, you wrote:

What you are talking about is the surface of the past lightcone—the green dots. The problem is that such a representation is not hyperbolic. It translates to C on your diagram

And you now appear to be saying that:

An orthogonal projection of a hyperbola would be your diagram A.

Have you switched from C to A?

If so, for what reason?

CC

modest · June 2, 2010

I just edited my last post then saw this one. So I'll post my edit here instead.

I also edited mine. I wrote "diagram A" when I meant C.

[edit]I see you caught that. I didn't notice it until I read your second to last post :shrug: [/edit]

Spacetime diagrams are explained: Space-Time Diagrams

I'd really like to first nail down what physical observations are represented in your diagram.

~modest

modest · June 2, 2010

I'm sorry, this is correct:

An orthogonal projection of a hyperbola would be your diagram A.

this was not correct:

the green dots would be projected onto the y axis (the time axis) orthogonally—which is just another way of saying that the y axis measures time. This would correspond to diagram A.

That's the one I changed. It should have been "diagram C"

An orthogonal projection of a hyperbola is definitely your diagram A. An orthogonal projection of the green dots on Ned Wright's spacetime diagram onto the y axis would be your diagram C.

~modest

coldcreation · June 2, 2010

I'd really like to first nail down what physical observations are represented in your diagram.

Excellent.

However, just before we proceed on that route (a very important consideration indeed), I'd like to make it perfectly clear what we are looking at. I think it is clear for you now, but for anyone else following or joining us it may not be.

In contrast then to the hyperbolic projection you posted above (from the Ned Wright tutorial) which is a cross section through the light cone, the image below has the observer looking through the actual continuum (the past light cone). Figure A through Figure 4A above are simply the curved line flatted out. The result, rather than seeing actual curvature, is that spatial and temporal increments (as measured from the observer's rest-frame) appear to increase nonlinearly, are dilated, respectively, in the look-back time with increasing distance. In effect, diagrams A through 4A are cross sections of the actual universe, set on a plane for simplified viewing. We can very easily, and intuitively, project the plane into a four-dimensional continuous manifold which would represent real-world observations (since the observer could be considered centered on spherical shells corresponding to the 'concentric' circles of Figure 4A).

Figure 5A

The hyperbolically curved line in Figure 5A is the image plane (represented in Figure 4A above). The observer is at the origin (the point where incoming rays converge). This is a reduced dimension illustration of a hyperbolically curved general relativistic spacetime manifold as viewed from any observer, and is consistent with, and has the same characteristic features as Figure A, 1A, 2A, 3A and 4A.

Now I will proceed to answer (in a forthcoming post) your query to nail down how physical observations are represented in the Figure A series. i.e., "How does one use a telescope or other instruments to plot data for your diagram?"

CC

modest · June 2, 2010

I'm sorry, CC. Nothing in the above post makes sense to me. I don't know what the image is supposed to represent. If the straight lines are light rays then the vertical axis is not time. If the vertical axis is not time then I have no idea what it would represent.

I don't think we're going to get anywhere until you can label a diagram and explain what physical measurements it corresponds to.

~modest

EDIT---> Here is another good link for spacetime diagrams: http://physics.syr.edu/courses/modules/LIGHTCONE/events.html

coldcreation · June 2, 2010

I'm sorry, CC. Nothing in the above post makes sense to me. I don't know what the image is supposed to represent. If the straight lines are light rays then the vertical axis is not time. If the vertical axis is not time then I have no idea what it would represent.

In the diagram you linked here, there is a cut plane through the past light cone. That is why spatiotemporal increments and intervals would appear to converge with increasing distance from the origin O.

The increments of the hyperboloid model (the curved line) shown in Figure 5A are preserved when flattened out into Figure 4A, and would remain consistent with real-world observations (at the telescope), since the linear transformation preserves the spacetime increments and intervals of the manifold.

The projective model shown as slice, or an "image plane," in your link above would not be consistent with empirical observations, as the curvature would not be preserved spatiotemporaly (since it cuts the manifold artificially resulting in false sense of isometry that tend to infinity: it is a projective linear isomorphism). Sure, every light ray projected from the hyperbola to the 'image plane' (at the crossing of the asymptotes) will fall on this parametrized Euclidean plane, but there is no physical reason why the plane needs to be there in the first place; giving the false impression of a closed spherical system.

That does not mean the projection is invalid (it is certainly a valid approach), it simply means that there is a better way to visualize 4-D hyperbolic spacetime curvature in reduced dimensions. As you can see from Figure 5D, it does not imply the visual 'shrinkage' of the projected objects with distance. Nor does it imply spatial increments become smaller with distance. Nor for that matter would Figure 5A imply that parallel line converge with distance. But that is exactly what 'appears' to happen in your link (as it appears to happen with Escher's hyperbolic spheres; based on the Poincaré model, and the Klein model) when the cut (the projection) is made through the surface.

In Figure 5A, that line (or plane) does not cut through the past light cone to arrive at a polar coordinate system of the type Figure 4A. Instead, the aim is to flatten the curved line in order to visualize the same spacetime geometry without distortion due to the cut. Figure 5A simply lets the observer look out into all directions without that artificial cut. There is no reason why that cut should be straight. It should be a curve.

That is the only difference between the two illustrations, aside from some artistic license. :thumbs_up The vertical axis along with the other lines (tangent vectors) that converge on the origin still represents time (the look-back time). The time is determined by how long it takes for a photon to arrive at the origin, relative to the constancy of the speed of light. If there is a deviation from the Minkowski (or special relativistic) expectation, the it can either be postulated that the universe is expanding, or the manifold is hyperbolically curved. One could not assume the photon is actually slowing down with distance, since relative to any observer along the wave-front path the speed of the photon (locally) will always be c. That would simply be time dilation in the look-back time (for both models).

Another simple way to visualize this, is to take a hyperbolic paraboloid (a saddle shape) and flatten it out. The edges would stretch outwards onto a plane, departing from the linearity near the outer edge, while towards the origin there would be less distortion, all the way to the origin where there would be virtually no deviation from linearity. That is why the local universe would appear quasi-Euclidean.

Take a piece of Play-Doh, or clay, and flatten it out to form a circular shape. Now, bend the edges both upwards and downwards to form a hyperbolic paraboloid. You will see that the outer edge (tending towards the interior) will have to be stretched, whereas the center will not.

All Figure A4 does is flatten a hyperbolic paraboloid, instead of cutting a plane through a hyperbolic surface. The pseudosphere is an example of an isometric embedding of a manifold with constant curvature (an increasing departure from linearity the greater the distance from the observer). Figure 4A is an important example to consider when trying to visualize a hyperbolic space. Intuitively, manifolds with hyperbolic curvature are difficult to fit inside a Euclidean space because volume grows too fast (exponentially) with distance from O. You can see that in Figure 4A, but the isometries of the space are clear enough (represented by 'concentric' circles, or spherical shells and in light year distances relative to O).

Notice what happens to the angles of a cosmic triangle on this manifold below. Compare the sum of the angles with those of identical triangles superimposed onto a Euclidean manifold (e.g., figure B above), and then on a manifold of the type figure C above. What conclusion would you arrive at, i.e., which manifold is geometrically hyperbolic and which is spherical?

Figure Y

I don't think we're going to get anywhere until you can label a diagram and explain what physical measurements it corresponds to.

I don't think we're going to get anywhere until it is understood what is represented in figure 4A. How to go about applying physical measurements will follow.

CC

modest · June 2, 2010

In the diagram you linked here, there is a cut plane through the past light cone.

There is no past lightcone in that image. That image,

I posted only to correct your statement that Escher's drawing was an orthogonal projection. Escher's drawing is a stereographic projection like the method shown in the image above.

A past lightcone is the surface between the blue and white area in this spacetime diagram:

That is the only difference between the two illustrations, aside from some artistic license. :thumbs_up The vertical axis along with the other lines (tangent vectors) that converge on the origin still represents time (the look-back time).

Both the distance along the vertical axis and the distance along the Hyperbola are lookback time. It makes no sense.

In a spacetime diagram the vertical axis is the time that a clock measures and the horizontal axis is the distance that a ruler measures. This is a spacetime diagram of a low density universe:

Vertical distance is the time that a clock moving along the center worldline measures. Horizontal distance is the distance that a ruler moving along the center worldline measures. The red lines are the past lightcone of an observer at the apex of that lightcone. The hyperbolas are slices of constant cosmic time—which is to say that any clock along any one of those lines will have measured the same amount of time since the big bang (or any other agreed upon cosmic event).

The look-back time is just the vertical distance between events along the lightcone (the vertical distance between green dots). A spacetime diagram therefore represents spacetime. Not for a lack of trying, but I don't know what your latest image represents.

~modest

coldcreation · June 3, 2010

I posted only to correct your statement that Escher's drawing was an orthogonal projection. Escher's drawing is a stereographic projection like the method shown in the image above.

Which drawing are you referring to?

You should read this post, if you haven't already.

Escher's drawings that I refer to are designed after the Poincaré hyperbolic disk; a two-dimensional space having hyperbolic geometry, such as this one: Circle Limit IV (Heaven and Hell), 1960

Look at the Poincaré Hyperbolic Disk. Clearly, it is an orthographic projection, just like Escher's famous hyperbolic drawings. The orthographic projection is a projection from infinity that preserves neither area nor angle.

The Poincaré disk is a model for hyperbolic geometry in which a line is represented as an arc of a circle whose ends are perpendicular to the disk's boundary (and diameters are also permitted). Two arcs which do not meet correspond to parallel rays, arcs which meet orthogonally correspond to perpendicular lines, and arcs which meet on the boundary are a pair of limits rays. The illustration above shows a hyperbolic tessellation similar to M. C. Escher's Circle Limit IV (Heaven and Hell) (Trott 1999, pp. 10 and 83).

Again, this type of projection serves little or no purpose for the present discussion about redshift z. Though interesting they are. :thumbs_up

Not for a lack of trying, but I don't know what your latest image represents.

Figure A, and many of the other diagrams I've drawn (including the latest one, Figure Y) are polar coordinate systems: two-dimensional coordinate systems in which each point on the plane is determined by a distance from a fixed point (the origin; analogous to the rest-frame of the observer) and an angle from a fixed direction.

What exactly do you not understand? It's a cross section of the visible universe, as if you drew a plane through it, and looked at it form above, from the pole.

Did you calculate the sum of the angles of the triangle in Figure Y yet?

CC

modest · June 3, 2010

Escher's drawing is a stereographic projection like the method shown in the image above.

Which drawing are you referring to?

Escher's drawing:

is a stereographic projection—done like this:

I posted the image directly above in order to correct this:

In brief, once again (and I know it's not easy), when you project orthographically a hyperbolic surface on to a flat plane, of the polar coordinate type, you end up with a projection that resembles Escher's famous hyperbolic sphere drawings

which I think is mistaken. I wasn't trying to compare the drawing directly above to a spacetime diagram or any other kind of physical situation. It was just the first image I found of a stereographic projection of a hyperbola. So, I don't really understand this:

In the diagram you linked
here
, there is a cut plane through the past light cone.

because there is nothing that I would understand as a past lightcone in that image.

Figure A, and many of the other diagrams I've drawn (including the latest on, Figure Y) are polar coordinate systems: two-dimensional coordinate systems in which each point on the plane is determined by a distance from a fixed point (the origin; analogous to the rest-frame of the observer) and an angle from a fixed direction.

What exactly do you not understand? It's a cross section of the visible universe, as if you drew a plane through it, and looked at it form above, from the pole.

If red is the observer and green is a galaxy that red observes,

My understanding of your explanation is that both the blue line and the yellow line are the look-back time. In other words, the length of the blue line and the length of the yellow line both correspond to the light travel time distance of that galaxy from our perspective. That doesn't make sense to me.

If the blue line along the hyperbola is the look-back time then I don't know what the y axis is for. The only way I think I'll understand what you're trying to show is if you explain what the distances, lines, and angles correspond to in terms of physical measurements. For example, if red has a clock, is there a distance on the diagram that corresponds to how much time that clock measures? That type of thing.

~modest

modest · June 3, 2010

I missed your edits...

You should read this post, if you haven't already.

Yes, I've read it.

Escher's drawings that I refer to are designed after the Poincaré hyperbolic disk; a two-dimensional space having hyperbolic geometry, such as this one: Circle Limit IV (Heaven and Hell), 1960

I am aware.

Look at the Poincaré Hyperbolic Disk. Clearly, it is an orthographic projection

No, for an orthographic projection the rays are parallel. Poincare's disk is a stereographic projection.

The Poincaré models

Main articles:
Poincaré disc model
,
Poincaré half-plane model
Another closely related pair of models of hyperbolic geometry are the Poincaré ball and Poincaré half-space models. The ball model comes from a
stereographic projection
of the hyperboloid in R
ⁿ⁺¹
onto the hyperplane {x
₀
= 0}.

Poincaré disk model

Relation to the hyperboloid model

The Poincaré disk model, as well as the Klein model, are related to the hyperboloid model projectively...

For Cartesian coordinates (t, xi) on the hyperboloid and (yi) on the plane, the conversion formulae are...

[math]y_i = \frac{x_i}{1 + t}[/math]

[math](t, x_i) = \frac {\left( 1+\sum{y_i^2},\, 2 y_i \right)} {1-\sum{y_i^2}}[/math]

Compare the formulae for
stereographic projection
between a sphere and a plane.

Orthogonal Projection

A projection of a figure by parallel rays...

Orthogonal Projection -- from Wolfram MathWorld

Projecting an hyperboloid directly onto a disk clearly does not have parallel rays of projection:

The vertical lines, you will notice, are not parallel. Here is another that should demonstrate the point,

...The diagram shows how the projection, viewed along the y-axis, of this surface from the "south pole" onto the horizontal plane gives a pleasant "model" for the hyperbolic plane. This is the model that was promoted by Henri Poincare.

Math 452, Classical Geometries--Introduction

The thick horizontal line is Poincare's disk (one dimension of it) and the hyperbola is Hyperbolic space (one dimension of it). The projecting rays are not parallel. It is not orthogonal.

An orthogonal projection onto an Euclidean plane would correspond to your diagram A and a stereographic projection would correspond to diagram C (eg Escher's drawings and Poincare's disc models).

Again, this type of projection serves little or no purpose for the present discussion about redshift z. Though interesting they are.

I believe the reason you brought up projections was to object to Ned Wright's spacetime diagram. A spacetime diagram is a normal and very well defined construct that has direct physical meaning—not to mention, it is very useful in visualizing spacetime. It is far from, as you say, "serving little or no purpose for the present discussion". It is exactly what cosmologists use when demonstrating curvature.

Your lengthy objection based on different kinds of projections feels like obfuscation.

Did you calculate the sum of the angles of the triangle in Figure Y yet?

A triangle in diagram A would have angles less than 180º. It is clear that diagram A is meant to represent hyperbolic, negatively curved, space. Nobody has assumed otherwise.

~modest

quantumtopology · June 3, 2010

I think you guys are mixing different types of diagrams and just need to pick one out and progress from there.

Escher drawings are obviously stereographic projections, the diagram Modest uses from Ned Wright's tutorial is a space-like slice hyperboloid from a minkowski spacetime (a manifold with null curvature only valid for Special relativity), so I don't think none of it can be compared with CC's manifold that is spacetime.

Hope this helps some, or at least it does not confuse things further.

Regards

modest · June 3, 2010

Ned Wright's tutorial is a space-like slice hyperboloid from a minkowski spacetime (a manifold with null curvature only valid for Special relativity)

If by "Ned Wright's tutorial" you mean this:

it is not a space-like slice from Minkowski spacetime (which would just be a hyperbola). It is a spacetime diagram of a low-density universe in special relativistic coordinates. The distance along the y axis is the proper time of an observer on the center world line.

One consequence of general relativity is that the curvature of space depends on the ratio of rho to rho(crit). We call this ratio Ω = rho/rho(crit). For Ω less than 1, the Universe has negatively curved or hyperbolic geometry. For Ω = 1, the Universe has Euclidean or flat geometry. For Ω greater than 1, the Universe has positively curved or spherical geometry. We have already seen that the zero density case has hyperbolic geometry, since the cosmic time slices in the special relativistic coordinates were hyperboloids in this model.

I agree that CC is presenting a diagram of some other nature, but it is not spacetime. I've been providing links to sites that explain spacetime diagrams and I think everyone would benefit from reading them.

~modest

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