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EntroPyrexoR
2009-Apr-16, 07:11 PM
Hi all,

This is my first post to these forums.

I would like to write a computer program that models uniform exansion. I'm clear on the idea, but not clear on the math.

Crawl before you walk...

Suppose I have 3 points (in only 2-D for now) A, B, and C. The relationship of distances is as follows:

A <--> B = 2
A <--> C = 3
B <--> C = 1

I understand that A will move "faster" away from C, while also moving away from B. So, let's pretend that after my next "time-step":

A <--> C = 6

How far is A <--> B then? 4?

If someone can shed some light on this, I would really appreciate it.

Thank You in advance,

EntroPyrexoR

EntroPyrexoR
2009-Apr-16, 07:16 PM
Better yet...

If A <--> C = 6

what are the other 2 relationships?

A <--> B = ?

B <--> C = ?

Hornblower
2009-Apr-16, 07:19 PM
Hi all,

This is my first post to these forums.

I would like to write a computer program that models uniform exansion. I'm clear on the idea, but not clear on the math.

Crawl before you walk...

Suppose I have 3 points (in only 2-D for now) A, B, and C. The relationship of distances is as follows:

A <--> B = 2
A <--> C = 3
B <--> C = 1

I understand that A will move "faster" away from C, while also moving away from B. So, let's pretend that after my next "time-step":

A <--> C = 6

How far is A <--> B then? 4?

If someone can shed some light on this, I would really appreciate it.

Thank You in advance,

EntroPyrexoR
If A <--> B is 4 at this point, I would call it a uniform expansion. Just imagine putting dots on a rubber band at these intervals and then stretching it.

EntroPyrexoR
2009-Apr-16, 07:28 PM
Thank You Hornblower.

So uniform expansion is linear? I was always under the impression that the rate of change (of the distance) increases as the distance increases, no?

If the relation ship goes from (1, 2, 3) to (2, 4, 6) after one interval, then the rate of expansion is equal equal all the way around.

I'm just confused because I thought:

(A <--> C)dx > (A <--> B)dx > (B <--> C)dx

antoniseb
2009-Apr-16, 07:30 PM
If A <--> B is 4 at this point, I would call it a uniform expansion. Just imagine putting dots on a rubber band at these intervals and then stretching it.

But take into account the speed of light, and Special Relativity's lack of constancy for objective lengths and times. What you are proposing will require that you specify certain points of view (frames of reference) that you will track things through.

Cougar
2009-Apr-16, 07:45 PM
Suppose I have 3 points (in only 2-D for now) A, B, and C. The relationship of distances is as follows:

A <--> B = 2
A <--> C = 3
B <--> C = 1

Unless you're using some very modern geometry, this describes a line. Lines are 1-D.

EntroPyrexoR
2009-Apr-16, 08:00 PM
Unless you're using some very modern geometry, this describes a line. Lines are 1-D.

Ah, Perhaps I wasn't clear enough.

1,2, and 3 describes the distances between the points in 2-D. I really don't care what their actual coordinates are. I just know that the distances between A, B, and C form some sort of triangle.

speedfreek
2009-Apr-16, 08:18 PM
Thank You Hornblower.

So uniform expansion is linear? I was always under the impression that the rate of change (of the distance) increases as the distance increases, no?

If the relation ship goes from (1, 2, 3) to (2, 4, 6) after one interval, then the rate of expansion is equal equal all the way around.

I'm just confused because I thought:

(A <--> C)dx > (A <--> B)dx > (B <--> C)dx

Try this for starters, with all objects in a straight line.

A-B-C-D-E are all 1 metre apart
A--B--C--D--E are all 2 metres apart
A---B---C---D---E are all 3 metres apart
A----B----C----D----E are all 4 metres apart
A-----B-----C-----D-----E are all 5 metres apart

This is uniform expansion.

A will find that B has moved from 1 metre away to 5 metres away in the same amount of time as it took E to move from 4 metres to 20 metres away. If each iteration took one second, then after 5 seconds A finds that B has moved away a distance 4 metres in that time, whilst E has receded by 16 metres. E is receding from A at 4 times the speed that B is receding from A! The further away something is, the faster it moves away.

Amber Robot
2009-Apr-16, 08:19 PM
I would like to write a computer program that models uniform exansion. I'm clear on the idea, but not clear on the math.

Suppose I have 3 points (in only 2-D for now) A, B, and C. The relationship of distances is as follows:

A <--> B = 2
A <--> C = 3
B <--> C = 1

I understand that A will move "faster" away from C, while also moving away from B. So, let's pretend that after my next "time-step":

A <--> C = 6

How far is A <--> B then? 4?

If someone can shed so

If it indeed *uniform* expansion, then yes, if AC distance doubles, then all distances double. Therefore AB goes from 2 to 4 and BC goes from 1 to 2.

As for the speed increasing with distance it obeys that two. AC has gone from 3 to 6, i.e. a change of 3, and BC has only gone from 1 to 2, i.e., a change of 1. Therefore, in the same amount of time, AC has expanded three times faster than BC.

EntroPyrexoR
2009-Apr-16, 08:30 PM
Ok, I think I've got it...

(Looking down on the triangle formed by vertices A, B, and C)

A is at point (0,0) B is at (0,2) and C is at (3,0). A will travel at a slope of -2/3...

...but only at the exact moment A is 2 units from B, and 3 units from C.

So how do you mathematically define the expansion of these points over time?

EntroPyrexoR
2009-Apr-16, 08:44 PM
As for the speed increasing with distance it obeys that two. AC has gone from 3 to 6, i.e. a change of 3, and BC has only gone from 1 to 2, i.e., a change of 1. Therefore, in the same amount of time, AC has expanded three times faster than BC.

I disagree. if all distances are multiplied by 2 (i,e. 1,2,3 becomes 2,4,6) AC has certainly not expanded three times faster than BC. AB, AC, and BC have all expanded at the same exact rate, 2.

What you've suggested is that after one interval (where all distances have doubled), you simply have a new, larger, yet similar triangle. This is not uniform expansion. Uniform expansion dicatates that each point A, B, and C are at the center of expansion (i,e. A,B,and C do not move away from the center of the triangle, they move away from each other).

So the direction and speed at which any given point is moving is dictated by its distance to the other points.

In my example, A will move at a slope of -2/3. However, as B and C also move, the direction A moves will approach -0 (or in other words, a slightly negative slope).

The purpose of my computer simulation is to answer less obvious questions like, will A ever move in a positive slope in the above example? Will C ever move in a more negative slope than A?

So obviously this a calculus problem for speed and direction. So my question is how do you define the rates of change in speed and direction for the 3 points in my example?

EntroPyrexoR
2009-Apr-16, 08:59 PM
Try this for starters, with all objects in a straight line.

A-B-C-D-E are all 1 metre apart
A--B--C--D--E are all 2 metres apart
A---B---C---D---E are all 3 metres apart
A----B----C----D----E are all 4 metres apart
A-----B-----C-----D-----E are all 5 metres apart

This is uniform expansion.

A will find that B has moved from 1 metre away to 5 metres away in the same amount of time as it took E to move from 4 metres to 20 metres away. If each iteration took one second, then after 5 seconds A finds that B has moved away a distance 4 metres in that time, whilst E has receded by 16 metres. E is receding from A at 4 times the speed that B is receding from A! The further away something is, the faster it moves away.

Ah thank you very much. I didn't see your post before. I will look at this and try to relate it to 2 and 3 dimensions. Nice example,

Thank you.

pzkpfw
2009-Apr-16, 09:04 PM
(Nitpick: [edit:typed before the post above]

I disagree. if all distances are multiplied by 2 (i,e. 1,2,3 becomes 2,4,6) AC has certainly not expanded three times faster than BC. AB, AC, and BC have all expanded at the same exact rate, 2.

I think Amber Robot possibly missed that you later clarified you meant a triangle in 2D, and based that comment on a line.

However, EntroPyrexoR, note:

Suppose I have 3 points (in only 2-D for now) A, B, and C. The relationship of distances is as follows:

A <--> B = 2
A <--> C = 3
B <--> C = 1

You described a line, not a triangle.

0--1--2--3--4--5
A-----B A to B is 2
A--------C A to C is 3
B--C B to C is 1

A-----B--C It's a line

Now double it.

0--1--2--3--4--5--6
A-----------B-----C

A-----------B A to B is 4
A-----------------C A to C is 6
B-----C B to C is 2

A to B increased by 2.
A to C increased by 3... so it did increase faster than A to B. (As Amber Robot correctly wrote.)
)

a1call
2009-Apr-16, 09:10 PM
I am not an expert by any stretch of imagination. Yet have hung around this board long enough to realize that you have to pay careful attention to post #5 in order to have a realistic model. Distances are not absolute. AC might be 3 meters from the vantage point of A at time t0 while/and AC might be 4 meters from the vantage point of B at time t0. And to make things even more fun neither A, B, nor C are required to agree on when t0 is. To have a true model you will have to have attributes for all points in different arrays for each point. In other words the observance of speeds/locations is relative and the binding rule is the constancy of speed of any/all beams of light from all vantage points. Do not think of location, speed or time as absolute, defined or definable parameters. :whistle::whistle::whistle:

Jeff Root
2009-Apr-16, 09:11 PM
What you've suggested is that after one interval (where all distances
have doubled), you simply have a new, larger, yet similar triangle.
This is not uniform expansion.
Sure it is.

Uniform expansion dicatates that each point A, B, and C are at the
center of expansion (i,e. A,B,and C do not move away from the center
of the triangle, they move away from each other).
The expansion has no center. Any point can be considered as
being at the center. A, B, C, or the center of the triangle can
each be considered as at the center of the expansion. Every
point, whether it is the corner of a triangle or not, moves away
from every other point at the same rate. That's uniform expansion.

I made these animations to illustrate uniform expansion. The first
is one-dimensional, the second is two-dimensional:

Uniform 1-D expansion (http://www.freemars.org/jeff2/expand3e.htm)
Uniform 2-D expansion (http://www.freemars.org/jeff2/expand5a.htm)

The second animation does not display very well in my browser,
but it displays ok in my animation-making program and in IrfanView.

-- Jeff, in Minneapolis

pzkpfw
2009-Apr-16, 09:17 PM
[Edit: gosh darn I type too slow (or Jeff Root types too fast...)]

Now a triangle - I'll use a nice 3-4-5 triangle.

A = (0,0), B = (4,0), C = (4,3)

B to C = 3
A to B = 4
A to C = 5

C
|
2
|
1
|
A--1--2--3--B

Double it (I'll just re-scale):

C
|
4
|
2
|
A--2--4--6--B

Now:

A = (0,0), B = (8,0), C = (8,6)

B to C = 6
A to B = 8
A to C = 10

Seems like the the distances do scale as we might expect.

The distance of A, B and C to the centre of that triangle (Z) is, I think, irrelevant.

...because you could simply build three triangles (Z,A,B), (Z,B,C) & (Z,C,A) - and they'd each behave as the example triangle (A,B,C).

Amber Robot
2009-Apr-16, 09:34 PM
So, am I wrong or right? I thought I had it right...

EntroPyrexoR
2009-Apr-16, 09:35 PM
I am not an expert by any stretch of imagination. Yet have hung around this board long enough to realize that you have to pay careful attention to post #5 in order to have a realistic model. Distances are not absolute. AC might be 3 meters from the vantage point of A at time t0 while/and AC might be 4 meters from the vantage point of B at time t0. And to make things even more fun neither A, B, nor C are required to agree on when t0 is. To have a true model you will have to have attributes for all points in different arrays for each point. In other words the observance of speeds/locations is relative and the binding rule is the constancy of speed of any/all beams of light from all vantage points. Do not think of location, speed or time as absolute, defined or definable parameters. :whistle::whistle::whistle:

Very true. I realized before posting here that this problem does not lend itself well to the procedural nature of a computer program. A might be 3 away from C and 2 away from B. OK, so move A so it's 6 away from C and 4 away from B. But during the same interval, where do B and C go? That's a head scratcher for sure.

Perhaps I was on the right track by proportioning out the slope (the direction a point will move) for all the points, and simply mvoing them in those directions over time.

Thank you all very much. You have been very helpful.

Amber Robot
2009-Apr-16, 09:37 PM
What you've suggested is that after one interval (where all distances have doubled), you simply have a new, larger, yet similar triangle. This is not uniform expansion.

Ok, I guess this is where we differ. I thought that was the definition of "uniform expansion". The uniform expansion of any shape will produce a larger, yet similar shape.

pzkpfw
2009-Apr-16, 09:46 PM
So, am I wrong or right? I thought I had it right...

I think you were right.

pzkpfw
2009-Apr-16, 09:51 PM
A might be 3 away from C and 2 away from B. OK, so move A so it's 6 away from C and 4 away from B. But during the same interval, where do B and C go? That's a head scratcher for sure.

I'd expect B-to-C to simply be twice whatever it was before (assuming Universal expansion).

See posts #15 and #16 (triangle expansion example).

speedfreek
2009-Apr-16, 10:03 PM
Ok, I guess this is where we differ. I thought that was the definition of "uniform expansion". The uniform expansion of any shape will produce a larger, yet similar shape.

I think you are right, too.

speedfreek
2009-Apr-16, 10:21 PM
I am not an expert by any stretch of imagination. Yet have hung around this board long enough to realize that you have to pay careful attention to post #5 in order to have a realistic model. Distances are not absolute. AC might be 3 meters from the vantage point of A at time t0 while/and AC might be 4 meters from the vantage point of B at time t0. And to make things even more fun neither A, B, nor C are required to agree on when t0 is. To have a true model you will have to have attributes for all points in different arrays for each point. In other words the observance of speeds/locations is relative and the binding rule is the constancy of speed of any/all beams of light from all vantage points. Do not think of location, speed or time as absolute, defined or definable parameters. :whistle::whistle::whistle:

This is definitely very important for the model, if you want to model anything other than arbitrary coordinates in uniform expansion. The OP never specifically mentioned anything about our universe, they just asked about uniform expansion. ;)

EntroPyrexoR
2009-Apr-17, 04:47 PM
Again thank you all for you responses.

Yes Amber Robot, you were correct. Thank you. I believe I was hung up on the "rate-of-change" which really is the same, however it's not the rate of change but simply the change in distance which is ocurring more rapidly, so yes you were right.

No I'm not trying to mdel the expansion of the universe, just uniform expansion in general. When I said computers didn't lend their hand to this problem well, it's because if I have 100 coordinates I initially have to run O(n^2) on the coordinates to figure out the direction and distance they need to move for each interval. This is a horrible efficiency.

Amber Robot
2009-Apr-17, 05:54 PM
No I'm not trying to mdel the expansion of the universe, just uniform expansion in general. When I said computers didn't lend their hand to this problem well, it's because if I have 100 coordinates I initially have to run O(n^2) on the coordinates to figure out the direction and distance they need to move for each interval. This is a horrible efficiency.

Can't you just multiply all the coordinates by a single value?

EntroPyrexoR
2009-Apr-17, 08:31 PM
Can't you just multiply all the coordinates by a single value?

Yea I guess you could, but to make sense of it you'd have to recenter the middle of the viewing area by multiplying it by the same number.

A's vertices are: (2,2) (2,4) (4,4) (4,2)

B's vertices are: (4,4) (4,8) (8,8) (8,4)

So yea it works.

Amber Robot
2009-Apr-17, 09:17 PM
Yea I guess you could, but to make sense of it you'd have to recenter the middle of the viewing area by multiplying it by the same number.

A's vertices are: (2,2) (2,4) (4,4) (4,2)

B's vertices are: (4,4) (4,8) (8,8) (8,4)

So yea it works.

You can shift to whatever center point, multiply by the expansion factor, and then shift back.

pzkpfw
2009-Apr-17, 11:06 PM
Now that uniform expansion is accepted, I'm not sure what you are going to achieve.

You may as well just draw a picture in MS Paint or some other tool, then use the zoom function to expand it. (Except your points would expand).

Modelling it, you could multiply each point by a number, to expand the Universe, then divide by the same number to re-scale the drawing and keep it visible on screen.

You'd see no change (the picture would seem static).

(If you were drawing objects (Galaxies) instead of just maintaining points, those objects would appear to shrink.)

Much much harder, would be accurately modelling general expansion of the Universe, while keeping Galaxies (or other gravitationally bound objects) together. e.g. if your points were stars, some in different Galaxies; with the Galaxies staying together, but getting further from other Galaxies.

That's not been covered in this thread.