Expansion of space

Agent Smith

Agent Smith

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Oct 18, 2023
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Space expansion is an accepted theory in cosmology. Lawrence Kraus illustrates this as points on a grid. From any point on that grid, it looks as though

1. Every other point is moving away from that point
2. n times as far away, n times greater the expansion

I've been learning matrices and am familiar with basic transformations of a given point/vector as the "weighted sum"

For a \(\displaystyle 2 \times 2\) e.g. \(\displaystyle \begin{bmatrix}a & b \\ c & d \end{bmatrix}\)transformation matrix applied to the vector/point \(\displaystyle \begin{bmatrix} x \\ y \end{bmatrix}\), we have the image at:
\(\displaystyle x \begin{bmatrix} a \\ c \end{bmatrix} + y \begin{bmatrix} b \\ d \end{bmatrix}\)

What kinda transformation matrix could be used to represent the expansion of space (cosmology)? Is there another (easier) way of doing this?
 
Agent Smith said:
What kinda transformation matrix could be used to represent the expansion of space (cosmology)? Is there another (easier) way of doing this?
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What else have you learned about matrices? E.g., can you write a matrix which multiplies every vector by 3 ?
 
blamocur said:
What else have you learned about matrices? E.g., can you write a matrix which multiplies every vector by 3 ?
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The simplified version of the real McCoy was 2 dimensional (thanks Lawrence Krauss for trying to make us understand). I have seen rather complicated transformations of the plane in videos (baffling).

As for your question, are you referring to scalar multiplication, looks like you are. The lessons I took are introductory, a lot of intermediate steps were skipped. Dunno if I can answer your question as a yes or no.
 
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Agent Smith said:
The simplified version of the real McCoy was 2 dimensional (thanks Lawrence Krauss for trying to make us understand). I have seen rather complicated transformations of the plane in videos (baffling).
Click to expand...
Where are the videos?

Agent Smith said:
The simplified version of the real McCoy was 2 dimensional (thanks Lawrence Krauss for trying to make us understand). I have seen rather complicated transformations of the plane in videos (baffling).

As for your question, are you referring to scalar multiplication, looks like you are. The lessons I took are introductory, a lot of intermediate steps were skipped. Dunno if I can answer your question as a yes or no.
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Agent Smith, I like you, but I can't understand most of your questions. The funny thing is that I am a Matrix expert and you are a Matrix beginner, but I can't understand your Matrix questions!

Let us assume that there is a Matrix that describes the expansion of Space. If you can't understand a matrix which multiplies every vector by 3, how the heck you will be able to understand the Space Matrix?
 
mario99 said:
Where are the videos?


Agent Smith, I like you, but I can't understand most of your questions. The funny thing is that I am a Matrix expert and you are a Matrix beginner, but I can't understand your Matrix questions!

Let us assume that there is a Matrix that describes the expansion of Space. If you can't understand a matrix which multiplies every vector by 3, how the heck you will be able to understand the Space Matrix?
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I was taught (regarding transformations of the plane) that if \(\displaystyle \begin{bmatrix}1 \\ 0 \end{bmatrix} \to \begin{bmatrix}3 \\ 2 \end{bmatrix} \wedge \begin{bmatrix} 0 \\ 1 \end{bmatrix} \to \begin{bmatrix}4 \\ 5 \end{bmatrix}\) (unit vectors and their images) then the image of the point/vector \(\displaystyle \begin{bmatrix}7 \\ 8 \end{bmatrix}\) is \(\displaystyle 7 \begin{bmatrix} 3 \\ 2 \end{bmatrix} + 8 \begin{bmatrix} 4 \\ 5 \end{bmatrix} \)

Texas university had an app that showed the changes to a pic of a dog when 2 × 2 matrix transformations were applied to it. It's no longer available. 🙃
 
If I wanted to do a reflection across the \(\displaystyle y\) axis then the transformation matrix would be \(\displaystyle [\begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}\). That's a start, oui? Assuming I'm correct
 
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I'm interested in space dilation and it seems for that, if I want a \(\displaystyle n \times\) dilation, the transformation matrix is \(\displaystyle \begin{bmatrix}n & 0 \\0 & n \end{bmatrix}\) for a point/vector \(\displaystyle \begin{bmatrix}a \\ b \end{bmatrix}\) and \(\displaystyle \begin{bmatrix}n & 0 \\ 0 & n \end{bmatrix} = n \begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix}\). Does that mean dilation is effected by multiplication by a scalar?
 
However the expansion of space dilation has another feature: take 2 points, A and B . If B is n times farther away than A (from the origin), B undergoes a dilation that's n times the dilation A undergoes. How do we encode this into a transformation matrix?
 
Agent Smith said:
I'm interested in space dilation and it seems for that, if I want a \(\displaystyle n \times\) dilation, the transformation matrix is \(\displaystyle \begin{bmatrix}n & 0 \\0 & n \end{bmatrix}\) for a point/vector \(\displaystyle \begin{bmatrix}a \\ b \end{bmatrix}\) and \(\displaystyle \begin{bmatrix}n & 0 \\ 0 & n \end{bmatrix} = n \begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix}\). Does that mean dilation is effected by multiplication by a scalar?
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Looks good.
Agent Smith said:
Does that mean dilation is effected by multiplication by a scalar?
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Not sure I understand this question, but linear dilation is multiplication by a scalar, or by an equivalent matrix which you've figured out earlier.
 
Agent Smith said:
However the expansion of space dilation has another feature: take 2 points, A and B . If B is n times farther away than A (from the origin), B undergoes a dilation that's n times the dilation A undergoes. How do we encode this into a transformation matrix?
Click to expand...
Unless I misunderstood this sounds like a non-linear dilation, which means it cannot be represented by a matrix -- or scalar -- multiplication.
 
blamocur said:
Unless I misunderstood this sounds like a non-linear dilation, which means it cannot be represented by a matrix -- or scalar -- multiplication.
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It's confusing. What does that statement actually mean?

For simplicity let's just stick to a 1D number line. There's a point 1 and a point 2 (the numbers 1 and 2). 2 is twice further from 0 than 1. I apply a \(\displaystyle 2 \times\) dilation. 1' = 2 and 2' = 4. So Did 2 undergo \(\displaystyle 2 \times\) the dilation of 1? No it didn't, right? If it did, 2' = (2 × 2) × 2 = 8. Do you mean to say we can't have a transformation matrix for the above, because it's nonlinear? Then how did Lawrence Kraus manage to create a CGI of space expansion in 2 D? 🤔
 
I suppose this is relative dilation. What happens to a point is relative to what happens to other points. Could we do this: Say we know the relative positions of 2 points A and B from the origin. B is n times further away than A. For any given dilation (k) for A, we could apply the dilation nk to the point B and all other points that are n times further away. We would need to compute the relative distances then for all points and adjust the dilator (nk). Je ne sais pas.
 
Agent Smith said:
There's a point 1 and a point 2 (the numbers 1 and 2). 2 is twice further from 0 than 1. I apply a 2×\displaystyle 2 \times2× dilation. 1' = 2 and 2' = 4. So Did 2 undergo 2×\displaystyle 2 \times2× the dilation of 1? No it didn't, right? If it did, 2' = (2 × 2) × 2 = 8.
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This is the way I understood your post #8, i.e. that '2' would dilate to '8'. But it seems that you meant constant dilation/scale factor, i.e. a linear transformation.
 
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