The Beginning of the Universe

[swansont]

3 hours ago, KJW said:

Time isn't a constant, but T is, by definition.

 

Then the proportionality only works at T, and if you are considering particles moving away from each other, more than one time is involved.

Why is it you can’t just admit the original statement was incorrect? You’ve already stated that the relative velocities are constant over time. Why the contortions to try and preserve the statement about varying with length? 

[Airbrush]

On 3/25/2024 at 11:22 AM, Mordred said:

An explosion has different dynamics that are measurable. You can easily confirm the difference yourself. Take a ruler from a central point at various different angles measure off 1 cm then 3 cm then 6 cm etc.  Now think of the ratio of change the result of radiating outward from the central point and measure the angles. You should notice angle changes as well as a preferred direction and location.

 

Now place dots on a balloon measure the angles prior to inflating (inflate just enough to get a sphere. Then inflate the balloon further. The angles do not change and the ratio of change in distance is identical between any two points. 

Thanks!  That is the best explanation why the expansion of space is not comparable to an explosion.

My favorite question is, what is the probability that the big bang expansion we can see, within a diameter of under 100 billion light-years, is likely to extend to infinity, assuming a flat, totally non-curved, space?  I would bet on a zero probability.

Edited March 27 by Airbrush

[Mordred]

Correct any finite portion will not extend to infinity. However we do not know if the entirety of universe beyond our Observational portion is infinite or finite. Our Observable universe portion will always remain finite. Had findings of our universe term been precisely flat at unity then we likely would have had an open infinite universe. However their is a slight curvature term that raises the possibility of a closed universe.

Edited March 27 by Mordred

[KJW]

On 3/27/2024 at 12:26 AM, swansont said:

Then the proportionality only works at T

T is arbitrary. Therefore, the proportionality works for all constant values of T (different proportionality for different T, but proportionality nevertheless).

 

 

On 3/27/2024 at 12:26 AM, swansont said:

if you are considering particles moving away from each other, more than one time is involved.

Not necessarily. My concern was about a snapshot in time, not time evolution.

 

 

On 3/27/2024 at 12:26 AM, swansont said:

Why is it you can’t just admit the original statement was incorrect?

My original statement was not incorrect. I was only concerned with the relationship between relative velocity and relative displacement for pairs of particles at a single time. It is you who was fixated on time evolution. When you pointed out that proportionality didn't apply over different times, I clarified what I meant.

 

 

On 3/27/2024 at 12:26 AM, swansont said:

You’ve already stated that the relative velocities are constant over time. Why the contortions to try and preserve the statement about varying with length? 

The entire point of what I wrote was to show that at any given instant in time, the relative velocity between any pair of particles is directly proportional to the relative displacement between that pair of particles, the same as for an expanding space. Your focus on time evolution was a distraction away from my consideration of the pairs of particles at a given time.

 

 

Edited March 29 by KJW

[swansont]

46 minutes ago, KJW said:

The entire point of what I wrote was to show that at any given instant in time, the relative velocity between any pair of particles is directly proportional to the relative displacement between that pair of particles, the same as for an expanding space. Your focus on time evolution was a distraction away from my consideration of the pairs of particles at a given time.

Oh, good grief. 

[KJW]

[math]\underline{\text{Non-relativistic:}}[/math]

[math]\text{For an explosion at time } t=0 \text{ at the origin of a rectangular coordinate system:}[/math]

[math]\text{Let } \textbf{x}_{0i} \text{ be the position vector of the } i \text{-th particle at time } t=T[/math]

[math]\text{Let } \textbf{v}_{0i} \text{ be the velocity vector of the } i \text{-th particle at time } t=T[/math]

[math]\text{Then: } \textbf{v}_{0i} = \dfrac{1}{T} \textbf{x}_{0i}[/math]

[math]\text{Similarly for the } j \text{-th particle at time } t=T \text{ : } \textbf{v}_{0j} = \dfrac{1}{T} \textbf{x}_{0j}[/math]

[math]\text{Then: } \textbf{v}_{0j} - \textbf{v}_{0i} = \dfrac{1}{T} (\textbf{x}_{0j} - \textbf{x}_{0i})[/math]

[math]\text{Let: } \textbf{x}_{ij} = \textbf{x}_{0j} - \textbf{x}_{0i} \ \ \text{and} \ \ \textbf{v}_{ij} = \textbf{v}_{0j} - \textbf{v}_{0i}[/math]

[math]\text{Then: } \textbf{v}_{ij} = \dfrac{1}{T} \textbf{x}_{ij}[/math]

[math]\text{From the perspective of the } i \text{-th particle, dropping the } i \text{ :}[/math]

[math]\textbf{v}_{j} = \dfrac{1}{T} \textbf{x}_{j}[/math]

Thus, all particles are on equal footing with regards to the observation of the other particles of the explosion.

[math]\underline{\text{Relativistic:}}[/math]

[math]\text{For an explosion at the origin of a Minkowskian coordinate system:}[/math]

[math]\text{Let } (x_i, y_i, z_i, t_i) \text{ be the coordinates of the } i \text{-th particle at proper time } \tau[/math]

[math]\text{Then: }c^2 t_i^2 - x_i^2 - y_i^2 - z_i^2 = c^2 \tau^2[/math]

[math]\text{The set of all particles at proper time } \tau \text{, taken as a continuum, defines a three-dimensional hyperboloid:}[/math]

[math]c^2 t^2 - x^2 - y^2 - z^2 = c^2 \tau^2[/math]

[math]\text{This is invariant to Lorentz transformations:}[/math]

[math]ct = \dfrac{c t' + \beta x'}{\sqrt{1 - \beta^2}}[/math]

[math]x = \dfrac{x' + \beta c t'}{\sqrt{1 - \beta^2}}[/math]

[math]y = y'[/math]

[math]z = z'[/math]

[math]\dfrac{(c t' + \beta x')^2}{1 - \beta^2} - \dfrac{(x' + \beta c t')^2}{1 - \beta^2} - y'^2 - z'^2 = c^2 \tau^2[/math]

[math]\dfrac{c^2 t'^2 + 2 c t' \beta x' + \beta^2 x'^2}{1 - \beta^2} - \dfrac{x'^2 + 2 x' \beta c t' + \beta^2 c^2 t'^2}{1 - \beta^2} - y'^2 - z'^2 = c^2 \tau^2[/math]

[math]\dfrac{(1 - \beta^2) c^2 t'^2 - (1 - \beta^2) x'^2}{1 - \beta^2} - y'^2 - z'^2 = c^2 \tau^2[/math]

[math]c^2 t'^2 - x'^2 - y'^2 - z'^2 = c^2 \tau^2[/math]

Although Lorentz transformations transform points of the three-dimensional hyperboloid to different points of the three-dimensional hyperboloid, the three-dimensional hyperboloid itself is invariant. Thus, all particles are on equal footing with regards to the observation of the other particles of the explosion. Note that Lorentz transformations take the role of relative velocities. By moving a selected particle to the spatial origin by the application of a Lorentz transformation, the velocities of all the other particles are now relative to the selected particle.

Similar to an expanding space, and contrary to what one might expect for an explosion, the space defined by constant age of the particles has no centre. Thus, each particle is able to regard itself as the centre of the space. In particular, the explosion itself has no identifiable location in the space. Because all the particles originated from the explosion, the explosion actually occurred everywhere in the space.

One thing that is notable from the above is that for a flat spacetime, the three-dimensional space of constant proper time from a single point is curved like a hyperboloid. For a flat three-dimensional space of constant proper time from a single point, the spacetime will be curved. And although I never dealt with a spherical three-dimensional space of constant proper time from a single point, I think one can extrapolate that the spacetime curvature will be larger than for the flat three-dimensional space.

 

Edited March 30 by KJW

[Mordred]

You do not show a metric for an explosion. An explosion radiates from starting point outward. So it is anisotropic.  The metric equations you have are essentially homogeneous and isotropic. You might want to consider the inner product of your vectors in terms of the symmetry relations of the Minkowski metric.

\[\mu\cdot\nu=\nu\cdot\mu\]

 describing an explosion of an ensemble of particles from an origin point would require angles.

for example take 3 particles on a sheet of graph paper. Set an origin point a 0,0. Now place a particle at 1.1 another at 3,5 the last at say 5,3. Now expansion the graph lines increase in scale. (using a draftman's scaling ruler would be useful simply set the new scale at 1:2 ratio and redraw your grid lines). None of the angles change between 1.1, 3,5 and 5,3. Those angles would change if those particles were radiating outward from the origin point. 

Edited March 30 by Mordred

[KJW]

24 minutes ago, Mordred said:

You do not show a metric for an explosion.

I'm not sure what you are referring to here. What I showed was the coordinates of each particle after a given proper time. Over a continuum of particles, this defines a three-dimensional space. I could've supplied a metric for this space, but did not.

 

 

24 minutes ago, Mordred said:

An explosion radiates from starting point outward. So it is anisotropic.  The metric equations you have are essentially homogeneous and isotropic.

Exactly. It was specifically my intention to prove that the three-dimensional space defined by the particles from the explosion is homogeneous and isotropic.

 

 

24 minutes ago, Mordred said:

You might want to consider the inner product of your vectors in terms of the symmetry relations of the Minkowski metric.

[math]/[\mu\cdot\nu=\nu\cdot\mu/][/math]

 describing an explosion of an ensemble of particles from an origin point would require angles

What angles are you referring to? The spacetime trajectories of all the particles are orthogonal to the three-dimensional space.

 

 

Edited March 30 by KJW

[Mordred]

17 minutes ago, KJW said:

I'm not sure what you are referring to here. What I showed was the coordinates of each particle after a given proper time. Over a continuum of particles, this defines a three-dimensional space. I could've supplied a metric for this space, but did not.

 

 

Exactly. It was specifically my intention to prove that the three-dimensional space defined by the particles from the explosion is homogeneous and isotropic.

 

 

What angles are you referring to? The spacetime trajectories of all the particles are orthogonal to the three-dimensional space.

 

 

The angles would be needed between any 2 points. In an explosion scenario the angles would vary between any two or more points. This would include a metric that is expanding outward from an origin point. (specifically if one wanted to have commoving coordinates radiating outward from an origin for an inhomogeneous and anistropic expansion ) A homogeneous and isotropic expansion no angles change including those of the metric points. That also includes curvature terms once k is set it doesn't change over time. If K=0 in the past it will remain k=o to the present.

With k=0 there is no time dilation with its uniform mass distribution. 

Edited March 30 by Mordred

[KJW]

36 minutes ago, Mordred said:

The angles would be needed between any 2 points. In an explosion scenario the angles would vary between any two or more points. This would include a metric that is expanding outward from an origin point. A homogeneous and isotropic expansion no angles change including those of the metric points

In the non-relativistic case, it is clear that angles do not change. I proved that in my first post in this thread. In the relativistic case, the curvature of the three-dimensional space led me to consider the invariance to Lorentz transformations as proof of homogeneous and isotropic expansion. I should point out that I'm not suggesting that an expanding hyperboloidal space is indistinguishable from an expanding flat space. But to distinguish between them is more advanced than the simple geometric arguments often presented against the explosion model of the big bang. I'm not advocating that the big bang was an explosion, btw. But I will suggest that arguments against it are not necessarily valid.

 

Edited March 30 by KJW

[Mordred]

the argument against an explosion is valid when you look at the angles and not just distance change between multiple points. That is literally the lesson the balloon analogy was originally designed to convey. Though a more accurate analogy for 3d is the raisin bread analogy

Edited March 30 by Mordred

[KJW]

14 minutes ago, Mordred said:

the argument against an explosion is valid when you look at the angles and not just distance change between multiple points. That is literally the lesson the balloon analogy was originally designed to convey. Though a more accurate analogy for 3d is the raisin bread analogy

But if one considers any three points forming a triangle, then if the length of each side changes by the same relative amount, the result is a similar triangle with unchanged angles.

 

[Mordred]

sigh use that same graph and draw a set three points to make a triangle in the explosion case draw a line from the origin to each corner of the triangle. Now move the triangle along the radial lines you just drew keeping the three points on those lines you drew from the center. Can you honestly tell me that the length of each side expands equally and has no angle change ?

 

An even simple example is any two points on the circumference of a circle. if the radius increases so to does the distance between the two points the rate of change isn't linear as the radius increases either. For example take a circle with radius line place two points on the circumference at two different angles. Now just to get a triangle place a point on one of those angles half of the circumference. Will all sides change equally  if you shift the triangle outward while preserving the same angle from the origin to each point of the triangle ?

the two points on the same radial angle will have no change in distance while the other point will increase in distance between the other two 

Edited March 30 by Mordred

[KJW]

27 minutes ago, Mordred said:

sigh use that same graph and draw a set three points to make a triangle in the explosion case draw a line from the origin to each corner of the triangle. Now move the triangle along the radial lines you just drew keeping the three points on those lines you drew from the center. Can you honestly tell me that the length of each side expands equally and has no angle change ?

I'm not sure I'm visualising your description correctly, but in my assessment of the explosion, triangles formed from any three particles will expand with unchanged angles over time. The above mathematics of the non-relativistic case makes this clear. And I'm quite confident, without explicitly checking, that the same is true for the relativistic case also, in spite of the curvature of the three-dimensional space.

 

Edited March 30 by KJW

[Mordred]

were using the triangle to help visualize coordinate changes remember ? It  is basic geometry and ratios of change in the case of expansion the distance between any ensemble of coordinates will change equally. That is not the case in an explosion scenario.

perhaps this will help lets apply actual coordinates. place a point at 0.0 on an x,y graph. Now place a point at 6,0 and 0,6. measure the distance between 0.6 and 6,0. Now change 0.6 and 6,0 to 12,0 and 0,12 and measure the change in distance between 0,12 and 12,0. Does that distance change by 6 which is the ratio of change along the x and y axis ?

 

the length of the Hypotenuse will have a different rate of change as per Pythagorus theorem using the x and y axis allows us to form a right angle triangle .

a^2+b^2=c^2

Edited March 30 by Mordred

[KJW]

Consider on the x,y-graph the following three points: (1,0), (0,2), and (3,4) at some time. At double the time, (1,0) will become (2,0) because its velocity is 1 along the x-axis; (0,2) will become (0,4) because its velocity is 2 along the y-axis; and (3,4) will become (6,8) because its velocity is 5 along the (3,4)-direction. The triangle is double the size and all the angles are the same.

 

 

[Mordred]

Are you going tell me you could not see that all sides did not change equally in the hypotenuse on a right angle triangle?

In the last example I gave above ?

[KJW]

8 minutes ago, Mordred said:

Are you going tell me you could not see that all sides did not change equally in the hypotenuse on a right angle triangle?

In the last example I gave above ?

I figure that because you are trying to correct my error that it would be better to use my example.

 

[Mordred]

Let me ask are you familiar enough with Dirac notation and triangle inequality  in vector space ? There is a few math expressions that beautifully express the distinction between triangle inequality and homogeneous and isotropic using vector spaces under Dirac notation is most commonly used.

the fact is if you change length A and length B by an equal amount for simplicity length C does not change by an equal amount without an change in angles. Pythagorus theorem clearly shows this with right angle triangles. A homogeneous and isotropic change all points must change equally in ratio in all cases on all sides of the triangle

 

Edited March 30 by Mordred

[KJW]

26 minutes ago, Mordred said:

Let me ask are you familiar enough with Dirac notation and triangle inequality in vector space ? 

I'm familiar with Dirac notation, though I'm not especially comfortable with it. And I am familiar with the triangle inequality. But I'm not seeing the relevance to this discussion. Do you see any error in the example I gave?

 

Edited March 30 by KJW

[Mordred]

the expression for the relation above is as follows without Dirac notation. Using metric space \(\mathcal{M}\) d is a distance function A metric space will satisfy three conditions

positivity ( simply sets the x and y axis positive norm so were ignoring any - axis.)

\(d(x,y\ge 0\) with iff \( "=" x=y\)

a Homogeneous and isotropic will be

\[d(x,y)=d(y,x)\]

triangle inequality as

\[d(x,z)\le d(x,y)+d(y,z)\]

there's triangle inequality in terms of distance. 

If your familiar with Pythagorus theorem and triangle inequality I really don't understand how you cannot see the difference between an expansion of a homogeneous and isotropic metric space and a metric space as an anistropic and inhomogeneous explosion.

 

 

 

 

Edited March 30 by Mordred

[KJW]

9 minutes ago, Mordred said:

I really don't understand how you cannot see the difference between an expansion of a homogeneous and isotropic metric space and a metric space as an anistropic and inhomogeneous explosion.

I think I need to point out that the mathematics I wrote says that in terms of geometry an explosion is not anisotropic and inhomogeneous from the perspective of any one of the particles from the explosion. Why hasn't the mathematics been considered during this discussion?

 

[Mordred]

The mathematics you posted isn't an explosion but a homogeneous isotropic space  is precisely why I posted the proof for a metric space. It is not a space resulting from an explosion. We aren't even considering the time dimension there is no time dilation involved in a uniform mass distribution so you don't require it. Stick with 3 dimensional space. If you expand a Euclidean space uniformly that's expansion however in an explosion the space further away from the point of origin will expand faster than the space near the origin point. 

You can easily test that by using a cone and slicing the cone at various distances from the origin and measuring its area at the end of the cone furthest from the origin point. The cone is your explosion geometry not Euclidean geometry. Though its correctly a sphere expanding outward from the origin however all particle and vectors will have a preferred direction (outward from origin). That isn't the case in our universe. In our universe there is no preferred direction no vectors has a preferred direction due to any previous force applied regardless if those vectors are coordinates of geometry change or particles. A kinetic style explosion certainly has a preferred direction (outward) from its origin.

A good way to model a direction preferred spacetime would be to apply spherical coordinates or even cylindrical coordinates such as the rotating Godel universe whose expansion was due to the outward force of rotation. There is a large list of spacetimes (mostly dealing with BH/WH scenarios). 

A particularly good example is this particular anistropic model

Is the Universe anisotropic right now? Comparing the real Universe with the Kasner's space-time

https://arxiv.org/pdf/2305.02726.pdf

Edited March 31 by Mordred

[Mordred]

 Lets try another thought experiment using space expanding outward from the explosion. Lets set rate of flow outward from origin. Lets have an increase of space set at 1 parsec per second why not. Now place an observer that flows outward from from the center and set the observer 1 sec after the BB. Simply so we have space ahead of that observer. As space is created outward from the origin the space between the observer and the origin increases but if the space 1 second previous to that observer is flowing outward. Then no new space is being formed ahead of that same observer. That isn't uniform expansion where the distance between every coordinate  expands uniformly. (ignore time dilation we are just dealing with how space increases in each case) 

In the FLRW metric I'm using the commoving observer or equivalent.

Edited March 31 by Mordred

[KJW]

19 hours ago, Mordred said:

The mathematics you posted isn't an explosion but a homogeneous isotropic space

It would appear that where we disagree is what an explosion is, and that we do agree that the three-dimensional hyperboloid that I described is indeed homogeneous and isotropic. Therefore, it suffices for me to prove that what I described is an explosion. If I succeed in doing that then this:

19 hours ago, Mordred said:

If you expand a Euclidean space uniformly that's expansion however in an explosion the space further away from the point of origin will expand faster than the space near the origin point. 

You can easily test that by using a cone and slicing the cone at various distances from the origin and measuring its area at the end of the cone furthest from the origin point. The cone is your explosion geometry not Euclidean geometry. Though its correctly a sphere expanding outward from the origin however all particle and vectors will have a preferred direction (outward from origin). That isn't the case in our universe. In our universe there is no preferred direction no vectors has a preferred direction due to any previous force applied regardless if those vectors are coordinates of geometry change or particles. A kinetic style explosion certainly has a preferred direction (outward) from its origin.

17 hours ago, Mordred said:

 Lets try another thought experiment using space expanding outward from the explosion. Lets set rate of flow outward from origin. Lets have an increase of space set at 1 parsec per second why not. Now place an observer that flows outward from from the center and set the observer 1 sec after the BB. Simply so we have space ahead of that observer. As space is created outward from the origin the space between the observer and the origin increases but if the space 1 second previous to that observer is flowing outward. Then no new space is being formed ahead of that same observer. That isn't uniform expansion where the distance between every coordinate  expands uniformly. (ignore time dilation we are just dealing with how space increases in each case)

is irrelevant because you have already indicated that the three-dimensional hyperboloid is homogeneous and isotropic.

To me, an explosion is simply a set of straight trajectories in spacetime that originate from a single point in spacetime. The trajectory of the [math]i[/math]-th particle, originating from the origin, and parametised by proper time [math]\tau[/math] is:

[math]x_i(\tau) = v_{xi} t_i(\tau)[/math]
[math]y_i(\tau) = v_{yi} t_i(\tau)[/math]
[math]z_i(\tau) = v_{zi} t_i(\tau)[/math]
[math]t_i(\tau) = \dfrac{c\ \tau}{\sqrt{c^2 - v_{xi}^2 - v_{yi}^2 - v_{zi}^2}}[/math]

where [math]v_{xi}, v_{yi}, v_{zi}[/math] are constant for the [math]i[/math]-th particle. But instead of considering discrete trajectories, one can consider a continuum of trajectories, parametised by [math](v_x, v_y, v_z)[/math], [math]v_x^2 + v_y^2 + v_z^2 \lt c^2[/math]:

[math]x(v_x, v_y, v_z, \tau) = v_x t(v_x, v_y, v_z, \tau)[/math]
[math]y(v_x, v_y, v_z, \tau) = v_y t(v_x, v_y, v_z, \tau)[/math]
[math]z(v_x, v_y, v_z, \tau) = v_z t(v_x, v_y, v_z, \tau)[/math]
[math]t(v_x, v_y, v_z, \tau) = \dfrac{c\ \tau}{\sqrt{c^2 - v_x^2 - v_y^2 - v_z^2}}[/math]

At a fixed value of [math]\tau[/math], [math](x(v_x, v_y, v_z, \tau), y(v_x, v_y, v_z, \tau), z(v_x, v_y, v_z, \tau), t(v_x, v_y, v_z, \tau))[/math] define the three-dimensional hyperboloid:

[math]c^2 t^2 - x^2 - y^2 - z^2 = c^2 \tau^2[/math]

Or alternatively, a single parameter [math]\tau[/math] family of three-dimensional hyperboloids are defined.

Note that [math](v_x, v_y, v_z)[/math] covers all values satisfying [math]v_x^2 + v_y^2 + v_z^2 \lt c^2[/math]. This means that a complete three-dimensional space is defined for all values of [math]\tau \gt 0[/math]. Also, only geometry is being considered, there is no consideration of matter distribution with respect to [math](v_x, v_y, v_z)[/math]. Finally, because the explosion being considered is cosmological, notions of realism that apply to real-world explosions do not apply.