Cont: Why James Webb Telescope rewrites/doesn't the laws of Physics/Redshifts (2)

[W.D.Clinger]

You continue to speak as though you were completely ignorant of the relevant mathematics and physics.

 

[Mike Helland]

You continue to speak as though you were completely ignorant of the relevant mathematics and physics.

You might be looking too much into some part of it.

You have a spacetime with one dimension of time and two dimensions of space, x, and y, You have a metric tensor g, which defines a field of tensors, ie. a tensor at every point (t, x, y). Each of those tensors has a tt component.

Now, for the purpose of investigating some of the properties of this setup, you can visualize the dataset consisting of [x, y, g_tt(t, x, y)] as a 3D shape. If g_tt is constant, it will have a uniform height.

I'm not saying g_tt in an actual dimension of reality if that's what's throwing you off.

 

[W.D.Clinger]

In this metric though:

http://latex.codecogs.com/gif.latex?ds^2 = (1-\frac{r}{R})^{-2} dt^2 - dx^2 - dy^2 - dz^2​

The height of the chart would be infinity nearly everywhere, except around the observer, where it dips down to 1.

Are we to assume r = sqrt (x² + y² + z²) ?

Or should we assume R = sqrt (x² + y² + z²) ?

Or none of the above?

ETA: (deleted an incorrect form of the example given in post #1045)

 

[Mike Helland]

Are we to assume r = sqrt (x² + y² + z²) ?

Yeah. Not sure why I mixed coordinates like that. That is confusing.

In any case, I'm pretty sure I get what you're saying.

Let's say there's an event at (t, x, y, z). The metric tensor, being a tensor field, places a tensor at the point (t, x, y, z).

Let's say we change coordinates, using a different origin, and now that event lies at new coordinates (t', x', y', z').

The metric tensor should place a tensor there, and am I right that it should be the same tensor (have the same scalar values) as before?

 

[W.D.Clinger]

Let's say there's an event at (t, x, y, z). The metric tensor, being a tensor field, places a tensor at the point (t, x, y, z).

Let's say we change coordinates, using a different origin, and now that event lies at new coordinates (t', x', y', z').

The metric tensor should place a tensor there, and am I right that it should be the same tensor (have the same scalar values) as before?

It should be the same tensor, but the part I highlighted in red doesn't make sense as written, and would be incorrect if rewritten to convey what I think you were trying to say.

The reason that highlighted phrase doesn't make sense is that it doesn't really make sense to ask whether a tensor has "the same scalar values". I suspect you were trying to say the numerical components of the tensor are unchanged by the coordinate transformation, but that is flat-out false. Although it's the same tensor as before, its coordinate-dependent components will be different in the new coordinate system.

In my previous message, I was going to give you an example of that, but I got a minus sign wrong so I deleted the example. Here is the example:

ds² = − (ρ + 1)² c² dt² + (ρ + 1)⁻⁴ dρ² + (ρ / (ρ + 1))² [ dθ² + sin² θ dφ² ]​

In that metric form, the radial coordinate ρ (which is the Greek letter rho) ranges over the non-negative real numbers.

Do you recognize the tensor field expressed by that metric form?

 

[Mike Helland]

It should be the same tensor, but the part I highlighted in red doesn't make sense as written, and would be incorrect if rewritten to convey what I think you were trying to say.

The reason that highlighted phrase doesn't make sense is that it doesn't really make sense to ask whether a tensor has "the same scalar values". I suspect you were trying to say the numerical components of the tensor are unchanged by the coordinate transformation, but that is flat-out false. Although it's the same tensor as before, its coordinate-dependent components will be different in the new coordinate system.

What if you kept the units and coordinate system (Cartesian or polar) and units, but just changing the origin.

I'm using the 2D space + 1D of time because it's easier to visualize than 4D spacetime and a bit richer than 1D space + 1D time.

So, I'll try to stick with radial coordinates, so we have (t, r, θ). Piece of paper on the table. Place a dot, that's r₀, where r=0. Now any point on the paper can be described by an r and θ using the point.

At all points on the paper there is a tensor with a g_tt component.

At r=0, g_tt = 1 (ignoring the -c² for now).

r₀ is the original origin. Now let's move it to a new point, say θ=0, r=0.5 from the original origin. That's r₁, and now r=0 there.

Now the metric tensor gives a different g_tt component (using the 1/(1-r) metric) for r₀.

Now the g_tt for r₀, which is now at r=0.5 from the new origin, comes out to g_tt = 2² (eta the square).

That's what you're saying you can't do. That's trying to change the territory because we've already established g_tt = 1 for that point.

Is that correct?

In my previous message, I was going to give you an example of that, but I got a minus sign wrong so I deleted the example. Here is the example:

ds² = − (ρ + 1)² c² dt² + (ρ + 1)⁻⁴ dρ² + (ρ / (ρ + 1))² [ dθ² + sin² θ dφ² ]​

In that metric form, the radial coordinate ρ (which is the Greek letter rho) ranges over the non-negative real numbers.

Do you recognize the tensor field expressed by that metric form?

Assuming ρ = z, that's FLRW,

I think.

 

[W.D.Clinger]

It should be the same tensor, but the part I highlighted in red doesn't make sense as written, and would be incorrect if rewritten to convey what I think you were trying to say.

The reason that highlighted phrase doesn't make sense is that it doesn't really make sense to ask whether a tensor has "the same scalar values". I suspect you were trying to say the numerical components of the tensor are unchanged by the coordinate transformation, but that is flat-out false. Although it's the same tensor as before, its coordinate-dependent components will be different in the new coordinate system.

What if you kept the units and coordinate system (Cartesian or polar) and units, but just changing the origin.

I'm using the 2D space + 1D of time because it's easier to visualize than 4D spacetime and a bit richer than 1D space + 1D time.

So, I'll try to stick with radial coordinates, so we have (t, r, θ). Piece of paper on the table. Place a dot, that's r₀, where r=0. Now any point on the paper can be described by an r and θ using the point.

At all points on the paper there is a tensor with a g_tt component.

At r=0, g_tt = 1 (ignoring the -c² for now).

r₀ is the original origin. Now let's move it to a new point, say θ=0, r=0.5 from the original origin. That's r₁, and now r=0 there.

Your discussion is insufficiently precise. It is unclear, for example, what you mean by moving the origin to a new point. Are you transforming θ as well as r? If so, then your metric form is going to be a lot more complicated in the new coordinates than in the original coordinates. If you aren't transforming θ, then you are continuing to calculate the angle θ with respect to the original origin, not with respect to the new origin.

I will give a more precise example below.

By the way, you are likely to confuse yourself by using r₀ and r₁ as names for events. You were using the letter r as a coordinate; indeed, you used the same letter r to name the radial coordinate of both your original coordinate system and your transformed coordinate system. That use of the same letter with two different meanings was already sloppy and confusing. Using the same letter to name events, even though you add subscripts, makes it triply confusing.

Now the metric tensor gives a different g_tt component (using the 1/(1-r) metric) for r₀.

Now the g_tt for r₀, which is now at r=0.5 from the new origin, comes out to g_tt = 2² (eta the square).

If the origin is moved via a coordinate transformation analogous to the transformation I will give below, the value of g_tt at the original origin remains unchanged.

Your description of your coordinate transformation was insufficiently precise for me to say your last sentence quoted above is wrong, but I'm inclined to regard your g_tt = 2² as an indication of befuddlement.

That's what you're saying you can't do. That's trying to change the territory because we've already established g_tt = 1 for that point.

Is that correct?

I doubt whether you understand what I was saying. I was certainly not saying "we've already established g_tt = 1 for that point." We had established that g_tt = 1 for that point in your original coordinates, and it is possible that g_tt = 1 for that point in the new coordinates as well, but whether that is true depends on the specific coordinate transformation, and your description of your coordinate transformation was insufficiently precise for anyone (including yourself) to draw a firm conclusion here.

Let's return to your metric form for the geocentric Helland universe you've been talking about for the past several days. You wrote that metric form using coordinates t, r, θ, and φ, where 0 ≤ r < 1.

Let's transform to a new coordinate system that uses exactly the same t, θ, and φ coordinates, but changes the radial coordinate to q = r − ½, where −½ ≤ q < ½.

The metric tensor field of your geocentric Helland universe is then expressed by both of these line elements:

ds² = − (1 − r)⁻² c² dt² + dr² + r² [ dθ² + sin² θ dφ² ]
= − (½ − q)⁻² c² dt² + dq² + (q + ½)² [ dθ² + sin² θ dφ² ]​

Letting E₀ be the event whose t, r, θ, and φ coordinates are (0, 0, 0, 0), it's easy to see that the coordinates of E₀ in this new coordinate system are (0, −½, 0, 0). The event E₁ whose t, r, θ, and φ coordinates are (0, ½, 0, 0) has coordinates (0, 0, 0, 0) in this new coordinate system.

It is therefore possible to say that the transformation from r to q moves the origin from E₀ to E₁, but I don't know whether that's what you meant by moving the origin to a different point. If it isn't what you meant, then this transformation from r to q should help you to understand why you need to be more precise when describing a coordinate transformation.

In my previous message, I was going to give you an example of that, but I got a minus sign wrong so I deleted the example. Here is the example:

ds² = − (ρ + 1)² c² dt² + (ρ + 1)⁻⁴ dρ² + (ρ / (ρ + 1))² [ dθ² + sin² θ dφ² ]​

In that metric form, the radial coordinate ρ (which is the Greek letter rho) ranges over the non-negative real numbers.

Do you recognize the tensor field expressed by that metric form?

Assuming ρ = z, that's FLRW,

I think.

Nope.

That metric form describes the tensor field of your geocentric Helland universe.

All three of the line elements in this message describe exactly the same tensor field, but their descriptions use different coordinate systems. Those coordinate systems differ only in their radial coordinate: r, q, or ρ.

I strongly recommend the following exercise: Prove that all three of those metric forms describe exactly the same tensor field.

 

[Mike Helland]

The metric tensor field of your geocentric Helland universe is then expressed by both of these line elements:

ds² = − (1 − r)⁻² c² dt² + dr² + r² [ dθ² + sin² θ dφ² ]
= − (½ − q)⁻² c² dt² + dq² + (q + ½)² [ dθ² + sin² θ dφ² ]​

Letting E₀ be the event whose t, r, θ, and φ coordinates are (0, 0, 0, 0), it's easy to see that the coordinates of E₀ in this new coordinate system are (0, −½, 0, 0). The event E₁ whose t, r, θ, and φ coordinates are (0, ½, 0, 0) has coordinates (0, 0, 0, 0) in this new coordinate system.

It is therefore possible to say that the transformation from r to q moves the origin from E₀ to E₁, but I don't know whether that's what you meant by moving the origin to a different point. If it isn't what you meant, then this transformation from r to q should help you to understand why you need to be more precise when describing a coordinate transformation.

Huh.

Well, I thought the issue was the metric was "geocentric". That there is a special place. Specifically, there is a special observer.

So I said, you just move the origin to a new observer at a new place, and now, r=0 there, so it's not special.

You seem to be talking about coordinate transformations in a more general sense, where I specifically had in mind changing frames of reference to another observer.

But it's the same problem either way.

It seems this would also be a problem for de Sitter's original coordinates:

http://latex.codecogs.com/gif.latex?ds^2 = -\left(1-\frac{r^2}{\alpha^2}\right)dt^2 + \left(1-\frac{r^2}{\alpha^2}\right)^{-1}dr^2 + r^2 d\Omega_{2}^2.

I assumed since this was valid (at least mathematically) then the "geocentrism" isn't a huge deal.

I think we have to go back to what the intention was. The intention of the metric is to describe the past light cone of an observer. The relevant "territory" in this case is not the universe as a whole, but the past light cone of an observer.

So, the territory of one observer differs from the territory of another observer.

That's fine. All that means is the past light cone for one observer is different than the past light cone of another observer. That's basically a given.

In which case, I suppose they both need to be thought of as sub-manifolds of a greater manifold.

It seems the technique would be to start at the t=0 (present) and r=0 (position of the observer of the light cone) and then we can work our way back in time along a null path.

And any given place, change direction in space and time. When you get back to t=0 you'll be at a different observer.

Essentially, the universe as a whole doesn't have a particular metric (as of yet), but each observer in the universe has a past light cone, and you can use where they intersect to move between observers.

I strongly recommend the following exercise: Prove that all three of those metric forms describe exactly the same tensor field.

Sounds good. Before I get too far... is that -4 supposed to be there?

 

[W.D.Clinger]

Less than a week ago, on 22 October, Mike Helland began to propose metric forms that describe geocentric universes. It took him a while to decide which of those metric forms actually describes his new Helland universe, which is radically different from the universes he had proposed previously. For the moment, he has settled upon the first of these three metric forms:

ds² = − (1 − r)⁻² c² dt² + dr² + r² [ dθ² + sin² θ dφ² ]
= − (½ − q)⁻² c² dt² + dq² + (q + ½)² [ dθ² + sin² θ dφ² ]
= − (ρ + 1)² c² dt² + (ρ + 1)⁻⁴ dρ² + (ρ / (ρ + 1))² [ dθ² + sin² θ dφ² ]​

All three of those metric forms describe exactly the same tensor field, but they use different coordinate systems. All three of those coordinate systems share the same t, θ, and φ coordinates, but differ in their radial coordinate: r, q, or the Greek letter ρ. The ρ coordinate ranges over all non-negative real numbers, while r and q have a more limited range: 0 ≤ r < 1 and −½ ≤ q < ½. The three radial coordinates are related by r = q + ½ = ρ / (ρ + 1).

In the new Helland universe, the event E₀ with coordinates (t=0, r=0, θ=0, φ=0) corresponds to our position in that (mythical!) universe at the current time. That Helland universe is geocentric in the following sense: The Helland universe is spherically symmetric around the spatial location of E₀, and is not spherically symmetric around any other spatial location.

Mike Helland has been denying the geocentrism of his Helland universe, and continues to do so in his most recent post, from which I quote below.

Huh.

Well, I thought the issue was the metric was "geocentric". That there is a special place. Specifically, there is a special observer.

So I said, you just move the origin to a new observer at a new place, and now, r=0 there, so it's not special.

You seem to be talking about coordinate transformations in a more general sense, where I specifically had in mind changing frames of reference to another observer.

But it's the same problem either way.

Moving the origin is a coordinate transformation. Coordinate transformations do not change the spacetime manifold. The map is not the territory.

If the Helland universe were spatially homogeneous and isotropic, then it would not be geocentric. But the Helland universe is neither homogeneous nor isotropic, as Mike Helland appears to have admitted in post #1001.

The absolute value of the g_tt component of the Helland universe's metric tensor field reaches its global minimum at r=0. That is suggestive but not conclusive, because tensor components are coordinate-dependent.

Mike Helland has also calculated and posted the Ricci curvature tensor for his Helland universe. The tt component of that Ricci tensor becomes infinite at r=0, for essentially the same reason that the Ricci curvature of a cone blows up at its vertex. There is only one spatial location with r=0, and that is the only location at which the Ricci curvature blows up.

Although individual components of the Ricci curvature tensor are coordinate-dependent, the Ricci scalar R = g^μν R_μν is an invariant, the same in all coordinates. The Ricci scalar R becomes infinite at r=0, and that is the only spatial location at which the Ricci scalar blows up.

The spatial location of E₀ is therefore the unique spatial location of the Helland universe at which the Ricci scalar blows up, and it is also the unique spatial location for which the surrounding Ricci scalar field is spherically symmetric.

Coordinate transformations cannot give those special properties to some other spatial location of the Helland universe. The map is not the territory.

It seems this would also be a problem for de Sitter's original coordinates:

http://latex.codecogs.com/gif.latex?ds^2 = -\left(1-\frac{r^2}{\alpha^2}\right)dt^2 + \left(1-\frac{r^2}{\alpha^2}\right)^{-1}dr^2 + r^2 d\Omega_{2}^2.

I assumed since this was valid (at least mathematically) then the "geocentrism" isn't a huge deal.

As Mike Helland has been told on several occasions, using that particular coordinate system to express the metric of de Sitter space is misleading. For one thing, those coordinates lead to an artificial horizon at r=α.

Because de Sitter space has constant positive curvature, that artificial horizon can be removed by a coordinate transformation that converts the metric to FLRW form, thereby revealing the homogeneity and isotropy of de Sitter space.

You can't play that game with the Helland universe, because the curvature at its central location is unique to that location.

I think we have to go back to what the intention was. The intention of the metric is to describe the past light cone of an observer. The relevant "territory" in this case is not the universe as a whole, but the past light cone of an observer.

The metric tensor field is a mathematical object. You can't change it by talking rubbish about its "intention".

Although the metric tensor field determines light cones, it also describes curvature. You can't change that curvature by pretending some particular light cone, or light cones in general, are the only relevant "territory".

So, the territory of one observer differs from the territory of another observer.

The Helland universe is the same for all observers. You can use different coordinate systems to describe it, such as the three coordinate systems I used near the beginning of this post, but all of those coordinate systems describe exactly the same territory.

That's fine. All that means is the past light cone for one observer is different than the past light cone of another observer. That's basically a given.

The metric tensor field defines the Helland universe and determines the light cones for all observers.

In which case, I suppose they both need to be thought of as sub-manifolds of a greater manifold.

It seems the technique would be to start at the t=0 (present) and r=0 (position of the observer of the light cone) and then we can work our way back in time along a null path.

And any given place, change direction in space and time. When you get back to t=0 you'll be at a different observer.

It's hard to read that as anything other than the confused rambling of someone who doesn't actually understand general relativity.

Essentially, the universe as a whole doesn't have a particular metric (as of yet) , but each observer in the universe has a past light cone, and you can use where they intersect to move between observers.

The highlighted phrase is nonsense, and the entire paragraph is risible.

Mike Helland did eventually settle on a metric form for his Helland universe, and that metric form defines the metric tensor field of the Helland universe as a whole.

Referring to the metric form as written with ρ as the radial coordinate:

Sounds good. Before I get too far... is that -4 supposed to be there?

Yep.

It is of course possible that I made a mistake.

But I don't think you need to worry about getting too far. I doubt whether you'll be able to get far enough to figure out whether that exponent is correct, even though you have been given the relationship between r and ρ.

 

[Mike Helland]

In the new Helland universe

"The universe" is your sacred cow.

When I began to question the expansion of the universe it had eventually occurred to me that without the expansion of the universe, and without the big bang, the universe doesn't have an age... it doesn't really have a size, or shape. It doesn't have a birthday, or fax number.

It's really a whole lot of nothing.

Here's Newton's principia.

https://www.gutenberg.org/cache/epub/28233/pg28233.txt

Ctrl+F, "universe", not found.

Well, the whole thing's in Latin. So that's why.

"universum"? Not found.

https://redlightrobber.com/red/links_pdf/Isaac-Newton-Principia-English-1846.pdf

He tosses the word "universe" in at the end when talking about the Greeks.

But here's the thing.

My model has a volume of space, and in it are planets, and stars, and galaxies, and galaxies, and super clusters.

What's in FLRW? None of those thing. But it's the whole universe! Yeah, it's a perfect fluid. It's the whole universe, minus everything we actually observe.

So I've sort of adopted the mindset that the atoms and rocks and trees and people and houses and planets and stars and galaxies and galaxy clusters and walls of galaxies are all real.

The universe seems ... meh. Like, we have "Nature" and "Reality" and "The World" and "The Cosmos"?

Do we need another nickname? We do we need "the universe" in between us an reality?

How many omnipresent grounds of being do we need? And what's with the "uni" bit? Sounds like monotheism when you really get into it.

Anyways, I realize everyone believes in the universe thing. It's, like super important and stuff. But it's actually kind of silly. But then I think birthday's are silly. So whatever.


Again. Intentions.

The Helland metric doesn't describe a universe. It's describes observable reality. It describes a past light cone. And a larger picture of reality can be constructed through a network of past of light cones.


You can essentially arrive that this conclusion by saying that the speed of light is constant for all that observe it, and time is observed to run slower at great distances.

If clocks are running slower at z>0, then observer's looking at those clocks have to measure the speed of light as moving at the same speed as we measure it. So if their clock is going at 1/(1+z), then the speed of light is c/(1+z) too.

That gives the light cones the shape they we have.

Note that even in FLRW the speed of light in proper coordinates is c/(1+z), which is c - H(z)r(z).

http://latex.codecogs.com/gif.latex?\frac{c - Hd}{c} = 1- \frac{Hd}{c}

 

[hecd2]

Mike are you feeling ok?

 

[Mike Helland]

Mike are you feeling ok?

Sure.

Hypothetical scenario for you or anyone else reading this that would like to take a crack at it.

Let's assume I spoke perfect English and had no significant mental impairments.

Let's say I told you, where I'm from, we believe in rocks, and we believe in clouds, and stars and galaxies, and that these things are real and in reality, and reality is real.

All of that is just to preface, this isn't solipsism, which most people just assume it is.

Then I told you, we don't believe in a universe.

Why should I?

Why should I believe, beyond or in addition to reality and the things in it, I should believe in this relatively modern idea that is "the universe", which is apparently all of existence as we know it... and it's a whopping 3.5 times older than the rock we live on?

Go ahead.

 

[steenkh]

Sure.

Hypothetical scenario for you or anyone else reading this that would like to take a crack at it.

Let's assume I spoke perfect English and had no significant mental impairments.

Let's say I told you, where I'm from, we believe in rocks, and we believe in clouds, and stars and galaxies, and that these things are real and in reality, and reality is real.

All of that is just to preface, this isn't solipsism, which most people just assume it is.

Then I told you, we don't believe in a universe.

Why should I?

Why should I believe, beyond or in addition to reality and the things in it, I should believe in this relatively modern idea that is "the universe", which is apparently all of existence as we know it... and it's a whopping 3.5 times older than the rock we live on?

Go ahead.

For a starter, I’d ask you to look through a telescope, then gradually build upon that going through the history of astronomy and cosmology without the detours, and end up with what we know today.

What reason would there be to think that the universe does not exist, when you can observe it in every possible wavelength?

 

[Mike Helland]

For a starter, I’d ask you to look through a telescope, then gradually build upon that going through the history of astronomy and cosmology without the detours, and end up with what we know today.

What reason would there be to think that the universe does not exist, when you can observe it in every possible wavelength?

When I look through a telescope I observe stars and galaxies.

What direction do I point the telescope to see "the universe" and what should I be looking for?

 

[steenkh]

When I look through a telescope I observe stars and galaxies.

What direction do I point the telescope to see "the universe" and what should I be looking for?

Stars and galaxies.

 

[Mike Helland]

https://www.scientificamerican.com/...-lead-to-a-new-understanding-of-the-universe/

"Several unexplained measurements are threatening to upend scientists’ understanding of the universe’s origin and fate"

 

[3point14]

When I look through a telescope I observe stars and galaxies.

What direction do I point the telescope to see "the universe" and what should I be looking for?

"What the hell is this 'forest' you keep going on about. All I can see is trees!"

 

[Mike Helland]

"What the hell is this 'forest' you keep going on about. All I can see is trees!"

There are these two young fish swimming along, and they happen to meet an older fish swimming the other way, who nods at them and says, “Morning, boys. How’s the water?” And the two young fish swim on for a bit, and then eventually one of them looks over at the other and goes, “What the hell is water?” - David Foster Wallace

I get your point.

But my name for the forest is "reality".

Adding "the universe" to the lexicon has provided nothing except a ridiculous origin story and the necessity of the multiverse.

 

[The Man]

There are these two young fish swimming along, and they happen to meet an older fish swimming the other way, who nods at them and says, “Morning, boys. How’s the water?” And the two young fish swim on for a bit, and then eventually one of them looks over at the other and goes, “What the hell is water?” - David Foster Wallace

I get your point.

But my name for the forest is "reality".

Adding "the universe" to the lexicon has provided nothing except a ridiculous origin story and the necessity of the multiverse.

While removing "the universe" from the lexicon neither eliminates the evidence of an origin (a very hot dense period of almost, if, no spatial extent) nor the possibility (not "necessity") of a multiverse, in reality. Getting rid of the word doesn't do what you want because simply having it didn't do what you just claimed it did. It's the other way 'round, that evidence supports a very hot dense period of almost, if, no spatial extent (an origin for the perceived universe) and having the mere possibility of a multiverse extending the possible universe beyond what we can directly perceive. The word didn't create the evidence or implications but simply came and is defined as a result of them.

Similarly, just getting rid of the word "cancer" doesn't eliminate a condition where cells reproduce rapidly and out of control perhaps spreading throughout the body. The word didn't cause or drive the condition, in reality the condition simply spawned the word.

 

[Mike Helland]

While removing "the universe" from the lexicon neither eliminates the evidence of an origin (a very hot dense period of almost, if, no spatial extent) nor the possibility (not "necessity") of a multiverse, in reality. Getting rid of the word doesn't do what you want because simply having it didn't do what you just claimed it did. It's the other way 'round, that evidence supports a very hot dense period of almost, if, no spatial extent (an origin for the perceived universe)

One interpretation of the evidence supports that. The interpretation that says there's a universe with a finite age.

and having the mere possibility of a multiverse extending the possible universe beyond what we can directly perceive. The word didn't create the evidence or implications but simply came and is defined as a result of them.

Cosmologists have adopted a type of multiverse closely intertwined with the inflation epoch:

https://en.wikipedia.org/wiki/Eternal_inflation

"According to eternal inflation, the inflationary phase of the universe's expansion lasts forever throughout most of the universe. Because the regions expand exponentially rapidly, most of the volume of the universe at any given time is inflating. Eternal inflation, therefore, produces a hypothetically infinite multiverse, in which only an insignificant fractal volume ends inflation. "​

Similarly, just getting rid of the word "cancer" doesn't eliminate a condition where cells reproduce rapidly and out of control perhaps spreading throughout the body. The word didn't cause or drive the condition, in reality the condition simply spawned the word.

True.

But cancer is something you can point to. It has a physical position relative to other things.

That's not true for the universe.

And like I said, we already had a name for the universe. Several, in fact. Reality, Nature, the World, Existence, Being.