Explanation of why expansion redshift optical illusion of a different kind of redshift

An object exhibiting a doppler redshift looks like this:


t=1
| | | | | | | | | |
| | | | | X | | | | |
| | | | | | | | | |

t=2
| | | | | | | | | |
| | | | | | X | | | |
| | | | | | | | | |

t=3
| | | | | | | | | |
| | | | | | | X | | |
| | | | | | | | | |


But an object exhibiting expansion redshift looks like this:

t=1
| | | | | | | | | |
| | | | | X | | | | |
| | | | | | | | | |

t=2
| | | | | | | | | |
| | | | | X | | | | |
| | | | | | | | | |

t=3
| | | | | | | | | |
| | | | | X | | | | |
| | | | | | | | | |


In t=3, X is farther away from the ends of the graph than it is in
t=1, because space expands over time.

If we consider X to be light, then it moves at a constant speed.

v = d/t

If v is constant, d and t need to increase proportionately.

But, if v isn't constant, but actually decreases over a few hundred
million light years, then t can increase while d doesn't change at
all.

So if you can visualize X moving at a constant speed on a grid that
expands;

And then if you can visualize X moving on a constant grid, but
decelerating over it;

Then you should be able to determine that both graphs (constant_speed/
expanding_grid versus decelerating_speed/constant_grid) are
mathematically identical.

Here are three steps of an object traveling left to right over a grid,
and the object slows down:

|X.....| (step 1)
|..X...| (step 2)
|...X..| (step 3)

Now the same three steps, played while you're zooming-in during each
step (more space) then you can zoom-in on the grid to give the
illusion that its traveling at a constant speed

|X.....|
| . .X . . . |
|. . . X . . |

So expansion redshift is an optical illusion, light slowing down in
static space.

We're seeing the limits of special relativity's speed of light
postulate when we observe expansion.

This isn't a full model, of course, but it is different that Tired
Light models that have been examined in the past.
.

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