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

On Apr 21, 9:33 am, Michael Helland <mobyd...@xxxxxxxxx> wrote:
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.

Excluded by observation. Do a basic literature search some time before
you rehash yet another failed theory.

[snip]
.

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