Re: Formula for Decelerating Light

On May 6, 8:32 pm, jjs...@xxxxxxxxx wrote:
On May 6, 8:03 pm, Michael Helland <mobyd...@xxxxxxxxx> wrote:



On May 5, 1:12 pm, jjs...@xxxxxxxxx wrote:

On May 5, 1:07 pm, Michael Helland <mobyd...@xxxxxxxxx> wrote:

Think of Hubble Law, and think of Hubble's parameter as a parameter of
expansion.

You can redefine the expansion parameter in to a deceleration
parameter, with the units km/s^2, then you have:

v = c - (t * H)

Read this:http://en.wikipedia.org/wiki/Hubble_parameter#Derivation_of_the_Hubbl...

Hubbles law and parameter are a bit more complex then that.

Point is Hubble's law calculates the recessional velocity.

It could be reinterpreted to calculate decelerated velocity, then
Hubble's parameter would be in the units of deceleration and something
like this would work:

v = c - (t * H)

You don't grok the math for cosmology as evidenced by your formula so
go away...


H has to be deceleration, d/t^2.

Then it works fine.

Starting here:

c = fw

expansion: f goes down, w goes up.
tired light: f goes down, w stays the same.
deceleration: f goes down, w stays the same, but c bends to
compensate.

So deceleration and expansion match the same data.

There's no reason Hubble's parameter can't be written units of
deceleration, except taboo.
.