Re: Formula for Decelerating Light

On May 3, 5:02 pm, jjs...@xxxxxxxxx wrote:
On May 3, 5:15 pm, Michael Helland <mobyd...@xxxxxxxxx> wrote:

On May 3, 1:26 pm, jjs...@xxxxxxxxx wrote:

On May 3, 5:58 am, MichaelHelland<mobyd...@xxxxxxxxx> wrote:

On May 2, 11:20 pm, jjs...@xxxxxxxxx wrote:

On May 3, 12:26 am, MichaelHelland<mobyd...@xxxxxxxxx> wrote:
<snip>

Show a derivation of your formula....

Instead of calculating a Hubble distance, calculate Hubble time
instead.

Expansion time and deceleration time are identical, because f = 1/t.

Unlike Uncle Al I am interested to know where you dug these numbers up
from
v = 1 - (t * (20* 1.05702341) * 10^-13) like the 20 and 1.05702341....

The Hubble's Parameter is about 20 km / Mly

Actually try 71 km/s/Mpc


71 for megaparsecs, 20 for million light years.


That's the 20, and the 1.057 * 10^-13 converts km into light years.

435 364.304 m / s is what it should be

1 kilometers = 1.05702341 × 10^-13 light years

20 km should be 20 * 1.05702341 × 10^-13 light years.


I think I fucked up, because it should be in million light years.

So that would really be 10^-19?

So actually your math is total wrong as are some of your units and
your numbers...


But the concept is solid.

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)
.