Re: Formula for Decelerating Light



Michael Helland wrote:
On May 2, 3:35 pm, Sam Wormley <sworml...@xxxxxxxxx> wrote:
Michael Helland wrote:
On May 2, 11:50 am, Sam Wormley <sworml...@xxxxxxxxx> wrote:
Michael Helland wrote:
On May 1, 11:44 pm, Sam Wormley <sworml...@xxxxxxxxx> wrote:
Michael Helland wrote:
v = 1 - (t * (20* 1.05702341) * 10^-13)
t is the duration of the photon's journey in million years
v is the lights velocity in percentage of c (if v=0.5, then it is
0.5c)
So, picture two galaxies 5 million light years apart in steady space.
Plugging in t=5 gives you: 0.999999999989430000
So the velocity of light after traveling for 5 million years is
0.999999999989430000 * c.
Given a constant wave length, you can use the formula c = fw to figure
the lights new frequency and thus redshift.
Since the constant in the formula is derived from Hubble's Parameter,
it should be pretty consistent with Hubble's Law.
But, the key feature of this interpretation is that there is no
expansion, the same effect is achieved through deceleration.
<laughing>
The formula predicts a redshift for an object 5 million years away.
What is the redshift for an object 5 million years away?
Use 71 km/s/Mpc and the equations of the Doppler effect.
Which is about 20 km/Mly, which is how I derived the constant in my
formula.
I was hoping you could tell me so we'll see that my equation predicts
the right value without me knowing the right value.
In any case, let's say light at a wavelength of 650 nm travels 5
million years.
According to my formula, it's velocity should be:
1 - (5 * (20* 1.05702341) * 10^-13)
0.999999999989430000
The result is the percent of c, so it's velocity is:
0.999999999989430000 * c
If we use:
c = f w
Then
c = 650 f
and
0.999999999989430000 * c = 650 f_2
Therefore, the new frequency of light that has traveled for 5 million
years, in steady space, decelerating at the rate given by my formula
should be:
f_2 = (0.999999999989430000 * c) / 650
That's my prediction.
It should check out with what is observed.
Use 71 km/s/Mpc and the equations of the Doppler effect.


I did.

But I interpreted it as deceleration instead of expansion.

Then I calculated the new frequency, which should be consistent with
observation.

Did you getr identical red shifts with the two calculations?
Please show your calculations.
.



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