On 12/10/2017 5:25 PM, Lawrence Crowell wrote:
On Sunday, December 10, 2017 at 5:13:38 PM UTC-6, agrays...@gmail.com
wrote:
On Sunday, December 10, 2017 at 10:54:11 PM UTC, Lawrence Crowell
wrote:
On Sunday, December 10, 2017 at 3:34:33 PM UTC-6,
agrays...@gmail.com wrote:
On Saturday, December 9, 2017 at 2:17:38 PM UTC, Lawrence
Crowell wrote:
On Saturday, December 9, 2017 at 7:34:29 AM UTC-6,
agrays...@gmail.com wrote:
I think you're making the unwarranted assumption
that the measured shift in H is not
effected by the cosmological red shift which
presumably shifts all wave lengths. AG
Of course it shifts all wavelengths by the same
factor. So the spectrum of atoms are shifted
accordingly. With v = Hd the red shift factor is z =
v/c = H(d/c). for H = 70km/s/Mpc for v = c we then
have that d = c/H = 3x10^{5}km/s/(70Mpc/km/s) =
4.3x10^3Mpc = 1.4x10^{10}ly. So at z = 1 there lies
the cosmological horizon. We now observe galaxies with
z = 8 and the CMB has z = 1100. One can however thing
of these photons as emitted prior to these systems
crossing the horizon.
LC
Since a parsec is about 3.26 LY and the SoL is about
300,000 km/sec, the event horizon should be about
300,000/70 * 3.26 * 10^6 = 13971 * 10^6 LY =~ 13971 MLY =
13.971 BLY. But this is a far cry from about 50 BLY, which
is what I think the true distance is to the event horizon.
I probably didn't account for the intervening expansion.
How is accurate calculation done? TIA, AG
That is about it. There is a bit with significant figures for
you might want to use c = 299800km/s.
LC
But isn't the event horizon much farther out, about 50 BLY? AG
No that is about where the CMB surface of last scatter lies.
To clarify, you mean where it lies "/now"/; and /"now" /means the
(universe wide) time at which the CMB is 2.7degK.
Brent
It has z = 1100 and is further out beyond the horizon.
LC
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