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
--
You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to everything-list+unsubscr...@googlegroups.com <mailto:everything-list+unsubscr...@googlegroups.com>. To post to this group, send email to everything-list@googlegroups.com <mailto:everything-list@googlegroups.com>.
Visit this group at https://groups.google.com/group/everything-list.
For more options, visit https://groups.google.com/d/optout.

--
You received this message because you are subscribed to the Google Groups 
"Everything List" group.
To unsubscribe from this group and stop receiving emails from it, send an email 
to everything-list+unsubscr...@googlegroups.com.
To post to this group, send email to everything-list@googlegroups.com.
Visit this group at https://groups.google.com/group/everything-list.
For more options, visit https://groups.google.com/d/optout.

Reply via email to