The previous blog text about the cosmic strings was here.

The talk was not only a very nice review of all the stuff that we've discussed previously, but it also presented a shocking surprise: Mark has identified the coordinates of CSL-1 - the Cosmic String Lensing candidate. So far the coordinates have been largely secret.

Imagine that you're in charge of a telescope that is comparable to the Hubble telescope, or you have just launched your personal space shuttle which carries a probe, telescope, or something like that - and you want to see whether there is a discontinuity near the object CSL-1.

How do you find the coordinates?

You open the PDF version of the paper by Alcala et al.

and you search for "double early type". As a consistency check, the redshift should be z=0.463. Try to contact all your astro-friends and give them these coordinates of CSL-1. Well, to simplify your work, the coordinates are:

- OACDF
- right ascension: 12 hours, 23 minutes 30.6 seconds (note that this angle is measured in hours, 1 hour = 15 degrees)
- declination: -12 degrees 38 minutes 57 seconds (southern celestial hemisphere, close to the equator)
- double early type, z=0.463, S/N 12

- 12h 23min 29.7sec, -12 deg 38' 27'', z=0.223 (?)

I'm sure Quantoken's theory predicts the position of this object.

ReplyDeleteObviously Lubos's string theory has predicted those coordinates first. So that's an area string theory is more useful than GUITAR. But then Lubos's crystal might be more effective than his theory :-)

ReplyDeleteI have now figured out the CORRECT relative He4 abundance in the universe. It was derived from my previous calculation of CMB temperature, which we know gives the exactly correct result of 2.7243K.

In my model, consider the critical density as one, then the baryon density is g, and the CMB energy density would be g^3/PI. So the ratio between radiation and baryon mass is g^2/PI.

With g = (2/PI)*sqrt(alpha). The ratio of radiation to baryon mass worked out to be:

4/PI^3 * alpha

(Please note the alpha above!!!)

Now let's see how much energy is released when 4 protons fuse into one He4 nuclear, through the proton-proton process.

Check out the periodic table of chemical elements:

http://www.webelements.com/index.html

H 1.00794

He 4.002602

calculate how much mass is lost when 4 proton turn into one He, divide that number by the atomic weight of He4, you find the number is STRIKINGLY close to alpha:

(H * 4 - He)/He

= (1.00794*4 - 4.002602)/4.002602

= 0.00729 = 1/137.17

So when 4 proton turn into one He4, about alpha portion of the mass is lost and converted into some other energy form. Assuming half of the enrgy lost turn into neitrino, and the other half turn into CMB energy, consulting the above formula which says

CMB/Baryon = 4/PI^3 * alpha

The relative abundance of He4 in the universe is then

4/PI^3 / (0.5)

= 8/PI^3

= 25.8%

That figure is in excellent agreement with the observed relative He4 abundance of 24% to 28%

The Big bang predicted a He4 abundance of about 25%, but that would be at the Beginning of the early universe. It does not match today's figure, once you consider that all stars are still burning Hydrogen and turn them into He4.

My calculation has shown that virtually all observed He4 are created by stars burning hydrogen for billions of years, which leaved no room for Big Bang to create any He4 at all to begin with.

Quantoken

I might also add that there is absolutely no adjustable parameter or any fudge factor in my model.

ReplyDeleteThe same g, which is fuxed by the value of alpha (g = (2/PI) * sqrt(alpha) ), is used to calculate the correct baryon density (g), the CMB energy and temperature (g^3/PI), the solar constant (result=1360W/m^2/sec), and the He abundance (2*g^2/(PI*alpha)). All leads to correct result within the argin of error of observed data.

Is that a coincidence you get 4 correct result out of one none-adjustable parameter?

Quantoken

Hi everbody,

ReplyDeleteJust a couple of comments.

Unless I am mistaken I think

all the objects of the Capodimonte Deep Field are located at negative declination which means that they are located on the southern hemisphere.

Also, how can you be sure that

this is the right object?. Its

magnitudes on different bands

do not seem to agree with the ones reported on the first paper

on the CSL-1.