Monday, September 21, 2009

Dark Energy's Demise?

From National Geographic News:

Dark Energy's Demise? New Theory Doesn't Use the Force

"Dark energy, a mysterious force proposed more than a decade ago to explain why the universe is flying apart at an increasingly faster clip, is no longer necessary.

That's the conclusion of a controversial new theory that shows how the accelerated expansion of the universe could be just an illusion."

The article refers to work by Blake Temple of UC Davis and Joel Smoller of University of Michigan. Their work suggests that our galaxy is in the middle of a density fluctuation that mimics the effects of an accelerating universe. This would explain supernova data without cosmic acceleration, but violate the Copernican Principle.

The inferrence of a repulsive "dark energy" has led science nowhere. Focus on DE is one more factor eroding the reputation of physics. These new theories are nails in the coffin of DE. As readers know, the data can be precisely matched by GM=tc^3 and a changing speed of light.

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Blogger Tim said...

Hi Louise, regaring the supernova eveidence you mention, I am still having trouble with your calculation of the redshift adjusted for varying c. Earlier you gave me this example:

"An object of redshift Z = 2.0 would be receding at 80% of today's speed of light co. We must divide by:

ci/co = sqrt(1 + 0.65) ~ 1.3

80% of today's co is only about 60% of ci at time supernova exploded. Apparent redshift is then 1.0, so we move the data point to the left."

I am using the relativistic redshift formula

1 + Z = sqrt((1+v/c0)/(1-v/c0))

and with Z=2 I get v/c0=0.8 ie. 80% as you do, but I don't see how you get that "80% of today's co is only about 60% of ci at time supernova exploded". In your paper you state that "When light of redshift Z was emitted, c was greater by factor sqrt(1 + Z)" ie.

ci = sqrt(1+Z) c0

So if we agree that an object of redshift Z=2 is receding at v/c0 = 0.8, then at the time of emission it is receding at

v/ci = v/(sqrt(1+Z)*c0)
= 0.8/sqrt(1+2)
= 0.46

or 46% of ci at the time of the explosion, not 60%. So accounting for varying-c gives an adjusted value for z of

z = sqrt((1+0.46)/(1-0.46)) - 1
= 0.64

rather than z=1 as you gave in your working.

To give an explicit example: suppose we measure an object at Z=2. It's apparent velocity is 0.8*3x10^8 = 2.4x10^8 m/s. However, at the time of emission

ci = sqrt(1+2)*3x10^8
= 5.2x10^8 m/s

so the velocity then is 2.4/5.2 = 0.46 or 46% of c at time of emission.

So is my calculation of the adjusted redshift z=0.64 correct?

3:44 PM  
Anonymous Bjoern said...

Hello again,

Sorry for bothering you again - but I haven't heard back from you for three months now. Did you miss my last mail, or do you still have too much to do for answering? Or don't you want to reply any more?

For the record, my questions are still about this calculation by you:

Why do you set z = 0 in that calculations? Einstein says clearly (pp. 116, 117) that z = 0
describes an Euclidean (i. e., flat) space. You have said that in your model, you consider a spherical space (S^3). According to Einstein, you'd have to use z = 1 for that case.

3:14 AM  
Blogger L. Riofrio said...

Sorry boys, really busy lately. See my August 26 comments about the high redshift data.

5:11 AM  
Anonymous Bjoern said...

On August 26, you only made a post about "Big Splash at Jupiter", nothing about high redshift data there. Did you mean another date?

But anyway: I don't see what high redshift data has to do with my question. Or was that addressed to Tim?

8:25 AM  
Blogger Tim said...

Hi Louise, actually my original comment was in response to the examples given in the post of Aug 26. I don't understand how you can justify plugging 0.65 into the expression

ci/co = sqrt(1 + 0.65) ~ 1.3

You seem to be saying that for large "observed" Z the speed of light tends to sqrt(1+0.65)*c0. This already contradicts the rest of your theory that c goes like t^(-1/3).

I've tried to give the theory a fair hearing, and looked carefully at your evidence, but at this point without some further input from you I would have to conclude that the supernova data clearly contradicts the theory. The straight line on your plot comes from calculating just two adjusted z's and magnitudes and drawing a line through them, whereas calculating further points shows that the line should actually be a curve which is very different from the data.

7:08 PM  
Blogger JJHB2 said...

you might want to check this DM related question and see if your work might give him another point of view

Jerry's an interesting person, to say the least, with broad experience and contacts, some of whixh I bet you share. I've known him casually since '77, mostly through my IT work.

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