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Previous Next Up Topic Cosmology / Alternative Cosmology / CMBR as equilibrium temperature of an eternal Universe (14859 hits)
By machian cosmologist Date 2013-06-22 10:56 Edited 2013-06-24 11:52
Hello!

I'd like to discuss a calculation I made recently:

Assume stars like the sun shined since forever in the Universe.
Further assume the sun is a perfect average star.
Then you can calculate the energy density due to star-light.

Here are the values I used for this calculation:

-- Luminosity of the Sun:  P = 3.846 * 10^26 W
-- Lifetime of the Sun:       dt = 10 * 10^9 yr
-- Mass of the Sun:           M =   2 * 10^30 kg
-- Density of luminous matter: rho =  5 * 10^-28 kg/m³

I calculate the energy density like this:

u = P * dt / (M/rho)

And the resulting temperature like this:

T = sqrt(sqrt(u/a)),    with the radiation constant a = 7.565767 J m^-3 K^-4

With the values given above I get:

T = 2.516 K

Pretty close to the CMBR temperature, isn't it?

What do you think?

(EDIT: Luminousity --> Luminosity)
By Mike Petersen Date 2013-06-22 12:45
Hi...

Sorry, but I can't agree with your initial assumptions.  The sun does not shine forever, and the sun is not the perfect, average star.

If you could come up with similar calculations using more realistic assumptions, I might tend to be intrigued.  Until then, this just looks like a mathematical oddity.

I mean no offense.  You have come up with an exceedingly clever point of view.  Convince me to see it!

Regards,
Mike Petersen
By machian cosmologist Date 2013-06-22 14:07
Hi Mike!

Mike Petersen said:

The sun does not shine forever, and the sun is not the perfect, average star.


Ok, maybe you got me wrong regarding your first point:

I don't assume that the sun shines forever but that stars like the sun have been shining forever.
So you as you see I prefer an eternal Universe without an absolute beginning.

Now to the second point. You say the sun is not the perfect, average star.
I can only rely on what I can find in the world wide web.
But maybe my arguments would be richer if I could explain it with the Hertzsprung-Russel-diagram...

You can find a lot of links if you search for "sun average star".
Here is one example with a quote:

http://www.cosmoquest.org/forum/showthread.php?38628-The-Sun-as-an-Average-Star&s=663c3bbf0a4e40f517a791a1c0e882be
("The Sun is also said to be average. Average for a star is based on its luminosity or brightness. In comparison to other stars in the galaxy, the Sun is said to be of a typical luminosity.")

Kind Regards,
S.E.
By Ari Jokimäki Date 2013-06-22 15:35
I tried to recreate this calculation, but with bad results. I also didn't get the units to match (I don't seem to be able to get rid of meters, to have only kelvins as a unit for the result). Perhaps you could show your calculation with numerical values and units.
By machian cosmologist Date 2013-06-22 16:24 Edited 2013-06-22 17:15
Hi Ari!

Ari Jokimäki said:

Perhaps you could show your calculation with numerical values and units.


Of course I can do that.

Here we go:

u = (3.846 * 10^26 W) * (10 * 10^9 * 365 * 24 * 3600 s)  * (5 * 10^-28 kg/m³) / (2 * 10^30 kg)   =   3.032 * 10^-14 J/m³

T = (u/a)^(1/4) = sqrt( sqrt( (3.032 * 10^-14 J/m³) / (7.565767 * 10^-16 J m^-3 K^-4) ) ) = 2.516 K

As you can see the meters and Joule cancel out due to the dimension of the radiation constant a.

Can you reproduce it now?

Kind regards,
S.E.
By machian cosmologist Date 2013-06-22 16:53
My source for the relation between u and T was this link:
http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/raddens.html
By Ari Jokimäki Date 2013-06-23 06:47
I see now where I went wrong. I was a bit careless and had M*rho instead of M/rho under the divider line.
By Mike Petersen Date 2013-06-23 11:52
Hi, S.E.

You quoted from another source, "the Sun is said to be of a typical luminosity."  Yes, I agree.  I've read that same (or similar) statement many times.  The H-R Diagram is extremely useful, but only in a qualitative way.  I don't agree that "typical" equates to "average."

If I may give an example.  I consider myself to be a "typical" American male (of my age group - don't ask! lol) but does that mean I am "average?"  No, I am a typical American male, but the only thing you could "average" about me could be my weight, my age, my height; in other words, quantifiable terms.  The same must be said of the Sun.  Yes, it is typical, but is it "average?"  For example, what if it were the median (rather than the mean), would that throw things off?

Okay, I don't want to beat the horse, especially because he's not dead yet.  I'm not arguing your point, I am just asking how you arrived at it quantitatively rather than qualitatively.

Thanks for listening.

Regards,
Mike
By machian cosmologist Date 2013-06-23 13:25
Hi Mike,

I'm happy to tell you that now I have a quantitative argument.
I made a calculation of the average luminosity with the 25 nearest stars to the sun.

Here is the data:
http://www.astro.wisc.edu/~dolan/constellations/extra/nearest.html

And here is the formula I used:
https://en.wikipedia.org/wiki/Luminosity#Magnitude_formulae

My result for the average luminosity of these stars is:  L = 1.32883596 * L_sun

If I put this into my calculation of the resulting energy density I get even 2.7 K for the temperature.

Regards,
S.E.
By machian cosmologist Date 2013-06-28 11:08
Update:

I thought it would be better to take also the average of the masses of 22* of the 26 nearest stars for my calculation.
M_avg = 0.5917 * M_sun

So, with this I get:

T = 3.08 K.

Still close to the measured CMB value, isn't it?

*) Unfortunately I found no values for the missing 4 stars on wikipedia
By Mike Petersen Date 2013-06-28 11:44
Boy, I really hate to be a spoil sport, but taking average luminosity of the 22 nearest stars (or even 26 for that matter) is hardly a reasonable selection.  There is nothing that states the 22 nearest stars are an average representative sample of ALL stars in the universe, which is what you are really saying here.  I'm sorry, but I just can't agree.  Your tiny sample probably has a HUGE margin of error compared the real (if unknown) value.

Now, if you could find another study somewhere that gives an average luminosity of a good representative sample of all the stars in the observable universe, then I would buy that.  Heck, I would even take an approximation.

Good luck with your calculations. 

Regards, Mike
By machian cosmologist Date 2013-06-28 16:18
Okay, I guess you're right.
At least assuming a 50% error in every value I use the most "bad" results for the energy density are only one order of magnitude away from the measured value.
And the resulting temperature is somewhere between   1.7 K   and    5.0 K .

If there is someone out there who has an idea how to get better values for average luminosity, lifetime and mass of the stars in the universe I would be very happy to hear about it!

Best regards,
S.E.
By Ari Jokimäki Date 2013-06-29 12:46
You could search Vizier for star catalogs. Some of them might even have the desired parameters already in the catalog, but you might have to calculate some of the parameters yourself.
By machian cosmologist Date 2013-06-29 16:08
Hi again!

First of all: Thank you Ari for your recommendation!

Now I want to tell you about an alternative calculation which is even more simple than the one I made before:

I just take a typical luminosity of a galaxy multiply it with 100 billion years and divide it by the volume available for one galaxy:

u = 10^37 W * 10^11 yr / (10^21 ly^3) = 3.724 * 10^-14  J / m^3
=> T = 2.65 K

Remember that I prefer an eternal and infinite Universe.
So there is enough time and space for an equilibrium radiation due to starlight.
By Ari Jokimäki Date 2013-06-30 07:39
Mike said:

Boy, I really hate to be a spoil sport, but taking average luminosity of the 22 nearest stars (or even 26 for that matter) is hardly a reasonable selection.  There is nothing that states the 22 nearest stars are an average representative sample of ALL stars in the universe, which is what you are really saying here.


I think that the argument here is that the measured background radiation could be a result from starlight. As the background radiation has been measured near Earth, one could argue that the stars generating that radiation should be the nearest stars to the Earth. This would mean that the background radiation we measure would not be universal, but local thing. Of course, it can be questioned if 22 stars is enough.
By machian cosmologist Date 2013-06-30 18:29
Ari Jokimäki said:

I think that the argument here is that the measured background radiation could be a result from starlight.


Absolutely right! :)
By machian cosmologist Date 2013-07-01 16:05
...but I prefer a little bit more a theory in which the origin of the cosmic background radiation is quite far away from us.
In this theory luminous star hosting galaxies are only one stage in a very long cycle that repeats again and again.
So maybe it makes more sense to look at galaxies as a whole as the energy producers.

I'm very interested if my theory and/or my point of view about the reason for the CMB seems plausible to you.
Please let me know!

Kind regards,
S.E. 
By Jade Annand Date 2013-07-01 22:24
Anyone know anything more recent about the old "failed shadow test"?
By Mike Petersen Date 2013-07-02 11:39
An extensive Google search only showed that Creationists jumped on the Sept. 2006 article right away:

Big Bang Fails Another Test

Figures doesn't it?  Cherry-picking at its best.

Regards, Mike

PS - You'll have to stop the self-serving video that obnoxiously comes up by hitting pause and scrolling down to the article.  Yuck.
By Jade Annand Date 2013-07-03 06:39
Ah, John Hartnett - I think I've encountered him before.

I guess the idea is that if you're a creationist, you can get varying scientific views to fight it out in a big gun battle, and then when everyone is "dead", you get to wander in and claim the whole place as your own. As a matter of fact, there's a telling quote in another article on there:

Sarfati said:

We have also pointed out that with rival evolutionary models, the proponents of one model shoot fatal wounds into the theories of the other, leaving all materialistic models mortally wounded.


Total fallacy of the excluded middle, though, of course. You don't actually "win by default" this way :)
By bangstrom Date 2013-07-05 07:48
This is an excellent article on the historical attempts to measure the “temperature of space” resulting from stellar radiation.
http://redshift.vif.com/JournalFiles/Pre2001/V02NO3PDF/V02N3ASS.PDF

“We present the history of estimates of the temperature of intergalactic space. We begin with the
works of Guillaume and Eddington on the temperature of interstellar space due to starlight belonging to our Milky Way galaxy. Then we discuss works relating to cosmic radiation, concentrating on Regener and Nernst. We also discuss Finlay-Freundlich’s and Max Born’s important research on this topic. Finally, we present the work of Gamow and collaborators. We show that the models based on a Universe in dynamical equilibrium without expansion predicted the 2.7 K temperature prior to and better than models based on the Big Bang”
By machian cosmologist Date 2013-07-05 09:46
Thank you very much for this hint, bangstrom!
I will read this article with great interest! :)
By machian cosmologist Date 2013-07-05 10:01
Assis and Neves said:

But there is a third model of the Universe which has
been developed in this century by several scientists including Nernst, Finlay-Freundlich, Max Born and Louis
de Broglie (1966). It is based on a Universe in dynamical
equilibrium without expansion and without continuous
creation of matter. We reviewed this subject in earlier papers (Assis 1992, 1993). Although it is not considered by
almost any textbook dealing with cosmology nowadays,
this third model proves to be the most important one of
all of them.


I think this is the most exciting part of the paper regarding an alternative explanation of the CMB and cosmological redshift.
By Ari Jokimäki Date 2013-07-05 12:53
There are some more references in this thread, including a paper on Nernst cosmology.
By bangstrom Date 2013-07-06 07:48
What impressed me most about the paper was Eddington's theoretical “temperature of space” where a thermometer in the deepest of space far from any stars (galaxies now) should register a temperature on the positive side of absolute zero because no material object can be so far removed from the distant stars that it not heated even a tiny bit by their presence. This means that any non-luminous matter in deep space should have a non-zero equilibrium temperature with its deep space environment. Anything with a temperature radiates as it gains and loses energy with other bits of matter so this process of radiation, absorption and re-radiation should give the universe a uniform background temperature. This temperature was measured to be about 3 K prior to Penzias and Wilson.

If this temperature is the afterglow of the big bang it raises the question of what happened to energy that originated with the stars.

In light of this history, the strongest “proof” of the Big Bang model turns out to be its strongest counter-evidence.- Amitabha Ghosh “Origin of Inertia”
Previous Next Up Topic Cosmology / Alternative Cosmology / CMBR as equilibrium temperature of an eternal Universe (14859 hits)

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