If I can derive Hubble's constant from nothing but c and the age of the universe, would that be important?

Faye Kane

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Jun 30, 2012, 1:58:15 AM6/30/12

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Please give me a serious answer, because I think I know how. But I don't want to bother doing the calculations if it's already been done.

Thank you,

-faye

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xray4abc

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Jun 30, 2012, 1:18:28 PM6/30/12

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Yes. It is a very good exercise.
(This is a very serious answer.:-))
Regards, LL

Faye Kane

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Jun 30, 2012, 2:46:06 PM6/30/12

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> Yes. It is a very good exercise.
> (This is a very serious answer.:-))

Remember, I'm talking about using *only* c and the universe age. Cosmology calculators I've seen also require a density parameter.

Would somebody please give me a straight answer (with no smileys)? I can't tell whether this guy means deriving H0 is a trivial exercise for astronomy undergrads or if it is obviously impossible and he's being sarcastic.

dlzc

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Jun 30, 2012, 2:49:18 PM6/30/12

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Dear Faye Kane:

On Friday, June 29, 2012 10:58:15 PM UTC-7, Faye Kane wrote:
> Please give me a serious answer, because I think
> I know how. But I don't want to bother doing the
> calculations if it's already been done.

Probably not. Hubble's constant was very large for 300,000 years, then became much smaller very rapidly (as in orders of magnitude less than today), then accelerated to what we see today.

Your bigger problem will be Hubble's constant *establishes* the age of the Universe in a roundabout way, so you will essentially derive Hubble's constant, given Hubble's constant.

David A. Smith

dlzc

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Jun 30, 2012, 5:27:00 PM6/30/12

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Dear Faye Kane:

On Saturday, June 30, 2012 11:46:06 AM UTC-7, Faye Kane wrote:
> > Yes. It is a very good exercise.
> > (This is a very serious answer.:-))
>
> Remember, I'm talking about using *only*
> c and the universe age. Cosmology
> calculators I've seen also require a
> density parameter.

You cannot get rid of the length unit, if you don't invoke more.

> Would somebody please give me a straight
> answer (with no smileys)? I can't tell
> whether this guy means deriving H0 is a
> trivial exercise for astronomy undergrads
> or if it is obviously impossible and he's
> being sarcastic.

He is serious. He has other ways of poking fun at someone.

My interest will be piqued, if your derivation has the observed behavior for H0 over 0 < t < 13.7 Gy... regardless of what units problems you may or may not have.

David A. Smith

Faye Kane

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Jul 1, 2012, 12:52:46 AM7/1/12

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well I had to find an arbitrary-precision calculator because the simplest, most understandable way to compute this involves dividing two 28-digit numbers that differ only in the last six digits. (it's the ratio of a megaparsec to a megaparsec after the universe expands for one second).

But the accuracy of the result I got (which I assumed would be wrong) shocked me so much that I'm writing out the steps one by one and showing the calculation results in each step, so people will have to tell me it's an amazing coincidence (which you will) rather than "you must have made an arithmetic mistake somewhere" or just assume I'm lying and not do the calculation. I'll post it as a new message.

Note that I'm not proposing a new "theory" that proves aether or einstien being wrong or anything else crank. I just take the most accurate age we have for the universe and the speed of light, and do a simple calculation involving ct². The result you get for H-zero is (71.4 km/s)/Mpc.

Stand by.

BTW, to duplicate the result, you'll need an arbitrary-precision calculator like this one, as neither windows calc or google calc is accurate enough:
http://www.karenware.com/powertools/ptcalc.asp
or you can use another one from somewhere else.

Koobee Wublee

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Jul 1, 2012, 1:41:05 AM7/1/12

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On Jun 29, 10:58 pm, Faye Kane wrote:
>
> Please give me a serious answer, because I think I know how. But I
> don't want to bother doing the calculations if it's already been done.

Hubble’s constant is strictly a conjecture, and that should answer
your question. It was decided on what it means during Hubble’s time
where he could only see up to 500k parsecs at best. <shrug>

SolomonW

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Jul 1, 2012, 3:09:10 AM7/1/12

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What Hubble said is his initial data showed that the theory of the
expansion of the universe was correct. However, if you look at Hubble’s
original released data, it does not really show the expansion of the
universe.

Faye Kane

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Jul 1, 2012, 3:30:37 AM7/1/12

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Wow, good thing I'm not relying on Hubble's original released data, but on the modern values of c, the age of the universe, and geometry.

SolomonW

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Jul 1, 2012, 9:30:17 AM7/1/12

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So you should.

dlzc

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Jul 1, 2012, 10:13:54 AM7/1/12

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Dear Faye Kane:

On Saturday, June 30, 2012 9:52:46 PM UTC-7, Faye Kane wrote:
...

> Note that I'm not proposing a new "theory"
> that proves aether or einstien being wrong or
> anything else crank. I just take the most
> accurate age we have for the universe and the
> speed of light, and do a simple calculation
> involving ct².

This has units of length * time. H0 has units of time^-1. So you are just playing number games.

> The result you get for H-zero is (71.4 km/s)/Mpc.

And using your result for different epochs, you get a parabola. What the Universe shows is a bathtub curve. Something like:
http://microblog.routed.net/wp-content/uploads/2006/08/bathtub.JPG

Hubble's constant, isn't constant over the history of the Universe.

Don't focus on today's value only...

David A. Smith

G=EMC^2

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Jul 1, 2012, 10:46:28 AM7/1/12

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Red shift showing more reddining with distance is a bad ruler. Its off
by 40% at best. Reality is we need a better way to measure. Clair
Patterson gave the Earth its age by using radio active decay.It is
only off by 3% I have my own way of measuring the age of the universe.
It gives it 22 billion years,and counting. It goes against 13.5 given
by our imperial thinkers,so even knowing I'm on the money I get
laughed at. They fudge I never fudge I add up time lapses they
rather sweep under the rug. TreBert

SolomonW

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Jul 2, 2012, 8:20:03 AM7/2/12

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Well, the oldest object found is about 13 billion years old. That puts a
limit on how young the universe can be.

xxein

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Jul 3, 2012, 7:28:56 PM7/3/12

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xxein: What's c? What's age of the universe?

c=f(tu)?

How many in your 3rd grade class think this is important?

Brad Guth

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Jul 7, 2012, 1:21:32 AM7/7/12

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If that oldest object found was a reflected image, then what?

David Fuller

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Feb 9, 2016, 1:27:03 PM2/9/16

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((2 * (10^17))^2) / (Mpc / (69 331.718 (m / s))) = 8.98755183 × 10^16 hertz

((4*10^34) / (Mpc / (69 331.718 (m / s))) = 8.98755183 × 10^16 hertz

David Fuller

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Feb 9, 2016, 1:35:59 PM2/9/16

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1/(1.616199 ×10^-35 * ((2 * (10^17))^2) = 1.54683921968

1 / (6.1873569 * (10^34)) = 1.616199e-35

((6.1873569 * (10^34)) / (Mpc / (69331.718 * (m / s)))) / (0.5 * pi) = 8.8504776 × 10^16 Hz

David Fuller

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Feb 9, 2016, 2:44:44 PM2/9/16

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On Saturday, June 30, 2012 at 12:58:15 AM UTC-5, Faye Kane wrote:

1 / (((13.7 billion light years) / Mpc) / c) = 71 373.2194 m / s

((13.7 billion light years) / Mpc) * 71 373.2194 = 299792457.817

((1.2961201e+26 m) / (3.08567758 * (10^22) m)) * 71 373.2194 = 299798867.081

1 / (((1.2961201e+26 m) / (3.08567758 * (10^22) m)) / c) = 71 371.6936 m / s

13.7 billion years * (71 373.2194 / 69 331.718) = 1.41034022 × 10^10 years

(13.7 * (10^9) years) / (14.1034022 * (10^9) years) = 0.971396817

David Waite

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Feb 10, 2016, 12:49:35 AM2/10/16

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No it doesn't. For example if the universe is open curved, or if as the evidence suggests, its open flat, then its "size" is infinite no matter how old it is. The age and Hubble's "constant" merely relate to the scale factor function.
For the OP, you can't get Hubble's constant from the age of the universe without making some other assumption whether you know you're making them or not, which realizing it or not is decided by a density function and the cosmological constant. My bet is you're setting up the solution with those assumptions imbedded without knowing it, by like assuming its flat and the value of the cosmological constant.

David Waite

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Feb 10, 2016, 1:12:20 AM2/10/16

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So for example you might assume its flat and that the cosmological constant is a constant which directly corresponds to a specific stress-energy tensor telling you what the energy density is as a function of time equation 8.3.9.
http://www.modernrelativitysite.com/chap8.htm#BM8_3
Having the resulting solution for the scale factor yielding the line element solution as 8.3.6 you can then solve for Hubble's "constant" as a function of time, the age of the universe and the cosmological constant as equation 8.3.12. The current value of Hubble's "constant" is then given as a function of the age of the universe and the cosmological constant as equation 8.3.13, at which point, given observational data for the current value of Hubble's constant and the age of the universe, you can then use a root finder to get the value of the cosmological constant yeilding its value as equation 8.3.16.