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An elegant formula describing the Universe
- Larry Burford
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21 years 2 months ago #6519
by Larry Burford
Replied by Larry Burford on topic Reply from Larry Burford
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>
[Mac] I'm afraid that doesn't clear it up in my mind though for a surface to purportedly have mass.
<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>
The ratio of two masses has the same units as the ratio of two of any other quantity (such as two areas): none.
His equation works, as far as that goes. There are other issues...
LB
[Mac] I'm afraid that doesn't clear it up in my mind though for a surface to purportedly have mass.
<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>
The ratio of two masses has the same units as the ratio of two of any other quantity (such as two areas): none.
His equation works, as far as that goes. There are other issues...
LB
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- Anthony Mai
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21 years 2 months ago #6205
by Anthony Mai
Replied by Anthony Mai on topic Reply from
Larry:
Thanks for clarifying my formula and testify that the formula indeed gives the correct answer.
But clearly it can not be purely coincidental. We are talking a dimentionless number of 10 to the 82th power. And it matches the correct answer to within 2 to 3% error, actually it matches perfectly within the uncertainty of the known value. Consider that the equation is so simple and so elegant, it simply can not be coincidental.
In the frame work of Big Bang, sure the age of the universe is not a constant, but why should G be a constant, then?
On another hand, assuming G is indeed constant, my formula clearly indicates that the age of the universe is a constant, no matter how much time has elapsed. Because otherwise it will be too much a coincident that we happen to live at a specific time of the universe which makes my equation true, and at any other time just a little bit before or after the current time, my equation will be off.
The implication of a constant age of the universe is that of a static and infinite universe that has lived forever, but somehow NO information can be stored for a time longer than the "age" of the universe, or can be transmitted over a distance longer than the "radius" of the universe, which equals to light speed times the "age".
In another word, from the specific location and time where we are located, there is a sphere of visibility in the spacetime. And the radius of that sphere is "age" of the universe if measured in time unit, or "radius" of the universe if measured in distance. We will be unable to detect any information coming from any space time point outside that sphere of visibility. Hence the age and radius of the universe we are talking about is merely the radius of the visible universe, as observed from our location.
So "age" and "radius" of the universe, the visible universe, is a constant, even though the whole universe is actually infinite both in time and in space.
Thanks for clarifying my formula and testify that the formula indeed gives the correct answer.
But clearly it can not be purely coincidental. We are talking a dimentionless number of 10 to the 82th power. And it matches the correct answer to within 2 to 3% error, actually it matches perfectly within the uncertainty of the known value. Consider that the equation is so simple and so elegant, it simply can not be coincidental.
In the frame work of Big Bang, sure the age of the universe is not a constant, but why should G be a constant, then?
On another hand, assuming G is indeed constant, my formula clearly indicates that the age of the universe is a constant, no matter how much time has elapsed. Because otherwise it will be too much a coincident that we happen to live at a specific time of the universe which makes my equation true, and at any other time just a little bit before or after the current time, my equation will be off.
The implication of a constant age of the universe is that of a static and infinite universe that has lived forever, but somehow NO information can be stored for a time longer than the "age" of the universe, or can be transmitted over a distance longer than the "radius" of the universe, which equals to light speed times the "age".
In another word, from the specific location and time where we are located, there is a sphere of visibility in the spacetime. And the radius of that sphere is "age" of the universe if measured in time unit, or "radius" of the universe if measured in distance. We will be unable to detect any information coming from any space time point outside that sphere of visibility. Hence the age and radius of the universe we are talking about is merely the radius of the visible universe, as observed from our location.
So "age" and "radius" of the universe, the visible universe, is a constant, even though the whole universe is actually infinite both in time and in space.
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- PoPpAScience
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21 years 2 months ago #6207
by PoPpAScience
Replied by PoPpAScience on topic Reply from
Hi Anyhony Mai;
Great posts. Even though with my grade 10 math I have to read it many times. So far I think I see a formula for the age and size of the universe today. I'll have to read it more.
Is there more I should see, or, do not understand yet?
Thanks; PoPpA
Great posts. Even though with my grade 10 math I have to read it many times. So far I think I see a formula for the age and size of the universe today. I'll have to read it more.
Is there more I should see, or, do not understand yet?
Thanks; PoPpA
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21 years 2 months ago #6208
by Jeremy
Replied by Jeremy on topic Reply from
I guess I will add a little skepticism to this formula. All I can say is that it is interesting but I am not bowled over by it. Be very careful of saying that something must be significant because it matches a value to a close percentage. I have read tons of articles with many cleverly devised relations that all make a similar kind of argument and are based on completely different assumptions. Read Arthur Eddington's fundamental theory if you want to see a whole book of such relations.
The close value you cite is to me misleading since the values you are plugging in are subject to wild variation depending on whose numbers you want to use. The mass of the Electron is known accurately but the mass of the universe could vary by orders of magnitude according to whose theory you want to believe in. The age of the universe has gone up and down by billions of years over the past 30 years, I don't expect the present pronouncements to last any longer.
The close value you cite is to me misleading since the values you are plugging in are subject to wild variation depending on whose numbers you want to use. The mass of the Electron is known accurately but the mass of the universe could vary by orders of magnitude according to whose theory you want to believe in. The age of the universe has gone up and down by billions of years over the past 30 years, I don't expect the present pronouncements to last any longer.
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21 years 2 months ago #6301
by Mac
Replied by Mac on topic Reply from Dan McCoin
I find the idea interesting, if not also confusing and inconclusive. I think it merits further study but keeping in mind the number of ways that "Pi" has been shown in the design of the Pyrimids, etc.
Knowing to believe only half of what you hear is a sign of intelligence. Knowing which half to believe can make you a genius.
Knowing to believe only half of what you hear is a sign of intelligence. Knowing which half to believe can make you a genius.
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- Larry Burford
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21 years 2 months ago #6209
by Larry Burford
Replied by Larry Burford on topic Reply from Larry Burford
Hello Anthony,
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>[Anthony]
Larry:
Thanks for ... testify that the formula indeed gives the correct answer.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>That's not *exactly* what I meant.
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>
So "age" and "radius" of the universe, the visible universe, is a constant, even though the whole universe is actually infinite both in time and in space.
<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>
The radius of the visible universe isn't constant - it grows every time we build a new telescope. Shouldn't your equation have a factor for telescope power?
Regards,
LB
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>[Anthony]
Larry:
Thanks for ... testify that the formula indeed gives the correct answer.<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>That's not *exactly* what I meant.
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>
So "age" and "radius" of the universe, the visible universe, is a constant, even though the whole universe is actually infinite both in time and in space.
<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>
The radius of the visible universe isn't constant - it grows every time we build a new telescope. Shouldn't your equation have a factor for telescope power?
Regards,
LB
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