View Full Version : Say gravity were larger, matter "burns" faster?
Spinnor
Oct31-09, 03:36 PM
Say we want a universe with a stronger gravitational interaction.
How does a stronger gravitational force effect star birth, life, and death?
Would matter in such a universe "burn" off faster in stars and black holes?
Could we still have a long lived universe like our present universe if "things" were fine tuned?
Thanks for any help.
DaleSpam
Oct31-09, 06:27 PM
You need to be a little more specific about what you mean by "a universe with a stronger gravitational interaction". I assume you mean that G is bigger, but G cannot change in isolation. So do you mean that G increases without any change in any of the dimensionless fundamental constants, or do you mean that G increases with a corresponding change in the dimensionless fundamental constants (e.g. the ratio of the proton mass to the Planck mass).
Spinnor
Oct31-09, 06:38 PM
We say the force of gravity is roughly 10^40 weaker then the other forces. Make it say ten times stronger then it is so it would be "only" 10^39 times weaker than the other forces.
Then the universe would act different, but my hope is "things" can be tweaked so long lived universes could exist.
Thanks for your help.
DaleSpam
Oct31-09, 06:52 PM
I'm sorry I don't understand what you mean by gravity is 10^40 times weaker than the other forces. The key question is how the dimensionless fundamental constants change.
See http://math.ucr.edu/home/baez/constants.html
Spinnor
Oct31-09, 07:34 PM
I'm sorry I don't understand what you mean by gravity is 10^40 times weaker than the other forces. The key question is how the dimensionless fundamental constants change.
See http://math.ucr.edu/home/baez/constants.html
See:
http://en.wikipedia.org/wiki/Fundamental_interaction
I must be tired, good night %^).
DaleSpam
Oct31-09, 08:13 PM
Sorry, there is not enough information there for me to tell what is implied. As Wiki says: Both magnitude ("relative strength") and "range", as given in the table, are meaningful only within a rather complex theoretical framework
I'm sorry I don't understand what you mean by gravity is 10^40 times weaker than the other forces. The key question is how the dimensionless fundamental constants change.
See http://math.ucr.edu/home/baez/constants.html
Note: big G isn't one of the constants that Baez discusses.
As for how things change inside stars, if G is bigger then the pressure is larger for a given mass since pressure is usually...
P = k. G. M2/R4 ...where k's value depends on the central concentration of mass.
Nuclear fusion rates change very, very sensitively with temperature and thus pressure (remember PV = nRT?) So a bigger G means more fusion for a star of the same mass in our Universe and smaller stars can initiate fusion. In Stephen Baxter's "Raft" he imagines a Universe in which G is a billion times stronger. Stars are only a mile across and burn for a year, fusing all the way to iron before they sputter out.
Dmitry67
Nov1-09, 02:06 AM
If gravity was stronger from the very beginning (= less weaker then other forces) then stars will be smaller and they will burn faster, so there will be less time for life to form. The number of protons in an average star is a function of the weakness of gravity, so it is a function of that 10^-39
If gravity was stronger from the very beginning (= less weaker then other forces) then stars will be smaller and they will burn faster, so there will be less time for life to form. The number of protons in an average star is a function of the weakness of gravity, so it is a function of that 10^-39
Quite so. But doubtless there are subtle points we might miss in making such a big generalisation. Higher G means the Helmholtz contraction phase is longer and brighter, perhaps partly compensating for a shorter Main Sequence?
Dmitry67
Nov1-09, 02:46 AM
May be. In any case, are there any good numerical simulations? It is very interesting in a context of a 'fine tuning' problem.
May be. In any case, are there any good numerical simulations? It is very interesting in a context of a 'fine tuning' problem.
Fred Adams wrote a paper recently about stars in other Universes which may well be relevant. Of course you could contemplate writing the paper yourself?
Stars in Other Universes (http://arxiv.org/abs/0807.3697)
DaleSpam
Nov1-09, 08:42 AM
Note: big G isn't one of the constants that Baez discusses.Exactly. G is not fundamental, only the dimensionless constants are. G describes our system of units, not the physics of the universe.
As for how things change inside stars, if G is bigger then the pressure is larger for a given mass since pressure is usually...This is only true if some of the dimensionless fundamental constants also change (e.g. the ratio of the proton mass to the Planck mass). If G changes without a corresponding change in the dimensionless constants then the pressure would not measurably change nor would the reaction rates nor the intensity nor anything else that we could measure.
Dmitry67
Nov1-09, 10:39 AM
Fred Adams wrote a paper recently about stars in other Universes which may well be relevant. Of course you could contemplate writing the paper yourself?
Stars in Other Universes (http://arxiv.org/abs/0807.3697)
Wonderful. Thank you.
It would be very interesting to make that research much wider:
1. Add all parameters of the Standard Model to the parameter space
2. Analyze not only stars, but also supernovas, planet formation, life chemistry etc.
Wonderful. Thank you.
It would be very interesting to make that research much wider:
1. Add all parameters of the Standard Model to the parameter space
2. Analyze not only stars, but also supernovas, planet formation, life chemistry etc.
Well that's a much bigger task and we really don't know why Standard Model parameters are what they are. Nor do we really know the intricacies of planet formation as well as we understand the stars.
As for life's chemistry... too hard!
Exactly. G is not fundamental, only the dimensionless constants are. G describes our system of units, not the physics of the universe.
I'm not sure that really follows. For some uses, like stellar interiors, its actual value matters. You're confusing underlying mathematical structure (which c = h = G = 1 preserves) for actual physical predictions (which c = h = G = 1 doesn't preserve.)
This is only true if some of the dimensionless fundamental constants also change (e.g. the ratio of the proton mass to the Planck mass). If G changes without a corresponding change in the dimensionless constants then the pressure would not measurably change nor would the reaction rates nor the intensity nor anything else that we could measure.
Why? You seem to be asserting rather than proving. I suspect you're wrong. Show me as you would a novice - because that's what our questioner is after all.
However changing G does change the Planck units and I'm unsure what that would do to large scale physics.
DaleSpam
Nov1-09, 02:39 PM
I'm not sure that really follows. For some uses, like stellar interiors, its actual value matters. You're confusing underlying mathematical structure (which c = h = G = 1 preserves) for actual physical predictions (which c = h = G = 1 doesn't preserve.)
You seem to be asserting rather than proving. I suspect you're wrong. Show me as you would a novice - because that's what our questioner is after all.It does follow; the only physically significant constants are the dimensionless ones. But you are quite correct, I have been asserting rather than proving and that is bad form on my part. This topic came up in great detail in a previous thread where I worked out a couple of examples in detail. It certainly is not a mathematically rigorous proof, but it was sufficient for me:
http://www.physicsforums.com/showpost.php?p=2011753&postcount=55
http://www.physicsforums.com/showpost.php?p=2015734&postcount=68
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