light elements abundance in a static toy universe

Ken G

I have no idea why you think it is considered that. It certainly isn't considered that by cosmologists.
As someone else said, if you change the physics, you can get any answer you want. But in this universe, H-->He is only reversible in the early minutes of the Big Bang, conditions that did not exist in your question. That's why the Big Bang model answers the H/He ratio-- it represents exactly the ratio of neutrons to protons one would expect to be "frozen in" from the reversible process p<-->n in the early minutes of the Big Bang, assuming expansion. The cores of stars tend to only result in p-->n.

TrickyDicky

You are right, that is because cosmologists generally deal with physically realistic scenarios, I'm having problems getting you people into the "thought experiment mode" here.

Again, "this" universe (ours) is not the one I'm talking about.
Yes, the cores of stars as isolated systems tend to p-->n, so in the hypothetical static spacetime some mechanism should be compensating this, I guess.

Ken G

Not true, we have no issue with thought experiments. We like thought experiments, we think they are a nice way to learn real physics. Not make believe physics, though. You just didn't like the correct answer for some reason.That's not a thought experiment, that's make believe. There's a difference.

TrickyDicky

What I cannot understand is why you won't concede that in a static spacetime there is time symmetry and therefore nuclear reactions would be reversible, so the "all iron" answer can never be the correct answer.

Ken G

I can't concede it because it's wrong, the physics of that claim is confused. The static character of the spacetime has nothing at all to do with the nuclear reactions possible. The latter depends, not on the spacetime (which simply defines the inertial paths, and asserts that they are always the same), but on the conditions of the matter (temperature, density, and so on), and the physical processes allowed in those conditions. The model would have reached a steady state if the age is effectively infinite, so all processes that can occur must balance their inverse process. That doesn't mean you have some known H/He ratio, it might just mean you don't have any of either H or He. I'm saying that is what you would indeed have, because the conditions one can assume for your static spacetime (given that they are unspecified, yet you asked your question anyway, we can assume you intended conditions of T and density like we find in the universe today), do not have a process for turning He back into H, so we are on a one-way street leading to iron. Hence the answer that you don't like. Now, obviously if you are allowed to invent imaginary physics, you can get any H/He you are more happy with, but then there is also no reason to pose your question here.

TrickyDicky

Ok, let's imagine this spacetime was a solution of the EFE, in that case the matter conditions would also be fixed by the RHS of the EFE.

Ken G

You think the H/He ratio, and the nucleosynthesis physics, shows up on the RHS of the EFE? What is actually there?

twofish-quant

That's because it's not clear what rules you are imposing. If you can state the rules of the game, we can figure out what goes on.

And once you specify that mechanism then you get whatever answer you want.

TrickyDicky

Stress-energy tensor.

Ken G

Yes... no physics about nucleosynthesis at all. You can kind of tell this, actually-- Einstein did have a cosmological model with a static spacetime. So why didn't he go ahead and try to answer the question from your OP? Because he knew it would not be possible to do, there's not enough information without additional assumptions. Now, of course Einstein didn't know squat about nucleosynthesis, but what we do know about it now is what gives the answer "all iron", so Einstein would have then known his static solution was wrong in the absence of some new physics (which is what we are telling you, also). So the bottom line is, as has often been repeated, there are only two possible answers to your question:
1) if no new physics: all iron
2) if new physics: anything you want
I wish I had just said that from the start, but then again, I think I basically did.

TrickyDicky

And in the simplified case of a static universe with a fluid in thermodynamical equilibrium the stress-energy tensor is proportional to the hydrostatic pressure and the inverse of the metric tensor.

TrickyDicky

What?? So in your opinion, what physics is the matter tensor related to?

Ken G

The stress-energy tensor is what it is-- there are many different things that can lead to the same stress-energy tensor. You seem to imagine that tensor completely describes everything that is happening, but this is incorrect. Consider this analogy. As I write this, everything happening in my head can be influencing in some way the words that appear, yet you cannot take those words and infer everything happening in my head. So it is for the stress-energy tensor, and so it was for Einstein and his static spacetime cosmology, and that is also why he knew he could not use that cosmology to infer H/He. Why else do you think Einstein could design a theory around the stress-energy tensor without even knowing that nucleosynthesis existed?

To repeat: Einstein could make a static cosmology. He could not infer H/He from that cosmology, because he did not know the physics of nucleosynthesis. We do, so we can get H/He, and it's all iron, unless you want to put in some additional unknown physics, in which case you can get any answer you like.

TrickyDicky

You are weirdly hung upon that Einstein thing, that has nothing to do with my questions.
No, I don't think the stress tensor describes what you are thinking.
Thanks for your valuable help.

Ken G

Yeah, why would the fact that your OP stipulated a static spacetime, which is just what Einstein had in his static spacetime cosmology, why would that be relevant? All you added was nucleosynthesis, as if knowledge of that would suddenly let H/He be calculated in Einstein's cosmology simply because our H/He "formed in equilibrium." So all equilibrium are exactly the same then, in your mind? No, they're not. But you can't get this so, this must conclude our conversation. There isn't much point in repeating further-- your question is answered: "all iron if you add nothing to your OP, or anything you want if you add some made up physics you did not stipulate."

TrickyDicky

Certainly. But equilibrium is certainly a good condition to start with. And a static fluid in equilibrium - therefore thermodynamic and hydrostatic equilibrium- is a particular equilibrium that simplifies the problem.
So that the ratio of protons and neutrons when they are allowed to freely and reversibly transform into each other (this particular equilibrium), I understand, is determined just by their relative masses. This seems to be the only stipulating that is needed to calculate a H/He ratio under the postulated conditions. But please correct me if this is not so.
Sorry for not making this stipulations clear in the OP.

Ken G

Not just that, also the temperature. Just saying you have an equilibrium is only the beginning, it means you have a temperature, but it doesn't tell you what the temperature is. You need that to get H/He in equilibrium between the processes that make He and those that make H. But as I said, simply having a static spacetime doesn't mean you have equilibrium everywhere, you can have stars forming and exploding and so on, and those are the processes that will turn everything into iron regardless of what is the average temperature. It sounds to me like you wanted not only a static spacetime, but also a homogeneous density, but that's generally not stable to gravity even locally (never mind the global instability that dooms static spacetimes). If we could have a stable static spacetime, that was also locally stable, so you have equilibrium everywhere at the same T, even then, you still need to know what that T is before you can know H/He. That's the crucial input from the Big Bang model-- it tells you what happens to T, and that is what inevitably gives you H/He ~ 4 (by mass).