Consequences of a fourth neutrino type

  • Thread starter Thread starter szcsongor
  • Start date Start date
  • Tags Tags
    Neutrino Type
AI Thread Summary
The discussion centers on the hypothetical consequences of a fourth neutrino type on the early universe's development. Initially, a participant speculated that a fourth neutrino could lead to increased production of photons and helium-4, but later reconsidered this view. It was concluded that introducing a fourth neutrino would likely result in a higher neutron-to-proton ratio due to faster expansion and less time for neutron decay. This would consequently lead to more helium-4 generation, suggesting that answer (c) is correct. The conversation highlights the complex interplay between neutrinos and baryonic matter in cosmic evolution.
szcsongor
Messages
3
Reaction score
0
Homework Statement
If there was a fourth neutrino type it would have affected Universe's development the following way...
Relevant Equations
none
Hi,

I got this question on a recent astronomy course but can't really find the right answer (rough translation from Swedish):

If there was a fourth neutrino type it would have affected Universe's development the following way:
a. more photons and more He-4 generated
b. less photons and less He-4
c. more neutrons and more He-4
d. more protons and more He-4

My guess would be answer (a) but I'm not sure.
 
Physics news on Phys.org
Relevant Equations
none
Are you sure you don't have equations describing the evolution of the early universe? How does its temperature change over time? How does that depend on neutrinos?
szcsongor said:
My guess would be answer (a) but I'm not sure.
Why do you guess (a)?
 
Hi,

Thanks for the quick reply. I was on holiday, so couldn't answer sooner.
I was thinking about this question but unfortunately I don't know how to start solving it. I guess there are some relations (equations?) to use here but I'm unable to see it :(

The temperature of the universe is decreasing constantly, but I don't see the relation with neutrinos.
My guess was a) because I was thinking that neutrinos are produced in the process of H-1 -> He-4 alongside with photons, so another type of neutrinos could be a consequence of more He-4 and photons produced. But it might be true only in stars... Or might be that I'm wrong completely :)
 
Think earlier. How did neutrinos influence the temperature evolution?
Electron neutrinos were important for the neutron to proton ratio, but a fourth generation neutrino shouldn't affect that directly as it doesn't couple to electrons.

If you prefer you can also look up how Planck measured the number of neutrino generations in the early universe. Their result was 3...
 
Thanks for trying to help.
I think I see now how I should proceed with this. I also found clues in the course material now when I read through it:

If we "put in" an other type of neutrino into the universe there should be less baryonic matter to comply with the principle of conservation of mass. But since neutrinos do not interact with matter (and also almost massless), the universe would have expanded faster. The consequence would have been more neutrons (higher neutron / proton ratio) because there would have been less time for the n -> p + e + v reaction. And more neutrons mean more He-4 generated. Therefore answer (c) seem to be the right one.
Am I right? :)
 
Sounds plausible.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top