What temperature would 1 keV neutrinos have today?

In summary, the temperature of 1 keV neutrinos today can be estimated based on the cosmic background radiation and the evolution of the universe. As the universe expands, the temperatures of particles decrease, and neutrinos decouple from matter at high temperatures. Today, 1 keV neutrinos would correspond to a much lower effective temperature due to this cooling process, which can be approximated using the cosmological parameters and the characteristics of neutrino interactions in the early universe.
  • #1
ergospherical
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Suppose there was a 4th generation of neutrino (X, say) with mass m ~ 1 keV.

The other three neutrino generations decouple at T ~ 1 MeV and are not heated during ##e^{\pm}## annihilation (whereas the plasma is heated, leading to a bookwork ##T_{\nu}/T_{\gamma}## factor due to conservation of ##g_{\star S} (aT)^3##).

This X species would remain coupled to the photons until a much later time (long after ##e^{\pm}## annihilation), so I would say ##T_X \sim T_{\gamma}## today?
 
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  • #2
There are very strong constraints on 4th generation neutrinos. They would certainly not remain coupled to photons longer than the standard model neutrinos. Do you have a reference supporting this claim?
 
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  • #3
It's just a thermodynamics question, so whether or not a 1 kev neutrino can exist is beyond the scope. The question is, what would be the contribution ##\Omega_X h^2## of a 1 kev neutrino to the present day energy density. So I need to figure out the temperature of this new species w.r.t. the photon temperature.
 
  • #4
If it would exist and stay in equilibrium longer than the regular neutrinos it would not be a neutrino. It would need stronger than weak interactions with the standard model. If it does it would be very likely decay long before the present time.

Assuming that it didn’t (essentially that it is completely self-contradictory) it would be far from relativistic, it would no longer have thermal distribution as it is decoupled and what redshifts is momentum, not energy. Its energy density would scale as matter, not radiation, at least for late times.
 
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  • #5
ergospherical said:
It's just a thermodynamics question, so whether or not a 1 kev neutrino can exist is beyond the scope. The question is, what would be the contribution ##\Omega_X h^2## of a 1 kev neutrino to the present day energy density. So I need to figure out the temperature of this new species w.r.t. the photon temperature.
How would you even answer that if you don't have an applicable cross section ?
The cross sections are used to determine [tex] N_{eff)[/tex]

What commands are now used for latex ? The older [tex] no longer works for this site.
 
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  • #6
If such a thing is not visible in the decay [itex]Z \rightarrow \nu_4 \overline{\nu}_4[/itex] and related processes, how does it get in thermal equilibrium anyway?
 
  • #7
Vanadium 50 said:
If such a thing is not visible in the decay [itex]Z \rightarrow \nu_4 \overline{\nu}_4[/itex] and related processes, how does it get in thermal equilibrium anyway?
It is much much worse than that. In order to stay in equilibrium much longer than the regular neutrinos, it needs to interact stronger than weakly. This is only going to get worse if you also by "much later time" means that the keV mass starts becoming comparable to the temperature of the SM sector and therefore the equilibrium density starts becoming Boltzmann suppressed. In short, this is a wholly unbelievable scenario.
 
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  • #8
What's the new latex format ? tex doesn't seem to work
 
  • #9
I got it confused initially. We're still supposed to assume that the decoupling is set by when the weak force timescale ~ the Hubble timescale, as is the case for the ordinary neutrinos (i.e. at 1 MeV, not at 1 keV when these hypothetical particles become non-relativistic).

So the reasoning is pretty similar to ordinary neutrinos, i.e. these new ones have temperature ##T_X = (4/11)^{1/3} T_{\gamma}## and number density ##n_X = \tfrac{3}{4} \tfrac{4}{11} n_{\gamma}## and energy density ##\rho_X = (1 \mathrm{keV}) \times n_X##.

Not sure the reason for the confusing premise. I wouldn't read too much into the feasibility as it is meant as a thermodynamics exercise.
 
  • #10
Is this homework? Then it should go in the homework forums.
 
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  • #11
Not homework, just found it looking through an old exam. Usually I will mark something as homework if it's a technical issue and in a general subforum if it's a conceptual issue, since that seems to work best on this website.
 
  • #12
$$\frac{1}{2}$$
ergospherical said:
I got it confused initially. We're still supposed to assume that the decoupling is set by when the weak force timescale ~ the Hubble timescale, as is the case for the ordinary neutrinos (i.e. at 1 MeV, not at 1 keV when these hypothetical particles become non-relativistic).

So the reasoning is pretty similar to ordinary neutrinos, i.e. these new ones have temperature ##T_X = (4/11)^{1/3} T_{\gamma}## and number density ##n_X = \tfrac{3}{4} \tfrac{4}{11} n_{\gamma}## and energy density ##\rho_X = (1 \mathrm{keV}) \times n_X##.

Not sure the reason for the confusing premise. I wouldn't read too much into the feasibility as it is meant as a thermodynamics exercise.
so taking $$n_{eff}(\frac{3}{4})(\frac{7}{8})$$ this is based on all neutrinos decoupling at the same time for the value of 3 if memory serves me correctly. If I recall that value is slightly off due to the different flavor oscillations of each neutrino type

$$N_{eff} $$ according to the following

https://arxiv.org/abs/hep-ph/0506164

the link goes into further detail in regards to neutrino mass terms via the CKMS mass mixing matrix with regards to the cross sections and how they relate
 
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  • #13
If it is actual homework or not is irrelevant. If it is something posted as a task for students or is similar to that, then it should be posted in the homework forums with the appropriate template and attempts.

In the technical forums a certain adherence to mainstream physics should be expected and what is proposed here is severely ruled out by constraints on fourth generation fermions.
 
  • #14
Mordred said:
$$\frac{1}{2}$$

so taking $$n_{eff}(\frac{3}{4})(\frac{7}{8})$$ this is based on all neutrinos decoupling at the same time for the value of 3 if memory serves me correctly. If I recall that value is slightly off due to the different flavor oscillations of each neutrino type

$$N_{eff} $$ according to the following

https://arxiv.org/abs/hep-ph/0506164
It is not the effect of oscillations that lead to a value different from three. In fact, the abstract of the paper you linked state this explicitly
We find that oscillations do not essentially modify the total change in the neutrino energy density
Instead, what is relevant is that ##N_{\rm eff}## is defined to be exactly 3 if there are 3 active neutrino species fully decoupled before ##e^\pm## annihilation. However, the neutrinos - in particular of electron type - only approximately have time to fully decouple. Hence the neutrino sector will have slightly higher energy density than if they fully decoupled.
 
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  • #15
Orodruin said:
If it is actual homework or not is irrelevant. If it is something posted as a task for students or is similar to that, then it should be posted in the homework forums with the appropriate template and attempts.

In the technical forums a certain adherence to mainstream physics should be expected and what is proposed here is severely ruled out by constraints on fourth generation fermions.

Fair enough 👍
 

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