CQW is an educational resource for theoretical physics and astrophysics, field theory, relativity and cosmology. we post a new question every wednesday for students to get and for teachers to stay in shape.

Wednesday, July 24, 2013

Planck's summer: observation of the CNB

fourth post in the "Planck's summer"-series by Youness

When did the cosmic neutrino background form, why is its temperature lower than that of the CMB and how could an experiment to detect it look like?

I'm a little bit late, but I guess you could do a very rough estimate of the decoupling by dimensional analysis. Considering freeze-out takes place when $ \Gamma \approx H $, and $ \Gamma \approx n \sigma v $. From the dimensionality, a number density scales as energy^3 (so T^3). We can estimate $ \sigma v \propto \frac{g^2}{m_Z^2} T^2 $; and the Hubble rate is $ \sqrt{ \frac{8 \pi}{3} G \rho } $ with $\rho \propto T^4 $. Putting everything together we can solve for the temperature, and this gives roughly 1 MeV.

Youness's answer: It formed at neutrino decoupling around $t \approx 1$s in cosmic time when the interaction rate with electrons due to the weak interaction could no longer compete with the expansion.

Let's briefly recall some basics about the weak interaction. At low energies, Fermi described it as a four-fermion interaction taking place at a point; whereas, fundamentally, the weak interaction is due to the exchange of W and Z bosons. The corresponding Feynman diagram implies an amplitude $\propto g \frac{1}{Q^2 + M_W^2} g$. At low energies, $Q^2 \ll M_W^2$, and we can approximate this as a point interaction with an amplitude $\propto \frac{g^2}{M_W^2}$. This is essentially Fermi's constant $G_F \approx 10^{-5} \text{GeV}^{-2}$; it carries two negative mass dimensions since we absorbed the W mass in it. The cross section scales with the squared amplitude and thus with $G_F^2$. From dimensional grounds, it's then suggestive to write $\sigma \propto G_F^2 E^2$ since the cross section is an area (two negative mass dimensions). As the interaction rate depends on the electron density, $\Gamma = n_e \sigma$, we get an overall dependence $\Gamma \propto T^5$. This eventually leads to the above statement, namely a decoupling at $t \approx 1$s.

The direct detection of the cosmic neutrino background is hard due to the scaling $\sigma \propto G_F^2 E^2$ and the low energy of the neutrino background. Any experiment to detect the cosmic neutrinos should thus employ a trick to get to higher energies. In principle, this is possible by large velocities. For example, if we could place a typical neutrino detector like the big water tank experiments on a spaceship and travel with it at extremely high velocities (very close to the speed of light), the cosmic neutrinos would be 'blue-shifted' in the laboratory frame and could reach sufficiently high energies to be detected in principle. In reality, we would probably need some astrophysical source of high-energy particles and hope to detect their interactions with the neutrinos.

I'm a little bit late, but I guess you could do a very rough estimate of the decoupling by dimensional analysis. Considering freeze-out takes place when $ \Gamma \approx H $, and $ \Gamma \approx n \sigma v $. From the dimensionality, a number density scales as energy^3 (so T^3). We can estimate $ \sigma v \propto \frac{g^2}{m_Z^2} T^2 $; and the Hubble rate is $ \sqrt{ \frac{8 \pi}{3} G \rho } $ with $\rho \propto T^4 $. Putting everything together we can solve for the temperature, and this gives roughly 1 MeV.

ReplyDeleteYouness's answer: It formed at neutrino decoupling around $t \approx 1$s in cosmic time when the interaction rate with electrons due to the weak interaction could no longer compete with the expansion.

ReplyDeleteLet's briefly recall some basics about the weak interaction. At low energies, Fermi described it as a four-fermion interaction taking place at a point; whereas, fundamentally, the weak interaction is due to the exchange of W and Z bosons. The corresponding Feynman diagram implies an amplitude $\propto g \frac{1}{Q^2 + M_W^2} g$. At low energies, $Q^2 \ll M_W^2$, and we can approximate this as a point interaction with an amplitude $\propto \frac{g^2}{M_W^2}$. This is essentially Fermi's constant $G_F \approx 10^{-5} \text{GeV}^{-2}$; it carries two negative mass dimensions since we absorbed the W mass in it. The cross section scales with the squared amplitude and thus with $G_F^2$. From dimensional grounds, it's then suggestive to write $\sigma \propto G_F^2 E^2$ since the cross section is an area (two negative mass dimensions). As the interaction rate depends on the electron density, $\Gamma = n_e \sigma$, we get an overall dependence $\Gamma \propto T^5$. This eventually leads to the above statement, namely a decoupling at $t \approx 1$s.

The direct detection of the cosmic neutrino background is hard due to the scaling $\sigma \propto G_F^2 E^2$ and the low energy of the neutrino background. Any experiment to detect the cosmic neutrinos should thus employ a trick to get to higher energies. In principle, this is possible by large velocities. For example, if we could place a typical neutrino detector like the big water tank experiments on a spaceship and travel with it at extremely high velocities (very close to the speed of light), the cosmic neutrinos would be 'blue-shifted' in the laboratory frame and could reach sufficiently high energies to be detected in principle. In reality, we would probably need some astrophysical source of high-energy particles and hope to detect their interactions with the neutrinos.

There is a known trick to get to higher energies by using (163)Ho as an target nuclius. You can find the paper here:

Deletehttp://arxiv.org/abs/1012.0760