Axial Symmetry Breaking in Self-Induced Neutrino Flavor Conversions

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#Today I Read#

Author: Lei Ma
Summary: Local symmetry can be broken by a mult-azimuth-angle instability.
Categories: { collective oscillations }
Tags: #collective oscillations #axial symmetry breaking

The authors consider collision less neutrino propagation. The convention for Hamiltonian is

\[\mathrm H = \frac{\mathrm M^2}{2E} + \sqrt{2}G_F\left( \mathrm N_l +\int_{-\infty}^\infty dE' \int \frac{d\mathbf v'}{(2\pi)^3} \rho' (1-\mathbf v \cdot \mathbf v') \right)\]

The idea is that axial symmetry can be spontaneously broken in multiangle approach or continuous angle distribution. That is to say, even we start from a axial symmetric configuration, the axial symmetric for flavor instabilities can be broken locally.

Comment: This result seems to be weird. We have axial symmetry of initial conditions. Meanwhile the equation of motion preserves the axial symmetry.

Notations

Here is a list of all the definitions and notations.

  1. The vacuum Hamiltonian

    \[\mathrm M^2\]
  2. Matter contribution to Hamiltonian

    \[\mathrm N_l\]
  3. $\theta_R$ position of emission point on neutrino sphere.
  4. $u=\sin^2\theta_R$ is a parameter that is used to work out the radial velocity $v_{r,u}=\sqrt{1-u R^2/r^2}$ and transverse velocity $\beta_{r,u}=\sqrt{u}R/r$.
  5. Flux matrix is related to density matrix

    \[\frac{\mathrm F(r,E,u,\phi)}{4\pi r^2} \frac{d E du d\phi}{v(u,r)} = \rho(r,\mathbf p) \frac{d^2\mathbf p}{(2\pi)^3}.\]

    Flux matrix can be used instead of density matrix to describe the dynamics,

    \[\partial_r \mathbf F = -i[\mathrm H,\mathrm F]\]
  6. Coherent scattering becomes

    \[\mathrm H_{\nu\nu} = \sqrt{2}G_F \int d\Gamma' \mathrm F' \frac{1 - v v' - \boldsymbol \beta \cdot \boldsymbol \beta'}{v v'}.\]
  7. Linearization is done using flux matrix

    \[\mathrm F = \frac{\mathrm{Tr}\mathrm F}{2} + \frac{F_{ee}-F_{xx}}{2} \begin{pmatrix} s & S \\ S^* & -s \end{pmatrix}.\]
  8. Matter potential

    \[\lambda = \sqrt{2}G_F ( n_e - n_{\bar e} ) \frac{R^2}{2r^2},\]

    and

    \[\bar \lambda = \lambda + \epsilon \mu.\]
  9. Normalized self-interaction potential

    \[\mu = \frac{\sqrt{2}G_F (F_{\bar \nu_e} (R ) - F_{\bar \nu_x}(R) ) }{4\pi r^2} \frac{R^2}{2r^2}.\]

Linearized Equation of Motion

Assuming the solution to perturbation $S$ has the form $S = Q_\Omega e^{i\Omega r}$, where $\Omega=\gamma + i \kappa$.

\[(\omega + \mu \bar \lambda-\Omega) Q _ \Omega=\mu \int d\Gamma' (u + u' -2\sqrt{u u'} \cos(\phi - \phi') ) g' Q'_ \Omega.\]

Question: What happens if \(\omega + \mu \bar \lambda -\Omega =0?\)

Then the equation of motion becomes the rhs integral being 0. Either the integrand is zero, or the integrand has positive and negative regions and they sum up to zero.

I don’t think we can solve this equation.

The integral on the right hand side should have the form

\[a + b u + \sqrt{u}(c\cos \phi + d\sin\phi).\]

The solution to $Q_\Omega$ should be of the form

\[Q _ \Omega= \frac{a + b u + \sqrt{u}(c\cos \phi + d\sin\phi)}{ \omega + \mu \bar\lambda - \Omega }.\]

Then we plug this result back into the equation of motion and find out a matrix equation for the unknown coefficients $a,b,c,d$. We require that the equation has nontrivial solutions. So the problem becomes an eigenvalue problem.

Axial Symmetric Emission

The spectrum $g$ doesn’t depend on $\phi$. But the solution to frequency can depend on $\phi$.

About Equation 13

In the beginning, it wasn’t so clear how could it be true.

Mathematically, with $I$ defined,

\[I_n = I \langle u^n \rangle.\]

In equation 12,

\[(I_1 - 1)^2 = I_0 I_2\]

becomes

\[(I \langle u \rangle -1 )^2 = I^2 \langle u^2 \rangle.\]

Move them to the same side, we have

\[\begin{align} (I \langle u \rangle -1 )^2 - (I \sqrt{\langle u^2 \rangle})^2 &= 0.\\ (I \langle u \rangle -1 - I \sqrt{\langle u^2 \rangle})(I \langle u \rangle -1 + I \sqrt{\langle u^2 \rangle}) &= 0, \end{align}\]

from which we solve $1/I$.

\[\begin{align} \langle u \rangle - \sqrt{\langle u^2 \rangle} &= 1/I \\ \langle u \rangle + \sqrt{\langle u^2 \rangle} &= 1/I . \end{align}\]

From the meaning of $a,b,c,d$ we know that $c,d$ are associated with angle $\phi$ so they are related to the MAA solution. We also know that the solution $I_1=-1$ is the one related to $c,d$ which indicates that this is the MAA solution.

The paper also says that

\[I_1 - 1 = \sqrt{I_0 I_2}\]

is bimodal solution while

\[I_1 - 1 = \sqrt{I_0 I_2}\]

is the MZA solution.

Not sure why.

Single Energy Model

Energy spectrum is

\[g(\omega) = - \delta(\omega + \omega_0) + (1+\epsilon) \delta(\omega-\omega_0).\]

The author solved system

  1. Inverted hierarchy ($\omega_0>0$), bimodal instability.
  2. NH ($\omega_0<0$), MZA & MAA instability.

Matter Effect

At large matter density, instability occurs at larger $\sim\lambda/\lvert q_j\rvert\sim \lambda$. Matter suppression of instability.

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