1.2.4. Dispersion Relation

Reference to Notes

Izaguirre, I., Raffelt, G., & Tamborra, I. (2016). Fast Pairwise Conversion of Supernova Neutrinos: Dispersion-Relation Approach, 21101(January), 1–6. https://doi.org/10.1103/PhysRevLett.118.021101

1.2.4.1. Polarization Tensor

In Raffelt’s paper, they defined the polarization tensor as

Πμν=ημν+dΩ4πG(θ,ϕ)vμvνkμvμ.

For numerical calculations, we lower the second index and multiply on both side ω,

ωΠνμ=ωδνμ+dΩ4πG(θ,ϕ)vμvν1kωk^v,

where

vμ=(1sinθcosϕsinθsinϕcosθ)kμ=(ωksinθkcosϕkksinθksinϕkkcosθk).

1.2.4.1.1. Parametrization of Polarization Tensor

Raffelt et al parametrize kμ(n) where n=k/ω. Then the polarization tensor is decomposed into two parts,

ημν

and

1ωNμν,

where

Nμν=dΓGvvμvν1nk^v,

with kμ=(ω,k).

Note to self

The actual wave vector that determines the instability is Kμ which is related to kμ,

kμ=Kμ(Λμ+Φμ).

Since Λμ and Φμ are real, imaginary part of ω (k) equal imaginary part of Ω (K). Thus we only discuss the dispersion relation of kμ.

Density matrix is written as

ρ=(1ϵϵ1).

The perturbation ϵ is assumed to have the form

ϵ=Q(Ω,K)ei(ΩtKx).

This assumption indicates that even though we find instabilities, a proper initial condition/boundary condition is required to stimulate this instability.

The polarization tensor is in fact

Πμν=η+1ωNμν.

The equation of motion becomes

vμΠμνaν=0,

where

aν=dΓvνGvQv.

Since vμ is (component) of a null dual vector, we require Πμνaν to be (component) proportional to vμ. Since we have a lot of directions, different vμ are independent of each other. So we require Πμνaν=0.

Then we need to find the solution to

Det(Πμν)=0,

which is simplified to

Det(ωημν+Nμν)=0.

We can also use the polarization tensor Πνμ

Det(ωδνμ+Nνμ)=0

is the determinant of a matrix

ωI+Nνμ.

Equivalently, we only need to find the eigenvalues of Nνμ then multiply on each of them by negative sign.

1.2.4.1.2. Examples of Parametrization

Raffelt et al proposed that can now solve the dispersion relation by finding the value of kμ(n) for each n. We make the plot ω vs k.

Here is an example that I calculated.

The axial symmetric system can be calculated easily using this method. The paper gave an example of two polar angle beams with axial symmetry.

../../_images/listpltOmegan1.png

Fig. 1.62 ω(n) for G=0.5δ(cosθ0.8)+0.5δ(cosθ+0.2).

../../_images/listpltDispersionRelationDecompose1.png

Fig. 1.63 Dispersion relation.

We can check what happens for multibeams. I can plot the dispersion relation for similar configuration but with different number of beams.

../../_images/listanimi1.png

Fig. 1.64 Animition of dispersion relation.

dataPltNBeamsPlt[Join[Table[1/beams, {n, 1, beams/2}],
Table[-1/beams, {n, 1, beams/2}]],
Table[Pi/3 + n Pi/2/(beams - 1), {n, 0, beams - 1}], {-10, 10}, 0.049, {{-10, 10}, {-10, 10}}]

I plot the ω(n) relation for different number of beams

../../_images/listpltOmegan12List-2.png ../../_images/listpltOmegan12List-4.png ../../_images/listpltOmegan12List-8.png ../../_images/listpltOmegan12List-10.png

Similar to the previous example, confining the range of n leads to only a partial patch of the dispersion relation.

../../_images/pltDiffBeamsConfined-n-in--1-to-1-beams-10.png

Fig. 1.65 The code for it

pltDiffBeamsConfined[beams_] := dataPltNBeamsPlt[
Join[Table[1/beams, {n, 1, beams/2}],
Table[-1/beams, {n, 1, beams/2}]],
Table[Pi/3 + n Pi/2/(beams - 1), {n, 0, beams - 1}], {-1, 1},
0.049, {{-10, 10}, {-10, 10}}]

This should be the continuous limit?

As a comparison, we can plot the dispersion relation in a larger range of n for 10 beams.

../../_images/listpltOmegan12List-10.png

Fig. 1.66 10 beams.

On the other hand, we can calculate the continuous limit for the same angle range.

../../_images/compare-continuous-and-10-beams-within-n-range--1-to-1.png

Fig. 1.67 Dispersion relation for 10 beams (n[1,1]), and continuous limit.

MEH

1.2.4.2. Analyze the Symmetries in Polarization Tensor

Vectors Using Spherical Harmonics

Four velocity can be expressed in terms of spherical harmonics.

In principle, solving the dispersion relation is not easy. Neverthless, symmetries would significantly simplify the problem.

Axial symmetry indicates that the integrals of first orders of sinϕcosϕ, sinϕ, and cosϕ are 0 on the range ϕ[0,2Π].

We denote the integral

dΩ4πG(θ,ϕ)vμvνωkk^v

as Pνμ. The polarization tensor becomes

Πμνμ=I+Pμνμ.

For axial symmetric emission, only terms P000,P030,P303,P333,P111,P222 are nonzero, given k in z direction, i.e., ϕk=θk=0.

To simplify the calcualtion, we denote n=|k|ω. We will NOT solve ω(n). Instead we write down the form of the eigenvalues of

Nμνμ=ωPμνμ,

which shows is an analytical expression of ω. We do not solve this relation. Instead, we plugin the definition n=|k|ω and find out the relation between ω and k=|k|

The four velocity is

vμ(1,sinθcosϕ,sinθsinϕ,cosθ)T.

We define

In(θ)=cosθ2cosθ1dcosθG(θ)cosnθ1ncosθ,

where θ1 and θ2 are

Since

02πdϕ=2π02πdϕcos2ϕ=02ϕdϕsin2ϕ=π,

the matrix Nμνμ is simplified,

Nμνμ=ωPμνμ(12I00012I1014(I0I2)000014(I0I2)012I10012I2).

The equation we are solving is

Det(ω+Nμνν)=0.

We find the solutions to omega,

(1.35)ω=14(I0I2),14(I0I2±(I02I1+I2)(I0+2I1+I2)).

Mathematica Code

The Mathematica code for solving ω as a function of n is

We plug in the definition n=k/ω then solve dispersion relation from each of the solutions in Eq. (1.35).

The questions are

  1. What does each of the solutions mean?

1.2.4.2.1. Eigenvalues and Axial Symmetry

By definition, the meaning of polarization tensor, Πνμaν=0 implies that a1 and a2 are the ϕ angle dependent components. To prove this, we rewrite Q,

Q=aμvμkνvν,

which clearly shows that the 1, and 2 component of aμ is related to the phi dependence of Q. a1=a2=0 indicates that Q has no ϕ dependence.

Is this related to eigenvectors?

The eigenvalue 14(I0I2) of matrix Nμνμ corresponds to eigenvectors (0,0,1,0) and (0,1,0,0).

I don’t think it is related to eigenvalues. However, eigenvalues set limit on the actual solution. When we write down the solution to aμ, the coefficients are related to each other because we have determinant of coefficient matrix being 0. There are degeneracies.

That is to say, the part

(14(I0I2)0014(I0I2))

are the only elements that determines the whether we have a ϕ dependence in Q, since this is the only part that needs to be acted on in Gaussian elimination method. It is obvious that we have

a1=a2=0.

In turn, it determines the angle dependence of Q,

Q=a0a1sinθcosϕa2sinθsinϕa3cosθkμvμ=Q0+Q3(θ).

We have no ϕ dependence in Q if we foce the emission to be axially symmetric.

1.2.4.2.2. Continuous Emission within Angle Range

In this case we have to calculate In specifically for the angle range, then plug in the expression n=k/ω to find the dispersion relation.

1.2.4.2.3. Discrete Emission Beams

For discrete emission G(θ)=iGiδ(cosθcosθi), we can define new quantities

I~n(θ)=iGicosnθi1ncosθi.

Thus

ω=14(I0I2)ω=14iGi1cos2θi1ncosθi.

For two sets of beams, we have

4=G11cos2θ1ωkcosθ1+G21cos2θ2ωkcosθ2,

which is a conic section. We have already used n=k/ω.

Hyperbola

For an quadratic equation [HyperbolaWikipedia]

Axxx2+2Axyxy+Ayyy2+2Bxx+2Byy+C=0,

it is hyperbola if

D:=|AxxAxyAxyAyy|<0.

Center of the hyperbola (xc,yc) is

xc=1D|BxAxyByAyy|yc=1D|AxxBxAxyBy|.

Principal axis is tilted away from x axis by angle β

tan2β=2AxyAxxAyy.
HyperbolaWikipedia

Hyperbola @ Wikipedia

We can prove that this is a hyperbola. Simplify the equation to standard form of conic sections

4cosθ1cosθ2k22×2(cosθ1+cosθ2)ωk+4ω2+2×G1(1cos2θ1)cosθ2+G2(1cos2θ2)cosθ12k2×G1(1cos2θ1)+G2(1cos2θ2)2ω=0.

The ω k plane

We use ω k plane, so that we map k to x and ω to y.

So the coefficients are defined as

The condition for it to be hyperbola is D<0, where

D=|4cosθ1cosθ22(cosθ1+cosθ2)2(cosθ1+cosθ2)4|=4(cosθ1cosθ2)2.

As long as we have different angles, D is always less than 0. We always have a hyperbola. The center of the hyperbola is

kc=1D|BkAkωBωAωω|=14(cosθ1cosθ2)2|G~1cosθ2+G~2cosθ122(cosθ1+cosθ2)G~1+G~224|ωc=1D|AkkBkAkωBω|=14(cosθ1cosθ2)2|4cosθ1cosθ2G~1cosθ2+G~2cosθ122(cosθ1+cosθ2)G~1+G~22|.

A special case for it is G1=±G2, however, the expression for the center doesn’t really simplify that much.

We are interested in gaps, so the asymptotic lines are the lines that we are interested in.

First of all, we need to find out the principal axis. The angle between the principal axis and x axis is defined to be β,

tan2β=2AkωAkkAωω=cosθ1+cosθ21cosθ1cosθ2.

Suppose we have angles θi=arctan(cosθi),

β=θ1+θ22.

This indicates that the angle β is always within range β[π/4,π/4].

The other solutions

For the solutions

ω=14(I0I2±(I02I1+I2)(I0+2I1+I2)),

it becomes much more complicated.

1.2.4.3. Why is it called Polarization Tensor

Why is aμ called polarization vector?

We kind of see why aμ is some kind of polarization given the definition

aμ=dΓvμG(θ,ϕ)Q(θ,ϕ).

In some sense it is a weighted average of Q. And S=Qeikμxμ is the “field” we are insterested in.

Comparing to electrodynamics, where we have the field Aμ which tells us about the polarization,

Polarization in Electrodynamics

We can assume that the four vector field is

We make sense of it by interpretating ϵμ as the polarization vector and a as the amplitude of the field strength. This solution is only for one particular case. We use it as an example because it is simple.

To calculate the electric field, Ei where i=1,2,3, we apply the definition of it

By working it out, we find ϵi, which is the spatial part of ϵμ, indeed plays a role in the direction of field.

1.2.4.4. Solving N Beams

For consistancy check, we parametrize ω(n). For MAA solution, we have

ω(n)=14iGi1cosnθi1ncosθi.
../../_images/DR-omega-n-2-beams-lr.png

Fig. 1.68 ω(n) with two angles cosθ=0.9,0.3 and G1=G2=2π/2. The grid lines are the values of n for singularity, n=1/0.3,1/0.9.

../../_images/DR-omega-k-2-beams-lr.png

Fig. 1.69 Same parameters as above. The orange dashed lines are the singularity lines.

Fig. 1.68 shows that the limit of n1/0.9,1/0.3 leads to ω infinities. These infinities also indicates that ω(k) at large |ω| are straight lines with slope n=1/0.9,1/0.3, as shown in Fig. 1.69.

Similar plots are made for 4,6,8 beams.

../../_images/DR-omega-n-2-4-6-8-beams-lr.png

Fig. 1.70 2, 4, 6, 8 beams with equal division of cosine of emission angle. For example 4 beams are emission at cosθ=0.9,0.7,0.5,0.3.

../../_images/DR-omega-k-2-4-6-8-beams-lr.png

Fig. 1.71 Same parameters as above.

We also calculated the homogeneous emssion.

../../_images/DR-omega-n-2-4-6-8-beams-homogeneous.png

Fig. 1.72 2, 4, 6, 8 beams for homogeneous G.

../../_images/DR-omega-k-2-4-6-8-beams-homogeneous.png

Fig. 1.73 Same parameters as above.

Gap

Whenever a gap in omega appears, we might find in the ω(n) plot that we have less solutions to n for some given ω values. It seems that this can only happen for ELN spectrum with crossing.

To illustrate this idea, I shaded the region that n(ω) has no solution for the case of two emission angle and crossing in Fig. 1.74.

../../_images/DR-omega-k-2-beams-lr-shaded-region.png

Fig. 1.74 The shaded regions are the region that n(ω) has no solution. It’s hard to analytically solve n(ω). But we can get an idea of this using ω(n) plot.

This provides an method to determine whether we have a gap in ω or k. For gap in k, we plot out k(n).

../../_images/DR-k-n-2-beams-lr-shaded-region.png

Fig. 1.75 The shaded region are the region where n(k) has no solution.

The behavor of ω(n) and k(n) are qualititatively the same.

../../_images/DR-k-n-2-beams-lr-and-homogeneous.png

Fig. 1.76 k(n) for non-crossing and crossing. The are qualitatively the same as ω(n). That being said, we always have the same upside down U-shape for spectrum with crossing for both k(n) and ω(n), if the region of emission angles θ satisfies cosθ>0.

1.2.4.5. Solving Continuous Emission

Suppose neutrinos are emitted within a angle range [θ1,θ2]. Using Mathematica, we find the three important integrals

I0=c2c1dx11kcosθ/ω=ωkln(ωc2kωc1k)I1=c2c1dxx1kcosθ/ω=ωk(c2c1+ωkln(ωc2kωc1k))I2=c2c1dxx21kcosθ/ω=ω2k(c2c1)(2ωk+c1+c2)+(ωk)3ln(ωc2kωc1k).

where c1=cosθ1 and c2=cosθ2.

1.2.4.5.1. Homogeneous Emission

Assuming G=1, the MAA solution is

ω=g4(I0I2),

which becomes

(1.36)ω=(c2c1)(1+(c1+c2)n/2)4n3+1n24n3ln(1c2n1c1n)

Meanwhile we could write down the MZA/bimodal solution in the form of ω(n).

For MAA and MZA we can plot ω as a function of n.

../../_images/DR-omega-k-direct-continuous-maa-no-crossing.png

Fig. 1.77 ω(n) for MAA solution. The spectrum is feeded in as {{cosθ1,cosθ2},G}={{0.9,0.3},1}.

../../_images/DR-omega-k-direct-continuous-mza-no-crossing.png

Fig. 1.78 ω(n) for MZA solution. The spectrum is feeded in as {{cosθ1,cosθ2},G}={{0.9,0.3},1}.

On the other hand, we know k=nω, so we have parametrized the dispersion relation using a parameter n.

../../_images/DR-omega-k-direct-continuous-dr-no-crossing.png

Fig. 1.79 Dispersion relation out of Eq. (1.36). For large k, the relation becomes proportional. The discontinuties are at location of limn{nω(n),ω(n)}. In this example, {0.184653,0} (MAA solution) and {0.729306,0} (MZA solution).

The Limits

There are several limits in the dispersion relation.

From the figure of ω, we notice the singularities at the two ends of the distribution, 1/c1 and 1/c2. At these points, a tiny change of n will cause a significant change in ω and k. In fact the relation between them becomes a proportional relation since n remains almost constant.

Another limit is n. Since ln(n) increases slower than n, we have

limnω(n)=limnI0I24=0.

We can calculate limn{nω(n),ω(n)}, for c1=0.9 and c1=0.3,

limn{nω(n),ω(n)}={0.184653,0.},for MAA solutionlimn{nω(n),ω(n)}={0.729306,0.},for MZA solution.

In general, we have the limits

limn{nω(n),ω(n)}={(c12c22+2ln|c2c1|)/8,0},for MAA solutionlimn{nω(n),ω(n)}={(c12c222ln|c2c1|)/4,0},for MZA solution.

1.2.4.5.2. Emission with Crossing

I have to break each of the integral into two parts. I calculate I0I2 for the first region then add to it the second region. Within a region [θ1,θ2] and

G={g1,θ1<θ<θ0g2,θ0<θ<θ2.

In other words, we have a box-like spetrum.

For MAA solution we define a function,

RHS(c1,c2,g)=I0I24=g4[(1n1n3)ln(1nc21nc1)c2c1n(c1+c22+1n)].

The dispersion relation is given by

ω=RHS(c1,c0,g1)+RHS(c0,c2,g2).

Then we parametrically plot {nω(n),ω(n)} to get the dispersion relation, for MAA solution. Similarly I can find that of MZA solution.

Limits

Before we do any numerical calculations, we can calculate the limits first.

limn±{nω(n),ω(n)}={(g1(c12c02)+g2(c02c22)+2g1lnc0c1+2g2lnc2c0)/8,0},for MAA solutionlimn{nω(n),ω(n)}={{twoverycomplicatedexpression,tryMathematica},0},for MZA solution.

For g1=g2=1 these limit match the homogeous result, which they should.

We also have the large k limit which are ω=1cik.

Mathematically, we also have

limn1/c1{nω(n),ω(n)}={DirectedInfinity(g1(1c12)),c1DirectedInfinity(Sign(g1(1+c12)))},for MAA solutionlimn1/c0{nω(n),ω(n)}={DirectedInfinity((g1g2)(1c02)),c0DirectedInfinity((g1g2)(1+c02))},for MAA solutionlimn1/c2{nω(n),ω(n)}={DirectedInfinity(g2(1c22)),c2DirectedInfinity(g2(1+c22))},for MAA solution.

For simplicity, we choose g2=g1=1.

../../_images/DR-omega-k-direct-continuous-two-regions-dr-crossing.png

Fig. 1.80 Dispersion relation for spectral crossing. The discontinuties are at {0.0944205,0} (MAA solution) and {0.098841,0} (MZA solution). The spectrum I use is {{{cosθ1,cosθ0},g1},{{cosθ0,cosθ2},g2}}={{{0.9,0.6},3},{{0.6,0.3},3}}.

I can also plot the MAA and MZA soltions for ω(n).

../../_images/DR-omega-k-direct-continuous-two-regions-maa-crossing.png

Fig. 1.81 ω(n) for MAA solution. The vertical grid lines are n=1/c1,1/c2. The spectrum I use is {{{0.9,0.6},3},{{0.6,0.3},3}}. At n±, both of the lines approaches 0.

../../_images/DR-omega-k-direct-continuous-two-regions-mza-crossing.png

Fig. 1.82 ω(n) for MZA solution. The vertical grid lines are n=1/c1,1/c2. The spectrum I use is {{{0.9,0.6},3},{{0.6,0.3},3}}. At n±, all lines approach 0. At n1/0.9, + solution becomes and th - solution becomes a complex number 0.63947436.1614i. (This should be irrelevant.) At n1/0.3, + solution approaches while - solution approaches 0.0445.

The reason we have no real values between 1/0.9 and 1/0.3 is because the argument of the ln function is negative within this regime.

../../_images/DR-omega-k-direct-continuous-two-regions-arg-ln.png

Fig. 1.83 The argument of the ln function. The vertical grid lines are n=1/0.3,1/0.6,1/0.9. Even n=1/0.6 gives us the zero argument, which means the ln function is infinite, I do not think this is some point that we can have a ω,k relation.

Some Discussions about ω(n)

It seems that if we plot ω(n) for MZA we will have some kind of similar plot compared to MAA. Well the singularities are all the same location. If we do that we will have the same |ω| for n very close to the sigularities thus same |k|. Of couse for MAA and MZA, ω have different signs.

In other words, the slopes of ω(k) will have the same value for MZA and MAA but in different quadrant.

1.2.4.6. General Discussions of Significance of Spectra

../../_images/DR-omega-of-n-for-different-spectra-table-0.9-0.6-0.3.png

Fig. 1.84 Function ω(n) for different g1,g2. The emission was set to G={g1,θ1<θ<θ0g2,θ0<θ<θ2 where cosθ1=0.9 and cosθ2=0.3 cosθ0=0.6.

../../_images/DR-k-of-n-for-different-spectra-table-0.9-0.6-0.3.png

Fig. 1.85 k(n) for the same parameters.

The corresponding dispersion relations are shown in Fig. 1.86

../../_images/DR-omega-k-continuous-table-0.9-0.6-0.3.png

Fig. 1.86 DR for spectra {{{0.9,0.6},g1},{{0.6,0.3},g2}}. Everytime we have g1 or g2 becomes 0, the results should be ignored.

../../_images/DR-omega-of-n-for-different-spectra-table-0.9-0.4-0.3.png

Fig. 1.87 Function ω(n) for different g1,g2. The emission was set to G={g1,θ1<θ<θ0g2,θ0<θ<θ2 where cosθ1=0.9 and cosθ2=0.3 cosθ0=0.4

../../_images/DR-k-of-n-for-different-spectra-table-0.9-0.4-0.3.png

Fig. 1.88 k(n) for the above case.

../../_images/DR-omega-k-continuous-table-0.9-0.4-0.3.png

Fig. 1.89 Dispersion relations for the above plot.

Why

It seems that crossing is important to a change in the number of solution to n(ω). If we have crossing, then the number of solutions to n(ω) will change for different values of ω. Otherwise, the number of solutions won’t change.

And a change of the number of solutions indicates a possible gap. I need some verification about the relation between such non-explicit gap and instabilities.

However, I also notice that the combination 2,1 also has such change of number of solutions. In this case, I can spot that for some branches we have complex k. What I don’t understand is that we always find real solutions to k for some other branches. Does it mean that only one such complex solutions is sufficient? And what exactly is the requirement?

Can I derive some expression for the ln function for a continuous distribution G(θ)?

1.2.4.6.1. Do we really need a gap?

As long as we have a point on ω(k) plane that ω(kb)= and ω(kb,)ω(kb,+)<0, we might obtain a complex k for some range of it.

Or I can simply consider k(n) and find whether we have a region of k(n)=0 and k(n)0 which generates different numbers of solutions in different region of k.

Imagary Part in k

For MAA solution, I can try to solve n given value of ω. If I obtain complex value of n, then I have complex value of k thus instability.

Didn’t find any weird numbers here.

Fig. 1.86 also indicates that crossing probably change the number of solutions to omega given different n thus change the number of solutions to k given different n. For example, the lower left panel shows that ω has 1 solutions within n[,1/0.9] but has 3 solutions within n[1/0.3,]. The lower right panel have 3 solutions whithin n[,1/0.9] and n[1/0.3,]. This might be important when identifying the possible gaps.

1.2.4.7. How to Use the Dispersion Relation

I can analyze some points on the ωk plane.

../../_images/lsa-DR-963-MZA.png

Fig. 1.90 MZA solution for spectrum {{{0.9,0.6},3},{{0.6,0.3},3}}. In principle I should have some real solutions fall onto the dispersion relation lines. Other points solved directely from linear stability analysis should be the unstable regions.

../../_images/lsa-963-kimag-omega-MZA-p.png

Fig. 1.91 for MZA+ solution for spectrum {{{0.9,0.6},3},{{0.6,0.3},3}}. The points for ω>0 are not real because these are values returned without convergence. The values on the left are returned with convergence.

../../_images/lsa-963-kimag-omega-MZA-m.png

Fig. 1.92 for MZA- solution. The points for ω<0 are not real because these are values returned without convergence.

../../_images/lsa-DR-93-MZA-Omega-Real.png

Fig. 1.93 MZA solution for spectrum {{{0.9,0.3},1}}. Solve complex k for real ω.

../../_images/lsa-DR-93-MZA-K-Real.png

Fig. 1.94 MZA for spectrum {{{0.9,0.3},1}}. Solve complex ω for real k.

1.2.4.8. Discrete Case and Continuous Case

Will we have a continuous case if the number of beams is infinite.

For discrete case

(1.37)I0I2=i=1NGi1cos2θiωkcosθi=iGi1ui2ωkui.

The continuous case is

(1.38)I0I2=dcosθG(cosθ)1cos2θωkcosθ=duG(u)1u2ωku.

We notice that Eq. (1.37) and Eq. (1.38) are the same when number of beams becomes large.

Gi is in fact Gi=giΔui, where Δui is the range of cosθi around θi.

1.2.4.9. Arbitrary Spectrum

../../_images/DR-continuous-spectrum-near-realistic.png

Fig. 1.95 Reproducing Fig. 3 of Raffelt’s paper.

../../_images/DR-continuous-spectrum-near-realistic-spectrum-created.png

Fig. 1.96 The spectrum I created and used. 11.2738u511.5219u4+9.1079u3+12.6683u2+4.51014u+2.50958.

The unstable regions can be calculated exactly by setting ω to be real.

../../_images/DR-continuous-spectrum-near-realistic-no-crossing-maa.png

Fig. 1.97 MAA

../../_images/DR-continuous-spectrum-near-realistic-no-crossing-mzap.png

Fig. 1.98 MZA+

../../_images/DR-continuous-spectrum-near-realistic-no-crossing-mzam.png

Fig. 1.99 MZA-


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