Final formatting edits for EPJ ST submission.

02z600StrangeHadrons.tex

@@ -253,9 +253,10 @@ \subsection{Strange hadron abundance in cosmic plasma}
 \end{align}
 Here the experimental cross-sections can be parameterized as 
 \begin{align}
-&\sigma_{K^-p\rightarrow \Lambda\pi^0}\!\!=\!\!\left(\begin{array}{l}\!\!1479.53\mathrm{mb}\!\cdot\!\exp{\left(\frac{-3.377\sqrt{s}}{\mathrm{GeV}}\right)},\; \mathrm{for}\,\sqrt{s_m}\!\!<\!\!\sqrt{s}\!<\!3.2\mathrm{GeV} \\ \\0.3\mathrm{mb}\!\cdot\!\exp{\left(\frac{-0.72\sqrt{s}}{\mathrm{GeV}}\right)},\; \mathrm{for}\sqrt{s}>3.2\mathrm{GeV}\end{array}\right.\,,
-\\
-&\sigma_{K^-n\rightarrow \Lambda\pi^-}\!\!=\!\!1132.27\mathrm{mb}\!\cdot\!\exp{\left(\frac{-3.063\sqrt{s}}{\mathrm{GeV}}\right)},\; \mathrm{for}\sqrt{s}>1.699\mathrm{GeV},
+&\sigma_{K^-p\rightarrow \Lambda\pi^0}\!\!=\!\!\left\{\begin{array}{l}\!\!1479.53\mathrm{mb}\!\cdot\!\displaystyle e^{\left(\frac{-3.377\sqrt{s}}{\mathrm{GeV}}\right)}\qquad   \mathrm{for\ } \sqrt{s_m}\!\!<\!\!\sqrt{s}\!<\!3.2\mathrm{GeV}\,, \\[0.5cm]
+0.3\mathrm{mb}\!\cdot\!\displaystyle e^{\left(\frac{-0.72\sqrt{s}}{\mathrm{GeV}}\right)}\qquad\qquad   \mathrm{for\ }\sqrt{s}>3.2\mathrm{GeV}\,,\end{array}\right.
+\\[0.4cm]
+&\sigma_{K^-n\rightarrow \Lambda\pi^-}\!\!=\!\!1132.27\mathrm{mb}\!\cdot\!e^{\left(\frac{-3.063\sqrt{s}}{\mathrm{GeV}}\right)} \qquad\   \mathrm{for\ }\sqrt{s}>1.699\mathrm{GeV}\,,
 \end{align}
 where $\sqrt{s_m}=1.473\GeV$.\\
 2) For the cross-section $\sigma_{\pi N\rightarrow K\Lambda}$ we use~\cite{Cugnon:1984pm}
@@ -264,7 +265,12 @@ \subsection{Strange hadron abundance in cosmic plasma}
 \end{align}
 The experimental $\sigma_{\pi p\rightarrow K^0\Lambda}$ can be approximated as follows
 \begin{align}
-\sigma_{\pi p\rightarrow K^0\Lambda}=\left(\begin{array}{l}\frac{0.9\mathrm{mb}\cdot\left(\sqrt{s}-\sqrt{s_0}\right)}{0.091\mathrm{GeV}},\; \mathrm{for} \sqrt{s_0}<\sqrt{s}<1.7\mathrm{GeV} \\ \\ \frac{90\mathrm{MeV\cdot mb}}{\sqrt{s}-1.6\mathrm{GeV}},\; \mathrm{for}\sqrt{s}>1.7\mathrm{GeV},\end{array}\right.
+\sigma_{\pi p\rightarrow K^0\Lambda}=\left\{
+\begin{array}{l}
+\displaystyle\frac{0.9\mathrm{mb}\cdot\left(\sqrt{s}-\sqrt{s_0}\right)}{0.091\mathrm{GeV}}\qquad  \mathrm{for\ } \sqrt{s_0}<\sqrt{s}<1.7\mathrm{GeV}\,, \\[0.5cm]
+\displaystyle\frac{90\mathrm{MeV\cdot mb}}{\sqrt{s}-1.6\mathrm{GeV}}\qquad\qquad \quad  \mathrm{for\
+ }\sqrt{s}>1.7\mathrm{GeV}\,,\end{array}\right.
+ \,,
  \end{align}
  with $ \sqrt{s_0}=m_\Lambda+m_K$. 
 

10appDNeutrinoScatteringIntegrals.tex

@@ -188,8 +188,8 @@ \subsection{Collision integral inner products}
 \end{align}
 where
 \begin{align}
-t(x)=&\frac{1}{4}((q^0)^2-r^2+((q^{\prime})^0)^2-(r^{\prime})^2-2q^0(q^{\prime})^0+2rr^{\prime}x),\\
-=&\frac{1}{4}((q^0-(q^{\prime})^0)^2-r^2-(r^{\prime})^2+2rr^{\prime}x)\,.\notag
+t(x)=&\frac{1}{4}((q^0)^2-r^2+((q^{\prime})^0)^2-(r^{\prime})^2-2q^0(q^{\prime})^0+2rr^{\prime}x), 
+= \frac{1}{4}((q^0-(q^{\prime})^0)^2-r^2-(r^{\prime})^2+2rr^{\prime}x)\,. 
 \end{align}
 
 %%%%%%%%%%%%%%%%%%%%%%%%
@@ -335,15 +335,15 @@ \subsection{Electron and neutrino collision integrals}\label{nu:matrix:elements}
 \end{table}
 Using this we find
  \begin{align}
-&\int_0^{2\pi} S |\mathcal{M}|^2 (s,t(\cos(\psi)\sqrt{1-y^2}\sqrt{1-z^2}+yz))d\psi\\
-=&\frac{\pi C}{16} s^2(3+4 yz-y^2-z^2+3y^2z^2)
-\equiv\frac{\pi C}{16} s^2q(y,z)\,\notag\\
-&K(s,p)=\frac{\pi^2C}{2}s^2\int_{-1}^1 \left[\int_{-1}^1q(y,z)G_{34}(p^0,-p y) dy\right] G_{12}(p^0,-p z)dz\,.\notag
+\int_0^{2\pi} S |\mathcal{M}|^2 (s,t(\cos(\psi)\sqrt{1-y^2}\sqrt{1-z^2}+yz))d\psi 
+=& \frac{\pi C}{16} s^2(3+4 yz-y^2-z^2+3y^2z^2)
+\equiv\frac{\pi C}{16} s^2q(y,z)\,, \\
+K(s,p)=&\frac{\pi^2C}{2}s^2\int_{-1}^1 \left[\int_{-1}^1q(y,z)G_{34}(p^0,-p y) dy\right] G_{12}(p^0,-p z)dz\,.\notag
 \end{align}
 Therefore
 \begin{align}\label{eq:M:nu:nubar}
-&M_{\nu\bar\nu\rightarrow\nu\bar\nu}=\frac{C}{2048(2\pi)^5 }T^8\!\!\!\int_0^\infty\!\!\!\!\int_0^\infty\!\!\! \tilde{s}^2\bigg[\int_{-1}^1\!\int_{-1}^1q(y,z)\tilde{G}_{34}(\tilde p^0,-\tilde{p} y)\notag\\
-&\hspace{58mm}\tilde{G}_{12}(\tilde p^0,-\tilde{p} z)dydz\bigg]\frac{\tilde{p}^2}{\tilde{p}^0}d\tilde{p}d\tilde{s}\,.\notag
+M_{\nu\bar\nu\rightarrow\nu\bar\nu}=\frac{C}{2048(2\pi)^5 }T^8\! \int_0^\infty\!\!\int_0^\infty\!\!\! \tilde{s}^2\bigg[\int_{-1}^1\!\int_{-1}^1q(y,z)\tilde{G}_{34}(\tilde p^0,-\tilde{p} y) 
+\tilde{G}_{12}(\tilde p^0,-\tilde{p} z)dydz\bigg]\frac{\tilde{p}^2}{\tilde{p}^0}d\tilde{p}d\tilde{s}\,.
 \end{align}
  If we want to emphasize the role of $C$ then we write $M_{\nu\bar\nu\rightarrow\nu\bar\nu}(C)$. Note that due to the polynomial form of the matrix element integral, the double integral in brackets breaks into a linear combination of products of one dimensional integrals, meaning that the nesting of integrals is again only three deep in  practice.
 
@@ -372,20 +372,20 @@ \subsection{Electron and neutrino collision integrals}\label{nu:matrix:elements}
 
 
 
-  The integral of each of these terms is
+The integral of each of these terms is
  \begin{align}
-&\int_0^{2\pi}\frac{(s+t(\psi)-m_e^2)^2}{4}d\psi=\frac{\pi}{16}s(3s-4m_e^2)+\frac{\pi}{4}s^{3/2}\sqrt{s-4m_e^2}yz\\
-&\qquad\qquad-\frac{\pi}{16}s(s-4m_e^2)(y^2+z^2)+\frac{3\pi}{16}s(s-4m_e^2)y^2z^2\,,\notag\\
-&\int_0^{2\pi} \frac{(m_e^2-t(\psi))^2}{4}d\psi=\frac{\pi}{16}s(3s-4m_e^2)-\frac{\pi}{4}s^{3/2}\sqrt{s-4m_e^2}yz\notag\\
-&\qquad\qquad-\frac{\pi}{16}s(s-4m_e^2)(y^2+z^2)+\frac{3\pi}{16}s(s-4m_e^2)y^2z^2\,,\notag\\
-&\int_0^{2\pi} m_e^2\frac{s}{2} d\psi=\pi m_e^2s\,.\notag
+\int_0^{2\pi}\frac{(s+t(\psi)-m_e^2)^2}{4}d\psi=&\frac{\pi}{16}s(3s-4m_e^2)+\frac{\pi}{4}s^{3/2}\sqrt{s-4m_e^2}yz
+-\frac{\pi}{16}s(s-4m_e^2)(y^2+z^2)+\frac{3\pi}{16}s(s-4m_e^2)y^2z^2\,,\\
+\int_0^{2\pi} \frac{(m_e^2-t(\psi))^2}{4}d\psi=&\frac{\pi}{16}s(3s-4m_e^2)-\frac{\pi}{4}s^{3/2}\sqrt{s-4m_e^2}yz
+-\frac{\pi}{16}s(s-4m_e^2)(y^2+z^2)+\frac{3\pi}{16}s(s-4m_e^2)y^2z^2\,,\notag\\
+\int_0^{2\pi} m_e^2\frac{s}{2} d\psi=&\pi m_e^2s\,.\notag
 \end{align}
 Therefore 
 \begin{align}
-&\int_0^{2\pi} S |\mathcal{M}|^2 (s,t(\psi))d\psi\\
-=&\frac{\pi}{16}s[3s(A+B)+4m_e^2(4C-A-B)]+\frac{\pi}{4}s^{3/2}\sqrt{s-4m_e^2}(A-B)yz\notag\\
-&-\frac{\pi}{16}s(s-4m_e^2)(A+B)(y^2+z^2)+\frac{3\pi}{16}s(s-4m_e^2)(A+B)y^2z^2\notag\\
-\equiv& \pi q(m_e,s,y,z)\,,\notag
+\int_0^{2\pi} S |\mathcal{M}|^2 (s,t(\psi))d\psi
+=&\frac{\pi}{16}s[3s(A+B)+4m_e^2(4C-A-B)]+\frac{\pi}{4}s^{3/2}\sqrt{s-4m_e^2}(A-B)yz\\
+&-\frac{\pi}{16}s(s-4m_e^2)(A+B)(y^2+z^2)+\frac{3\pi}{16}s(s-4m_e^2)(A+B)y^2z^2
+\equiv \pi q(m_e,s,y,z)\,,\notag
 \end{align}
 and hence
 \begin{align}
@@ -548,8 +548,8 @@ \subsection{Electron and neutrino collision integrals}\label{nu:matrix:elements}
 
 Finally, we point out how the vanishing of these inner products is a reflection of certain conservation laws. From \req{n:div}, \req{collision:integrals}, and the fact that $\hat\psi_0,\hat\psi_1$ span the space of polynomials of degree $\leq 1$, we have the following expressions for the change in number density and energy density of a massless particle
 \begin{align}
-\frac{1}{a^3} \frac{d}{dt}(a^3n)=&\frac{g_p}{2\pi^2}\int \frac{1}{E}C[f]p^2dp=c_0 R_0\,,\\
-\frac{1}{a^4}\frac{d}{dt}(a^4\rho)=&\frac{g_p}{2\pi^2}\int C[f] p^2dp=d_0R_0+d_1R_1\,,\notag
+\frac{1}{a^3} \frac{d}{dt}(a^3n)= \frac{g_p}{2\pi^2}\int \frac{1}{E}C[f]p^2dp=c_0 R_0\,,\qquad
+\frac{1}{a^4}\frac{d}{dt}(a^4\rho)= \frac{g_p}{2\pi^2}\int C[f] p^2dp=d_0R_0+d_1R_1\,, 
 \end{align}
 for some $c_0,d_0,d_1$. Therefore, the vanishing of $R_0$ is equivalent to conservation of comoving particle number.  The vanishing of $R_0$ and $R_1$ implies $\rho\propto 1/a^4$ i.e. that the reduction in energy density is due entirely to redshift; energy is not lost from the distribution due to scattering.  These findings match the situations above where we found one or both of $R_0=0$, $R_1=0$.  $R_0$ vanished for all kinetic scattering processes and we know that all such processes conserve comoving particle number.  Both $R_0$ and $R_1$ vanished for a distribution scattering from itself and in such a process  there is no energy loss energy  from the distribution by scattering; energy is only redistributed among the particles corresponding to that distribution.
 
@@ -609,27 +609,27 @@ \subsection{Comparison with an alternative method for computing scattering integ
 These expressions are symmetric under $1\leftrightarrow 2$ and $3\leftrightarrow 4$ and so without loss of generality we can assume $p_1\geq p_2$, $p_3\geq p_4$. We require $p_1\leq p_2+p_3+p_4$ (and cyclic permutations) by conservation of energy.  In the case where the above conditions all hold, we separate the computation into four additional cases in which the integrals can be evaluated analytically, as  in \cite{Dolgov:1997mb,Dolgov:1998sf}:\\
 ${\bf p_1+p_2>p_3+p_4\text{, \hspace{1mm} }p_1+p_4>p_2+p_3}${\bf :}
 \begin{align}
-D_1=&\frac{\pi}{8}(p_2+p_3+p_4-p_1)\,,\\
-D_2=&\frac{\pi}{48}((p_1-p_2)^3+2(p_3^3+p_4^3)-3(p_1-p_2)(p_3^2+p_4^2)\,,\notag\\
+D_1=&\frac{\pi}{8}(p_2+p_3+p_4-p_1)\,,\qquad
+D_2=\frac{\pi}{48}((p_1-p_2)^3+2(p_3^3+p_4^3)-3(p_1-p_2)(p_3^2+p_4^2)\,, \\
 D_3=&\frac{\pi}{240}(p_1^5-p_2^5+5p_2^3(p_3^2+p_4^2)-5p_1^3(p_2^2+p_3^2+p_4^2)-(p_3+p_4)^3(p_3^2-3p_3p_4+p_4^2)\notag\\
 &\hspace{7mm}+5p_2^2(p_3^3+p_4^3)+5p_1^2(p_2^3+p_3^3+p_4^3))\,.\notag
 \end{align}
 ${\bf p_1+p_2<p_3+p_4\text{, \hspace{1mm} }p_1+p_4>p_2+p_3}${\bf :}
 \begin{align}
-D_1=&\frac{\pi }{4}p_2\,,\\
-D_2=&\frac{\pi }{24}p_2(3(p_3^2+p_4^2-p_1^2)-p_2^2)\,,\notag\\
-D_3=&\frac{\pi}{120}p_2^3(5(p_1^2+p_3^2+p_4^2)-p_2^2)\,.\notag
+D_1= \frac{\pi }{4}p_2\,,\qquad
+D_2= \frac{\pi }{24}p_2(3(p_3^2+p_4^2-p_1^2)-p_2^2)\,,\qquad 
+D_3= \frac{\pi}{120}p_2^3(5(p_1^2+p_3^2+p_4^2)-p_2^2)\,. 
 \end{align}
 ${\bf p_1+p_2>p_3+p_4\text{, \hspace{1mm} }p_1+p_4<p_2+p_3}${\bf :}
 \begin{align}
-D_1=&\frac{\pi }{4}p_4\,,\\
-D_2=&\frac{\pi}{12} p_4^3\,,\notag\\
-D_3=&\frac{\pi }{120}p_4^3(5(p_1^2+p_2^2+p_3^2)-p_4^2)\,.\notag
+D_1= \frac{\pi }{4}p_4\,,\qquad 
+D_2= \frac{\pi}{12} p_4^3\,,\qquad 
+D_3= \frac{\pi }{120}p_4^3(5(p_1^2+p_2^2+p_3^2)-p_4^2)\,. 
 \end{align}
 ${\bf p_1+p_2<p_3+p_4\text{, \hspace{1mm} }p_1+p_4<p_2+p_3}${\bf :}
 \begin{align}
-D_1=&\frac{\pi}{8}(p_1+p_2+p_4-p_3)\,,\\
-D_2=&\frac{\pi}{48}(-(p_1+p_2)^3-2p_3^3+2p_4^3+3(p_1+p_2)(p_3^2+p_4^2))\,,\notag\\
+D_1=&\frac{\pi}{8}(p_1+p_2+p_4-p_3)\,,\qquad
+D_2= \frac{\pi}{48}(-(p_1+p_2)^3-2p_3^3+2p_4^3+3(p_1+p_2)(p_3^2+p_4^2))\,, \\
 D_3=&\frac{\pi}{240}(p_3^5-p_4^5-(p_1+p_2)^3(p_1^2-3p_1p_2+p_2^2)+5(p_1^3+p_2^3)p_3^2-5(p_1^2+p_2^2)p_3^3\notag\\
 &\hspace{7mm}+5(p_1^3+p_2^3-p_3^3)p_4^2+5(p_1^2+p_2^2+p_3^2)p_4^3)\,.\notag
 \end{align}