S H I N E

Saturday, September 22, 2018

Cosmic rays propagation equation

the particles number through energy boundary in $dt$ is $dN$

(E+dE)-------->|(E)-------->(E-dE)

boundary

\[ dN = \int_E^{E+dE} dN/dE(E) dE = dN/dE(E_0)dE \approx dN(E)/dEdE = \frac{dN(E)}{dE}\frac{dE}{dt} dt\]

the number of particles that pass through the energy boundary per unit of time,
\[ J_E = \frac{dN}{dt} = \frac{dN}{dE}\frac{dE}{dt} \]

----------------->[(E+dE)------------------](E)--------------->

the change of particle number due to energy variation within energy interval $[E,E+dE]$ per unit time if $dE/dt<0$.

\begin{equation}
\begin{split}
\frac{dN}{dt} & = dN/dE(E+dE)d(E+dE)/dt - dN/dE(E)dE/dt \\
& = [dN/dE(E)+\frac{d^2N}{dE^2}(E) dE] \cdot [dE/dt + \frac{d^2E}{dEdt}(E)dE ] - dN/dE(E)dE/dt \\
& = \frac{d}{dE}(\frac{dN}{dE}(E)\frac{dE}{dt}) \cdot dE
\end{split}
\end{equation}

thus, \[ \frac{dN}{dEdt} = \frac{d}{dE}(\frac{dN}{dE}(E)\frac{dE}{dt}) = \frac{dJ_E}{dE} \]

the change of particle number due to motion

Some operation about vector and tensor

1 derivative of a vector

\[ Y = (y_1(x_1,x_2,x_3) ,y_2(x_1,x_2,x_3),y_3(x_1,x_2,x_3)) \]
\[ dy_1 = dy_1/dx_1 \cdot dx_1+ \ldots = \sum_i^3 \frac{dy_1}{dx_i} \cdot dx_i\]
\[ dy_i = \sum_i^3 \frac{dy_1}{dx_i} dx_i \]

\begin{align}
dy_i = \begin{pmatrix}
dy_i/dx_1 &dy_i/dx_2 &dy_i/dx_3
\end{pmatrix} \,
\begin{pmatrix}
dx_1 \\
dx_2 \\
dx_3 \\
\end{pmatrix}
\end{align}

\begin{align}
\begin{pmatrix}
dy_1 \\
dy_2 \\
dy_3 \\
\end{pmatrix} =
\begin{pmatrix}
dy_1/dx_1 &dy_1/dx_2 &dy_1/dx_3 \\
dy_2/dx_1 &dy_2/dx_2 &dy_2/dx_3 \\
dy_3/dx_1 &dy_3/dx_2 &dy_3/dx_3 \\
\end{pmatrix} \,
\begin{pmatrix}
dx_1 \\
dx_2 \\
dx_3 \\
\end{pmatrix}
\end{align}

\begin{align}
\frac{d\vec{y}}{d\vec{x}} =
\begin{bmatrix}
dy_1/dx_1 &dy_1/dx_2 &dy_1/dx_3 \\
dy_2/dx_1 &dy_2/dx_2 &dy_2/dx_3 \\
dy_3/dx_1 &dy_3/dx_2 &dy_3/dx_3 \\
\end{bmatrix}
\end{align}

2 double dot product of two tensor
\[\vec{T}:\vec{U} = \sum_i\sum_j T_{ij}U_{ji}\]

3 the dot product of a tensor with a vector
\[ \vec{T}\cdot \vec{v} = \sum_i \vec{\delta}_i (\sum_j T_{ij}v_j) \]

4 the dot product of a vector with a tensor
\[\vec{v}\cdot{T} = \sum_i \vec{\delta}_i (\sum_j T_{ji}) v_j \]

5. unit tensor

\[ \nabla \cdot \vec{I} = 0 \]

\[ \vec{v} \cdot \vec{I} = \vec{v} \]

6. vector dot gradient of tensor

\[ \vec{A} \cdot \nabla \cdot \vec{T} = \nabla \cdot( \vec{A} cdot \vec{T}) - \vec{T}:\nabla \vec{A} \]

REFERENCE:
http://cs231n.stanford.edu/vecDerivs.pdf
http://www.polymerprocessing.com/notes/root92a.pdf
difference between matrix,bmatrix,pmatrix,vmatix,Vmatrix
https://math-linux.com/latex-26/faq/latex-faq/article/how-to-write-matrices-in-latex-matrix-pmatrix-bmatrix-vmatrix-vmatrix

Resource
The Poor Man’s Introduction to Tensors

Tuesday, September 18, 2018

First order partial different equation

\[\frac{\partial u}{\partial t} + c(x,t)\frac{\partial u}{\partial x} = 0\]
find a curve that the u is constant along it.

\[ du = \frac{\partial u}{\partial t}dt + \frac{ \partial u}{\partial x}dx \]
\[ t = t(r),\qquad x=x(r) \]
\[ dt = \frac{dt}{dr} dr,\qquad dx = \frac{dx}{dr} dr\]

\[\frac{du}{dr} = \frac{\partial u}{\partial t}\frac{dt}{dr} + \frac{ \partial u}{\partial x}\frac{dx}{dr} \]

we hope $du=0$,then along the curve, the u is constant.
$\frac{du}{dr} = 0 = \frac{\partial u}{\partial t} + c(x,t)\frac{\partial u}{\partial x}$
we get \[\frac{dt}{dr}=1\],and \[\frac{dx}{dr}=c(x,t)\].
they determine the characteristic curve.
\[u(t(r1),x(r1) = u(t(r2),x(r2))\]

Guide to chromebook

screen ctrl + switch window -> full screenshot
ctrl shift switch -> select area

shortcuts of chromebook
https://support.google.com/chromebook/answer/183101?hl=en

latex on chromebook
https://github.com/macbuse/Chromebook/blob/master/LaTeX.md

chrome as text editor
data:text/html, <body contenteditable style="font: 1.5rem/1.5 monospace;max-width:60rem;margin:0 auto;padding:4rem;">

rotate screen
ctrl shift refresh

Friday, September 14, 2018

Force-field solution to solar modulation

propagation equation

\begin{equation}
\frac{\partial f}{\partial t} + \nabla \cdot (C \vec{V}_{sw} f - K \cdot \nabla f) - \frac{1}{3p^2} \frac{\partial (p^3V_{sw}\cdot \nabla f)}{\partial p} =0 \label{eq1}
\end{equation}

$C = -1/3\frac{\partial lnf}{\partial lnp}$

stream current density
\begin{equation}
J = C V_{sw} f - K \cdot \nabla f \label{eq:stream}
\end{equation}

\begin{equation}
K = \begin{bmatrix}
k_{rr} &k_{r\theta} &k_{r\phi} \\
k_{\theta r} &k_{\theta\theta} &k_{\theta\phi} \\
k_{\phi r} &k_{\phi\theta} &k_{\phi\phi}
\end{bmatrix} \nonumber
\end{equation}

$k_{rr} = k_{\parallel}\cos^2(\psi) + k_{\perp,r}\sin^2(\psi)$
$\tan\psi = r*\Omega/V_{sw}$

$\nabla = \hat r \frac{\partial}{\partial r} + \frac{\hat \theta}{r} \frac{\partial}{\partial \theta} + \frac{\hat \phi}{r \sin(\theta)} \frac{\partial}{\partial \phi}$
The dot product of a tensor with a vector is:
$A\cdot B = i(A_{11} B_1 + A_{12}B_2 + A_{13}B_3) + j(A_{21}B_1 + A_{22}B_2 + A_{23}B_3) + k(A_{31}B_1 + A_{32}B_2 + A_{33}B_3)$

One example for syntax hightlight in blogspot

Follow the guide https://eric0806.blogspot.com/2014/04/blogger-google-code-prettify.html .

#include&ltstdio.h&gt
int main (void)
{
    print("Hello World");
    return 0;
}

$E=mC^2$

xxx