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:
$A\cdot B = i(A_{11} B_1 + A_{12}B_2 + A_{13}B_3) + j(A_{21}B_1 + A_{22}B_2 + A_{23}B_3) + k(A_{31}B_1 + A_{32}B_2 + A_{33}B_3)$

force-field solution

Assuming the particle distribution function f is isotropic and the solar wind velocity is $\vec{V}_{sw} = V_{sw}\hat{r}$.

$$\nabla f = [\frac{\partial f}{\partial r}\hat{r},0,0]$$
$$K \cdot \nabla f = \hat{r}k_{rr} \frac{\partial f}{\partial{r}}$$

if $k_{\parallel} = k_{\perp,r}=k$, then $k_{rr}=k$.

let J in eq $\ref{eq:stream}$ equals zero,
$$\begin{equation} %\Rightarrow 1/3pV_{sw}\frac{\partial f}{\partial p} + k\frac{\partial f}{\partial r} = 0 \end{equation}$$

solution of this equation

characteristic equation:
$$\frac{dp}{1/3pV_{sw}} = \frac{dr}{k}$$ .
f is constant along the trajectories in the phase space which satisfy the characteristic equation.

assuming $k = \beta k_1(r)k_2(p)$
$$\begin{equation} \int \frac{\beta k_2(p) dp}{p} = \int \frac{V_{sw}dr}{3k_1(r)} \label{gff}. \end{equation}$$

$\beta = v = p/E = \sqrt{(E^2-m^2)/(E^2)} = \sqrt{(1-m^2/E^2)}$
when $E = T + m \gg m$ or $T\gg 0$, $\beta =1$.

example: $T = m$ $\Rightarrow$ $\beta = \sqrt{(1-{\frac{1}{2}}^2)} \approx 0.866$.



if $k_2(p) = p$ and $T>m$, eq $\ref{gff}$ becomes
$$\begin{equation}\label{eq:ceq} \int dp = \int \frac{V_{sw} dr}{3 k_1(r)} \equiv \phi . \end{equation}$$
curve: $$p +c = \phi$$
$$f = F(c)=F(p-\phi) ??$$
$$f = F(p_1-\phi_1) = F(p_2-\phi_2) ??$$

f is constant along the curve. This indicates that $$p_1-\phi_1(r_1) = c,p_2-\phi_2(r_2) =c$$ .

Finally we get
$$p_2 = p_1 + \phi_2(r_2) - \phi_1(r_1) .$$
$$p_2 = p_1 + \phi(r_2) - \phi(r_1) .$$

redefine $\phi = \int_0^{r} \frac{V_{sw} dr}{3 k_1(r)}$ and $f(r=0,p) = f(r,p_2)$

$f = \frac{dN}{d^3r d^3p} = \frac{dN}{dV p^2dp d\Omega} = \frac{dN}{dA v dt dT d\Omega p^2 dp/dT} = J/(p^2)$
$dp/dT = 1/v$

$J(T) = p^2 f$

$$\begin{equation} \frac{J(T1)}{p_1^2} = \frac{J(T2)}{p_2^2} = \frac{J(T2)}{(p_1+\phi)^2} \label{eq:solu} \end{equation}$$

Finally, we get the force-field solution $\ref{eq:solu}$.



We can also directly integrate the $\ref{eq:ceq}$
$$\int_{p_1}^{p_2} dp = \int_{r_1}^{r_2} \frac{V_{sw} dr}{3 k_1(r)} \equiv \phi.$$

REFERENCE

Spherical Coordinates
www.cpp.edu/~ajm/materials/delsph.pdf
Tensor
www.polymerprocessing.com/notes/root92a.pdf
force-field modulation
A MONTE CARLO APPROACH FOR THE SOLAR MODULATION