next up previous contents
Next: Adding high energy moments Up: The algorithm Previous: Choice of frame   Contents

The normalised non-linear system

We normalise the unknown moments in terms of the measured ones as:
\begin{displaymath}
V_e^{\prime} = \frac{\vert{\bf V}_e\vert}{V_{m}}
\end{displaymath} (4.1)


\begin{displaymath}
V_{Te}^\prime = \frac{V_{Te}}{V_{Tm}}
\end{displaymath} (4.2)


\begin{displaymath}
\eta = \frac{1}{\sqrt{\pi}}\frac{N_e/N_m}{V_e^\prime V_{Te}^\prime }.
\end{displaymath} (4.3)

Where $V_{Te}$ is the thermal speed, from which the temperature can be inferred:
\begin{displaymath}
T_e = \frac{m_eV_{Te}^2}{2k}
\end{displaymath} (4.4)

The inputs to the solver are the normalised quantities
\begin{displaymath}
\zeta_{sc} = \frac{V_{m}}{V_{Tm}}
\end{displaymath} (4.5)


\begin{displaymath}
\epsilon_{sc} = \frac{\mathcal{E}}{V_{Tm}^2}
\end{displaymath} (4.6)


\begin{displaymath}V_{L,U} = \frac{v_{L,U}}{V_{Tm}}\end{displaymath} (4.7)

For the inputs $v_L=0$, $v_U = \infty$, $\mathcal{E} = 0$, the exact solution yields unity for equations 4.5 & 4.6 and $1/\sqrt{\pi}$ for equation 4.7.

The set of non-linear equations which must be inverted are:
\begin{displaymath}g_1(N_e,V_e,T_e) - N_m = 0 \end{displaymath} (4.8)


\begin{displaymath}g_2(N_e,V_e,T_e) - N_mV_{m} = 0 \end{displaymath} (4.9)


\begin{displaymath}g_3(N_e,V_e,T_e) - 3N_mkT_m - m_eN_mV_m^2 = 0 \end{displaymath} (4.10)

That is, the measured moments are functions of the real moments. We recast to a normalised system such that we wish to find the triplet ($\eta$, $V_e^\prime$, $V_{Te}^\prime$), given the normalised inputs ($\zeta_{sc}$, $\epsilon_{sc}$, $V_{L,U}$). The equations to be solved are
\begin{displaymath}
1-\frac{\eta}{\zeta_{sc}}\int_{V_L}^{V_U} \sqrt{V^2-\epsilon_{sc}}\left(E^- - E^+\right){\rm d}V = 0
\end{displaymath} (4.11)


$\displaystyle \zeta_{sc}^2-\eta\int_{V_L}^{V_U} (V^2-\epsilon_{sc}) \left[E^- + E^+
-\frac{V_{Te}^{\prime 2}(E^-E^+)}{2VV_e^\prime\zeta_{sc}}\right]{\rm d}V = 0$     (4.12)


$\displaystyle \frac{3}{2} + \zeta_{sc}^2 - \frac{\eta}{\zeta_{sc}}\int_{V_L}^{V_U} (V^2-\epsilon_{sc}) ^ {3/2}
\times(E^- - E^+){\rm d}V = 0$     (4.13)

where
\begin{displaymath}
E^{\pm}=\exp{\left[-\left(\frac{V\pm V_e^\prime\zeta_{sc}}{V_{Te}^\prime}\right)^2\right]}
\end{displaymath} (4.14)


next up previous contents
Next: Adding high energy moments Up: The algorithm Previous: Choice of frame   Contents
Steve Schwartz 2005-03-26