# Diff of /trunk/doc/cookbook/example10.tex

revision 3233 by jfenwick, Mon Oct 4 00:54:26 2010 UTC revision 3308 by jfenwick, Tue Oct 26 03:24:54 2010 UTC
# Line 29  where $\gamma$ is the gravitational cons Line 29  where $\gamma$ is the gravitational cons
29  Consider now the \esc general form, which can simply be related to  Consider now the \esc general form, which can simply be related to
30  \autoref{eqn:poisson} using two coefficients. The result is  \autoref{eqn:poisson} using two coefficients. The result is
31
32  -\left(A\hackscore{jl} u\hackscore{,l} \right)\hackscore{,j} = Y  -\left(A_{jl} u_{,l} \right)_{,j} = Y
33
34  one recognises that the LHS side is equivalent to  one recognises that the LHS side is equivalent to
35   \label{eqn:ex10a}   \label{eqn:ex10a}
36  -\nabla A \nabla u  -\nabla A \nabla u
37
38  If $A=\delta\hackscore{jl}$ then \autoref{eqn:ex10a} is equivalent to  If $A=\delta_{jl}$ then \autoref{eqn:ex10a} is equivalent to
39  \begin{equation*}  \begin{equation*}
40  -\nabla^2 u  -\nabla^2 u
41  \end{equation*}  \end{equation*}
42  and thus Poisson's \autoref{eqn:poisson} satisfies the general form when  and thus Poisson's \autoref{eqn:poisson} satisfies the general form when
43
44  A=\delta\hackscore{jl} \text{ and } Y= 4\pi\gamma\rho  A=\delta_{jl} \text{ and } Y= 4\pi\gamma\rho
45
46  At least one boundary point must be set for the problem to be solvable. For this  At least one boundary point must be set for the problem to be solvable. For this
47  example we have set all of the boundaries to zero. The normal flux condition is  example we have set all of the boundaries to zero. The normal flux condition is
# Line 73  The potential $U$ is related to the grav Line 73  The potential $U$ is related to the grav
73
74  \vec{g} = \nabla U  \vec{g} = \nabla U
75
76  This for example as a vertical component $g\hackscore{z}$ where  This for example as a vertical component $g_{z}$ where
77
78  g\hackscore{z}=\vec{g}\cdot\hat{z}  g_{z}=\vec{g}\cdot\hat{z}
79
80  Finally, there is the magnitude of the vertical component $g$ of  Finally, there is the magnitude of the vertical component $g$ of
81  $g\hackscore{z}$  $g_{z}$
82
83  g=|g\hackscore{z}|  g=|g_{z}|
84
85  These values are derived from the \esc solution \verb!sol! to the potential $U$  These values are derived from the \esc solution \verb!sol! to the potential $U$
86  using the following commands  using the following commands

Legend:
 Removed from v.3233 changed lines Added in v.3308