User:Egm6321.f12.team6.taylor.bt

=Report 7=

Problem Statement: The Laplacian in Spherical Coordinates Using the Math/Physics Convention

 * Express the Laplacian Operator in spherical coordinates using the Math Physics Convention.

Given

 * The Laplacian in General Curvilinear Coordinates;

Nomenclature

 * The Math Physics convention for spherical coordinates expresses three coordinate axes:
 * $$ (r, \bar \theta, \varphi) = (\xi_1, \xi_2, \xi_3)$$
 * In relation to the cartesian coordinate axes the Math Physics axes are measured as demonstrated in the figure below;


 * The vector, $$ \mathbf g_i $$, is a vector tangent to the curvilinear coordinate axis.

Solution

 * Step 1 -
 * The first objective is to relate the coordinates from the Cartesian coordinate system, to the Spherical system. This is done by projecting the vector formed from the Cartesian point, $$ (x, y, z) $$, in terms of the Spherical system, $$ (r, \bar \theta, \varphi) $$, onto each axis.  Inspection of Figure 1 yields the following results:


 * Step 2 -
 * The next step is to find the vectors tangent to the spherical coordinate axes. We can interpret equations, 6.3, 6.4 and 6.5 as functions such that, $$ x, \; y, \; z $$, are functions of $$ r, \; \bar \theta, \; \varphi $$. This can be expressed as $$ x(r, \bar \theta, \varphi), \; y(r, \bar \theta, \varphi), \; z(r, \bar \theta) $$. We can now infer that the derivative of equations 6.3, 6.4 and 6.5 define the tangent vector, $$ \mathbf g_i $$. Therefore the tangent vectors will have three components that define them. These components are in terms of the spherical coordinate axis, such that;


 * There are nine separate derivations that need to be performed. The results are found to be;


 * Step 3 -
 * Now that we have found the tangent vectors, we can find, $$ h_i $$, as defined from equation 6.2.
 * First we solve for $$ h_1 $$;


 * $$ h_1= \parallel \mathbf g_1 \parallel $$
 * $$ = \sqrt {\sin^2 \bar \theta \cos^2 \varphi + \sin^2 \bar \theta \sin^2 \varphi + \cos^2 \bar \theta} $$
 * $$ = \sqrt {\sin^2 \bar \theta ( \cos^2 \varphi + \sin^2 \varphi) + \cos^2 \bar \theta} $$
 * $$ = \sqrt {\sin^2 \bar \theta (1) + \cos^2 \bar \theta} $$
 * $$ = \sqrt {\sin^2 \bar \theta + \cos^2 \bar \theta} $$
 * $$ = \sqrt {1} $$


 * Next we solve for $$ h_2 $$;
 * $$ h_2= \parallel \mathbf g_2 \parallel $$
 * $$ = \sqrt {r^2 \cos^2 \bar \theta \cos^2 \varphi + r^2 \cos^2 \bar \theta \sin^2 \varphi + r^2 \sin^2 \bar \theta} $$
 * $$ = \sqrt {r^2 [\cos^2 \bar \theta ( \cos^2 \varphi + \sin^2 \varphi) + \sin^2 \bar \theta]} $$
 * $$ = \sqrt {r^2 [\cos^2 \bar \theta (1) + \sin^2 \bar \theta]} $$
 * $$ = \sqrt {r^2 (1)} $$


 * Last we solve for $$ h_3 $$;
 * $$ h_3= \parallel \mathbf g_3 \parallel $$
 * $$ = \sqrt {r^2 \sin^2 \bar \theta \sin^2 \varphi + r^2 \sin^2 \bar \theta \cos^2 \varphi} $$
 * $$ = \sqrt {r^2 \sin^2 \bar \theta ( \sin^2 \varphi + \cos^2 \varphi)} $$
 * $$ = \sqrt {r^2 \sin^2 \bar \theta (1)} $$


 * Step 4 -
 * The next operation is to solve the Laplacian operator in equation 6.1.
 * It will be convienient to solve for;


 * The summation in equation 6.1 will expand 3 times. We will solve for Laplacian for each expansion of the summation. Note that we take use of a nomenclature change in that;
 * $$ (r, \bar \theta, \varphi) = (\xi_1, \xi_2, \xi_3)$$


 * For the summation index, $$ i=1 $$, we have;


 * For the summation index, $$ i=2 $$, we have;
 * $$ \Delta u_2 = \frac{1}{r^2 \sin \bar \theta} \frac{\partial }{\partial \bar \theta} \left ( \frac{r^2 \sin \bar \theta}{r^2} \frac{\partial u}{\partial \bar \theta} \right )$$


 * For the summation index, $$ i=3 $$, we have;
 * $$ \Delta u_3 = \frac{1}{r^2 \sin \bar \theta} \frac{\partial }{\partial \varphi} \left ( \frac{r^2 \sin \bar \theta}{r^2 \sin^2 \bar \theta} \frac{\partial u}{\partial \varphi} \right )$$


 * We now can formulate our final equation;
 * $$ \Delta u = \frac{1}{r^2 \sin \bar \theta} \frac{\partial }{\partial r} \left ( r^2 \sin \bar \theta \frac{\partial u}{\partial r} \right ) + \frac{1}{r^2 \sin \bar \theta} \frac{\partial }{\partial \bar \theta} \left ( \sin \bar \theta \frac{\partial u}{\partial \bar \theta} \right ) + \frac{1}{r^2 \sin \bar \theta} \frac{\partial }{\partial \varphi} \left ( \frac{1}{ \sin \bar \theta} \frac{\partial u}{\partial \varphi} \right ) $$


 * {| style="width:100%" border="0"|-


 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * $$ \displaystyle \Delta u= \frac{1}{r^2 \sin \bar \theta}\left ( \frac{\partial }{\partial r} \left ( r^2 \sin \bar \theta \frac{\partial u}{\partial r} \right ) + \frac{\partial }{\partial \bar \theta} \left ( \sin \bar \theta \frac{\partial u}{\partial \bar \theta} \right ) + \frac{\partial }{\partial \varphi} \left ( \frac{1}{ \sin \bar \theta} \frac{\partial u}{\partial \varphi} \right ) \right )$$


 * (6.15)
 * }