User:Egm6321.f2010.team5.riveros/hw6p6

= Problem 6 - Spherical Coordinate Problem = From (meeting 35 page 4)

Given
We are given the following relations,
 * $$\displaystyle x_1 = r \sin \bar{\theta} \cos \phi =  \xi_1 \sin \xi_2 \cos \xi_3 $$
 * $$\displaystyle x_2 = r \sin \bar{\theta} \sin \phi =  \xi_1 \sin \xi_2 \sin \xi_3 $$
 * $$\displaystyle x_3 = r \cos \bar{\theta} =  \xi_1 \cos \xi_2$$

Noting that $$\displaystyle \xi_1=r$$, $$\displaystyle \xi_2=\bar{\theta}$$, and $$\displaystyle \xi_3=\phi$$

Find
We are to derive the laplacian, $$\Delta\Psi$$, in spherical coordinates using the math/physics convention in which,
 * $$\bar\theta=\frac{\pi}{2}-\theta$$

Solution
We first calculate the total derivatives, $$dx_i$$, in spherical coordinates using the given relations. We have,
 * $$\displaystyle dx_1=\frac{\partial x_1}{\partial \xi_1}d \xi_1 + \frac {\partial x_1}{\partial \xi_2} d \xi_2 + \frac{\partial x_1}{\partial\xi_3}d\xi_3=sin\xi_2 cos \xi_3 d\xi_1 + \xi_1 cos \xi_2 cos \xi_3 d \xi_2 - \xi_1 sin \xi_2 sin \xi_3 d \xi_3 $$
 * $$\displaystyle dx_2=\frac{\partial x_2}{\partial \xi_1}d\xi_1+\frac{\partial x_2}{\partial \xi_2}d\xi_2+\frac{\partial x_2}{\partial \xi_3}d\xi_3=dx_2=sin\xi_2 sin\xi_3 d\xi_1 + \xi_1 cos\xi_2 sin\xi_3 d\xi_2 + \xi_1 sin\xi_2 cos\xi_3 d\xi_3 $$
 * $$\displaystyle dx_3=\frac{\partial x_3}{\partial\xi_1}d\xi_1+\frac{\partial x_3}{\partial \xi_2}d\xi_2+\frac{\partial x_3}{\partial \xi_3}d\xi_3=dx_3=cos\xi_2 d\xi_1 - \xi_1 sin\xi_2 d\xi_2 + 0 $$

Next, $$(ds)^2$$ is found by similarly. We have,
 * $$\displaystyle (ds)^2=\sum (dx_i)^2 $$

By substituting the given relations and using the first Pythagorean identity, $$(ds)^2$$ simplifies to,
 * $$\displaystyle ds^2=d\xi_1^2+\xi_1^2d\xi_2^2+\xi_1^2sin^2\xi_2d\xi_3^2$$

We note that $$h_i^2$$ is the coefficient of $$\xi^2_i$$. Listing the coefficients,
 * $$\displaystyle h_1=1$$
 * $$\displaystyle h_2=\xi_1$$
 * $$\displaystyle h_3=\xi_1 sin\xi_2$$

The Laplacian for orthogonal curvilinear coordinates can be expressed as
 * $$\displaystyle \Delta\Psi=\frac{1}{h_1h_2h_3}\sum_{i=1}^{3}\frac{\partial}{\partial \xi_i}\left(\frac {h_1h_2h_3}{(h_i)^2}\cdot\frac{\partial\Psi}{\partial\xi_i}\right)$$

We substitute the values for $$h_i$$ found above to yield
 * $$\displaystyle \Delta\Psi=\frac{1}{\xi^2_1sin\xi_2}\left[\frac{\partial}{\partial\xi_1}\left(\xi^2_1sin\xi_2\frac{\partial\Psi}{\partial \xi_1}\right)

+\frac{\partial}{\partial\xi_2}\left(sin\xi_2\frac{\partial\Psi}{\partial \xi_2}\right) +\frac{\partial}{\partial\xi_3}\left(\frac{1}{sin\xi_2}\frac{\partial\Psi}{\partial \xi_3}\right)\right]$$ We expand the above relation to yield the Laplacian in spherical coordinates,
 * $$\Delta\Psi =\frac {1}{r^2 sin \bar\theta} \left[ \frac {\partial}{\partial r} \left(r^2 sin \bar\theta \frac {\partial\Psi}{\partial r} \right)

+ \frac {\partial}{\partial \bar\theta} \left(sin \bar\theta \frac {\partial \Psi}{\partial \bar\theta} \right) + \frac {\partial}{\partial \phi} \left(\frac {1}{sin \bar\theta} \frac {\partial \Psi}{\partial \phi}\right) \right] $$

Author and proof-reader
[Author]
 * Raul Riveros

[Proof-reader]

=Problem 5: Derive the Laplacian in Cylindrical Coordinates=

Given
From the lecture notes p 35-3
 * {| style="width:100%" border="0"

{x}_{1}=x= { \xi}_{1} \cos { \xi}_{2} $$ $$
 * $$\displaystyle (Eq 5.1)
 * style= |
 * }


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

{x}_{2}=y= { \xi}_{1} \sin { \xi}_{2} $$ $$
 * $$\displaystyle (Eq 5.2)
 * style= |
 * }


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

{x}_{3}=z= { \xi}_{3} $$ $$
 * $$\displaystyle (Eq 5.3)
 * style= |
 * }

Find
$$ 1) $$ Find $$ \left\{  {dx}_{i}\right\}=  \left\{  {dx}_{1}, {dx}_{2}, {dx}_{3}\right\} $$ in terms of $$  \left\{  {\xi}_{j}\right\}=  \left\{  {\xi}_{1}, {\xi}_{2}, {\xi}_{3}\right\} $$ and $$ \left\{  {d\xi}_{k}\right\} $$

$$ 2) $$ Find $${ds}^{2} = \sum_{i} { \left(d {x}_{i} \right)}^{2}=  \sum_{i}  {\left(  {h}_{i}\right)}^{2}  { \left( {d\xi}_{i} \right)}^{a}$$ Identify $$ \left\{  {h}_{i}\right\} $$ in terms of $$ \left\{  {\xi}_{i}\right\} $$

$$ 3) $$ Find $$ \Delta \Psi $$ in cylindrical coordinates

$$ \left\{ {dx}_{i}\right\} $$
To solve we find the total derivatives of $$ \left\{ {dx}_{1}, {dx}_{2}, {dx}_{3}\right\} $$


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

{dx}_{1} = \frac{\partial  {x}_{1}}{\partial  {\xi}_{1}} {d\xi}_{1}+\frac{\partial  {x}_{1}}{\partial  {\xi}_{2}} {d\xi}_{2}+\frac{\partial  {x}_{1}}{\partial  {\xi}_{3}} {d\xi}_{3} $$ $$ Substituting $$\displaystyle (Eq 5.1) $$ into $$\displaystyle (Eq 5.4) $$ we get
 * $$\displaystyle (Eq 5.4)
 * style= |
 * }


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

{dx}_{1} = \frac{\partial }{\partial  {\xi}_{1}} \left(  {\xi}_{1} \cos {\xi}_{2}\right) {d\xi}_{1} + \frac{\partial }{\partial  {\xi}_{2}} \left(  {\xi}_{1} \cos {\xi}_{2}\right) {d\xi}_{2} +\cancelto{0}\frac{\partial }{\partial  {\xi}_{3}} \left(  {\xi}_{1} \cos {\xi}_{2}\right){d\xi}_{3} $$
 * 
 * style= |
 * }


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

$$ {dx}_{1} =  \cos {\xi}_{2}{d\xi}_{1} - {\xi}_{1} \sin {\xi}_{2} {d\xi}_{2} $$ $$
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 * $$\displaystyle (Eq 5.5)
 * style= |
 * }

Similarly,


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

{dx}_{2} = \frac{\partial }{\partial  {\xi}_{1}} \left(  {\xi}_{1} \sin {\xi}_{2}\right) {d\xi}_{1} + \frac{\partial }{\partial  {\xi}_{2}} \left(  {\xi}_{1} \sin {\xi}_{2}\right) {d\xi}_{2} +\cancelto{0}\frac{\partial }{\partial  {\xi}_{3}} \left(  {\xi}_{1} \sin {\xi}_{2}\right){d\xi}_{3} $$
 * 
 * style= |
 * }


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

$$ {dx}_{2} =  \sin {\xi}_{2}{d\xi}_{1} + {\xi}_{1} \cos {\xi}_{2} {d\xi}_{2} $$ $$
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 * $$\displaystyle (Eq 5.6)
 * style= |
 * }

Since there is no projection on $$ {x}_{3}$$ from $$\displaystyle (Eq 5.4) $$


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

$$ {dx}_{3} =  {d\xi}_{3} $$ $$
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 * $$\displaystyle (Eq 5.7)
 * style= |
 * }

$$ {\left(ds\right)}^{2} $$
To solve we use the relationship


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

{ds}^{2} = { \left(d {x}_{1} \right)}^{2}+ { \left(d {x}_{2} \right)}^{2}+ { \left(d {x}_{3} \right)}^{2} $$
 * 
 * style= |
 * }

And substituting equations 5.5 through 5.7 we obtain the following


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

{ds}^{2} = { \left(\cos {\xi}_{2}{d\xi}_{1} - {\xi}_{1} \sin {\xi}_{2} {d\xi}_{2} \right)}^{2}+ { \left(\sin {\xi}_{2}{d\xi}_{1} + {\xi}_{1} \cos {\xi}_{2} {d\xi}_{2} \right)}^{2}+ { \left(d {\xi}_{3} \right)}^{2} $$ $$
 * $$\displaystyle (Eq 5.8)
 * style= |
 * }

Expanding equation 5.8 and combining like terms gives us


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

{ds}^{2} = \cancelto{1}{\left( {\cos {\xi}_{2}}^{2}+ { \sin {\xi}_{2}}^{2} \right)} {\left( {d\xi}_{1} \right)}^{2}+ \cancelto{{{\xi}_{1}}^{2}}{\left({{\xi}_{1}}^{2}{\cos {\xi}_{2}}^{2}+ {  {{\xi}_{1}}^{2}\sin {\xi}_{2}}^{2} \right)}  {\left( {d\xi}_{2} \right)}^{2}+  {\left( {d\xi}_{3} \right)}^{2} $$ $$
 * $$\displaystyle (Eq 5.9)
 * style= |
 * }


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

$$ {ds}^{2}={\left( {d\xi}_{1} \right)}^{2}+ {{\xi}_{1}}^{2}{\left( {d\xi}_{2} \right)}^{2}+ {\left( {d\xi}_{3} \right)}^{2} $$ $$
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 * $$\displaystyle (Eq 5.10)
 * style= |
 * }

Now by inspection we see that


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

$$ {h}_{1}= 1 ; {h}_{2} =  {\xi}_{1} ; {h}_{3}= 1 $$ $$
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 * $$\displaystyle (Eq 5.11)
 * style= |
 * }

$$\Delta \Psi $$
To obtain the Laplacian in cylindrical coordinates we use the following equation


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

\Delta \Psi = \frac{1}{ {h}_{1}{h}_{2} {h}_{3}} \sum_{i=1}^{3}  \frac{\partial }{\partial  {\xi}_{i}} \left(  \frac{{h}_{1}{h}_{2} {h}_{3}}{ {\left( {h}_{i} \right)}^{2}} \frac{\partial  \Psi}{\partial {\xi}_{i}}\right) $$ $$
 * $$\displaystyle (Eq 5.12)
 * style= |
 * }

Now we plug in our values for $$ {h}_{i}$$ and expand the summation


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

\Delta \Psi =  \frac{1}{ {\xi}_{1}}  \frac{\partial }{\partial  {\xi}_{1}} \left(  \frac{1}{{\xi}_{1}} \frac{\partial  \Psi}{\partial {\xi}_{1}}\right)+ \cancelto{1}\frac{{\xi}_{1}}{ {\xi}_{1}} \frac{\partial {}^{2} \Psi}{\partial  {\xi}_{2}^{2}} + \cancelto{1}\frac{{\xi}_{1}}{ {\xi}_{1}} \frac{\partial {}^{2} \Psi}{\partial  {\xi}_{3}^{2}} $$ $$
 * $$\displaystyle (Eq 5.13)
 * style= |
 * }

Finally we replace the $$ \xi s $$ with the cylindrical coordinate convention to obtain


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

$$ \Delta \Psi =  \frac{1}{r}  \frac{\partial }{\partial  r} \left(  \frac{1}{r} \frac{\partial  \Psi}{\partial r}\right)+ \frac{\partial {}^{2} \Psi}{\partial  { \theta}^{2}} + \frac{\partial {}^{2} \Psi}{\partial  {z}^{2}} $$ $$
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 * <p style="text-align:right;">$$\displaystyle (Eq 5.14)
 * style= |
 * }

Contributing Members
Solved and Posted by--Egm6321.f10.team3.cook 02:45, 16 November 2010 (UTC)
 * Proofread by Egm6321.f10.team03.sigillo 02:58, 16 November 2010 (UTC)

= Problem 6 The Laplace Equation in Cylindrical Coordinates =

Given
Converting from rectangular to spherical coordinates:

$$\displaystyle x_1 = x = r sin \bar\theta cos \phi $$

$$\displaystyle x_2 = y = r sin \bar\theta sin \phi $$

$$\displaystyle x_3 = z = r cos\bar\theta $$

$$\displaystyle \xi_1 = r $$

$$\displaystyle \xi_2 = \bar\theta $$

$$\displaystyle \xi_3 = \phi $$

where $$\displaystyle \bar\theta $$ is equal to $$\displaystyle \pi / 2 - \theta $$ by the math/physics convention, as opposed to the astronomic convention

Find
the Laplacian, $$\displaystyle \Delta \Psi $$, in terms of spherical coordinates.

Solution
$$\displaystyle dx_1 = \frac {\partial x_1}{\partial xi_1} d \xi_1 + \frac {\partial x_1}{\partial xi_2} d \xi_2 + \frac {\partial x_1}{\partial xi_3} d \xi_3 $$

$$\displaystyle dx_1 = sin \xi_2 cos \xi_3 d \xi_1 + \xi_1 cos \xi_2 cos \xi_3 d \xi_2 - \xi_1 sin \xi_2 sin \xi_3 d \xi_3 $$

$$\displaystyle dx_2 = \frac {\partial x_2}{\partial xi_1} d \xi_1 + \frac {\partial x_2}{\partial xi_2} d \xi_2 + \frac {\partial x_2}{\partial xi_3} d \xi_3 $$

$$\displaystyle dx_2 = sin\xi_2 sin\xi_3 d\xi_1 + \xi_1 cos\xi_2 sin\xi_3 d\xi_2 + \xi_1 sin\xi_2 cos\xi_3 d\xi_3 $$

$$\displaystyle dx_3 = \frac {\partial x_3}{\partial xi_1} d \xi_1 + \frac {\partial x_3}{\partial xi_2} d \xi_2 + \frac {\partial x_3}{\partial xi_3} d \xi_3 $$

$$\displaystyle dx_3 = cos\xi_2 d\xi_1 - \xi_1 sin\xi_2 d\xi_2 + 0 $$

$$\displaystyle ds^2 = \sum dx_i^2 $$

$$\displaystyle $$

Using the trigonometric identity

$$\displaystyle cos^2\alpha + sin^2\alpha = 1 $$

and regrouping terms yields

$$\displaystyle ds^2 = d\xi_1^2 + \xi_1^2 d\xi_2^2 + \xi_1^2 sin^2 \xi_2 d\xi_3^2 $$

where $$\displaystyle h_i^2 $$ are the coefficients of each term in the above expression.

$$\displaystyle h_1 = 1 $$

$$\displaystyle h_2 = \xi_1 = r $$

$$\displaystyle h_3 = \xi_1 sin\xi_2 = r sin\bar\theta $$

For any set of orthogonal curvilinear coordinates, the Laplacian is given by

$$\displaystyle \Delta \Psi = \frac {1}{h_1h_2h_3} \sum \frac {\partial}{\partial \xi_i} (\frac {h_1h_2h_3}{h_i^2} \frac {\partial \Psi}{\partial \xi_i}) $$

for $$\displaystyle i = 1 $$

$$\displaystyle = \frac {1}{\xi_1} ( \frac {\partial}{\partial \xi_1} \frac {\xi_1}{h_1^2} \frac {\partial \Psi}{\partial \xi_1} ) $$

$$\displaystyle = \frac {1}{r^2 sin\bar\theta} \frac {\partial}{\partial r} (r^2 sin\bar\theta \frac {\partial \Psi}{\partial r} $$

for $$\displaystyle i = 2 $$

$$\displaystyle = \frac {1}{\xi_1} (\frac {\partial}{\partial \xi_2} \frac {\xi_1}{h_2^2} \frac {\partial \Psi}{\partial \xi_2}) $$

$$\displaystyle = \frac {1}{r^2 sin\bar\theta} \frac{\partial}{\partial \bar\theta} (\frac {r^2 sin \bar\theta}{r^2} \frac {partial \Psi}{\partial \bar\theta}) $$

for $$\displaystyle i = 3 $$

$$\displaystyle = \frac {1}{\xi_1} (\frac {\partial}{\partial \xi_3} \frac {\xi_1}{h_3^2} \frac {\partial \Psi}{\partial \xi_3}) $$

$$\displaystyle = \frac {1}{r^2 sin\bar\theta} \frac{\partial}{\partial phi} (\frac {r^2 sin\bar\theta}{r^2 sin^2 \bar\theta} \frac {\partial \Psi}{\partial \phi}) $$

Therefore, the expression for the Laplacian in terms of spherical coordinates becomes

128.227.51.176 17:51, 12 November 2010 (UTC)Egm6321.f10.team4.osentowski - Author