User:Egm6321.f10.team3.cook/HW6

=Problem 5: Derive the Laplacian in Cylindrical Coordinates=

Given
From the lecture notes p 35-3
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{x}_{1}=x= { \xi}_{1} \cos { \xi}_{2} $$ $$
 * $$\displaystyle (Eq 5.1)
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{x}_{2}=y= { \xi}_{1} \sin { \xi}_{2} $$ $$
 * $$\displaystyle (Eq 5.2)
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{x}_{3}=z= { \xi}_{3} $$ $$
 * $$\displaystyle (Eq 5.3)
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Find
$$ 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 $$ \left\{ {dx}_{1}, {dx}_{2}, {dx}_{3}\right\} $$


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{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
 * $$\displaystyle (Eq 5.4)
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{dx}_{1} = \frac{\partial }{\partial  {\xi}_{1}} \left(  {\xi}_{1} \cos {\xi}_{2}\right) {d\xi}_{1} + \frac{\partial }{\partial  {\xi}_{2}} \left(  {\xi}_{1} \cos {\xi}_{2}\right) {d\xi}_{2} +\cancelto{0}\frac{\partial }{\partial  {\xi}_{3}} \left(  {\xi}_{1} \cos {\xi}_{2}\right){d\xi}_{3} $$
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$$ {dx}_{1} =  \cos {\xi}_{2}{d\xi}_{1} - {\xi}_{1} \sin {\xi}_{2} {d\xi}_{2} $$ $$
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 * $$\displaystyle (Eq 5.5)
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Similarly,


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{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} $$
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$$ {dx}_{2} =  \sin {\xi}_{2}{d\xi}_{1} + {\xi}_{1} \cos {\xi}_{2} {d\xi}_{2} $$ $$
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 * $$\displaystyle (Eq 5.6)
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Since there is no projection on $$ {x}_{3}$$ from $$\displaystyle (Eq 5.4) $$


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$$ {dx}_{3} =  {d\xi}_{3} $$ $$
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 * $$\displaystyle (Eq 5.7)
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$$ {\left(ds\right)}^{2} $$
To solve we use the relationship


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{ds}^{2} = { \left(d {x}_{1} \right)}^{2}+ { \left(d {x}_{2} \right)}^{2}+ { \left(d {x}_{3} \right)}^{2} $$
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And substituting equations 5.5 through 5.7 we obtain the following


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{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)
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Expanding equation 5.8 and combining like terms gives us


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{ds}^{2} = \left( {\cos {\xi}_{2}}^{2}+ { \sin {\xi}_{2}}^{2} \right) {\left( {d\xi}_{1} \right)}^{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} $$ $$ Now we use the property $$ {\sin}^2+{\cos}^2 = 1$$ we obtain
 * $$\displaystyle (Eq 5.9)
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$$ {ds}^{2}={\left( {d\xi}_{1} \right)}^{2}+ {{\xi}_{1}}^{2}{\left( {d\xi}_{2} \right)}^{2}+ {\left( {d\xi}_{3} \right)}^{2} $$ $$
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 * $$\displaystyle (Eq 5.10)
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Now by inspection we see that


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$$ {h}_{1}= 1 ; {h}_{2} =  {\xi}_{1} ; {h}_{3}= 1 $$ $$
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 * $$\displaystyle (Eq 5.11)
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$$\Delta \Psi $$
To obtain the Laplacian in cylindrical coordinates we use the following equation


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\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) $$ $$
 * $$\displaystyle (Eq 5.12)
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Now we plug in our values for $$ {h}_{i}$$ and expand the summation


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\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}} $$ $$
 * $$\displaystyle (Eq 5.13)
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Finally we replace the $$ \xi s $$ with the cylindrical coordinate convention to obtain


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$$ \Delta \Psi =  \frac{1}{r}  \frac{\partial }{\partial  r} \left(  \frac{1}{r} \frac{\partial  \Psi}{\partial r}\right)+ \frac{\partial {}^{2} \Psi}{\partial  { \theta}^{2}} + \frac{\partial {}^{2} \Psi}{\partial  {z}^{2}} $$ $$
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 * <p style="text-align:right;">$$\displaystyle (Eq 5.14)
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Contributing Members
--Egm6321.f10.team3.cook 02:45, 16 November 2010 (UTC)