Base locale sferica

Ricaviamo la base locale sferica in uno spazio tridimensionale. La figura mostra i versori della base diretti lungo le direzioni, in ogni punto ortogonali in coppia alle superfici e tangenti alle linee coordinate.

base sferica locale in uno spazio tridimensionale

Partiamo dalle espressioni delle trasformazioni sferiche: $$ \Phi: \begin{cases} x = \rho sin\theta cos\phi \\ y = \rho sin\theta sin\phi \\ z = \rho cos\theta \end{cases} $$ Sappiamo che il vettore \( r\) può essere espresso come somma dei suoi vettori componenti, quindi possiamo scrivere:

$$ r = x\hat x + y\hat y + z \hat z = (\rho sin\theta cos\phi) \hat x + (\rho sin\theta sin\phi) \hat y + (\rho cos\theta)\hat z $$ $$ r = x\hat x + y\hat y + z \hat z = (\rho sin\theta cos\phi) \hat x + (\rho sin\theta sin\phi) \hat y + (\rho cos\theta)\hat z $$ $$ r = x\hat x + y\hat y + z \hat z = \begin{align}(\rho sin\theta cos\phi) \hat x + \\ (\rho sin\theta sin\phi) \hat y + \\ (\rho cos\theta)\hat z \end{align} $$

Calcoliamo i vettori tangenti alle linee coordinate, che in questo caso sono la \( \theta\)-curva, la \( \phi\)-curva e la \( r\)-curva:

$$ \begin{cases} {\partial r \over \partial \rho} = (sin\theta cos\phi) \hat x + (sin\theta sin\phi) \hat y + (cos\theta)\hat z \\ {\partial r \over \partial \theta} = (\rho cos\theta cos\phi) \hat x + (\rho cos\theta sin\phi) \hat y - (\rho sin\theta)\hat z \\ {\partial r \over \partial \phi} = -(\rho sin\theta sin\phi) \hat x + (\rho sin\theta cos\phi) \hat y \end{cases} $$ $$ \begin{cases} {\partial r \over \partial \rho} = (sin\theta cos\phi) \hat x + (sin\theta sin\phi) \hat y + (cos\theta)\hat z \\ {\partial r \over \partial \theta} = (\rho cos\theta cos\phi) \hat x + (\rho cos\theta sin\phi) \hat y - (\rho sin\theta)\hat z \\ {\partial r \over \partial \phi} = -(\rho sin\theta sin\phi) \hat x + (\rho sin\theta cos\phi) \hat y \end{cases} $$ $$ \small \begin{cases} {\partial r \over \partial \rho} = (sin\theta cos\phi) \hat x + (sin\theta sin\phi) \hat y + (cos\theta)\hat z \\ {\partial r \over \partial \theta} = (\rho cos\theta cos\phi) \hat x + (\rho cos\theta sin\phi) \hat y - (\rho sin\theta)\hat z \\ {\partial r \over \partial \phi} = -(\rho sin\theta sin\phi) \hat x + (\rho sin\theta cos\phi) \hat y \end{cases} $$

osservate come nell'ultima derivata, la componente lungo \( z\) va via essendo costante rispetto a \( \phi\). Calcoliamoci ora le norme dei vettori tangenti (fattori di scala)

$$ \begin{align}h_u = \begin{Vmatrix} {\partial r \over \partial u} \end{Vmatrix} = \sqrt{(sin\theta cos\phi)^2+ (sin\theta sin\phi)^2 + (cos\theta)^2} = \\ \sqrt{ sin^2\theta cos^2\phi+ sin^2\theta sin^2\phi + cos^2\theta } = \\ \sqrt{ sin^2\theta \bigl(cos^2\phi+ sin^2\phi\bigr) + cos^2\theta } = \\ \sqrt{ sin^2\theta + cos^2\theta } = \sqrt{1} = 1 \end{align} $$

$$ \begin{align}h_v = \begin{Vmatrix} {\partial r \over \partial v} \end{Vmatrix} = \sqrt{ (\rho cos\theta cos\phi)^2 + (\rho cos\theta sin\phi)^2 + (-\rho sin\theta)^2} = \\ \sqrt{ \rho^2 cos^2\theta cos^2\phi + \rho^2 cos^2\theta sin^2\phi + \rho^2 sin^2\theta } = \\ \rho \sqrt{ cos^2\theta \bigl(cos^2\phi + sin^2\phi \bigr) + sin^2\theta } = \\ \rho \sqrt{ cos^2\theta + sin^2\theta } = \rho \end{align} $$

$$ \begin{align}h_w = \begin{Vmatrix} {\partial r \over \partial w} \end{Vmatrix} = \sqrt{(-\rho sin\theta sin\phi)^2 + (\rho sin\theta cos\phi)^2 } = \\ \sqrt{ \rho^2 sin^2\theta sin^2\phi + \rho^2 sin^2\theta cos^2\phi } = \\ \rho \sqrt{ sin^2\theta \bigl(sin^2\phi + cos^2\phi\bigr) } = \\ \rho \sqrt{ sin^2\theta } = \rho sin\theta \end{align} $$

$$ \begin{align}h_u = \begin{Vmatrix} {\partial r \over \partial u} \end{Vmatrix} = \sqrt{(sin\theta cos\phi)^2+ (sin\theta sin\phi)^2 + (cos\theta)^2} = \\ \sqrt{ sin^2\theta cos^2\phi+ sin^2\theta sin^2\phi + cos^2\theta } = \\ \sqrt{ sin^2\theta \bigl(cos^2\phi+ sin^2\phi\bigr) + cos^2\theta } = \\ \sqrt{ sin^2\theta + cos^2\theta } = \sqrt{1} = 1 \end{align} $$

$$ \begin{align}h_v = \begin{Vmatrix} {\partial r \over \partial v} \end{Vmatrix} = \sqrt{ (\rho cos\theta cos\phi)^2 + (\rho cos\theta sin\phi)^2 + (-\rho sin\theta)^2} = \\ \sqrt{ \rho^2 cos^2\theta cos^2\phi + \rho^2 cos^2\theta sin^2\phi + \rho^2 sin^2\theta } = \\ \rho \sqrt{ cos^2\theta \bigl(cos^2\phi + sin^2\phi \bigr) + sin^2\theta } = \\ \rho \sqrt{ cos^2\theta + sin^2\theta } = \rho \end{align} $$

$$ \begin{align}h_w = \begin{Vmatrix} {\partial r \over \partial w} \end{Vmatrix} = \sqrt{(-\rho sin\theta sin\phi)^2 + (\rho sin\theta cos\phi)^2 } = \\ \sqrt{ \rho^2 sin^2\theta sin^2\phi + \rho^2 sin^2\theta cos^2\phi } = \\ \rho \sqrt{ sin^2\theta \bigl(sin^2\phi + cos^2\phi\bigr) } = \\ \rho \sqrt{ sin^2\theta } = \rho sin\theta \end{align} $$

$$ h_u = \begin{Vmatrix} {\partial r \over \partial u} \end{Vmatrix} $$ $$ \downarrow $$ $$ \small \begin{align}\sqrt{(sin\theta cos\phi)^2+ (sin\theta sin\phi)^2 + (cos\theta)^2} = \\ \sqrt{ sin^2\theta cos^2\phi+ sin^2\theta sin^2\phi + cos^2\theta } = \\ \sqrt{ sin^2\theta \bigl(cos^2\phi+ sin^2\phi\bigr) + cos^2\theta } = \\ \sqrt{ sin^2\theta + cos^2\theta } = \sqrt{1} = 1 \end{align} $$

$$ h_v = \begin{Vmatrix} {\partial r \over \partial v} \end{Vmatrix} $$ $$ \downarrow $$ $$ \small \begin{align}\sqrt{ (\rho cos\theta cos\phi)^2 + (\rho cos\theta sin\phi)^2 + (-\rho sin\theta)^2} = \\ \sqrt{ \rho^2 cos^2\theta cos^2\phi + \rho^2 cos^2\theta sin^2\phi + \rho^2 sin^2\theta } = \\ \rho \sqrt{ cos^2\theta \bigl(cos^2\phi + sin^2\phi \bigr) + sin^2\theta } = \\ \rho \sqrt{ cos^2\theta + sin^2\theta } = \rho \end{align} $$

$$ h_w = \begin{Vmatrix} {\partial r \over \partial w} \end{Vmatrix} $$ $$ \downarrow $$ $$ \small \begin{align}\sqrt{(-\rho sin\theta sin\phi)^2 + (\rho sin\theta cos\phi)^2 } = \\ \sqrt{ \rho^2 sin^2\theta sin^2\phi + \rho^2 sin^2\theta cos^2\phi } = \\ \rho \sqrt{ sin^2\theta \bigl(sin^2\phi + cos^2\phi\bigr) } = \\ \rho \sqrt{ sin^2\theta } = \rho sin\theta \end{align} $$

Riassumendo, i fattori di scala cercati sono i seguenti: $$ h_u = \begin{Vmatrix} {\partial r \over \partial u} \end{Vmatrix} = 1 \hspace{6mm} h_v = \begin{Vmatrix} {\partial r \over \partial v} \end{Vmatrix} = \rho \hspace{6mm} h_w = \begin{Vmatrix} {\partial r \over \partial w} \end{Vmatrix} = \rho sin\theta $$ $$ h_u = \begin{Vmatrix} {\partial r \over \partial u} \end{Vmatrix} = 1 \hspace{6mm} h_v = \begin{Vmatrix} {\partial r \over \partial v} \end{Vmatrix} = \rho \hspace{6mm} h_w = \begin{Vmatrix} {\partial r \over \partial w} \end{Vmatrix} = \rho sin\theta $$ $$ h_u = \begin{Vmatrix} {\partial r \over \partial u} \end{Vmatrix} = 1 \hspace{6mm} h_v = \begin{Vmatrix} {\partial r \over \partial v} \end{Vmatrix} = \rho \\ h_w = \begin{Vmatrix} {\partial r \over \partial w} \end{Vmatrix} = \rho sin\theta $$

Possiamo ricavare la base locale, in particolare i versori della base locale dividendo i vettori tangenti per i fattori di scala rispettivamente:


Base locale sferica

$$ \begin{cases} \hat u = {{\partial r \over \partial u} \over \begin{Vmatrix} {\partial r \over \partial u} \end{Vmatrix}} = (sin\theta cos\phi) \hat x + (sin\theta sin\phi) \hat y + (cos\theta)\hat z \\ \hat v = {{\partial r \over \partial v} \over \begin{Vmatrix} {\partial r \over \partial v} \end{Vmatrix}} = {{\partial r \over \partial v} \over \rho} = (cos\theta cos\phi) \hat x + (cos\theta sin\phi) \hat y - (sin\theta)\hat z \\ \hat w = {{\partial r \over \partial w} \over \begin{Vmatrix} {\partial r \over \partial w} \end{Vmatrix}} = {{\partial r \over \partial w} \over \rho sin\theta} = -(\rho sin\phi) \hat x + (\rho cos\phi) \hat y \end{cases} $$ $$ \begin{cases} \hat u = {{\partial r \over \partial u} \over \begin{Vmatrix} {\partial r \over \partial u} \end{Vmatrix}} = (sin\theta cos\phi) \hat x + (sin\theta sin\phi) \hat y + (cos\theta)\hat z \\ \hat v = {{\partial r \over \partial v} \over \begin{Vmatrix} {\partial r \over \partial v} \end{Vmatrix}} = {{\partial r \over \partial v} \over \rho} = (cos\theta cos\phi) \hat x + (cos\theta sin\phi) \hat y - (sin\theta)\hat z \\ \hat w = {{\partial r \over \partial w} \over \begin{Vmatrix} {\partial r \over \partial w} \end{Vmatrix}} = {{\partial r \over \partial w} \over \rho sin\theta} = -(\rho sin\phi) \hat x + (\rho cos\phi) \hat y \end{cases} $$ $$ \begin{cases} \hat u = {{{\partial r \over \partial u} \over \begin{Vmatrix} {\partial r \over \partial u} \end{Vmatrix}} = (sin\theta cos\phi) \hat x + \\ + (sin\theta sin\phi) \hat y + (cos\theta)\hat z} \\ \hat v = {{{\partial r \over \partial v} \over \begin{Vmatrix} {\partial r \over \partial v} \end{Vmatrix}} = {{\partial r \over \partial v} \over \rho} = (cos\theta cos\phi) \hat x + \\ + (cos\theta sin\phi) \hat y - (sin\theta)\hat z} \\ \hat w = {{{\partial r \over \partial w} \over \begin{Vmatrix} {\partial r \over \partial w} \end{Vmatrix}} = {{\partial r \over \partial w} \over \rho sin\theta} = -(\rho sin\phi) \hat x + \\ + (\rho cos\phi) \hat y } \end{cases} $$

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