La divergenza del rotore

Consideriamo un campo vettoriale \( \mathbb V\) e calcoliamone il rotore \( \mathbb{rotV}\). L'espressione del rotore, l'abbiamo calcolata più volte (riporto solo le componenti) è: $$ \begin{vmatrix} \widehat{i} & \widehat{j} & \widehat{k} \\ \frac{\partial}{\partial{x}} & \frac{\partial}{\partial{y}}& \frac{\partial}{\partial{z}} \\ \mathrm V_x & \mathrm V_y & \mathrm V_z \end{vmatrix} = \left( \frac{\partial \mathrm V_z}{\partial y} - \frac{\partial \mathrm V_y}{\partial z} \right)\hat{i} + \left( \frac{\partial \mathrm V_x}{\partial z} - \frac{\partial \mathrm V_z}{\partial x} \right)\hat{j} + \left( \frac{\partial \mathrm V_y}{\partial x} - \frac{\partial \mathrm V_x}{\partial y} \right)\hat{k} $$ $$ \begin{vmatrix} \widehat{i} & \widehat{j} & \widehat{k} \\ \frac{\partial}{\partial{x}} & \frac{\partial}{\partial{y}}& \frac{\partial}{\partial{z}} \\ \mathrm V_x & \mathrm V_y & \mathrm V_z \end{vmatrix} = \left( \frac{\partial \mathrm V_z}{\partial y} - \frac{\partial \mathrm V_y}{\partial z} \right)\hat{i} + \left( \frac{\partial \mathrm V_x}{\partial z} - \frac{\partial \mathrm V_z}{\partial x} \right)\hat{j} + \left( \frac{\partial \mathrm V_y}{\partial x} - \frac{\partial \mathrm V_x}{\partial y} \right)\hat{k} $$ $$ \begin{vmatrix} \widehat{i} & \widehat{j} & \widehat{k} \\ \frac{\partial}{\partial{x}} & \frac{\partial}{\partial{y}}& \frac{\partial}{\partial{z}} \\ \mathrm V_x & \mathrm V_y & \mathrm V_z \end{vmatrix} $$ $$ \downarrow $$ $$ \left( \frac{\partial \mathrm V_z}{\partial y} - \frac{\partial \mathrm V_y}{\partial z} \right)\hat{i} + \\ + \left( \frac{\partial \mathrm V_x}{\partial z} - \frac{\partial \mathrm V_z}{\partial x} \right)\hat{j} + \\ + \left( \frac{\partial \mathrm V_y}{\partial x} - \frac{\partial \mathrm V_x}{\partial y} \right)\hat{k} $$

Ora prendiamo il rotore e mandiamolo "in pasto" come input all'operatore di divergenza. Andiamo a calcolare quindi: $$ \mathrm{div}\begin{pmatrix} \frac{\partial \mathrm V_z}{\partial y} - \frac{\partial \mathrm V_y}{\partial z} \\ \frac{\partial \mathrm V_x}{\partial z} - \frac{\partial \mathrm V_z}{\partial x} \\ \frac{\partial \mathrm V_y}{\partial x} - \frac{\partial \mathrm V_x}{\partial y} \end{pmatrix} = \mathrm{div}\begin{pmatrix} \mathrm{rot}_x \\ \mathrm{rot}_y \\ \mathrm{rot}_z \end{pmatrix} $$

La divergenza di un vettore, ricordate, è uno scalare e si ottiene andando a sommare le derivate parziali di ogni componente calcolate rispetto alla medesima componente. $$ \mathrm{div(rot(V))} = \sum_{j=1}^3 {\partial \mathrm V_j \over \partial x_j} $$ $$ \downarrow $$ $$ {\partial \over \partial x} \left[ \frac{\partial \mathrm V_z}{\partial y} - \frac{\partial \mathrm V_y}{\partial z} \right] + {\partial \over \partial y} \left[ \frac{\partial \mathrm V_x}{\partial z} - \frac{\partial \mathrm V_z}{\partial x} \right] + {\partial \over \partial z} \left[ \frac{\partial \mathrm V_y}{\partial x} - \frac{\partial \mathrm V_x}{\partial y} \right] = $$ $${\small = {\partial \over \partial x}\frac{\partial \mathrm V_z}{\partial y} - {\partial \over \partial x}\frac{\partial \mathrm V_y}{\partial z} + {\partial \over \partial y}\frac{\partial \mathrm V_x}{\partial z} - {\partial \over \partial y}\frac{\partial \mathrm V_z}{\partial x} + {\partial \over \partial z}\frac{\partial \mathrm V_y}{\partial x} - {\partial \over \partial z}\frac{\partial \mathrm V_x}{\partial y} } $$ $$ {\partial \over \partial x} \left[ \frac{\partial \mathrm V_z}{\partial y} - \frac{\partial \mathrm V_y}{\partial z} \right] + {\partial \over \partial y} \left[ \frac{\partial \mathrm V_x}{\partial z} - \frac{\partial \mathrm V_z}{\partial x} \right] + {\partial \over \partial z} \left[ \frac{\partial \mathrm V_y}{\partial x} - \frac{\partial \mathrm V_x}{\partial y} \right] = $$ $${\small = {\partial \over \partial x}\frac{\partial \mathrm V_z}{\partial y} - {\partial \over \partial x}\frac{\partial \mathrm V_y}{\partial z} + {\partial \over \partial y}\frac{\partial \mathrm V_x}{\partial z} - {\partial \over \partial y}\frac{\partial \mathrm V_z}{\partial x} + {\partial \over \partial z}\frac{\partial \mathrm V_y}{\partial x} - {\partial \over \partial z}\frac{\partial \mathrm V_x}{\partial y} } $$ $$ {\partial \over \partial x} \left[ \frac{\partial \mathrm V_z}{\partial y} - \frac{\partial \mathrm V_y}{\partial z} \right] + \\ + {\partial \over \partial y} \left[ \frac{\partial \mathrm V_x}{\partial z} - \frac{\partial \mathrm V_z}{\partial x} \right] + \\ + {\partial \over \partial z} \left[ \frac{\partial \mathrm V_y}{\partial x} - \frac{\partial \mathrm V_x}{\partial y} \right] = $$ $$ \downarrow $$ $$ = {\partial \over \partial x}\frac{\partial \mathrm V_z}{\partial y} - {\partial \over \partial x}\frac{\partial \mathrm V_y}{\partial z} + \\ + {\partial \over \partial y}\frac{\partial \mathrm V_x}{\partial z} - {\partial \over \partial y}\frac{\partial \mathrm V_z}{\partial x} + \\ + {\partial \over \partial z}\frac{\partial \mathrm V_y}{\partial x} - {\partial \over \partial z}\frac{\partial \mathrm V_x}{\partial y} $$

Per il Teorema di Schwartz le derivate seconde miste opposte in segno si possono elidere quindi il risultato è \( 0 \)

$$ {\Large \mathrm{div(rot(V))} = 0 } \hspace{3cm} {\Large \nabla \cdot \nabla\times\mathrm V = 0} $$ $$ {\Large \mathrm{div(rot(V))} = 0 } \hspace{3cm} {\Large \nabla \cdot \nabla\times\mathrm V = 0} $$ $$ \large \mathrm{div(rot(V))} = 0 $$ $$ \Large \nabla \cdot \nabla\times\mathrm V = 0 $$

$$ \diamond $$
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