Seja f um campo escalar e \boldsymbol{u} um campo vetorial, determinar \boldsymbol{\nabla} f, \boldsymbol{\nabla} \cdot \boldsymbol{u}, \boldsymbol{\nabla}^{2} f e \boldsymbol{\nabla} \times \boldsymbol{u} para o sistema de coordenadas cilíndrico tridimensional.

Passo 1

O enunciado nos pede para escrever as expressões de \boldsymbol{\nabla}f, \boldsymbol{\nabla}\cdot\boldsymbol{u}, \boldsymbol{\nabla}^{2}f e \boldsymbol{\nabla}\times\boldsymbol{u} no sistema de coordenadas cilíndricas (r,\varphi,z). Vamos usar a teoria de operadores diferenciais em coordenadas curvilíneas ortonormais, onde os fatores de escala são h_{r}=1, h_{\varphi}=r, h_{z}=1, e os versores \hat{\mathbf{e}}_{r}, \hat{\mathbf{e}}_{\varphi}, \hat{\mathbf{e}}_{z} dependem de \varphi.

Passo 2

Para o gradiente de um campo escalar f(r,\varphi,z) em coordenadas com fatores de escala h_{i}, vale
\boldsymbol{\nabla}f=\sum_{i}\frac{1}{h_{i}}\frac{\partial f}{\partial q_{i}}\hat{\mathbf{e}}_{i}.
No caso cilíndrico (q_{1}=r,q_{2}=\varphi,q_{3}=z) com h_{r}=1, h_{\varphi}=r, h_{z}=1, obtemos
\boldsymbol{\nabla}f =\hat{\mathbf{e}}_{r}\frac{\partial f}{\partial r} +\hat{\mathbf{e}}_{\varphi}\frac{1}{r}\frac{\partial f}{\partial \varphi} +\hat{\mathbf{e}}_{z}\frac{\partial f}{\partial z}.

Passo 3

Para a divergência de um campo vetorial \boldsymbol{u}=u_{r}\hat{\mathbf{e}}_{r}+u_{\varphi}\hat{\mathbf{e}}_{\varphi}+u_{z}\hat{\mathbf{e}}_{z}, o operador geral em coordenadas curvilíneas é
\boldsymbol{\nabla}\cdot\boldsymbol{u} =\frac{1}{h_{1}h_{2}h_{3}} \sum_{i}\frac{\partial}{\partial q_{i}} \Bigl(h_{1}h_{2}h_{3}\,\frac{u_{i}}{h_{i}}\Bigr).
Substituindo h_{1}h_{2}h_{3}=r e cada h_{i}, temos
\boldsymbol{\nabla}\cdot\boldsymbol{u} =\frac{1}{r}\frac{\partial}{\partial r}(r\,u_{r}) +\frac{1}{r}\frac{\partial u_{\varphi}}{\partial \varphi} +\frac{\partial u_{z}}{\partial z}.

Passo 4

O laplaciano de um escalar f é \nabla^{2}f=\boldsymbol{\nabla}\cdot(\boldsymbol{\nabla}f). Aplicando a divergência do passo anterior ao gradiente de f, obtemos
\boldsymbol{\nabla}^{2}f =\frac{1}{r}\frac{\partial}{\partial r}\Bigl(r\frac{\partial f}{\partial r}\Bigr) +\frac{1}{r^{2}}\frac{\partial^{2}f}{\partial \varphi^{2}} +\frac{\partial^{2}f}{\partial z^{2}}.

Passo 5

Para o rotor (curl) de \boldsymbol{u}, em coordenadas curvilíneas ortonormais usa-se a fórmula
(\boldsymbol{\nabla}\times\boldsymbol{u})_{i} =\frac{1}{h_{i}} \sum_{j,k}\epsilon_{ijk}\, \frac{1}{h_{j}h_{k}}\, \frac{\partial}{\partial q_{j}}(h_{k}\,u_{k}),
onde \epsilon_{ijk} é o símbolo de Levi-Civita. Fazendo as contas para (r,\varphi,z), resulta
\boldsymbol{\nabla}\times\boldsymbol{u} =\hat{\mathbf{e}}_{r}\,\frac{1}{r}\Bigl(\frac{\partial u_{z}}{\partial \varphi}-\frac{\partial}{\partial z}(r\,u_{\varphi})\Bigr) +\hat{\mathbf{e}}_{\varphi}\,\Bigl(\frac{\partial u_{r}}{\partial z}-\frac{\partial u_{z}}{\partial r}\Bigr) +\hat{\mathbf{e}}_{z}\,\frac{1}{r}\Bigl(\frac{\partial}{\partial r}(r\,u_{\varphi})-\frac{\partial u_{r}}{\partial \varphi}\Bigr).

Resposta

\displaystyle \boldsymbol{\nabla}f =\hat{\mathbf{e}}_{r}\frac{\partial f}{\partial r} +\hat{\mathbf{e}}_{\varphi}\frac{1}{r}\frac{\partial f}{\partial \varphi} +\hat{\mathbf{e}}_{z}\frac{\partial f}{\partial z}. \displaystyle \boldsymbol{\nabla}\cdot\boldsymbol{u} =\frac{1}{r}\frac{\partial}{\partial r}(r\,u_{r}) +\frac{1}{r}\frac{\partial u_{\varphi}}{\partial \varphi} +\frac{\partial u_{z}}{\partial z}. \displaystyle \boldsymbol{\nabla}^{2}f =\frac{1}{r}\frac{\partial}{\partial r}\Bigl(r\frac{\partial f}{\partial r}\Bigr) +\frac{1}{r^{2}}\frac{\partial^{2}f}{\partial \varphi^{2}} +\frac{\partial^{2}f}{\partial z^{2}}. \displaystyle \boldsymbol{\nabla}\times\boldsymbol{u} =\hat{\mathbf{e}}_{r}\,\frac{1}{r}\Bigl(\frac{\partial u_{z}}{\partial \varphi}-\frac{\partial}{\partial z}(r\,u_{\varphi})\Bigr) +\hat{\mathbf{e}}_{\varphi}\,\Bigl(\frac{\partial u_{r}}{\partial z}-\frac{\partial u_{z}}{\partial r}\Bigr) +\hat{\mathbf{e}}_{z}\,\frac{1}{r}\Bigl(\frac{\partial}{\partial r}(r\,u_{\varphi})-\frac{\partial u_{r}}{\partial \varphi}\Bigr).

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