Duvida Float
Author:
Luiz Votto
Last Updated:
5 years ago
License:
Creative Commons CC BY 4.0
Abstract:
Sobre a fórmula do laplaciano
\begin
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Sobre a fórmula do laplaciano
\begin
Discover why 18 million people worldwide trust Overleaf with their work.
\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{wasysym}
\title{Dúvida do Float (Laplaciano em coordenadas cilíndricas)}
\author{Luiz Votto}
\date{Novembro 2019}
\begin{document}
\maketitle
Em coordenadas cartesianas, o divergente de um campo vetorial $\mathbf{A}:\mathbb{R}^3 \rightarrow \mathbb{R}^3$ é
\begin{align}
\nabla\cdot\mathbf{A} = \partial_x A_x + \partial_y A_y + \partial_z A_z.
\end{align}
Entretanto, em coordenadas cilíndricas, os versores $\hat{\rho}$ e $\hat{\phi}$ dependem da coordenada $\phi$:
\begin{align}
\partial_\phi \hat{\rho} & = \partial_\phi (\cos\phi \hat{\mathbf{x}} + \sin\phi \hat{\mathbf{y}}) = -\sin\phi\hat{\mathbf{x}} + \cos\phi\hat{\mathbf{y}} = \hat{\phi},\\
\partial_\phi \hat{\phi} & = \ldots = -\hat{\rho}.
\end{align}
As outras derivadas de versores são nulas. Tendo isto em mente, fazemos o divergente lembrando de tomar as derivadas \textbf{antes} de multiplicar. Assim,
\begin{align}
\nabla\cdot\mathbf{A} & = \left(\hat\rho\partial_\rho + \hat{\phi}\frac{1}{\rho}\partial_\phi + \hat{z} \partial_z\right) \cdot \left(\hat{\rho} A_\rho + \hat{\phi} A_\phi + \hat{z} A_z\right) \\
& = \hat{\rho}\cdot\left( \partial_\rho(\hat{\rho}A_\rho) + \partial_\rho(\hat{\phi}A_\phi) + \partial_\rho(\hat{z} A_z) \right)\\
& + \frac{\hat{\phi}}{\rho} \cdot \left( \partial_\phi(\hat{\rho}A_\rho) + \partial_\phi(\hat{\phi}A_\phi) + \partial_\phi(\hat{z} A_z) \right)\\
& + \hat{z} \cdot \left(\partial_z(\hat{\rho}A_\rho) + \partial_z(\hat{\phi}A_\phi) + \partial_z(\hat{z} A_z) \right).
\end{align}
Lembrando que a derivada numa coordenada $v$ do produto de duas funções $f$ e $g$ que dependem de $v$ é
\begin{align}
\partial_v(f g) = \partial_v (f) g + \partial_v(g) f = g \partial_v f + f \partial_v g,
\end{align}
logo
\begin{align}
\nabla \cdot \mathbf{A} & = \hat{\rho} \cdot \left( \hat\rho\partial_\rho A_\rho + \hat\phi \partial_\rho A_\phi + \hat z \partial_\rho A_z \right) \\
& + \frac{\hat\phi}{\rho} \cdot \left( \left( A_\rho \partial_\phi\hat\rho + \hat\rho \partial_\phi A_\rho \right) + \left( A_\phi \partial_\phi\hat\phi + \hat\phi \partial_\phi A_\phi \right) + \hat z \partial_\phi A_z \right) \\
& + \hat z \cdot \left( \hat\rho \partial_z A_\rho + \hat\phi \partial_z A_\phi + \hat z \partial_z A_z \right)\\
& = \left(\partial_\rho A_\rho + \frac{1}{\rho} A_\rho \right) + \frac{1}{\rho} \partial_\phi A_\phi + \partial_z A_z\\
& = \frac{1}{\rho}\partial_\rho(\rho A_\rho) + \frac{1}{\rho}\partial_\phi A_\phi + \partial_z A_z.
\end{align}
Como o laplaciano de um campo escalar $\varphi:\mathbb{R}^3\rightarrow\mathbb{R}$ é dado por
\begin{align}
\nabla^2\varphi &= \nabla\cdot\nabla\varphi\\
&= \nabla\cdot\left(\hat\rho\partial_\rho\varphi + \hat\phi\frac{1}{\rho}\partial_\phi\varphi + \hat z \partial_z\varphi\right)\\
& = \frac{1}{\rho}\partial_\rho(\rho \partial_\rho\varphi) + \frac{1}{\rho^2}\partial_\phi^2 \varphi + \partial_z^2 \varphi.
\end{align}
\\
\centering
O raciocínio para coordenadas esféricas é análogo \\\textbf{E FICA COMO EXERCÍCIO} \footnote{É brincadeira. \smiley{}}.
\end{document}