% vim: tw=50 % 21/02/2022 10AM \begin{corollary*} For a scalar field $\phi$ \[ \int_V \nabla \phi \dd V = \int_{\partial V} \phi \bf{\dd s} \] \end{corollary*} \begin{proof} Use divergence theorem with \[ \bf{F} = \phi \bf{a} \] with $\bf{a}$ a constant. \[ \int_V \nabla \cdot (\phi \bf{a}) \dd V = \int_{\partial V} \phi \bf{a} \cdot \bf{\dd s} \] \[ \implies \bf{a} \cdot \left[ \int_V \nabla \phi \dd V - \int_{\partial V} \phi \bf{\dd s} \right] = 0 \] But true $\forall\,\,\bf{a}$ implies the result. \end{proof} \subsection{An Application: Conservation Laws} Many things are conserved, for example energy, momentum, angular momentum, electric charge. Importantly, all of these are conserved \emph{locally}. This means that stuff moves continuously to nearby points. \myskip Let $\rho(\bf{x}, t)$ be the density of the conserved object, for example electric charge. Then \[ Q = \int_V \rho \dd V \] is the charge in a region $V$. Conservation of $Q$ means that there exists a vector $\bf{J}(\bf{x}, t)$, known as a \emph{current density} such that \[ \pfrac{\rho}{t} + \nabla \cdot \bf{J} = 0 \] This is the \emph{continuity equation}. \myskip The change of $Q$ in some fixed region $V$ is \begin{align*} \dfrac{Q}{t} &= \int_V \pfrac{\rho}{t} \dd V \\ &= -\int_V \nabla \cdot \bf{J} \dd V \\ &= -\int_S \bf{J} \cdot \bf{\dd s} \end{align*} ($\bf{J}$ is current flowing in/out of $V$). If $\bf{J}(\bf{x}) = 0$ on $S$ then $\dot{Q} = 0$. Often consider $V = \RR^3$ and $\dot{Q} = 0$ provided that $J(\bf{x}) \to 0$ suitably quickly as $|\bf{x}| \to \infty$. \begin{example*} A fluid has mass density $\rho(\bf{x}, t)$ and mass current $\bf{J} = \rho \bf{u}$ with $\bf{u}(\bf{x}, t)$ the velocity field. \\ Mass is conserved \[ \implies \pfrac{\rho}{t} + \nabla \cdot \bf{J} = 0 \] But many fluids are incompressible with $\rho = \text{ constant}$. Then continuity equation gives \[ \nabla \cdot \bf{u} = 0 \] \end{example*} \begin{example*}[Diffusion] Consider a gas with energy density $\eps(\bf{x}, t)$. Energy is conserved \[ \implies \pfrac{\eps}{t} + \nabla \cdot \bf{J} = 0 \] ($\bf{J}$ is \emph{heat current}) \\ \ul{Fact 1}: \[ \eps(\bf{x}, t) = c_v T(\bf{x}, t) \] where $c_v$ is heat capacity and $T$ is temperature. \\ \ul{Fact 2}: (Fick's law) Heat current is due to temperature differences \[ \bf{J} = -\kappa \nabla T \] ($\kappa$ is thermal conductivity) \[ \implies \pfrac{T}{t} = D\nabla^2 T \] i.e. the heat equation with $D = \frac{\kappa}{c_v}$. \end{example*}