RelationBetweenForceAndPotentialEnergy 2006-07-14 10:32:00 2009-03-06 18:08:20 Definition derivation is true if and only if you impose (from the starting!) that the field's force $\mathbf{F}$ is irrotational, i.e. $\nabla\times\mathbf{F}=\marhbf{0}$, that is, $\nabla\times\mathbf{F}=\marhbf{0} \Leftrightarrow \marhbf{F}=-\nabla U$. In another words, the field's force is conservative if and only if it is irrotational. So the conseravation of mechanical energy $dE/dt=d(T+U)/dt=0$ is a consequence of that theorem. Once you impose $\nabla\times\mathbf{F}=\marhbf{0}$, then you are proving the necessary condition for $\marhbf{F}=-\nabla U$. No problem about that. Another consequence about that theorem is that the ``work'' of the field's force is independent of the path described by the particle in its motion. That is, if $\Gamma_1$ and $\Gamma_2$ are two different paths, described by the particle, and joininig its initial and end position on the time interval $[t_1,t_2]$, then the line integrals $\int_{\Gamma_1}\mathbf{F}\cdot d\mathbf{r}= \int_{\Gamma_2}\mathbf{F}\cdot d\mathbf{r}$ must be equal and hence the work of the field's force, as the particle describes a closed path, must be zero, i.e. $\oint\mathbf{F}\cdot d\mathbf{r}=0$. % this is the default PlanetPhysics preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here \textit{This is a contributed entry} \section{Potential Energy and Force acting on a Particle} The relation between the force, $\mathbf{F}$, acting on a particle, and the potential energy, $U$ of that particle is \begin{equation} \mathbf{F} = -\nabla U, \end{equation} where $\nabla$ is the gradient operator. \subsection{Derivation} The above relationship can be derived from the conservation of energy. Let $T$ denote the kinetic energy of a particle, and $U$ its potential energy, with $E$ the total energy, given by $E=T+U$. Take the total time derivative of $E$, giving \begin{equation} \frac{dE}{dt} = \frac{dT}{dt} + \frac{dU}{dT} \end{equation} The kinetic energy of a particle is expressed as $T=\frac{1}{2}mv^{2}$, where $m$ is the mass of the particle, and $v$ is the magnitude of the particle's velocity. Recall that by Newton's second law, $\mathbf{F} = md\mathbf{v}/dt$, where $\mathbf{v}$ is the velocity vector. Consider, next, the quantity $\mathbf{F}\cdot d\mathbf{r}$, where $\mathbf{r}$ is the position vector of the particle. Expanding $\mathbf{F}$ in terms of Newton's second law, it is seen that \begin{eqnarray*} \mathbf{F}\cdot d\mathbf{r} & = & m\frac{d\mathbf{v}}{dt}\cdot\frac{d\mathbf{r}}{dt}dt\\ & = & m\frac{d\mathbf{v}}{dt}\cdot\mathbf{v}dt\\ & = & \frac{1}{2}m\frac{d}{dt}(\mathbf{v}\cdot\mathbf{v})dt\\ & = & d(\frac{1}{2}mv^{2}) = dT. \end{eqnarray*} Therefore, $dT/dt = \mathbf{F}\cdot d\mathbf{r}/dt$. It is assumed that the potential energy is a function of time and space \emph{i.e.} $U=U(x_{1},x_{2},x_{3},t)$. The time derivative of the potential energy can be expanded through the chain rule as \begin{equation} \frac{dU}{dt} = \frac{\partial U}{\partial t} + \frac{\partial U}{\partial x_{1}}\frac{\partial x_{1}}{\partial t} + \frac{\partial U}{\partial x_{2}}\frac{\partial x_{2}}{\partial t} + \frac{\partial U}{\partial x_{3}}\frac{\partial x_{3}}{\partial t}. \end{equation} Notice that \begin{equation} \frac{\partial U}{\partial x_{1}}\frac{\partial x_{1}}{\partial t} + \frac{\partial U}{\partial x_{2}}\frac{\partial x_{2}}{\partial t} + \frac{\partial U}{\partial x_{3}}\frac{\partial x_{3}}{\partial t} = \nabla U\cdot\mathbf{v}, \end{equation} and substitute this result, as well as the expression for the time derivative of kinetic energy back into the original equation for the time derivative of the total energy, \begin{eqnarray} \frac{dE}{dt} & = & \frac{dT}{dt} + \frac{dU}{dt}\\ & = & \mathbf{F}\cdot\frac{d\mathbf{r}}{dt} + \frac{\partial U}{\partial t} + \nabla U\cdot\mathbf{v}\\ & = & (\mathbf{F} + \nabla U)\cdot\mathbf{v} + \frac{\partial U}{\partial t} \end{eqnarray} If the potential has no explicity time dependence \emph{i.e.} it is dependent upon position, which is dependent on time, then $dU/dt=0$, and the above becomes \begin{equation} \frac{dE}{dt} = (\mathbf{F} + \nabla U)\cdot\mathbf{v} = 0, \end{equation} where $dE/dt=0$ arises because of the conservation of energy within a closed system \emph{i.e.} energy does not enter or leave the system. Therefore, it follows that under the conservation of energy, and the time independence of potential energy, $\mathbf{F} + \nabla U = 0$, which can be rewritten as \begin{equation} \mathbf{F} = -\nabla U, \end{equation} which is the desired relation between the force acting on a particle, and its potential energy.