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<record version="1" id="198">
 <title>relation between force and potential energy</title>
 <name>RelationBetweenForceAndPotentialEnergy</name>
 <created>2006-07-14 10:32:00</created>
 <modified>2006-07-14 10:32:00</modified>
 <type>Definition</type>
 <creator id="139" name="mdo"/>
 <modifier id="139" name="mdo"/>
 <comment>Changes for correction #33 ('just spelling').</comment>
 <author id="139" name="mdo"/>
 <classification>
	<category scheme="msc" code="45.50.-j"/>
 </classification>
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 <content>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} &amp; = &amp; m\frac{d\mathbf{v}}{dt}\cdot\frac{d\mathbf{r}}{dt}dt\\
&amp; = &amp; m\frac{d\mathbf{v}}{dt}\cdot\mathbf{v}dt\\
&amp; = &amp; \frac{1}{2}m\frac{d}{dt}(\mathbf{v}\cdot\mathbf{v})dt\\
&amp; = &amp; 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} &amp; = &amp; \frac{dT}{dt} + \frac{dU}{dt}\\
&amp; = &amp; \mathbf{F}\cdot\frac{d\mathbf{r}}{dt} + \frac{\partial U}{\partial t} + \nabla U\cdot\mathbf{v}\\
&amp; = &amp; (\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 opun 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.

\textit{This entry will be expanded, and the definitions relevant to it added to the encyclop\ae{}dia as time progresses}</content>
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