# PlanetPhysics/Relation Between Force and Potential Energy

This is a contributed entry

## Potential Energy and Force acting on a Particle

Let us asssume from the start that the field's force $\mathbf {F}$ is irrotational, i.e. $\nabla \times \mathbf {F} =\mathbf {0}$, that is, $\nabla \times \mathbf {F} =\mathbf {0} \leftrightarrow \mathbf {F} =-\nabla U$. 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 one imposes $\nabla \times \mathbf {F} =\mathbf {0}$, then one is proving that the necessary condition is: $\mathbf {F} =-\nabla U$. Another consequence about the 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 </math>\int_{\Gamma_1}\mathbf{F}\cdot d\mathbf{r}= \int_{\Gamma_2}\mathbf{F}\cdot d\mathbf{r}$mustbeequalandhencethe[[../Work/|work]]ofthefield'sforce,astheparticledescribesaclosedpath,mustbezero,i.e.<math>\oint \mathbf {F} \cdot d\mathbf {r} =0$.
The relation between the force, $\mathbf {F}$, acting on a particle, and the potential energy, $U$ of that particle is:
   $\mathbf {F} =-\nabla U,$
where $\nabla$ is the gradient operator.

### 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
   ${\frac {dE}{dt}}={\frac {dT}{dt}}+{\frac {dU}{dT}}$
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
   ${\begin{matrix}\mathbf {F} \cdot d\mathbf {r} &=&m{\frac {d\mathbf {v} }{dt}}\cdot {\frac {d\mathbf {r} }{dt}}dt\\&=&m{\frac {d\mathbf {v} }{dt}}\cdot \mathbf {v} dt\\&=&{\frac {1}{2}}m{\frac {d}{dt}}(\mathbf {v} \cdot \mathbf {v} )dt\\&=&d({\frac {1}{2}}mv^{2})=dT.\end{matrix}}$
Therefore, $dT/dt=\mathbf {F} \cdot d\mathbf {r} /dt$.
It is assumed that the potential energy is a function of time and space 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
   ${\frac {dU}{dt}}={\frac {\partial U}{\partial t}}+{\frac {\partial U}{\partial x_{1}}}{\frac {\partial x_{1}}{\partial t}}+{\frac {\partial U}{\partial x_{2}}}{\frac {\partial x_{2}}{\partial t}}+{\frac {\partial U}{\partial x_{3}}}{\frac {\partial x_{3}}{\partial t}}.$
Notice that
   ${\frac {\partial U}{\partial x_{1}}}{\frac {\partial x_{1}}{\partial t}}+{\frac {\partial U}{\partial x_{2}}}{\frac {\partial x_{2}}{\partial t}}+{\frac {\partial U}{\partial x_{3}}}{\frac {\partial x_{3}}{\partial t}}=\nabla U\cdot \mathbf {v} ,$
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,
   ${\begin{matrix}{\frac {dE}{dt}}&=&{\frac {dT}{dt}}+{\frac {dU}{dt}}\\&=&\mathbf {F} \cdot {\frac {d\mathbf {r} }{dt}}+{\frac {\partial U}{\partial t}}+\nabla U\cdot \mathbf {v} \\&=&(\mathbf {F} +\nabla U)\cdot \mathbf {v} +{\frac {\partial U}{\partial t}}\end{matrix}}$
If the potential has no explicity time dependence i.e. it is dependent upon position, which is dependent on time, then $dU/dt=0$, and the above becomes
   ${\frac {dE}{dt}}=(\mathbf {F} +\nabla U)\cdot \mathbf {v} =0,$
where $dE/dt=0$ arises because of the conservation of energy within a closed system 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
   $\mathbf {F} =-\nabla U,$
which is the desired relation between the force acting on a particle and the the particle's potential energy in the presence of the force acting upon it.