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relation between force and potential energy (Definition)

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 $\int_{\Gamma_1}\mathbf{F}\cdot d\mathbf{r}= \int_{\Gamma_2}\mathbf{F}\cdot d\mathbf{r}$ must be equal and hence the work of the field's force, as the particle describes a closed path, must be zero, i.e. $\oint\mathbf{F}\cdot d\mathbf{r}=0$.

The relation between the force, $\mathbf{F}$, acting on a particle, and the potential energy, $U$ of that particle is:

$\displaystyle \mathbf{F} = -\nabla U,$ (1.1)
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

$\displaystyle \frac{dE}{dt} = \frac{dT}{dt} + \frac{dU}{dT}$ (1.2)

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

$\displaystyle \mathbf{F}\cdot d\mathbf{r}$ $\displaystyle =$ $\displaystyle m\frac{d\mathbf{v}}{dt}\cdot\frac{d\mathbf{r}}{dt}dt$  
  $\displaystyle =$ $\displaystyle m\frac{d\mathbf{v}}{dt}\cdot\mathbf{v}dt$  
  $\displaystyle =$ $\displaystyle \frac{1}{2}m\frac{d}{dt}(\mathbf{v}\cdot\mathbf{v})dt$  
  $\displaystyle =$ $\displaystyle d(\frac{1}{2}mv^{2}) = dT.$  

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

$\displaystyle \frac{dU}{dt} = \frac{\partial U}{\partial t} + \frac{\partial U}... ...artial t} + \frac{\partial U}{\partial x_{3}}\frac{\partial x_{3}}{\partial t}.$ (1.3)
Notice that
$\displaystyle \frac{\partial U}{\partial x_{1}}\frac{\partial x_{1}}{\partial t... ... U}{\partial x_{3}}\frac{\partial x_{3}}{\partial t} = \nabla U\cdot\mathbf{v},$ (1.4)
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,
$\displaystyle \frac{dE}{dt}$ $\displaystyle =$ $\displaystyle \frac{dT}{dt} + \frac{dU}{dt}$ (1.5)
  $\displaystyle =$ $\displaystyle \mathbf{F}\cdot\frac{d\mathbf{r}}{dt} + \frac{\partial U}{\partial t} + \nabla U\cdot\mathbf{v}$ (1.6)
  $\displaystyle =$ $\displaystyle (\mathbf{F} + \nabla U)\cdot\mathbf{v} + \frac{\partial U}{\partial t}$ (1.7)

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
$\displaystyle \frac{dE}{dt} = (\mathbf{F} + \nabla U)\cdot\mathbf{v} = 0,$ (1.8)
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
$\displaystyle \mathbf{F} = -\nabla U,$ (1.9)
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.



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"relation between force and potential energy" is owned by bci1. [ full author list (2) | owner history (1) ]

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Attachments:
1D example of the relation between force and potential energy (Example) by bloftin
lamellar field (Topic) by pahio

Cross-references: system, closed system, function, position vector, vector, velocity, magnitude, mass, kinetic energy, gradient operator, relation, work, position, motion, theorem, energy, field's

This is version 10 of relation between force and potential energy, born on 2006-07-14, modified 2009-03-06.
Object id is 198, canonical name is RelationBetweenForceAndPotentialEnergy.
Accessed 5750 times total.

Classification:
Physics Classification: 45.50.-j (Dynamics and kinematics of a particle and a system of particles)
Pending Errata and Addenda
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Discussion
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Force and U, excellent start by bloftin on 2006-07-14 14:02:57
Mdo,

Great start on the article, it read very well.  Something for future additions to the article, you might want to consider putting in the term conservative force (ie. where you mention position dependence) since this would be a great topic for a link to a future entry.  Also maybe a mention of the 1D case, I always have mixed feelings about showing things like this in the main article and maybe it can be addressed in an attached example.  Thanks for your contribution.

Ben
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