relation between force and potential energy

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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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As of this snapshot date, this entry was owned by bci1.