PlanetPhysics/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: $$ \mathbf{F} = -\nabla U, $$ 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.