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 [[../CosmologicalConstant/|field's]] [[../Thrust/|force]] $𝐅$ is irrotational, i.e. $\mathrm{\nabla }\times 𝐅=\mathrm{𝟎}$, that is, $\mathrm{\nabla }\times 𝐅=\mathrm{𝟎}\leftrightarrow 𝐅=-\mathrm{\nabla }U$. In another words, the field's force is conservative if and only if it is irrotational. So the conseravation of mechanical [[../CosmologicalConstant2/|energy]] $dE/dt=d(T+U)/dt=0$ is a consequence of that [[../Formula/|theorem]]. Once one imposes $\mathrm{\nabla }\times 𝐅=\mathrm{𝟎}$, then one is proving that the necessary condition is: $𝐅=-\mathrm{\nabla }U$. Another consequence about the theorem is that the "work" of the field's force is independent of the path described by the [[../Particle/|particle]] in its [[../CosmologicalConstant/|motion]]. That is, if ${\mathrm{\Gamma }}_1$ and ${\mathrm{\Gamma }}_2$ are two different paths, described by the particle, and joininig its initial and end [[../Position/|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}$mustbeequalandhencethe[[../Work/|work]]ofthefiel{d}^{\prime }sforce,astheparticledescribesaclosedpath,mustbezero,i.e.<math>\oint 𝐅\cdot d𝐫=0$.

The [[../Bijective/|relation]] between the force, $𝐅$, acting on a particle, and the potential energy, $U$ of that particle is:

$𝐅=-\mathrm{\nabla }U,$

where $\mathrm{\nabla }$ is the [[../Gradient/|gradient operator]].

Derivation

The above relationship can be derived from the conservation of energy. Let $T$ denote the [[../KineticEnergy/|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 $T=\frac{1}{2}m{v}^2$, where $m$ is the [[../Mass/|mass]] of the particle, and $v$ is the [[../AbsoluteMagnitude/|magnitude]] of the particle's [[../Velocity/|velocity]]. Recall that by Newton's second law, $𝐅=md𝐯/dt$, where $𝐯$ is the velocity [[../Vectors/|vector]]. Consider, next, the quantity $𝐅\cdot d𝐫$, where $𝐫$ is the [[../PositionVector/|position vector]] of the particle. Expanding $𝐅$ in terms of Newton's second law, it is seen that

$\begin{array}{ccc}𝐅\cdot d𝐫& =& m\frac{d𝐯}{dt}\cdot \frac{d𝐫}{dt}dt\\ & =& m\frac{d𝐯}{dt}\cdot 𝐯dt\\ & =& \frac{1}{2}m\frac{d}{dt}(𝐯\cdot 𝐯)dt\\ & =& d(\frac{1}{2}m{v}^2)=dT.\end{array}$

Therefore, $dT/dt=𝐅\cdot d𝐫/dt$.

It is assumed that the potential energy is a [[../Bijective/|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

$\frac{\partial U}{\partial x_1}\frac{\partial x_1}{\partial t}+\frac{\partial U}{\partial x_2}\frac{\partial x_2}{\partial t}+\frac{\partial U}{\partial x_3}\frac{\partial x_3}{\partial t}=\mathrm{\nabla }U\cdot 𝐯,$

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,

$\begin{array}{ccc}\frac{dE}{dt}& =& \frac{dT}{dt}+\frac{dU}{dt}\\ & =& 𝐅\cdot \frac{d𝐫}{dt}+\frac{\partial U}{\partial t}+\mathrm{\nabla }U\cdot 𝐯\\ & =& (𝐅+\mathrm{\nabla }U)\cdot 𝐯+\frac{\partial U}{\partial t}\end{array}$

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

$\frac{dE}{dt}=(𝐅+\mathrm{\nabla }U)\cdot 𝐯=0,$

where $dE/dt=0$ arises because of the conservation of energy within a [[../ThermodynamicLaws/|closed system]] i.e. energy does not enter or leave the [[../GenericityInOpenSystems/|system]]. Therefore, it follows that under the conservation of energy, and the time independence of potential energy, $𝐅+\mathrm{\nabla }U=0$, which can be rewritten as

$𝐅=-\mathrm{\nabla }U,$

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.