User:Prof McCarthy/Potential Energy

Derivable from a potential
In this section the relationship between work and potential energy is presented in more detail. The line integral that defines work along curve C takes a special form if the force F is related to a scalar field φ(x) so that
 * $$ \mathbf{F}={\nabla \varphi} = \left ( \frac{\partial \varphi}{\partial x}, \frac{\partial \varphi}{\partial y}, \frac{\partial \varphi}{\partial z} \right ). $$

In this case, work along the curve is given by
 * $$W =\int_{C} \mathbf{F} \cdot \mathrm{d}\mathbf{x}=\int_{C} \nabla \varphi\cdot \mathrm{d}\mathbf{x},$$

which can be evaluated using the gradient theorem to obtain
 * $$ W= \varphi(\mathbf{x}_B) - \varphi(\mathbf{x}_A).$$

This shows that when forces are derivable from a scalar field, the work of those forces along a curve C is computed by evaluating the scalar field at the start point A and the end point B of the curve. This means the work integral does not depend on the path between A and B and is said to be independent of the path.

Potential energy U=-φ(x) is traditionally defined as the negative of this scalar field so that work by the force field decreases potential energy, that is
 * $$ W= U(\mathbf{x}_A) - U(\mathbf{x}_B).$$

In this case, the application of the del operator to the work function yields,
 * $$ {\nabla W} = -{\nabla U} = -\left ( \frac{\partial U}{\partial x}, \frac{\partial U}{\partial y}, \frac{\partial U}{\partial z} \right ) = \mathbf{F},$$

and the force F is said to be "derivable from a potential." This also necessarily implies that F must be a conservative vector field. The potential U defines a force F at every point x in space, so the set of forces is called a force field.

Computing potential energy
Given a force field F(x), evaluation of the work integral using the gradient theorem can be used to find the scalar function associated with potential energy. This is done by introducing a parameterized curve γ(t)=r(t) from γ(a)=A to γ(b)=B, and computing,
 * $$\begin{align}

\int_{\gamma} \nabla\varphi(\mathbf{r}) \cdot  d\mathbf{r} &=\int_a^b \nabla\varphi(\mathbf{r}(t))  \cdot  \mathbf{r}'(t)dt, \\ &=\int_a^b \frac{d}{dt}\varphi(\mathbf{r}(t))dt =\varphi(\mathbf{r}(b))-\varphi(\mathbf{r}(a))=\varphi\left(\mathbf{x}_B\right)-\varphi\left(\mathbf{x}_A\right). \end{align} $$

For the force field F, let v= dr/dt, then the gradient theorem yields,
 * $$\begin{align}

\int_{\gamma} \mathbf{F} \cdot d\mathbf{r} &=\int_a^b \mathbf{F}  \cdot  \mathbf{v}dt, \\ &=-\int_a^b \frac{d}{dt}U(\mathbf{r}(t))dt =U\left(\mathbf{x}_A\right)- U\left(\mathbf{x}_B\right). \end{align} $$

The power applied to a body by a force field is obtained from the gradient of the work, or potential, in the direction of the velocity v of the point of application, that is
 * $$P(t) = -{\nabla U} \cdot \mathbf{v} = \mathbf{F}\cdot\mathbf{v}.$$

Examples of work that can be computed from potential functions are gravity and spring forces.

Gradient theorem
The gradient theorem, also known as the fundamental theorem of calculus for line integrals, says that a line integral through a gradient field can be evaluated by evaluating the original scalar field at the endpoints of the curve.

Let $$ \varphi : U \subseteq \mathbb{R}^n \to \mathbb{R}$$. Then


 * $$ \varphi\left(\mathbf{q}\right)-\varphi\left(\mathbf{p}\right) = \int_{\gamma[\mathbf{p},\,\mathbf{q}]} \nabla\varphi(\mathbf{r})\cdot d\mathbf{r}. $$

It is a generalization of the fundamental theorem of calculus to any curve in a plane or space (generally n-dimensional) rather than just the real line.

The gradient theorem implies that line integrals through gradient fields are path independent. In physics this theorem is one of the ways of defining a "conservative" force. By placing φ as potential, ∇φ is a conservative field. Work done by conservative forces does not depend on the path followed by the object, but only the end points, as the above equation shows.

If φ is a differentiable function from some open subset U (of Rn) to R, and if r is a differentiable function from some closed interval [a,b] to U, then by the multivariate chain rule, the composite function φ ∘ r is differentiable on (a, b) and


 * $$\frac{d}{dt}(\varphi \circ \mathbf{r})(t)=\nabla \varphi(\mathbf{r}(t)) \cdot \mathbf{r}'(t)$$

for all t in (a, b). Here the ⋅ denotes the usual inner product.

Now suppose the domain U of φ contains the differentiable curve γ with endpoints p and q, (oriented in the direction from p to q). If r parametrizes γ for t in [a, b], then the above shows that


 * $$\begin{align}

\int_{\gamma} \nabla\varphi(\mathbf{u}) \cdot  d\mathbf{u} &=\int_a^b \nabla\varphi(\mathbf{r}(t))  \cdot  \mathbf{r}'(t)dt \\ &=\int_a^b \frac{d}{dt}\varphi(\mathbf{r}(t))dt =\varphi(\mathbf{r}(b))-\varphi(\mathbf{r}(a))=\varphi\left(\mathbf{q}\right)-\varphi\left(\mathbf{p}\right) \end{align} $$

where the definition of the line integral is used in the first equality, and the fundamental theorem of calculus is used in the third equality.

Potential energy is associated with forces that are derivable from a potential. These forces are obtained as the gradient of a scalar field U(x), given by
 * $$ {\nabla W} = -{\nabla U} = -\left ( \frac{\partial U}{\partial x}, \frac{\partial U}{\partial y}, \frac{\partial U}{\partial z} \right ) = \mathbf{F},$$

work of forces F on a point traveling along a curve C=x(t) is computed by line integral,
 * $$W =\int_{C} \mathbf{F} \cdot \mathrm{d}\mathbf{x} $$


 * $$\int_C \mathbf{F}(\mathbf{r})\cdot\,d\mathbf{r} = \int_a^b \mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)\,dt = \int_a^b \frac{dG(\mathbf{r}(t))}{dt}\,dt = G(\mathbf{r}(b)) - G(\mathbf{r}(a)).$$

Again using the above definitions of F, C and its parametrization, we construct the integral from a Riemann sum. Partition the interval [a,b] into n intervals of length Δt = (b − a)/n. Letting ti be the ith point on [a,b], then r(ti) gives us the position of the ith point on the curve. However, instead of calculating up the distances between subsequent points, we need to calculate their displacement vectors, Δri. As before, evaluating F at all the points on the curve and taking the dot product with each displacement vector gives us the infinitesimal contribution of each partition of F on C. Letting the size of the partitions go to zero gives us a sum

By the mean value theorem, we see that the displacement vector between adjacent points on the curve is
 * $$\Delta\mathbf{r}_i = \mathbf{r}(t_i+\Delta t)-\mathbf{r}(t_i)\approx\mathbf{r}'(t_i)\Delta t$$

Substituting this in the above Riemann sum yields


 * $$I = \lim_{\Delta t \rightarrow 0} \sum_{i=1}^n \mathbf{F}(\mathbf{r}(t_i)) \cdot \mathbf{r}'(t_i)\Delta t$$

which is the Riemann sum for the integral defined above.

Path independence
If a vector field F is the gradient of a scalar field G (i.e. if F is conservative), that is,


 * $$\nabla G = \mathbf{F},$$

then the derivative of the composition of G and r(t) is


 * $$\frac{dG(\mathbf{r}(t))}{dt} = \nabla G(\mathbf{r}(t)) \cdot \mathbf{r}'(t) = \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t)$$

which happens to be the integrand for the line integral of F on r(t). It follows that, given a path C , then


 * $$\int_C \mathbf{F}(\mathbf{r})\cdot\,d\mathbf{r} = \int_a^b \mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)\,dt = \int_a^b \frac{dG(\mathbf{r}(t))}{dt}\,dt = G(\mathbf{r}(b)) - G(\mathbf{r}(a)).$$

In other words, the integral of F over C depends solely on the values of G in the points r(b) and r(a) and is thus independent of the path between them.

For this reason, a line integral of a conservative vector field is called path independent.

Applications
The line integral has many uses in physics. For example, the work done on a particle traveling on a curve C inside a force field represented as a vector field F is the line integral of F on C.