Work and Potential Energy as a Gradient in Conservative Force Fields

Energy methods are the shortest route through most of mechanics, and they rest on one idea from Chapter 1: for a certain class of force fields, the work done is the difference of a scalar function between the endpoints, so the force is the (negative) gradient of that scalar. Such forces are called conservative, the scalar is the potential energy, and the statement $\vec{F} = -\nabla U$ is the bridge between the vector language of forces and the scalar language of energy.

This chapter makes that bridge rigorous. It defines work as a line integral, proves the work–energy theorem from Newton's second law, gives four equivalent characterisations of whether a force is conservative, and then shows how a one-dimensional potential-energy curve $U(x)$ encodes the entire qualitative dynamics — equilibria, stability, turning points, oscillation frequencies — before a single equation of motion is solved. The reading is Morin and the early chapters of Goldstein.

Why "as a gradient" is in the title. The high-school picture of potential energy as "stored energy" is fine as far as it goes, but the operational definition you need is geometric: $U$ is a scalar field, its level surfaces are surfaces of constant energy, and $\vec{F} = -\nabla U$ points downhill on that field, perpendicular to the level surfaces, steepest where they are closest together.

1. Work as a Line Integral

The work done by a force $\vec{F}$ on a particle moving along a path $\mathcal{C}$ from point $A$ to point $B$ is

where $d\vec{\ell}$ is the infinitesimal tangent vector along the path. Only the component of $\vec{F}$ along the motion contributes: a force perpendicular to the velocity (the normal force on a slope, the magnetic force, the tension in a whirling string) does zero work. In general this integral depends on the whole path $\mathcal{C}$, not merely on $A$ and $B$ — friction is the standard example, doing more negative work along a longer route.

2. The Work–Energy Theorem

Take the total force on a particle of mass $m$ and dot Newton's second law with the velocity:

Integrating over time from $A$ to $B$, and using $\vec{v}\,dt = d\vec{\ell}$ on the left,

The net work done by all forces equals the change in kinetic energy. This holds for any force whatsoever — conservative or not, constant or not — because it is just Newton's second law integrated. It is the workhorse for speed-at-a-point problems.

3. Path Independence and Conservative Forces

A force field $\vec{F}(\vec{r})$ is conservative if the work it does between any two points is independent of the path taken. Equivalently, the work around every closed loop is zero:

The physical content: a conservative force cannot be used to extract energy by running a particle around a cycle. Gravity, the electrostatic force and ideal spring forces are conservative; friction, air drag and the force from a person's muscles are not (they are dissipative or non-conservative).

4. Four Equivalent Statements

On a simply connected region, the following four statements about a force field $\vec{F}$ are all equivalent — any one implies the other three:

1. $\vec{F}$ is conservative (work is path-independent).
2. $\displaystyle\oint \vec{F}\cdot d\vec{\ell} = 0$ around every closed loop.
3. $\nabla\times\vec{F} = \vec{0}$ everywhere (the field is irrotational).
4. $\vec{F} = -\nabla U$ for some single-valued scalar field $U$ (a potential energy exists).

The chain of implications: $2 \Leftrightarrow 1$ by definition; $2 \Rightarrow 3$ by shrinking the loop and invoking the curl-as-circulation-density interpretation, and $3 \Rightarrow 2$ by Stokes' theorem; $1 \Rightarrow 4$ by defining $U(\vec{r}) = -\int_{\vec{r}_0}^{\vec{r}} \vec{F}\cdot d\vec{\ell}$ (well-defined precisely because the integral is path-independent), and $4 \Rightarrow 3$ because the curl of a gradient is identically zero.

Test 3 is the practical one. Given a force field, compute its curl. If it vanishes identically, the force is conservative and a potential exists; if any component is non-zero anywhere, it is not.

5. Potential Energy and $\vec{F} = -\nabla U$

When a potential energy exists, the work done by the conservative force is

by the gradient theorem. Combining with the work–energy theorem $W_{AB} = K_B - K_A$ (when only conservative forces act):

— conservation of mechanical energy. The relation $\vec{F} = -\nabla U$ inverts as $U(\vec{r}) = -\displaystyle\int_{\vec{r}_0}^{\vec{r}} \vec{F}\cdot d\vec{\ell}$, defined up to the arbitrary additive constant set by the choice of reference point $\vec{r}_0$.

Worked example — three standard potentials.

• Uniform gravity: $\vec{F} = -mg\,\hat{z}$. Then $\nabla\times\vec{F} = \vec{0}$, and $U = mgz$ (taking $U = 0$ at $z = 0$), since $-\partial U/\partial z = -mg$.
• Ideal spring: $F = -kx$ in 1D. Then $U = \tfrac{1}{2}kx^2$, since $-dU/dx = -kx$.
• Inverse-square (gravitational/Coulomb): $\vec{F} = -\dfrac{GMm}{r^2}\,\hat{r}$. Using $\nabla g(r) = g'(r)\hat{r}$ backwards, $U(r) = -\dfrac{GMm}{r}$ (with $U \to 0$ as $r \to \infty$), since $-\dfrac{dU}{dr} = -\dfrac{GMm}{r^2}$. Its curl is zero everywhere it is defined, so it is conservative.

6. Worked Example: Testing a Field and Building Its Potential

Consider $\vec{F} = (2xy + z^3)\,\hat{x} + x^2\,\hat{y} + 3xz^2\,\hat{z}$ (units: newtons, with coordinates in metres). Is it conservative, and if so what is $U$?

Curl test.

The curl vanishes identically, so the force is conservative.

Constructing $U$. From $F_x = -\partial U/\partial x$, integrate: $U = -\int(2xy + z^3)\,dx = -x^2 y - x z^3 + g(y,z)$. Then $-\partial U/\partial y = x^2 - \partial g/\partial y$ must equal $F_y = x^2$, so $\partial g/\partial y = 0$. And $-\partial U/\partial z = 3xz^2 - \partial g/\partial z$ must equal $F_z = 3xz^2$, so $\partial g/\partial z = 0$. Hence $g$ is a constant and

7. Energy Diagrams and Equilibrium

For one-dimensional motion, plotting $U(x)$ against $x$ gives the qualitative dynamics at a glance, because $E = \tfrac{1}{2}m\dot{x}^2 + U(x)$ is constant and $\tfrac{1}{2}m\dot{x}^2 \ge 0$:

• Allowed region: motion is confined to $x$ where $U(x) \le E$. Boundaries where $U(x) = E$ are turning points ($\dot{x} = 0$, then reverses).
• Equilibrium: points where $F = -dU/dx = 0$, i.e. extrema of $U$. A minimum of $U$ is stable (a displacement produces a restoring force); a maximum is unstable; a flat region is neutral.
• Small oscillations: near a stable equilibrium $x_0$, Taylor-expand $U(x) \approx U(x_0) + \tfrac{1}{2}U''(x_0)(x - x_0)^2$. The force is then $F \approx -U''(x_0)(x - x_0)$, simple harmonic motion with

Worked example. A particle of mass $m$ moves in $U(x) = U_0\left[\left(\dfrac{a}{x}\right)^{12} - 2\left(\dfrac{a}{x}\right)^{6}\right]$ (the Lennard-Jones form). Setting $dU/dx = 0$: $-12\dfrac{a^{12}}{x^{13}} + 12\dfrac{a^6}{x^7} = 0$, giving $x_0 = a$. There $U(a) = U_0[1 - 2] = -U_0$ (the well depth). Differentiating again and evaluating at $x_0 = a$ gives $U''(a) = \dfrac{72\,U_0}{a^2}$, so small vibrations about the bottom of the well have angular frequency $\omega = \sqrt{\dfrac{72\,U_0}{m a^2}}$. A particle with energy $-U_0 < E < 0$ is bound and oscillates between two turning points; a particle with $E \ge 0$ is unbound and escapes to $x \to \infty$.

8. Common Pitfalls

• Applying energy conservation when a non-conservative force acts. If friction or drag is present, use $W_{\text{net}} = \Delta K$ with all forces, or $\Delta E_{\text{mech}} = W_{\text{non-cons}}$.
• Forgetting the minus sign in $\vec{F} = -\nabla U$. The force points toward decreasing potential energy — downhill on the $U$ landscape.
• Assuming zero curl guarantees a potential on any domain. It does so only where the region is simply connected; a field like $\dfrac{-y\,\hat{x} + x\,\hat{y}}{x^2 + y^2}$ is irrotational but has no single-valued $U$ on a loop around the $z$-axis.
• Confusing a turning point with an equilibrium. At a turning point $\dot{x} = 0$ but $F \ne 0$ (the particle accelerates away); at an equilibrium $F = 0$.
• Reading stability off the value of $U$ rather than its curvature. Stability depends on the sign of $U''$, not on whether $U$ is positive or negative.
• Dropping the arbitrary constant carelessly. $U$ is defined only up to an additive constant; differences $U(A) - U(B)$ and the equilibrium/oscillation analysis are what carry physical meaning.

Summary

• Work is the line integral $W = \int_{\mathcal{C}} \vec{F}\cdot d\vec{\ell}$; only the tangential component of $\vec{F}$ contributes, and in general the value depends on the path.
• Work–energy theorem: $W_{\text{net}} = \Delta K$, true for every force, being Newton's second law dotted with $\vec{v}$ and integrated.
• A conservative force does path-independent work, equivalently zero work around every closed loop.
• Four equivalent tests (simply connected domain): path-independence $\Leftrightarrow$ $\oint\vec{F}\cdot d\vec{\ell} = 0$ $\Leftrightarrow$ $\nabla\times\vec{F} = \vec{0}$ $\Leftrightarrow$ $\vec{F} = -\nabla U$. The curl test is the practical one.
• $\vec{F} = -\nabla U$ gives $W_{AB} = U(A) - U(B)$ and hence $E = K + U = \text{const}$ when only conservative forces act. $U = -\int \vec{F}\cdot d\vec{\ell}$, up to a constant.
• Standard potentials: $U = mgz$ (uniform gravity), $U = \tfrac{1}{2}kx^2$ (spring), $U = -GMm/r$ (inverse square).
• Energy diagrams: motion confined to $U(x) \le E$; turning points where $U = E$; equilibria at extrema of $U$ (minimum stable, maximum unstable); small oscillations at $\omega = \sqrt{U''(x_0)/m}$.