Conservative Forces and Potential Energy

Derivation

Conservative Forces and Potential Energy

D-023 Home PU-101 Threads force · energy · fields Depends on The Work-Energy Theorem, curl-and-path-independence

Statement

For a force field \(\vec{F}(\vec{r})\) that depends on position alone and is continuously differentiable on a simply connected region of \(\mathbb{R}^3\), the following three statements are equivalent: (i) the work \(\int_{\mathcal{C}}\vec{F}\cdot d\vec{r}\) depends only on the endpoints of the path \(\mathcal{C}\), equivalently \(\oint\vec{F}\cdot d\vec{r}=0\) around every closed loop; (ii) the field is curl-free, \(\nabla\times\vec{F}=\vec{0}\) everywhere; and (iii) there exists a single-valued scalar field \(U(\vec{r})\), the potential energy, with \(\vec{F}=-\nabla U\). A force satisfying any (hence all) of these is called conservative, and the work it does is \(W_{A\to B}=U(A)-U(B)\).

Why it matters

Conservativeness is the exact condition under which a moving body's history can be discarded: rather than integrating force along whatever trajectory the particle actually took, one reads off a single number \(U\) attached to each point of space. This is what converts Newton's second law into the conservation of mechanical energy \(E=T+U=\text{const}\), the single most useful bookkeeping identity in mechanics.

The same three-way equivalence is the mechanical prototype for every potential in physics: the gravitational potential, the electrostatic potential, and the thermodynamic potentials all inherit their existence from this curl-free / path-independent / gradient structure. Recognising when a force fails to be conservative (friction, magnetic forces, induced fields) is equally important, because it marks precisely when energy bookkeeping must be enlarged.

Assumptions

The force depends only on position, \(\vec{F}=\vec{F}(\vec{r})\).If \(\vec{F}\) also depends on velocity (drag, the magnetic force \(q\vec{v}\times\vec{B}\)) or on time explicitly, then \(\int\vec{F}\cdot d\vec{r}\) is not a property of the path alone and no position-only potential \(U(\vec{r})\) can reproduce it.

The field is continuous with continuous first derivatives (\(C^1\)).Curl and gradient are built from partial derivatives; if \(\vec{F}\) has discontinuities or kinks, \(\nabla\times\vec{F}\) is undefined there and Stokes' theorem cannot be carried across, breaking the curl-to-loop link.

The domain is simply connected (every closed loop contracts to a point).On a region with a hole a field can be curl-free everywhere yet still have \(\oint\vec{F}\cdot d\vec{r}\neq 0\) around a loop encircling the hole, so (ii) no longer implies (i) and the potential becomes multi-valued.

The work–energy theorem holds for the test particle.Without \(W_{\text{net}}=\Delta T\) the bridge between the field statements (i)–(iii) and the physical payoff (mechanical-energy conservation) is severed, though the purely geometric equivalence survives.

Derivation

We establish the cycle (i)\(\Rightarrow\)(ii)\(\Rightarrow\)(iii)\(\Rightarrow\)(i), so the three statements are mutually equivalent, then add the energy corollary.

\[ W_{\mathcal{C}} = \int_{\mathcal{C}}\vec{F}\cdot d\vec{r} \]

Definition of work as the line integral of force along the path \(\mathcal{C}\). A

\[ W_{\mathcal{C}_1}-W_{\mathcal{C}_2} = \int_{\mathcal{C}_1}\vec{F}\cdot d\vec{r} - \int_{\mathcal{C}_2}\vec{F}\cdot d\vec{r} = \oint_{\mathcal{C}_1-\mathcal{C}_2}\vec{F}\cdot d\vec{r} \]

Two paths \(\mathcal{C}_1,\mathcal{C}_2\) sharing endpoints \(A,B\); reversing \(\mathcal{C}_2\) flips the sign of its integral and joining the two makes a single closed loop. B

\[ W \text{ path-independent} \iff \oint_{\mathcal{C}}\vec{F}\cdot d\vec{r}=0 \ \text{ for every closed } \mathcal{C} \]

From step 2 the difference of any two same-endpoint works equals a loop integral; requiring path-independence for all pairs is exactly requiring every closed-loop integral to vanish. This is statement (i). B

\[ \oint_{\partial S}\vec{F}\cdot d\vec{r} = \iint_{S}(\nabla\times\vec{F})\cdot d\vec{S} \]

Stokes' theorem (prior result: curl-and-path-independence), valid because \(\vec{F}\) is \(C^1\); here \(S\) is any surface whose boundary \(\partial S\) is the loop, which exists inside the domain precisely because it is simply connected. B

\[ \iint_{S}(\nabla\times\vec{F})\cdot d\vec{S}=0 \ \ \forall S \quad\iff\quad \nabla\times\vec{F}=\vec{0} \]

If the flux of \(\nabla\times\vec{F}\) vanishes through every surface it must vanish pointwise: shrink \(S\) to a point at any location and orientation \(\hat{n}\), forcing each curl component to zero. This is statement (ii). C

\[ U(\vec{r}) \equiv -\int_{\vec{r}_0}^{\vec{r}}\vec{F}\cdot d\vec{r}' \]

Given path-independence (equivalent to (ii) here), the integral from a fixed reference \(\vec{r}_0\) depends only on the endpoint \(\vec{r}\), so this defines a single-valued function of position. B

\[ U(\vec{r}+d\vec{r}) - U(\vec{r}) = -\int_{\vec{r}}^{\vec{r}+d\vec{r}}\vec{F}\cdot d\vec{r}' = -\vec{F}(\vec{r})\cdot d\vec{r} \]

Move the endpoint by an infinitesimal \(d\vec{r}\); path-independence lets us take the straight segment, along which \(\vec{F}\) is constant to first order. C

\[ dU = \nabla U\cdot d\vec{r} \quad\Rightarrow\quad \nabla U\cdot d\vec{r} = -\vec{F}\cdot d\vec{r} \quad \forall\, d\vec{r} \]

The left equality is the total differential of a scalar field (definition of the gradient); equate with step 7, which holds for every direction \(d\vec{r}\). C

\[ \vec{F} = -\nabla U \]

Two vectors giving equal dot products with every \(d\vec{r}\) must be equal. This is statement (iii), closing (ii)\(\Rightarrow\)(iii). B

\[ W_{\mathcal{C}} = \int_{\mathcal{C}}(-\nabla U)\cdot d\vec{r} = -\int_{\mathcal{C}} dU = U(A)-U(B) \]

Converse (iii)\(\Rightarrow\)(i): the gradient theorem for line integrals collapses the integral to endpoint values, manifestly path-independent. The cycle is complete. B

\[ \nabla\times\vec{F} = \nabla\times(-\nabla U) = \vec{0} \]

A direct check that (iii)\(\Rightarrow\)(ii): the curl of any gradient vanishes identically because mixed partials commute for \(C^2\) \(U\). A

\[ W_{\text{net}}=\Delta T \ \wedge\ W=U(A)-U(B) \ \Rightarrow\ \Delta(T+U)=0 \]

If \(\vec{F}\) is the only force, insert the work–energy theorem (prior result) into step 10: \(T_B-T_A=U(A)-U(B)\), i.e. \(T+U\) is conserved. B

Result

\[ \oint_{\mathcal{C}}\vec{F}\cdot d\vec{r}=0 \iff \nabla\times\vec{F}=\vec{0} \iff \vec{F}=-\nabla U,\qquad W_{A\to B}=U(A)-U(B) \]

Reading. A force is conservative when, and only when, it curls nowhere; equivalently when its work around any closed trip is exactly zero; equivalently when it is the downhill gradient of a potential-energy landscape \(U(\vec{r})\). The minus sign encodes that a body is pushed toward lower potential energy. Combined with the work–energy theorem, the height in this landscape plus the kinetic energy is a constant of the motion.

Units check. \(U\) has units of energy, \(\mathrm{J}\). Its gradient \(\nabla U\) carries \(\mathrm{J\,m^{-1}}=\mathrm{N}\), matching a force. The loop integral \(\oint\vec{F}\cdot d\vec{r}\) has \(\mathrm{N\cdot m}=\mathrm{J}\) (a work, correctly zero). The curl \(\nabla\times\vec{F}\) has \(\mathrm{N\,m^{-1}}\); its surface integral, \(\mathrm{N\,m^{-1}\cdot m^2}=\mathrm{J}\), returns the loop work via Stokes.

Limiting cases

One dimension: curl is vacuous, so every \(C^1\) force \(F(x)\) is automatically conservative with \(U(x)=-\int F\,dx\). The distinction only bites in two or more dimensions.
Uniform field \(\vec{F}=\vec{F}_0\) (near-surface gravity \(-mg\hat{z}\)): trivially curl-free, \(U=-\vec{F}_0\cdot\vec{r}\); for gravity \(U=mgz\), a linear ramp.
Central field \(\vec{F}=f(r)\hat{r}\): always curl-free wherever \(f\) is \(C^1\), with \(U(r)=-\int f(r)\,dr\) depending on radius alone. Covers Newtonian gravity and the Coulomb force.
Additive constant: \(U\to U+C\) leaves \(\vec{F}=-\nabla U\) and every work difference unchanged, so the potential is defined only up to a reference.

Breaks when

Non-simply-connected domain. Remove a line or point and a curl-free field can still circulate: the vortex \(\vec{F}=\dfrac{k}{x^2+y^2}(-y,\,x,\,0)\) has \(\nabla\times\vec{F}=\vec{0}\) off the \(z\)-axis, yet \(\oint\vec{F}\cdot d\vec{r}=2\pi k\neq 0\) around a loop enclosing the axis. No single-valued \(U\) exists; the potential winds like an angle.
Velocity- or time-dependent forces. Friction \(\vec{F}=-b\vec{v}\) and the magnetic force \(\vec{F}=q\vec{v}\times\vec{B}\) are not fields of position, so \(\nabla\times\vec{F}\) is ill-posed and \(\int\vec{F}\cdot d\vec{r}\) depends on how fast and when the path is traversed; energy is dissipated (friction) or reshuffled without a potential.
Explicit time dependence. An induced field with \(\nabla\times\vec{E}=-\partial\vec{B}/\partial t\neq\vec{0}\) (Faraday's law) does nonzero work around a loop; no static potential energy exists.
Non-conservative force superposed. When a conservative force acts alongside a dissipative one, \(\Delta(T+U)=W_{\text{non-cons}}\neq 0\); the potential of the conservative part still exists but mechanical energy is not conserved.

Failure modes

Sign slip: writing \(\vec{F}=+\nabla U\). The physical force points down the potential gradient; the minus sign is not optional and flips every stability conclusion (minima of \(U\) become maxima).
Curl-free \(\Rightarrow\) conservative, always: forgetting the simply-connected proviso and declaring the vortex field conservative because its curl vanishes; the global loop integral is the real test.
Absolute potential energy: treating a specific numeric \(U\) as physical rather than the choice of reference \(\vec{r}_0\); only differences \(\Delta U\) and gradients \(\nabla U\) are measurable.
Energy conservation with friction present: using \(T+U=\text{const}\) when a non-conservative force acts; the correct statement is \(\Delta(T+U)=W_{\text{non-cons}}\).
Confusing potential with potential energy: for gravity \(U=m\phi\) and for electrostatics \(U=qV\); dropping the coupling factor \(m\) or \(q\) scrambles units and magnitudes.
Path-dependent bookkeeping: laboriously integrating \(\vec{F}\cdot d\vec{r}\) along a chosen path for a force already known to be conservative, instead of using \(W=U(A)-U(B)\).

Discussion

The deepest content of the theorem is that a vector field with three components collapses, when conservative, to a single scalar function of position. That is an enormous compression: instead of specifying \(F_x,F_y,F_z\) at every point subject to hidden consistency, you specify one number \(U\) and differentiate. The curl-free condition is precisely the integrability condition (\(\partial F_x/\partial y=\partial F_y/\partial x\) and cyclically) that lets the three components descend from one potential.

Physically, the equivalence is what makes the potential-energy landscape a legitimate picture. Equilibria are the stationary points \(\nabla U=\vec{0}\); their stability is read from the curvature — the Hessian of \(U\) — a minimum being stable and a saddle or maximum unstable. Small oscillations about a minimum are governed by the local quadratic \(U\approx \tfrac{1}{2}k(\Delta x)^2\), which is why simple harmonic motion is universal near any smooth potential well: the effective spring constant is just \(U''\) evaluated at the minimum.

The result also draws a sharp line through all of physics between forces that store energy and forces that dissipate or merely transport it. Gravity and electrostatics are conservative and get potentials; friction converts organized kinetic energy into heat and cannot; the magnetic force sits in a third category, always perpendicular to velocity so it does zero work yet is not derivable from a scalar potential energy at all. Recognising a force's category tells you in advance whether energy methods will short-cut the dynamics, and \(U\) itself becomes the potential term in the Lagrangian \(L=T-U\) and the Hamiltonian \(H=T+U\).

Mathematically, "curl-free \(\Rightarrow\) gradient" is the Poincaré lemma, and its failure on multiply connected domains is the first whisper of de Rham cohomology: the obstruction is measured by the loop integrals around the holes, topological invariants of the domain rather than local properties of the field. In the language of forms, \(\vec{F}\cdot d\vec{r}\) is a 1-form \(\omega\); curl-free means \(\omega\) is closed (\(d\omega=0\)), and having a potential means \(\omega\) is exact (\(\omega=-dU\)). The vortex counterexample is a generator of \(H^1(\mathbb{R}^2\setminus\{0\})\), the same structure that reappears as the Aharonov–Bohm phase, where a curl-free vector potential outside a solenoid nonetheless produces observable interference because the enclosing loop cannot be contracted.

Common misconceptions. "Zero net force means conservative" — no; conservative is a property of the whole field's spatial structure, not of the force at one instant. "Path-independence is obvious for any smooth force" — it is exactly what fails for the vortex and for friction. "A conservative force does no work" — it does work, but that work is fully recoverable and equals \(-\Delta U\).

Worked examples

Example 1 — Test a field and build its potential. Take \(\vec{F}=c\,(2xy\,\hat{x}+x^2\,\hat{y})\) with \(c=1\,\mathrm{N\,m^{-2}}\) and \(x,y\) in metres.

\[ (\nabla\times\vec{F})_z = \frac{\partial F_y}{\partial x}-\frac{\partial F_x}{\partial y} = c(2x)-c(2x)=0 \]

Compute the only nonzero curl component in the plane before touching numbers; it vanishes, and the plane is simply connected, so \(\vec{F}\) is conservative. B

\[ -\frac{\partial U}{\partial x}=2cxy \ \Rightarrow\ U=-cx^2y+g(y) \]

Integrate the \(x\)-component of \(\vec{F}=-\nabla U\); the integration "constant" may depend on \(y\). B

\[ -\frac{\partial U}{\partial y}=cx^2-g'(y)=cx^2 \ \Rightarrow\ g'(y)=0 \ \Rightarrow\ g=\text{const} \]

Match the \(y\)-component; the residual freedom is the arbitrary reference. Take \(U=0\) at the origin. B

\[ W_{(0,0)\to(1,2)} = U(0,0)-U(1,2) = 0-[-(1)(1)^2(2)] = 2\ \mathrm{J} \]

Endpoint values only (path-independent); with \(x,y\) in metres and \(c=1\,\mathrm{N\,m^{-2}}\) the result is in joules. A

\[ U(x,y)=-(1\,\mathrm{N\,m^{-2}})\,x^2 y,\qquad W=+2\ \mathrm{J} \]

Reading. The field is conservative; moving from the origin to \((1\,\mathrm{m},2\,\mathrm{m})\) it does \(2\,\mathrm{J}\) of work regardless of route.

Units check. \(c\,x^2 y=(\mathrm{N\,m^{-2}})(\mathrm{m^2})(\mathrm{m})=\mathrm{N\cdot m}=\mathrm{J}\).

Example 2 — Escape speed from Earth via the gravitational potential. A body of mass \(m\) sits on a planet of mass \(M=5.97\times10^{24}\,\mathrm{kg}\), radius \(R=6.37\times10^{6}\,\mathrm{m}\). Find the minimum launch speed to reach infinity.

\[ \vec{F}=-\frac{GMm}{r^2}\hat{r} \ \Rightarrow\ U(r)=-\int_{\infty}^{r}\vec{F}\cdot d\vec{r}' = -\frac{GMm}{r} \]

The central force is curl-free, hence conservative; integrate radially with the conventional reference \(U(\infty)=0\). B

\[ \tfrac{1}{2}mv_{\text{esc}}^2 - \frac{GMm}{R} = \tfrac{1}{2}m(0)^2 + 0 \]

Apply energy conservation (valid because gravity is the only, conservative, force). The threshold trajectory just reaches infinity with zero speed. A

\[ v_{\text{esc}}=\sqrt{\frac{2GM}{R}} \]

Solve for the speed; the test mass \(m\) cancels — symbols rearranged before numbers. A

\[ v_{\text{esc}}=\sqrt{\frac{2(6.674\times10^{-11})(5.97\times10^{24})}{6.37\times10^{6}}} \approx 1.12\times10^{4}\ \mathrm{m\,s^{-1}} \]

Insert \(G=6.674\times10^{-11}\,\mathrm{N\,m^2\,kg^{-2}}\) and the given data. A

\[ v_{\text{esc}}\approx 11.2\ \mathrm{km\,s^{-1}} \]

Reading. Escape depends only on the depth of the potential-energy well per unit mass, not on the launch direction — a direct payoff of path-independence.

Units check. \(\sqrt{\mathrm{N\,m^2\,kg^{-2}}\cdot\mathrm{kg}/\mathrm{m}}=\sqrt{\mathrm{J\,kg^{-1}}}=\sqrt{\mathrm{m^2\,s^{-2}}}=\mathrm{m\,s^{-1}}\).

Problems

A 5.0 kg block is carried up 2.0 m, across 3.0 m horizontally, then down 0.5 m. Taking \(g=9.81\,\mathrm{m\,s^{-2}}\), find the work done by gravity.
Solution
Gravity is conservative, so only net vertical displacement matters: \(\Delta y = 2.0-0.5 = 1.5\,\mathrm{m}\) upward. \(W_{\text{grav}}=-mg\,\Delta y = -(5.0)(9.81)(1.5) = \mathbf{-73.6\ J}\). The horizontal leg contributes nothing since \(\vec{F}\perp d\vec{r}\) there.
Test whether \(\vec{F}=(yz,\ xz,\ xy)\,\mathrm{N}\) (coordinates in m, unit constant suppressed) is conservative, and if so find \(U\).
Solution
Curl: \(\big(\partial_y(xy)-\partial_z(xz),\ \partial_z(yz)-\partial_x(xy),\ \partial_x(xz)-\partial_y(yz)\big) = (x-x,\ y-y,\ z-z)=\vec{0}\), so conservative. Integrate \(\vec{F}=-\nabla U\): \(-\partial_x U=yz\Rightarrow U=-xyz+g(y,z)\); matching \(-\partial_y U=xz\) and \(-\partial_z U=xy\) forces \(g=\text{const}\). Hence \(\boxed{U=-xyz+C}\).
A force \(\vec{F}=-k\vec{r}\) (with \(\vec{r}\) the position vector, \(k=8.0\,\mathrm{N\,m^{-1}}\)) acts on a particle. Find \(U(\vec{r})\) and the work done by the force moving from the origin to \((0.3,0.4,0)\,\mathrm{m}\).
Solution
\(\vec{F}=-k(x,y,z)\Rightarrow U=\tfrac12 k(x^2+y^2+z^2)=\tfrac12 kr^2\) (central, curl zero). Endpoint \(r^2=0.3^2+0.4^2=0.25\,\mathrm{m^2}\). \(U(\text{end})=\tfrac12(8.0)(0.25)=1.0\,\mathrm{J}\), \(U(0)=0\). \(W=U(0)-U(\text{end})=\mathbf{-1.0\ J}\).
An ideal spring exerts \(F=-kx\) with \(k=200\,\mathrm{N\,m^{-1}}\). Show the force is conservative, write \(U(x)\), and find the work the spring does as it is stretched from 0 to 0.10 m.
Solution
In 1D every \(C^1\) force is conservative (curl vacuous). \(U=-\int F\,dx=\int kx\,dx=\tfrac12 kx^2\). Work by the spring \(=U(0)-U(0.10)=-\tfrac12(200)(0.10)^2=\mathbf{-1.0\ J}\); the energy is stored, so the spring does negative work on whatever stretches it.
Consider \(\vec{F}=\kappa(-y,\ x,\ 0)\) with \(\kappa=3.0\,\mathrm{N\,m^{-2}}\). Compute \(\nabla\times\vec{F}\), evaluate \(\oint\vec{F}\cdot d\vec{r}\) around the unit circle \(x^2+y^2=1\,\mathrm{m^2}\), and state whether \(\vec{F}\) is conservative.
Solution
\((\nabla\times\vec{F})_z=\partial_x(\kappa x)-\partial_y(-\kappa y)=\kappa+\kappa=2\kappa=6.0\,\mathrm{N\,m^{-2}}\neq 0\). By Stokes, \(\oint\vec{F}\cdot d\vec{r}=\iint 2\kappa\,dA=2\kappa(\pi R^2)=2(3.0)\pi(1)^2=6\pi\approx\mathbf{18.8\ J}\). The nonzero loop integral means \(\vec{F}\) is not conservative; no potential energy exists for this rotational field.