Effective Potentials & Equilibrium

A graph you can roll a ball down

In the last lesson you tracked energy sloshing between kinetic and potential in a mass on a spring, and the total stayed fixed. That worked because you already knew the potential: $U = \tfrac12 kx^2$, a parabola, and a parabola gives simple harmonic motion.

Most systems are not springs. An atom bound to another atom, a bead threaded on a bent wire, a marble in a bowl, a satellite in an elliptical orbit: none of them has an obvious spring in it. And yet all of them oscillate, and all of them oscillate at a frequency you can write down in about three lines. The reason is that the potential energy curve, whatever its shape, looks like a parabola if you get close enough to the bottom of a dip in it.

That is the whole lesson. Draw $U$ against the coordinate, find the dips, and measure how sharp each dip is. Sharpness gives you the frequency. Everything else here is bookkeeping about what counts as "the coordinate" and what counts as "the mass".

Force is minus the slope

Start with something you already have: for motion along one coordinate $x$ with no friction, the work done by the force as the particle moves a small distance $dx$ is $F\,dx$, and that work comes out of the potential energy, so $F\,dx = -dU$. Divide by $dx$:

$F(x) = -\frac{dU}{dx}$

Read this as a picture. Plot $U$ against $x$. Where the curve slopes up to the right, $dU/dx$ is positive, so $F$ is negative and the particle is pushed left: downhill. Where the curve slopes down to the right, the particle is pushed right. Also downhill. A particle on a potential energy graph behaves exactly like a ball rolling on that graph under gravity. That analogy is not a coincidence, and it is worth trusting.

Equilibrium means no net force, so it means no slope:

$\frac{dU}{dx} = 0$

Flat spots on the graph. Nothing more.

Stable, unstable, and why the second derivative decides

A flat spot alone does not tell you whether a system will sit there. A marble balanced on top of a hill has zero net force too. What separates the hilltop from the valley is the curvature: the second derivative.

Suppose $U'(x_0) = 0$ and you push the particle a small distance $\delta$ to the right. The slope there is approximately $U'(x_0 + \delta) \approx U''(x_0)\,\delta$, since the slope starts at zero and changes at rate $U''$. So the force is

$F \approx -U''(x_0)\,\delta$

If $U''(x_0) > 0$, the force is opposite to the displacement: push it right and it pushes back left. It returns. That is stable equilibrium, and on the graph it is a minimum, a valley.

If $U''(x_0) < 0$, the force points the same way as the displacement, the displacement grows, and the particle runs away. That is unstable equilibrium, a maximum. If $U'' = 0$ as well, the test says nothing and you have to look at the next non-zero derivative or just sketch the curve.

Worked example 1. A particle moves in $U(x) = 2x^3 - 9x^2 + 12x$, with $U$ in joules and $x$ in metres. Find the equilibrium points and say which is stable.

Differentiate:

$U'(x) = 6x^2 - 18x + 12 = 6(x^2 - 3x + 2) = 6(x-1)(x-2)$

So $U' = 0$ at $x = 1$ m and $x = 2$ m. Now the second derivative:

$U''(x) = 12x - 18$

At $x = 1$: $U'' = 12 - 18 = -6$ J/m², negative, so that is the hilltop: unstable. At $x = 2$: $U'' = 24 - 18 = +6$ J/m², positive, so $x = 2$ m is the stable one. Figure 1 is this exact potential, so you can check the algebra against the picture.

The frequency hiding in the curvature

Now the payoff. Near a stable point, write $U$ as a Taylor expansion about $x_0$: this is the tool that says any smooth curve equals its value, plus its slope times the offset, plus half its curvature times the offset squared:

$U(x) \approx U(x_0) + U'(x_0)(x-x_0) + \tfrac12 U''(x_0)(x-x_0)^2$

The first term is a constant, and a constant in $U$ changes no force. The second term is zero, because $x_0$ is an equilibrium. What survives is

$U(x) \approx U(x_0) + \tfrac12 U''(x_0)\,(x-x_0)^2$

Compare that to the spring potential $\tfrac12 k x^2$ from the last lesson. It is the same expression with the displacement measured from $x_0$ and with

$k_{\text{eff}} = U''(x_0)$

So the particle does not merely return: it does simple harmonic motion, with the frequency you already know:

$\omega = \sqrt{\frac{U''(x_0)}{m}}$

and period $T = 2\pi/\omega$. The units check: $U''$ is joules per metre squared, which is newtons per metre, which is a spring constant. Every stable equilibrium in physics is a harmonic oscillator if you look closely enough. That one sentence is why this technique appears everywhere.

Worked example 2. A 0.50 kg particle sits at the stable equilibrium of the same $U(x) = 2x^3 - 9x^2 + 12x$ and is nudged. Find $\omega$.

From example 1 the stable point is $x_0 = 2$ m and $U''(2) = 6$ N/m. So

$\omega = \sqrt{\frac{6}{0.50}} = \sqrt{12} = 3.46\ \text{rad/s}$

The amplitude never entered. That is the signature of harmonic motion, and it is only true while the amplitude is small enough that the parabola is a good fit.

When the coordinate is not a straight line

Often the moving thing is not a particle on the $x$-axis. A bead slides on a bent wire, a ball rolls in a bowl, a pendulum swings through an angle. The fix is small: write the kinetic energy in terms of your chosen coordinate $q$, and whatever multiplies $\tfrac12\dot q^2$ plays the role of the mass. Call it the effective mass $m_{\text{eff}}$. Then

$\omega = \sqrt{\frac{U''(q_0)}{m_{\text{eff}}}}$

Worked example 3. A bead slides without friction on a wire bent into the parabola $y = cx^2$ in a vertical plane, with $c = 0.80$ m⁻¹. Take $g = 9.8$ m/s². Find $\omega$ for small oscillations about the bottom.

Potential energy first: $U = mgy = mgcx^2$. Then $U' = 2mgcx$, zero at $x = 0$, and $U'' = 2mgc$, a constant.

Kinetic energy: the bead moves along the wire, so its speed has both components, $v^2 = \dot x^2 + \dot y^2$. Since $y = cx^2$, $\dot y = 2cx\dot x$, giving $v^2 = \dot x^2(1 + 4c^2x^2)$. Near the bottom $x$ is small, so that bracket is 1 and $m_{\text{eff}} = m$. Then

$\omega = \sqrt{\frac{2mgc}{m}} = \sqrt{2gc} = \sqrt{2(9.8)(0.80)} = \sqrt{15.68} = 3.96\ \text{rad/s}$

The mass cancelled, as it must for anything driven purely by gravity.

Now a case where the effective mass does not cancel. A uniform solid ball of radius $r = 2.0$ cm rolls without slipping near the bottom of a spherical bowl of radius $R = 20.0$ cm. Its centre stays on a circle of radius $R - r$, so if $\theta$ is the angle of the centre from the bottom, $U = mg(R-r)(1-\cos\theta)$. For small $\theta$, $1 - \cos\theta \approx \theta^2/2$, so $U \approx \tfrac12 mg(R-r)\theta^2$ and $U''(0) = mg(R-r)$.

The kinetic energy has two pieces because the ball spins as it goes. Its centre moves at $v = (R-r)\dot\theta$, and rolling without slipping means it spins at $\omega_{\text{spin}} = v/r$. With $I = \tfrac25 mr^2$,

$K = \tfrac12 mv^2 + \tfrac12\left(\tfrac25 mr^2\right)\frac{v^2}{r^2} = \tfrac{7}{10}mv^2 = \tfrac12\left[\tfrac75 m(R-r)^2\right]\dot\theta^2$

So $m_{\text{eff}} = \tfrac75 m(R-r)^2$, and

$\omega = \sqrt{\frac{mg(R-r)}{\tfrac75 m(R-r)^2}} = \sqrt{\frac{5g}{7(R-r)}} = \sqrt{\frac{5(9.8)}{7(0.180)}} = \sqrt{38.89} = 6.24\ \text{rad/s}$

A sliding marble in the same bowl would give $\sqrt{g/R}$, faster. The rolling ball carries extra inertia in its spin, so it is lazier.

Terms that are not really potential energy

The word effective in "effective potential" usually means you have folded something that is not a potential energy into $U$, because it acts like one.

The common case is rotation. A puck on a frictionless table is tethered by a string through a hole, so it orbits with angular momentum $L$ about the hole, and $L$ is conserved. Its kinetic energy splits into radial and tangential parts:

$K = \tfrac12 m\dot r^2 + \tfrac12 m(r\dot\theta)^2 = \tfrac12 m\dot r^2 + \frac{L^2}{2mr^2}$

using $L = mr^2\dot\theta$. The second term depends only on $r$, so as far as the radial motion is concerned it behaves exactly like a potential energy. Move it across:

$E = \tfrac12 m\dot r^2 + U_{\text{eff}}(r), \qquad U_{\text{eff}}(r) = U(r) + \frac{L^2}{2mr^2}$

Now the radial motion is a one-dimensional problem with the machinery you already have. That extra term is the centrifugal barrier. It blows up as $r \to 0$, which is the statement that a body with angular momentum cannot reach the centre. A circular orbit is just an equilibrium of $U_{\text{eff}}$, and its stability and radial frequency come from $U_{\text{eff}}''$.

Rotating frames work the same way. A bead on a wire spun about a vertical axis at rate $\Omega$ feels a centrifugal term $-\tfrac12 m\Omega^2 x^2$ that you add to the gravitational $U$ before differentiating. For the parabolic wire $y = cx^2$ spun at $\Omega$,

$U_{\text{eff}} = mgcx^2 - \tfrac12 m\Omega^2 x^2, \qquad \omega = \sqrt{2gc - \Omega^2}$

Spin it fast enough and $2gc - \Omega^2$ goes negative: the bottom stops being a minimum, and the bead runs outward instead of oscillating.

Worked example 4. A spacecraft moves in Earth's field with energy per unit mass $E = -1.42\times10^{7}$ J/kg and angular momentum per unit mass $\ell = 6.47\times10^{10}$ m²/s. At a turning point all the radial speed is gone, so $E = U_{\text{eff}}$:

$E = \frac{\ell^2}{2r^2} - \frac{GM}{r}$

Find the closest approach, with $GM = 3.986\times10^{14}$ m³/s².

Multiply through by $r^2$ and collect:

$Er^2 + GM\,r - \frac{\ell^2}{2} = 0$

Put the numbers in. $\ell^2 = (6.47\times10^{10})^2 = 4.186\times10^{21}$, so $\ell^2/2 = 2.093\times10^{21}$. Multiplying by $-1$ to make the leading coefficient positive:

$1.42\times10^{7}\,r^2 - 3.986\times10^{14}\,r + 2.093\times10^{21} = 0$

Quadratic formula. The discriminant is

$(3.986\times10^{14})^2 - 4(1.42\times10^{7})(2.093\times10^{21}) = 1.589\times10^{29} - 1.189\times10^{29} = 4.00\times10^{28}$

whose square root is $2.00\times10^{14}$. So

$r = \frac{3.986\times10^{14} \pm 2.00\times10^{14}}{2(1.42\times10^{7})}$

The two roots are $r = 2.10\times10^{7}$ m and $r = 6.99\times10^{6}$ m. Two turning points, because the orbit is bound: the far one is apogee, the near one is perigee. Closest approach is $6.99\times10^{6}$ m.

There is a useful check built into that answer. Two real roots means bound. One root, where the discriminant vanishes, means the energy sits exactly at the minimum of $U_{\text{eff}}$: a circular orbit. No real roots would mean that energy is not available at that angular momentum at all.

How far the parabola can be trusted

The frequency $\sqrt{U''/m}$ is an approximation, good while the displacement is small enough that the cubic and quartic parts of the Taylor series stay negligible. Push the amplitude up and the period starts to depend on amplitude.

For $U(x) = \tfrac12 kx^2 + \beta x^4$ the standard first correction is

$\omega \approx \omega_0\left(1 + \frac{3\beta A^2}{2k}\right)$

with $\omega_0 = \sqrt{k/m}$ and $A$ the amplitude. Take $m = 0.20$ kg, $k = 50$ N/m, $\beta = 4.0\times10^{3}$ J/m⁴ and $A = 3.0$ cm. Then $\omega_0 = \sqrt{50/0.20} = \sqrt{250} = 15.81$ rad/s, and the correction factor is

$1 + \frac{3(4.0\times10^{3})(0.030)^2}{2(50)} = 1 + \frac{10.8}{100} = 1.108$

so $\omega = 15.81 \times 1.108 = 17.5$ rad/s. A positive $\beta$ makes the well steeper than a parabola, so the oscillation is faster. That sign is worth remembering as a sanity check.

One more thing the higher derivatives are good for. The force is $-U'$, so the force is largest in magnitude where $U'' = 0$: the inflection point. For the Lennard-Jones potential

$U(r) = 4\varepsilon\left[\left(\frac{\sigma}{r}\right)^{12} - \left(\frac{\sigma}{r}\right)^{6}\right]$

the minimum sits where $U' = 0$, which gives $r_0 = 2^{1/6}\sigma$, about $1.122\sigma$; with $\sigma = 0.340$ nm that is 0.382 nm. The inflection, where the attraction is strongest and beyond which the bond gives way, is at $r = (26/7)^{1/6}\sigma \approx 1.244\sigma = 0.423$ nm. Same function, three different questions, three different derivatives.

Recognising the technique

Reach for this whenever a problem hands you an energy rather than a force. The tells:

• The problem states $U(x)$, $U(r)$ or $U(\theta)$ explicitly and asks where something sits, or how fast it wiggles. Set $U' = 0$, then use $U''$.
• The words small oscillations, slightly displaced, nudged, or near the bottom appear. That is a request for $\sqrt{U''/m_{\text{eff}}}$ and nothing else.
• Something rotates at fixed angular momentum, or the frame spins at fixed $\Omega$. Build $U_{\text{eff}}$ first, with the $L^2/2mr^2$ or $-\tfrac12 m\Omega^2 x^2$ term inside it, then treat $r$ or $x$ as an ordinary one-dimensional coordinate.
• You are asked for a turning point, a closest approach, or the range of motion at a given energy. Set $E = U_{\text{eff}}$ and solve; the number of real roots tells you whether the motion is bound.
• The system is not a particle on a line: a rolling ball, a bead on a wire, a rigid body on a pivot. Write $K$ in your coordinate, read off $m_{\text{eff}}$, and carry on.

The one habit to build is drawing the curve. Sketch $U$, mark the flat spots, decide which hold water, and only then start differentiating. Half the errors in this topic are solving correctly for the wrong root.

Next you take the same idea of a quantity varying with position and apply it to something that fills space rather than sitting at a point, in Pressure & Buoyancy.