Drop a marble into a steel katori from a little above the rim, just inside the edge. It rolls down, runs up the far side, stops and rolls back, again and again. Flick it harder and it jumps over the rim and away. Balance it on an upturned katori and it rolls off at the lightest touch. One graph explains all three.
The curve and the line
An energy diagram is a graph of the potential energy \(U\) of a particle against its position \(x\). Draw on it a horizontal line at the total mechanical energy \(E\) of the particle. If no non-conservative force does work, \(E\) is constant, so the line stays level while the particle moves. The curve belongs to the forces; the line belongs to this particular particle.
At any position, \(E = K + U\), so \(K = E - U\). On the diagram that is the vertical gap between the line and the curve. Where the gap is large, the particle moves fast. Where the gap closes, the particle stops. And from the last lesson, the force at each point is minus the slope of the curve. So the diagram shows speed, force and direction together.
Set \(E = 9\,\text{J}\) and press play. The particle slides into the well, fastest at its bottom, where the gap is \(9 - 2 = 7\,\text{J}\). It climbs the right side, slows, and stops at \(x = 5.1\,\text{m}\). Then it returns and stops again at \(x = 1.7\,\text{m}\). It swings between these two points for ever.
In the sim, set \(E = 10\,\text{J}\). The mass is \(1\,\text{kg}\) and \(U = 2\,\text{J}\) at the bottom of the well. What is the particle's speed as it passes the bottom of the well, in m/s?
Turning points and forbidden regions
Kinetic energy \(\tfrac12 mv^2\) can never be negative. So the particle can be only where \(K = E - U \ge 0\), that is, where the curve lies on or below the line. Where the curve rises above the line, \(K\) would have to be negative. Such a region is a forbidden region: the particle can never be there.
A point where the line meets the curve is a turning point. There \(U = E\), so \(K = 0\) and the particle stops for an instant. The curve there is usually steep, so the force is not zero. The force pushes the particle back, and it reverses. In the sim, the markers \(x_1\) and \(x_2\) are the two turning points.
A turning point is a position where the potential energy equals the total energy, \(U(x) = E\). There the kinetic energy is zero, the particle stops for an instant, and the force turns it back.
Now set \(E = 13\,\text{J}\). The turning points move apart, to \(1.4\,\text{m}\) and \(6.2\,\text{m}\). Look past the hump: the line meets the curve again near \(7.8\,\text{m}\), and beyond that the curve is below the line once more. That region is allowed, since \(K\) would be positive there. Yet the particle released from the left wall can never reach it, because the hump between is forbidden. Allowed is not the same as reachable.
On an energy diagram the total-energy line meets the potential energy curve at \(x = x_1\). What is happening to the particle there?
Trapped or free
While \(E\) is below the top of the hump, the particle has two turning points and swings between them for ever. This is bound motion. Raise the line above the hump, to \(16\,\text{J}\), and the right turning point disappears. The particle crosses the hump, where it slows but does not stop, and runs away down the far side. With nothing to turn it back, it never returns. This is unbound motion.
The boundary is exactly the height of the hump. At \(E = 14\,\text{J}\) the particle creeps towards the top, slowing all the way, and ends balanced on the crest. Whether a particle is trapped is not a property of the curve alone. It depends on where its energy line sits on the curve. The same idea decides whether a rocket escapes the Earth, in the chapter on Gravitational Energy and Orbits.
In the sim, a \(1\,\text{kg}\) particle sits at the bottom of the well, where \(U = 2\,\text{J}\). The top of the hump is at \(14\,\text{J}\). What is the least kinetic energy it must be given at the bottom to escape over the hump, in joules?
Where a particle can rest
A particle is in equilibrium where the force on it is zero. Since \(F = -dU/dx\), equilibrium is wherever the curve is flat: \(dU/dx = 0\). The well and the hump of the sim both qualify. But the two are very different places to rest, and the difference shows as soon as the particle is nudged.
At a minimum of \(U\), the curve rises on both sides. A small displacement puts the particle on an upward slope, and the force points back towards the minimum. The particle returns, overshoots, and swings about the minimum. This is stable equilibrium: a marble at the bottom of a katori. At a maximum, the curve falls on both sides. A small displacement puts the particle on a downward slope, and the force pushes it further away. This is unstable equilibrium: a marble on an upturned katori.
Where \(U\) is flat over a whole region, a displacement leads to no force at all. The particle stays wherever it is put. This is neutral equilibrium: a marble on a level table. The test is the curvature of the curve, the second derivative \(d^2U/dx^2\). Positive curvature, bending upward like a cup, means stable. Negative curvature, bending downward like a dome, means unstable.
A particle moves along the \(x\) axis with potential energy \(U = x^3 - 3x\), with \(U\) in joules and \(x\) in metres. Find its positions of equilibrium and decide whether each is stable or unstable.
Equilibrium needs \(dU/dx = 0\). Stability is decided by the sign of \(d^2U/dx^2\).
\(dU/dx = 3x^2 - 3\) and \(d^2U/dx^2 = 6x\).
\(3x^2 - 3 = 0\) gives \(x = +1\,\text{m}\) and \(x = -1\,\text{m}\). At \(x = +1\), \(d^2U/dx^2 = +6\), positive. At \(x = -1\), \(d^2U/dx^2 = -6\), negative.
The particle has a stable equilibrium at \(x = 1\,\text{m}\) and an unstable one at \(x = -1\,\text{m}\).
A particle has potential energy \(U = x^2 - 4x + 7\), with \(U\) in joules and \(x\) in metres. Where is it in equilibrium, and of what kind?
A marble is balanced exactly on the crest of a potential energy hill. Is it in equilibrium?
Near a stable equilibrium, any smooth curve looks like a small cup: \(U \approx U_0 + \tfrac12 k_{\text{eff}}(x - x_0)^2\), with \(k_{\text{eff}} = d^2U/dx^2\). So a particle nudged a little from any stable equilibrium moves like a block on a spring of constant \(k_{\text{eff}}\). This is the start of simple harmonic motion, the subject of the unit on Oscillations and Waves.
Return to the marble in the katori. The katori is a well, and the marble's energy line sits in it: the marble swings between two turning points on the sides. A hard flick raises the line above the rim, so the right turning point disappears and the marble escapes. On the upturned katori the marble sits on a hump, an unstable equilibrium, and the lightest touch sends it away.
Two atoms that form a molecule, such as the two hydrogen atoms of \(\text{H}_2\), have a potential energy curve with a deep well. Far apart, they attract; pushed too close, they repel strongly. The bottom of the well sets the bond length, about \(0.074\,\text{nm}\) for hydrogen. The depth of the well is the energy needed to break the bond. At room temperature the molecule's energy line sits low in the well, so the two atoms stay bound and vibrate about the bottom.
In his Analytical Mechanics of 1788, Lagrange stated that a system is in stable equilibrium where its potential energy is a minimum. He gave an argument, but it had a gap. In 1846 the German mathematician Dirichlet supplied a short, complete proof. It uses the idea of this lesson: near a minimum, a small energy keeps the system inside a small cup of the curve. The result is now called the Lagrange-Dirichlet theorem.
- Taking a turning point to be an equilibrium.At a turning point the speed is zero but the force is not. Equilibrium needs a flat curve, \(dU/dx = 0\).
- Deciding stability from the value of \(U\).Use the shape: a minimum (\(d^2U/dx^2 > 0\)) is stable, a maximum is unstable.
- Assuming a particle can reach any region where \(U < E\).It must also get there without crossing a region where \(U > E\).