Chapter 5 Energy Diagrams, Stability and Phase Portraits
Chapter 5
Energy Diagrams, Stability and Phase Portraits
Theory, worked examples and 16 previous-year questions with full solutions — IIT JAM, GATE, JEST and CSIR NET-JRF, 2020 onwards.
Theory
A particle moving along a straight line under a conservative force is the simplest dynamical system, yet it already displays most of the qualitative behaviour met in mechanics: oscillation, escape, reflection, equilibrium that is stable or unstable, and the delicate motion that separates one kind of behaviour from another. In this chapter we learn to read all of this from the potential energy alone, without solving the equation of motion. We first use energy diagrams to locate the regions in which motion is possible, then classify the points of equilibrium, then find the frequency of small oscillations about a stable equilibrium, and finally draw the complete family of motions as curves in the plane of position and momentum.
5.1 Energy diagrams
Consider a particle of mass \(m\) moving along the \(x\)-axis under a force \(F(x)\) that depends on position alone. Such a force is always conservative in one dimension, because the work \(\int F\,dx\) between two points depends only on the end-points. We define the potential energy by
where \(x_0\) is an arbitrary reference point. Multiplying Newton’s equation \(m\ddot x=-dV/dx\) by \(\dot x\) gives \(\frac{d}{dt}\left(\frac12m\dot x^2+V\right)=0\), so the total mechanical energy
\[\boxed{E=\frac12m\dot x^2+V(x)}\]
is a constant of the motion (Chapter 4). This single first integral contains a great deal of information. Since the kinetic energy \(T=E-V(x)\) can never be negative, the particle is confined to the regions in which
\[\boxed{V(x)\le E}.\]
These are the classically allowed regions; the regions with \(V(x)>E\) are classically forbidden. The points at which \(V(x)=E\) are the turning points: there the kinetic energy and the velocity vanish, and the particle reverses its direction of motion unless the force there is also zero. In an allowed region the speed follows directly from the energy,
so the particle moves fastest where the potential energy is lowest.
These facts are displayed in an energy diagram, in which the potential energy \(V(x)\) is plotted against \(x\) and the total energy \(E\) appears as a horizontal straight line, because it does not change during the motion. At any position the kinetic energy \(T=E-V\) is the vertical distance from the curve up to the line, and the allowed regions are those in which the line lies above the curve.
(a) Bounded motion
The energy diagram of a harmonic oscillator, \(V=\frac12kx^2\), is shown in Figure 5.1. The potential energy is a parabola centred on the origin, and the line \(E\) cuts it at the two turning points \(x_1=-\sqrt{2E/k}\) and \(x_2=+\sqrt{2E/k}\). Motion is confined to the shaded region \(x_1\le x\le x_2\), where \(E\ge V\).
Figure 5.1 Energy diagram of a harmonic oscillator. The motion is confined to the shaded region between the turning points \(x_1\) and \(x_2\); the kinetic energy \(T=E-V\) is the height of the line \(E\) above the curve.
The diagram tells the whole story of the motion. The kinetic energy is greatest at the origin. As the particle moves away from the origin in either direction it is slowed by the spring, and it comes momentarily to rest at a turning point. It then moves back towards the origin with increasing kinetic energy, passes through it, and is brought to rest at the other turning point, so that the cycle repeats indefinitely. This is bounded motion: as \(E\) increases the turning points move farther apart, but the particle can never escape to infinity. As \(E\) decreases the amplitude decreases, until for \(E=0\) the particle lies at rest at \(x=0\).
For any bounded motion between two turning points \(x_1\) and \(x_2\) the time taken to go from one to the other and back follows from the expression for \(\dot x\). Writing \(dt=dx/\lvert\dot x\rvert\), the period is
For the harmonic oscillator the integral gives \(\tau=2\pi\sqrt{m/k}\), independent of the energy; for any other potential the period depends in general on the amplitude.
(b) Unbounded motion
Figure 5.2 shows a repulsive potential \(V=A/r\) with \(A>0\), as for two particles carrying charges of the same sign, plotted for \(r>0\). For any energy \(E>0\) the line \(E\) cuts the curve at a single point, the distance of closest approach\(r_{\min}=A/E\). A particle sent in towards the origin loses kinetic energy as it climbs the potential, comes momentarily to rest at \(r_{\min}\), and is then pushed back out to infinity. The motion is unbounded, because the allowed region \(r\ge r_{\min}\) extends to infinity. Since \(E\) is conserved, the speed at any distance is the same on the way out as on the way in; the encounter merely reverses the velocity. For a steep potential the distance of closest approach changes only slowly as \(E\) is increased, while the kinetic energy \(E-V\) grows rapidly as the particle recedes.
Figure 5.2 Energy diagram for the repulsive potential \(V=A/r\). The particle approaches to the distance \(r_{\min}\), where \(V=E\), and is then turned back.
Example 5.1 The potential energy of a particle moving along the \(x\)-axis has the piecewise constant form shown in Figure 5.3, with
and \(V=0\) everywhere else. Find the kinetic energy in every region when the total energy is (a) \(E_1=10\) J, (b) \(E_2=5\) J.
Figure 5.3 A piecewise constant potential and two energy levels \(E_1\) and \(E_2\).
Solution. In each region the kinetic energy is \(T=E-V\), provided this is positive; if \(V>E\) the region is forbidden.
For \(E=E_1=10\) J the line \(E_1\) lies above the potential everywhere, so every region is allowed. For \(x<a\), \(T=10-0=10\) J; for \(a<x<b\), \(T=10-8=2\) J; for \(b<x<c\), \(T=10-3=7\) J; for \(c<x<d\), \(T=10-6=4\) J; for \(d<x<e\), \(T=10-(-4)=14\) J; and for \(x>e\), \(T=10\) J again. A particle travelling from left to right therefore slows down over the barrier \(V_1\), speeds up over the step \(V_2\), slows again over \(V_3\), moves fastest in the well \(V_4\), and emerges with its original speed.
For \(E=E_2=5\) J the regions \(a<x<b\) (where \(V_1=8\) J) and \(c<x<d\) (where \(V_3=6\) J) have \(V>E\) and are classically forbidden. In the allowed regions, \(T=5\) J for \(x<a\), \(T=5-3=2\) J for \(b<x<c\), \(T=5-(-4)=9\) J for \(d<x<e\), and \(T=5\) J for \(x>e\). The allowed regions are now separated from one another by the forbidden ones: a particle in \(b<x<c\) bounces back and forth between the walls at \(b\) and \(c\), a particle in \(x<a\) is reflected at \(a\), and a particle in \(x>d\) is reflected at \(d\). Which of these motions occurs depends on where the particle starts.
5.2 Equilibrium and stability in one dimension
A point \(x_0\) is a point of equilibrium if the force on the particle vanishes there,
\[F(x_0)=-\left.\frac{dV}{dx}\right|_{x_0}=0,\]
so that a particle placed at rest at \(x_0\) remains at rest. Equilibrium points are the stationary points of the potential energy: its minima, maxima and points of inflection with horizontal tangent. Their character is decided by what happens when the particle is displaced slightly from \(x_0\). Expanding the force about \(x_0\),
If \(V''(x_0)>0\), the force is opposite to the displacement \(\xi\) and pushes the particle back: the equilibrium is stable, and \(x_0\) is a minimum of the potential. If \(V''(x_0)<0\), the force is along the displacement and drives the particle farther away: the equilibrium is unstable, and \(x_0\) is a maximum. Thus
A stable point acts as an attractive centre for nearby motion and an unstable point as a repulsive one. Since \(F=-V'\), the same test can be stated in terms of the force: an equilibrium is stable where the graph of \(F(x)\) crosses zero with negative slope, and unstable where it crosses with positive slope.
If \(V''(x_0)=0\) the second-derivative test fails and we must look at the first derivative \(V^{(n)}(x_0)\) that does not vanish. If \(n\) is even and \(V^{(n)}(x_0)>0\), the point is a minimum and the equilibrium is stable, as for \(V=cx^4\) with \(c>0\). If \(n\) is odd, the point is a point of inflection, such as the origin for \(V=cx^3\); the force has the same sign on both sides, and a displacement in one direction is not reversed, so the equilibrium is unstable. If \(V\) is constant over a range of \(x\), every point of the range is an equilibrium and a displaced particle stays where it is put; this is called neutral equilibrium.
The examples below show how the potential, the force and the equilibrium points fit together. In each case the curve \(F(x)=-V'(x)\) crosses the \(x\)-axis at the equilibrium points, and the extreme values of \(F\) occur at the points of inflection of \(V\), where \(F'=-V''=0\).
Example 5.2 A particle moves in the potential \(V(x)=-\dfrac{x^2}{2}+\dfrac{x^4}{4}\), \(-\infty<x<\infty\). (a) Find the points where the potential vanishes. (b) Find the points of equilibrium. (c) Determine which are stable and which unstable. (d) Sketch \(V(x)\). (e) Sketch the force \(F(x)\).
Solution. (a) \(V=\dfrac{x^2}{4}\left(x^2-2\right)=0\) gives \(x=0\) and \(x=\pm\sqrt2\).
\(V'(x)=-x+x^3=x(x^2-1)=0\) gives the three equilibrium points \(x=0\), \(x=1\) and \(x=-1\).
\(V''(x)=-1+3x^2\). At \(x=\pm1\), \(V''=2>0\), so these are stable equilibria (minima), with \(V(\pm1)=-\frac14\). At \(x=0\), \(V''=-1<0\), so the origin is an unstable equilibrium (a local maximum), with \(V(0)=0\).
The potential is even in \(x\) and rises without limit as \(\lvert x\rvert\to\infty\). It therefore has the shape of a symmetric double well, with minima of depth \(-\frac14\) at \(x=\pm1\) separated by a hump at the origin (Figure 5.4).
Figure 5.4 The double-well potential \(V=-x^2/2+x^4/4\).
The force is \(F(x)=-V'(x)=x-x^3\), which vanishes at \(x=0,\pm1\). Its stationary points follow from \(F'(x)=1-3x^2=0\), that is \(x=\pm1/\sqrt3\). Since \(F''(x)=-6x\), the point \(x=1/\sqrt3\) is a maximum and \(x=-1/\sqrt3\) a minimum of the force, with
The force is odd in \(x\), tends to \(-\infty\) as \(x\to+\infty\) and to \(+\infty\) as \(x\to-\infty\), and crosses zero with negative slope at the stable points \(x=\pm1\) and with positive slope at the unstable point \(x=0\) (Figure 5.5).
Figure 5.5 The force \(F=x-x^3\) derived from the potential of Figure 5.4.
Example 5.3 A particle moves in the potential \(V(x)=\dfrac{x^2}{2}-\dfrac{x^3}{3}\), \(-\infty<x<\infty\). Find and classify the equilibrium points, and sketch the potential and the force.
Solution. The potential \(V=\dfrac{x^2}{6}(3-2x)\) vanishes at \(x=0\) and \(x=\frac32\). Equilibrium requires \(V'(x)=x-x^2=x(1-x)=0\), so \(x=0\) or \(x=1\). Since \(V''(x)=1-2x\), we have \(V''(0)=1>0\), a stable equilibrium at the minimum \(V(0)=0\), and \(V''(1)=-1<0\), an unstable equilibrium at the maximum \(V(1)=\frac12-\frac13=\frac16\)
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