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I am having trouble understanding the following diagram from pg. 18 of Arnold's book:

enter image description here

I am unable to see why local maxima of the potential energy correspond to unstable equilibria (and, reciprocally, why local minima correspond to stable equilibria).

My reasoning leads me to the opposite conclusion - for a local maxima $x_M$, $f'(x) > 0$ for $x < x_M$, so that things should increase to $x_M$, whereas $f'(x) < 0$ for $x > x_M$, so that things should decrease to $x_M$. However, I am being imprecise when I say "things", since I am not sure whether what I refer to is the same "thing" which is considered in the phase portrait.

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Force is equal to minus the potential gradient, ie $F=-\dfrac {dU}{dx}$.

Local minimum at $x_{\rm min}$

For $x<x_{\rm min}$, $\dfrac {dU}{dx}<0$, ie $F>0$ ie in positive x-direction

For $x>x_{\rm min}$, $\dfrac {dU}{dx}>0$, ie $F<0$ ie in negative x-direction

Thus a small displacement from $x_{\rm min}$ will result in a restoring force back to $x_{\rm min}$, ie stable equilibrium.

Local maximum at $x_{\rm max}$

For $x<x_{\rm max}$, $\dfrac {dU}{dx}>0$, ie $F<0$ ie in negative x-direction

For $x>x_{\rm max}$, $\dfrac {dU}{dx}<0$, ie $F>0$ ie in positive x-direction

Thus a small displacement from $x_{\rm max}$ will result in a restoring force away from $x_{\rm max}$, ie unstable equilibrium.

In terms of the phase diagram just consider what happens if a little more energy is added to the system.
Inside the thick black line "contour" the trajectory will still be around the position of the local minimum.
At the thick black line "contour" the trajectory will no longer be just around one of the local minima.

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  • $\begingroup$ Thank you for your response! If small displacements from $x_{min}$ result in restoring force back to $x_{min}$, then intuitively for me, trajectories should spiral into $x_{min}$ (rather than orbiting around it). Why doesn't this occur? $\endgroup$
    – algebroo
    Jun 5 at 7:56
  • $\begingroup$ I am sorry that I cannot give you a mathematical reason but I think that you are missing the point about the difference between the two representations. In the top graphical representation which you would have met first you can imagine an oscillation about the $x_{\rm min}$ position and passing through $x_{\rm min}$. In the bottom graphical representation the contour represents constant energy with energy interchanging between potential and kinetic.as one moves around the contour. $\endgroup$
    – Farcher
    Jun 5 at 8:04
  • $\begingroup$ Then note that there is a position of the contour when $x = x_{\rm min}$ and $|\dot x|$ is a maximum which corresponds to a maximum kinetic energy minimum potential energy state. Any "spiraling"towards the local minimum would indicate a loss of total energy to the system. $\endgroup$
    – Farcher
    Jun 5 at 8:07
  • $\begingroup$ That makes a lot of sense! Thank you. $\endgroup$
    – algebroo
    Jun 5 at 8:09
  • $\begingroup$ @algebroo Regarding "If small displacements from $x_\text{min}$ result in restoring force back to$x_\text{min}$, then intuitively for me, trajectories should spiral into $x_\text{min}$ (rather than orbiting around it)". Think of a pendulum: the restoring force always points towards the equilibrium point but the pendulum always bypasses it. Because it has momentum. The average position is at the equilibrium point, but the pendulum never stops there. Also, the moon doesn't fall on Earth – same reason. $\endgroup$
    – Themis
    Jun 12 at 0:12

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