# Why is F=dU/dx=0 either side of inflexion point?

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In Tipler & Mosca 5th edition p173 it defines neutral equilibrium as a point in a U-x curve where ##\frac{dU}{dx}=0## and also ##\frac{dU}{dx}=0## for a small displacement either side of the point. However I do not understand why ##\frac{dU}{dx}## remains ##0## either side of the inflexion point. Surely the gradient of an inflexion point is just ##0## at the inflexion point itself, and either side of it the gradient is nonzero. If this is indeed the case, then when a particle is displaced either side of an inflexion point, even by a very small amount, there will immediately be a nonzero force on it and hence it will immediately lose equilibrium, meaning that ##\frac{dU}{dx}\ne 0##.

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In Tipler & Mosca 5th edition p173 it defines neutral equilibrium as a point in a U-x curve where ##\frac{dU}{dx}=0## and also ##\frac{dU}{dx}=0## for a small displacement either side of the point. However I do not understand why ##\frac{dU}{dx}## remains ##0## either side of the inflexion point. Surely the gradient of an inflexion point is just ##0## at the inflexion point itself, and either side of it the gradient is nonzero. If this is indeed the case, then when a particle is displaced either side of an inflexion point, even by a very small amount, there will immediately be a nonzero force on it and hence it will immediately lose equilibrium, meaning that ##\frac{dU}{dx}\ne 0##.

I suspect they are looking at a Taylor expansion of ##U## about an inflexion point. Remember that for an inflexion point the second derivative is zero, hence ##U## is approx constant (to second order in ##x##). Hence ##\frac{dU}{dx}## is apprx zero to first order in ##x##.

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In Tipler & Mosca 5th edition p173 it defines neutral equilibrium as a point in a U-x curve where ##\frac{dU}{dx}=0## and also ##\frac{dU}{dx}=0## for a small displacement either side of the point. However I do not understand why ##\frac{dU}{dx}## remains ##0## either side of the inflexion point.

The equilibrium point is a point where the first derivative is zero. An inflection point is a point where the second derivative is zero.

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In Tipler & Mosca 5th edition p173 it defines neutral equilibrium as a point in a U-x curve where ##\frac{dU}{dx}=0## and also ##\frac{dU}{dx}=0## for a small displacement either side of the point. However I do not understand why ##\frac{dU}{dx}## remains ##0## either side of the inflexion point. Surely the gradient of an inflexion point is just ##0## at the inflexion point itself, and either side of it the gradient is nonzero. If this is indeed the case, then when a particle is displaced either side of an inflexion point, even by a very small amount, there will immediately be a nonzero force on it and hence it will immediately lose equilibrium, meaning that ##\frac{dU}{dx}\ne 0##.

That's a definition of neutral equilibrium.

If you displace the system slightly but there is a restoring force which brings it back to the equilibrium point, that's stable equilibrium. Most equilibrium situations of interest have this character, for instance simple harmonic motion.

If you displace the system slightly and the resulting force takes it farther away from equilibrium, that's unstable equilibrium. For instance, something you've managed to balance on the top of a sphere.

If you displace the system slightly and there is no force at all, so the new point is also an equilibrium point, that's neutral equilibrium. The above is the definition of this condition. It doesn't say all equilibria are of this nature, it's defining a particular situation. An example is an object experiencing neutral buoyancy. You can move it around and it still stays in the same location. An even simpler example is an object sitting on the ground.

Surely the gradient of an inflexion point is just ##0## at the inflexion point itself, and either side of it the gradient is nonzero.

You are restricting your thinking to a very narrow class of functions. Here's a counter example. Take a parabola. Slice off the bottom of it, replacing the curved portion with a horizontal line.

That function has the property that there is a finite region over which all derivatives are 0. If it represents a potential, there are no forces anywhere in that flat region.

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