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 $\frac{dU}{dx}\ne 0$.


The stable or unstable equilibrium is the point at which $dU/dx=0$. Let's use:

$$ F = -\frac{dU}{dx} \tag{1} $$

to write it as $F=0$. Now suppose we move a small distance $\delta x$ from this position, then to first order the change in the force $F$ is:

$$ \delta F = \frac{dF}{dx} \delta x $$

and substituting using equation (1) we get:

$$ \delta F = \frac{d^2U}{dx^2} \delta x \tag{2} $$

For a neutral equilibrium we require $\delta F = 0$ and hence from (2) we get:

$$ \frac{d^2U}{dx^2} = 0 $$

which is of course the condition for a point of inflexion. The condition given in Tipler and Mosca:

$\frac{dU}{dx}=0$ for a small displacement either side of the point

is just the condition that $d^2U/dx^2 = 0$.

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