0
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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 $d^2U/dx^2 = 0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.