Why is a 1d conservative force approximately linearly related to position?

[zenterix]

Homework Statement

I'd like to understand a simple derivation I found in a book that aims to show that a conservative 1d force is approximately linearly related to position.

Relevant Equations

The derivation uses a Taylor series of a potential function for the force.

There are a few statements and calculations which I don't fully understand.

Everything below is from this (free) book about Vibrations and Waves from MIT OCW's 8.03 course.

Consider a conservative force $F=-V'(x)$.

At a point of equilibrium, $x_0$, the force vanishes.

$$F=-V'(x_0)=0$$

Let $x_0=0$. We can write this because we are free to redefine the coordinate system.

Expand the force in Taylor series

$$F(x)=-V'(x)=-V'(0)-xV''(0)-\frac{1}{2}x^2V'''(0)+\ldots\tag{1}$$

$$=xV''(0)-\frac{1}{2}x^2V'''(0)+\ldots\tag{2}$$

The equilibrium is stable if $V''(x_0)>0$ as this means $x_0$ is a local minimum of potential energy.

For sufficiently small $x$, the second term in (2), and all subsequent terms will be much smaller than the first term in (2).

The third term is negligible if

$$|xV'''(0)|<<V''(0)\tag{3}$$

Typically, each extra derivative will bring with it a factor of $1/L$ where $L$ is the distance over which the potential energy changes by a large fraction.

Then, (3) becomes

$$x<<L\tag{4}$$

My questions are

1) How do we derive (3) exactly?

2) What does the quote above about the factor $1/L$ mean?

3) I think with answers to 1) and 2) I can answer this: where does (4) come from?

The point of the derivation above is to show that for small displacements from equilibrium, we obtain an approximately linear relationship between force and displacement, ie we get Hooke's law.

 

[zenterix]

Here are my thoughts about my questions.

We have a term $-\frac{1}{2}x^2V'''(0)$ and a term $-xV''(0)$.

The absolute values are

$$\left |\frac{1}{2}x^2V'''(0)\right |\tag{5}$$

$$|xV''(0)|\tag{6}$$

If (5) is smaller than (6) then

$$\left |\frac{1}{2}xV'''(0)\right |<|V''(0)|\tag{7}$$

By assumption, $V''(0)>0$ and so

$$\left |\frac{1}{2}xV'''(0)\right |<V''(0)\tag{8}$$

which looks like the expression I am asking about but there is an extra factor of $1/2$.

The symbol $<<$ is used instead of $<$ in the book.

$$\left |\frac{1}{2}xV'''(0)\right |<<V''(0)\tag{9}$$

means the lhs is way smaller than the rhs.

 

[jbriggs444]

For sufficiently small $x$, the second term in (2), and all subsequent terms will be much smaller than the first term in (2).

Pedant point...

This is not true for all possible force functions. For instance, there are pathological functions where the radius of convergence of the Taylor expansion is zero.

For instance, consider the potential $f(x) = e^{-1/x^2}$ for $x \ne 0$ and $f(x) = 0$ for $x = 0$. This function has a minimum at $x=0$.

Expand a Taylor series about that point and you will find that $0 = f(0) = f'(0) = f''(0) = f'''(0), \text{ etc}$ i.e. this function is not analytic.

 

[zenterix]

@jbriggs444 This is mentioned in the text

As you said, the exceptions are pathological functions. Would it not be the case that disregarding the argument being made about approximate linearity by demonstrating such exceptions is pedant?

My questions are about about the calculations for the non-pathological cases.