Understanding the generic nature of linearity in Physics

[zenterix]

Homework Statement

I'd like to understand an argument presented in the book "The Physics of Waves" by Howard Georgi.

Relevant Equations

The link to a free online version is below.

Here is a link to the book. This question is about the section between the end of page 6 to the start of page 9. That section discusses the "generic nature of linearity".

Let me go through the reasoning.

Suppose there is a particle moving along the $x$-axis with potential energy $V(x)$.

The gradient (in our case, just the simple derivative) of $V(x)$ is a conservative force.

$$F(x)=-\frac{d}{dx}V(x)$$

Suppose the force vanishes at a point of equilibrium $x_0=0$.

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

Expand the force in a Taylor series.

$$F(x)=-V'(x)=-V'(0)-xV''(0)-\frac{1}{2}V'''(0)x^2+\ldots\tag{1.23}$$

The first term, $-V'(0)$ vanishes by our earlier assumption.

The second term looks like Hooke's law with

$$K=V''(0)\tag{1.24}$$

The equilibrium is stable if the second derivative of the potential energy is positive, so that $x=0$ is a local minimum of potential energy..

The important point is that for sufficiently small $x$, the third term and all subsequent terms will be much smaller than the second.

The third term is negligible if

$$|xV'''(0)|<<V''(0)\tag{1.25}$$

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 (1.25) becomes

$$x<<L$$

Question 1: What is this $L$ that is mentioned?

There are only two ways that a force derived from a potential energy can fail to be approximately linear for sufficiently small oscillations about stable equilibrium.

1) Potential is not smooth (first or second derivative of potential not well defined at equilibrium point).

2) Even if the derivatives exist, it may happen that $V''(0)=0$. In this case, to have a stable equilibrium we must have $V'''(0)=0$ as well, otherwise a small displacement in one direction or the other would grow with time. Then the next term in the Taylor expansion dominates at small $x$, giving a force proportional to $x^3$.

Question 2: I don't really understand this point 2). We have $V'(0)=V''(0)=0$ and so the Taylor expansion becomes

$$F(x)=-\frac{1}{2}x^2V'''(0)-\frac{1}{6}x^3V^{(4)}(x)+\ldots$$

Suppose $V'''(0)> 0$.

Suppose that $x<0$, that is we are at a position below the equilibrium position $x=0$. Then, the force will be negative which means negative acceleration and the particle will move further and further away from equilibrium.

If $V'''(0)=0$ and we have the fourth term non-zero, $-\frac{1}{6}V^{(4)}(x)x^3$ then we don't have the same issue because if $x<0$ then force is positive and if $x>0$ then force is negative. Ie, the force is restoring to equilibrium.

I think this answers my question 2.

 

[zenterix]

I have another question about the example given to illustrate the reasoning above.

For example, consider the following potential energy

$$V(x)=E\left (\frac{L}{x}+\frac{x}{L}\right )\tag{1.27}$$

This is shown in the figure below

First of all, it is not clear to me what $E$ means in (1.27).

The minimum (at least for positive $x$) occurs at $x=L$ so we first redefine $x=X+L$ so that

$$V(X)=E\left (\frac{L}{X+L}+\frac{X+L}{L}\right )$$

The corresponding force is

$$F(X)=E\left (\frac{L}{(X+L)^2}-\frac{1}{L}\right )\tag{1.29}$$

Shouldn't the negative sign in (1.29) be a positive sign?

We can look near $X=0$ and expand in a Taylor series

$$F(X)=-2\frac{E}{L}\left (\frac{X}{L}\right )+3\frac{E}{L}\left (\frac{X}{L}\right )^2+\ldots$$

Now the ratio of the first nonlinear term to the linear term is

$$\frac{3X}{2L}$$

which is small if $X<<L$.

 

[haruspex]

First of all, it is not clear to me what $E$ means in (1.27).

Just some constant energy. In the context of SHM it would be related to the amplitude.

Shouldn't the negative sign in (1.29) be a positive sign?

No, it is right. Try it again.