What’s the Shape of a Hanging Chain?

The principle of least action stands superior to both [conservation of energy and conservation of momentum], even when considered together, and it appears to govern all the reversible processes of Nature — Max Planck, The Principle of Least Action, 1915.

Consider a chain of constant length {\ell} and linear mass density {\rho}, suspended at its endpoints {(x_0,y_0)} and {(x_1,y_1)} in a homogeneous gravitational field. What function {y(x)} has a graph that looks like this hanging chain?

A hanging chain
A hanging chain

There are many ways to answer this question, some elementary, others quite sophisticated. Here we solve it using Lagrange multipliers and the principle of least action. Along the way we will encounter a few important notions of theoretical physics. The whole idea of the solution presented below can be traced back to the following hint: “The chain’s center of mass is as low as possible.”

Let {g} denote the gravitational acceleration. As a first step, we derive a continuum version of the well-known potential energy formula

\displaystyle U=(\text{mass})\times\text{(gravitational acceleration)}\times\text{(height)}.

For this let’s chop up the chain into {n} little pieces, i.e. take a partition of the interval {[x_0,x_1]}:

\displaystyle x_0=:x^0<x^1<\dots<x^{n-1}<x^n:=x_1.

Each piece can be approximated by a line segment, hence we get a nice estimation of their masses

\displaystyle m_i\approx\rho\sqrt{(\Delta x_i)^2+(\Delta y_i)^2},

where {\Delta x_i=x^i-x^{i-1}} and {\Delta y_i=y(x^i)-y(x^{i-1})}, {i=1,\ldots,n}. Now, if we put each piece’s mass into its, say, right endpoint {h_i:=y(x^i)} we get the potential energy of our discretized chain

\displaystyle U[y]\approx\sum_{i=1}^nm_igh_i=\sum_{i=1}^ng\rho\,y(x^i)\sqrt{1+\bigg(\frac{\Delta y_i}{\Delta x_i}\bigg)^2}\Delta x_i.

By sending the mesh of partition to zero, that is taking the limit {\max\{\Delta x_i\}\rightarrow 0}, we arrive at the integral

\displaystyle U[y]=\int_{x_0}^{x_1}g\rho\,y(x)\sqrt{1+(y'(x))^2}\,dx.

This functional is the potential energy of the chain. Our goal is to minimize {U} under the constraint of fixed length, i.e.

\displaystyle \int_{x_0}^{x_1}\sqrt{1+(y'(x))^2}\,dx=\ell.

The method of Lagrange multipliers comes to mind in which (instead of {U}) another functional is introduced with the constraint built into it using a real parameter the so-called multiplier. The extrema of this new functional is a solution of the original (constrained) problem. The new functional has the form

\displaystyle S[y,\lambda]:=U[y]+\lambda\bigg(\int_{x_0}^{x_1}\sqrt{1+(y'(x))^2}\,dx-\ell\bigg),

or equivalently

\displaystyle S[y,\lambda]=\int_{x_0}^{x_1}\bigg[\big(g\rho\,y(x)+\lambda\big)\sqrt{1+(y'(x))^2}\bigg]\,dx-\lambda\ell.

The variation of the action functional {S} is zero along the true evolution of a physical system. This is Hamilton’s principle and in this context

\displaystyle L(y,y',x,\lambda):=\big(g\rho\,y+\lambda\big)\sqrt{1+(y')^2}

is what’s called the Lagrangian. The corresponding Euler-Lagrange equation can be written as

\displaystyle \frac{d}{dx}\frac{\partial L}{\partial y'} = \frac{\partial L}{\partial y}.

The Lagrangian {L} does not depend on {x} explicitly, hence Noether’s theorem tells us that

\displaystyle Q:=\frac{\partial L}{\partial y'}y'-L

is a conserved quantity — a Noether charge. Indeed, it can be easily checked that

\displaystyle \frac{d}{dx}Q=0.

Therefore the variation of {y} leads us to

\displaystyle Q=\frac{\partial L}{\partial y'}y'-L=(-c),

with some constant {c}. Differentiating {L} with respect to {y'} we get the following differential equation

\displaystyle \frac{g\rho y+\lambda}{\sqrt{1+(y')^2}}=c.

Since {c=0} is a special case of {y} being constant, we can assume that {c} is non-zero. Then the above equation can be rearranged as

\displaystyle \frac{Y}{\sqrt{1+\bigg(\dfrac{c}{g\rho}Y'\bigg)^2}}=1,\quad\text{with}\ Y(x):= \frac{g\rho y(x)+\lambda}{c}.

Recalling some well-known identities of the hyperbolic functions {\sinh} and {\cosh}, viz.,

\displaystyle \cosh^2x=1+\sinh^2x,\qquad (\cosh x)'=\sinh x,

the general solution can be written as

\displaystyle Y(x)=\cosh\bigg(\frac{g\rho}{c}x+c_1\bigg),

where {c_1} is the constant of integration. Therefore we find that the shape the hanging chain is given by

\displaystyle \boxed{y(x)=\frac{c}{g\rho}\cosh\bigg(\frac{g\rho}{c}x+c_1\bigg)-\frac{\lambda}{g\rho}}

This curve is called the catenary (latin for chain).
Given that the constants {g} and {\rho} are known, there are {3} parameters, namely {c}, {c_1}, and {\lambda}, to be determined. These parameters are uniquely fixed by the boundary conditions (suspension points, length of chain), although cannot be expressed by explicit formulae. Why not?

Well, notice that we have the freedom of choosing the origin of our coordinate system. Assume it can be positioned such a way that the chain’s lowest point has coordinates {(0,cg/\rho)}. Then the solution takes the somewhat simpler form

\displaystyle y(x)=\frac{c}{g\rho}\cosh\bigg(\frac{g\rho}{c}x\bigg),

where {c} is the only unknown parameter. The boundary conditions {y(x_0)=y_0}, {y(x_1)=y_1} give us the following equation

\displaystyle y_1-y_0=\frac{c}{g\rho}\bigg[\cosh\bigg(\frac{g\rho}{c}x_1\bigg)-\cosh\bigg(\frac{g\rho}{c}x_0\bigg)\bigg] \ \ \ \ \ (*)

while the length constraint turns into

\displaystyle \ell=\frac{c}{g\rho}\bigg[\sinh\bigg(\frac{g\rho}{c}x_1\bigg)-\sinh\bigg(\frac{g\rho}{c}x_0\bigg)\bigg]. \ \ \ \ \ (**)

The quantity {\sqrt{\ell^2-(y_1-y_0)^2}} can be expressed using (*), (**), and some additional identities of the hyperbolic functions

\displaystyle \sqrt{\ell^2-(y_1-y_0)^2}=\frac{2c}{g\rho}\sinh\bigg(\frac{g\rho}{2c}(x_1-x_0)\bigg).

This is a transcendental equation, that is {c} cannot be found algebraically. However, some elementary calculus shows us that there is a unique {c>0} solution, thus numerical methods can be applied to get a satisfying approximate value.

The parameter c can be “read off”.

There are lots of simulations of catenaries on the internet. Here is two of them: A Java Applet and a Wolfram Demonstration

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s