# 16.2: The Continuous Uniform Linear Chain

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The Lagrangian for the discrete lattice chain, for longitudinal modes, is given by equation $$(14.10.3)$$ to be

$L = \frac{1}{2} \sum^{n+1}_{j=1} \left( m\dot{q}^2_j - \kappa (q_{j-1} – q_j)^2\right)\label{16.1}$

where the $$n$$ masses are attached in series to $$n+1$$ identical springs of length $$d$$ and spring constant $$\kappa$$. Assume that the spring has a uniform cross-section area $$A$$ and length $$d$$. Then each spring volume element $$\Delta \tau = Ad$$ has a mass $$m$$, that is, the volume mass density $$\rho = \frac{m }{\Delta \tau}$$ or $$m = \rho \Delta \tau$$. Chapter $$16.5$$ will show that the spring constant $$\kappa = \frac{EA}{ d}$$ where $$E$$ is Young’s modulus, $$A$$ is the cross sectional area of the chain element, and $$d$$ is the length of the element. Then the spring constant can be written as $$\kappa = \frac{E\Delta \tau }{ d^2}$$. Therefore Equation \ref{16.1} can be expressed as a sum over volume elements $$\Delta \tau = Ad$$

$L = \frac{1}{2} \sum^{n+1}_{j=1} \left( \rho \dot{q}^2_j − E \left(\frac{q_{j-1} -q_j}{d} \right)^2 \right) \Delta \tau \label{16.2}$

In the limit that $$n \rightarrow \infty$$ and the spacing $$d = dx \rightarrow 0$$, then the summation in Equation \ref{16.2} can be written as a volume integral where $$x = jd$$ is the distance along the linear chain and the volume element $$\boldsymbol{\Delta} \boldsymbol{\tau} \rightarrow \mathcal{0}$$. Then the Lagrangian can be written as the integral over the volume element $$d\tau$$ rather than a summation over $$\Delta \tau$$. That is,

$L = \frac{1}{ 2} \int \left( \rho \dot{q}^2 − E \left(\frac{dq(x, t)}{ dx }\right)^2 \right) d\tau\label{16.3}$

The discrete-chain coordinate $$q(t)$$ is assumed to be a continuous function $$q(x, t)$$ for the uniform chain. Thus the integral form of the Lagrangian can be expressed as

$L = \frac{1}{2} \int \left( \rho \dot{q}^2 − E \left( \frac{dq(x, t) }{dx } \right)^2 \right) d\tau = \int \mathfrak{L}d\tau \label{16.4}$

where the function $$\mathfrak{L}$$ is called the Lagrangian density defined by

$\mathfrak{L} \equiv \frac{1}{2} \left( \rho \dot{q}^2 − E \left(\frac{dq(x, t) }{dx }\right)^2 \right) \label{16.5}$

The variable $$x$$ in the Lagrangian density is not a generalized coordinate; it only serves the role of a continuous index played previously by the index $$j$$. For the discrete case, each value of $$j$$ defined a different generalized coordinate $$q_i$$. Now for each value of $$x$$ there is a continuous function $$q(x, t)$$ which is a function of both position and time.

Lagrange’s equations of motion applied to the continuous Lagrangian in Equation \ref{16.4} gives

$\rho \frac{ d^2q}{ dt^2} − E \frac{d^2q}{ dx^2} = 0 \label{16.6}$

This is the familiar wave equation in one dimension for a longitudinal wave on the continuous chain with a phase velocity

$v_{phase} = \sqrt{\frac{E}{ \rho}} \label{16.7}$

The continuous linear chain also can exhibit transverse modes which have a Lagrangian density were the Young’s modulus $$E$$ is replaced by the tension $$\tau$$ in the chain, and $$\rho$$ is replaced by the linear mass density $$\mu$$ of the chain, leading to a phase velocity for a transverse wave $$v_{phase} = \sqrt{\frac{\tau}{\mu}}$$.

This page titled 16.2: The Continuous Uniform Linear Chain is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Douglas Cline via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.