Consider a one dimensional [[Harmonic Oscillator chain]] as shown above. Such a system has a characteristic potential $V(q_{i+1}-q_i)=\frac{k}{2}(d_{i+1}-d_i)^2$ that corresponds to the pair-wise bonding between any two points on the chain. Here along the chain we have points with mass $m$ that may move left or right when subject to a perturbation. Otherwise the energy of the system is $0$ when the points are in their equilibirum positions - i.e. at distances $\Delta q^0$ from each other. Here we assume periodic boundary conditions (i.e. the chain is infinite or it forms a ring). # Perturbed Lagrangian We desire a Lagrangian corresponding to a perturbed mass on the chain and we take both the kinetic and potential energies at the equilibrium to be $0.$ Thus, reducing the model to a sequence of [1D harmonic oscillator](1D%20Harmonic%20Oscillator.md)s. The The kinetic energy of the perturbed system follows from the general formula for kinetic energy. Thus $T=\sum_i \frac{m}{2}\dot{q}_i^2 = \sum_i \frac{m}{2}\dot{d}_i^2$ and accounting only for the quadratic term due to the perturbation, the potential energy of the whole system is $V=\sum_i \frac{k}{2}(d_{i+1}-d_i)^2.$ This gives us the following Lagrangian (where we account for both nearest neighbors): $\mathcal{L}=\sum_i\bigg(\frac{m}{2}\dot{d}_i^2-\frac{k}{2}(d_{i+1}-d_i)^2+\frac{k}{2}(d_i-d_{i-1})^2\bigg),$ which should look like a the Lagrangian for a sum of 1D Harmonic oscillators. We may re-write the Lagrangian in terms of the equilibrium length, $\Delta q^0:$ $\mathcal{L} = \frac{1}{2}\sum_i \Delta q^0 \bigg[\frac{m}{\Delta q^0}\dot{d}_i^2-k\Delta q^0\bigg(\frac{d_{i+1}-d_i}{\Delta q^0}\bigg)^2+k\Delta q^0\bigg(\frac{d_i-d_{i-1}}{\Delta q^0}\bigg)^2\bigg]$$=\sum_i \Delta q^0\mathcal{L}_i$ We can define $\mu = k\Delta q^0$ since we notice that this is [Young's modulus](Young's%20modulus.md) for a [[continuous rod]] and rewrite the Lagrangian as $\mathcal{L} = \frac{1}{2}\sum_i \Delta q^0 \bigg[\frac{m}{\Delta q^0}\dot{d}_i^2-\mu\bigg(\bigg(\frac{d_{i+1}-d_i}{\Delta q^0}\bigg)^2+\bigg(\frac{d_i-d_{i-1}}{\Delta q^0}\bigg)^2\bigg)\bigg]$ ## Equations of motion The corresponding [Euler-Lagrange equation](Euler-Lagrange%20equation.md) at some point $i$ along the chain are $\frac{m}{\Delta q^0}\ddot{d}_i - \mu\bigg(\frac{d_{i+1}-d_i}{(\Delta q^0)^2}\bigg)+\mu\bigg(\frac{d_{i}-d_{i-1}}{(\Delta q^0)^2}\bigg)=0$ where $\mu = k\Delta q^0$ is [Young's modulus](Young's%20modulus.md). If we were to ignore $\mu$ and keep $k\Delta q^0$ in its place notice that the lengths $q^0$ cancel out and the equation of motion reduces to $m\ddot{d}_i=k(d_{i+1}-d_i)-k(d_i-d_{i-1})$ This means that for a given spring constant that is independent of $\Delta q^0$, the equations of motion are also independent of $\Delta q^0.$ #Mechanics #Mechanics/WaveMechanics #PhysicalExamples/PhysicalExamplesInMechanics