I have been looking at taking the continuum limit for a linear elastic rod of length $l$ modeled by a series of masses each of mass $m$ connected via massless springs of spring constant $k$. The distance between each mass is $\Delta x$ which we use to express the total length as $l=(n+1)\Delta x$. The displacement from the equilibrium position is given by $\phi(x,t)$.
The discrete Lagrangian in terms of the $i$th particle $\mathscr L$ is composed as follows,
\begin{equation} \mathscr L=\frac{1}{2}\sum _{i=1}^{n}m\dot \phi _i^2-\frac{1}{2}\sum ^n _{i=0}k(\phi_{i+1}-\phi _i)^2 \end{equation}
At this point we take the continuum limit such that the number of masses in the fixed length of rod tends to infinity and correspondingly the inter-particle distance tends to zero. It is fruitful to multiply top and bottom by $\Delta x$ such that we can define two quantities that remain constant during this limit namely the linear density ($\mu=m/\Delta x$) and the elastic modulus ($\kappa=k\Delta x$).
\begin{equation} \mathscr L=\frac {1}{2} \sum _{i=1}^{n}\Delta x\bigg(\frac{m}{\Delta x}\bigg)\dot {\phi} _i^2-\frac {1}{2} \sum _{i=0}^{n}\Delta x(k\Delta x)\bigg(\frac{\phi _{i+1}-\phi _i}{\Delta x}\bigg)^2 \end{equation}
It is easy to see why the linear density remains constant since both the number of masses per unit length increases while simultaneously the unit length decreases.
However my question is regarding the elastic modulus, I fail to see how it remains constant in this limit.
The argument goes as follows; Since the extension of the rod per unit length is directly proportional to the force exerted on the rod the elastic modulus being the constant of proportionality. The force between two discreet particles is $F_i=k(\phi _{i+1}-\phi _i)$, the extension of the inter particle spacing per unit length is $(\phi _{i+1}-\phi _i)/\Delta x$. Therefore (HOW) $\kappa=k\Delta x$ is constant. Its easy to relate them but why is it constant!?!
Answer
Honestly, I think this is one of those cases where you should just accept it and push on. This 'derivation' is really nothing more than a pedagogical device to make field theory seem somewhat natural to students with a background in classical mechanics.
What we are trying to do is to take the continuum i.e. $N\to \infty$ limit of the following Lagrangian:
$$L_N=\frac{1}{2} \Biggl(\sum_{i=1}^N\Delta x \frac{m}{\Delta x} \dot{\phi_i}^2-\sum_{i=1}^{N-1}\Delta x\ k\Delta x \biggl[\frac{\phi_{i+1}-\phi_i}{\Delta x}\biggr]^2\Biggr) $$
define $\mu=\frac{m}{\Delta x}$ and $Y=k\Delta x$
Clearly, for a continuum limit, we get infinitely many particles, so the total kinetic energy of the system should diverge... unless we impose (or put in by hand, as they call it), that $\mu$ remains constant, not $m$. Similarly, it is obvious that the equilibrium force of each spring $F=k \Delta x$ should vanish... unless we impose that $k\Delta x$ is constant when we take our limit. With these ad-hoc assumptions, and replacing the discrete index $i$ with a continuous spatial coordinate, we get
$$L\equiv \lim_{N\to \infty}L_N=\frac{1}{2}\int_0^l \mathrm{d}x \biggl(\mu\dot \phi^2 -Y(\nabla\phi)^2\biggr)$$ This gives us the right action for a free, massless, scalar field \begin{align*}S[\phi]&=-\frac{Y}{2}\int_0^l \mathrm{d}x\ \mathrm{d}t \biggl(-\frac{\mu}{Y}\dot \phi^2+(\nabla \phi)^2\biggr)\\ &=-\frac{\mu c^2}{2} \int_0^l \mathrm{d}x\ \mathrm{d}t \biggl(-\frac{1}{c^2}(\partial_t\phi)^2+(\nabla\phi)^2\biggr) \hspace{2cm}c=\sqrt{\frac{Y}{\mu}}\\ &=-\mu c^2\int_0^l\mathrm{d}^2x\ \frac{1}{2}\eta^{\mu\nu}\partial_\mu\phi\partial_\nu\phi\end{align*}
The definition of $c$ is the standard one for the speed of longitudinal waves, and as one can see this Lagrangian is also reminiscent of the action for a relativistic point particle (especially the prefactor). This is, of course, a very nice result, so we can be happy about the way we took our limit, even if we had to make some ad-hoc assumptions.
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