I am currently reading these Weigand's QFT lecture notes. But already in the beginning I dont understand one step. Given is the energy $E$ of a string of length $L$ of $N$ connected particles that oscillate in the transverse direction. The Amplitude of the oscillation of the $r^\text{th}$ particle is given by $q_r(t)$. $$\tag{1.4} E=\sum_{r=0}^N\frac12m\left(\frac{dq_r(t)}{dt}\right)^2+\frac TL(q_r^2-q_rq_{r-1}) $$ In the limit of large $N$ the $q_1(t),...,q_N(t)$ can be replaced by a single function $\phi(x,t)$ whose value is equal to the amplitude of the particle at $x$.The energy after the transformation becomes $$\tag{1.5} E=\int_0^L\frac12\frac ML\left(\frac{\partial\phi(x,t)}{\partial t}\right)^2+\frac12\frac MLc^2\left(\frac{\partial\phi(x,t)}{\partial x}\right)^2dx $$ The function that gets integrated over can be seen as an energy density.
I get the transformation of the first term with the time derivative. You simply have $m\simeq\frac ML\mathrm dx$ and $\phi(\frac rNL,t)=q_r(t)$.
But I have trouble understanding the transformation of the second term. Here are my thoughts so far
$$\tag1
\frac{\partial\phi(x,t)}{\partial x}\simeq\frac{\phi(x,t)-\phi(x-\delta x,t)}{\delta x}\simeq\frac{q_r-q_{r-1}}{\delta x}
$$
$$\tag2
\Rightarrow\frac12\frac MLc^2\left(\frac{\partial\phi(x,t)}{\partial x}\right)^2\mathrm dx\simeq\frac12\frac MLc^2\frac{q_r^2-2q_rq_{r-1}+q_{r-1}^2}{(\delta x)^2}\mathrm dx
$$
Now my thoughts are that the $\mathrm dx$ can get canceled by one of the $\delta x$. And at first I thought that I can approximate $q_{r-1}\approx q_r$ such that $\frac 12q_r^2+q_{r_1}^2\approx q_r^2$. But that would also mean that the off diagonal term would need to be rewritten $q_rq_{r-1}\approx q_r^2$. But this term needs to stay as it is. I hope someone has an idea of how this transformation can be justified.
