# Taylor expansion in classical 1D harmonic chain (classical field theory)

I'm struggling with something apparently really easy but not entirely straightforward. In "Condensed Matter Field Theory" of Altland and Simons the classical 1D harmonic chain is treated as introductory example to classical field theory. Consider an harmonic chain of N particles on a periodic 1D lattice (N+1=1). The position of the I-th particle is $R_I=Ia$. The Lagrangian reads

$L=\sum_{I=1}^N[\frac{m}{2}\dot{R_{I}^2}-\frac{k_s}{2}(R_{I+1}-R_{I}-a)^2]$

Now he parametrizes $R(t)=R_I+\phi(t)$ where the deviation from equilibrium is small, so that

$L=\sum_{I=1}^N[\frac{m}{2}\dot{\phi_{I}^2}-\frac{k_s}{2}(\phi_{I+1}-\phi_{I})^2]$

Now he goes to the continuum limit, that is, he introduces continuum degrees of freedom $\phi(x)$ and applies a Taylor expansion such that:

$\phi_I\rightarrow a^{1/2}\phi(x)\left.\right|_{x=Ia} \hspace{30pt} \phi_{I+1}-\phi_I\rightarrow a^{3/2}\partial_x\phi(x)\left.\right|_{x=Ia}$

I can't figure out how this substitution and Taylor expansion works. If you look to the first, you should get for the second one:

$\phi_{I+1} - \phi_I\rightarrow a^{1/2}\left[\phi(x+a)-\phi(x)\right]\left.\right|_{x=Ia}$

How is this now equal to the derivative, should i take a limit for $a\rightarrow 0$? And where does the prefactor $a^{3/2}$ come from?

Cheers

• Remember the finite difference definition of a derivative: $df/dx = \lim_{\epsilon\to 0} (f(x+\epsilon)-f(x))/\epsilon$. – Michael Brown Sep 23 '13 at 9:16