Physics Asked by George Dixon on December 7, 2020

I have been given the following Lagrangian of a series of $N$ one-dimensional coupled oscillators, with distance a. I have also been given the boundary conditions: $y_0=0=y_{N+1},$ but have been given little explanation on where this came from.

$$L=sum_{i=1}^{N}frac{1}{2}mdot{y_i}^2-sum_{i=0}^Nfrac{1}{2}Tleft(frac{y_{i+1}-y_i}{a}right)^2,$$

I am having trouble understanding how this lagrangian has been formed, in particular; why there is an a on the denominator of the second squared term.

I understand that the Lagrangian is expressed:

$$L=KE-U,$$

The kinetic energy contribution of each of the $N$ coupled oscillators is simply:

$$KE=frac{1}{2}mdot{y_i}^2,$$

Where $y_i$ is the displacement of the $i$th oscillator.

Assuming all of the oscillators have equal mass, and none are fixed we should have the total kinetic energy:

$$KE=sum_{i=0}^{N}frac{1}{2}mdot{y_i}^2,$$

Generally, the potential energy of a spring is:

$$U=frac{1}{2}kx^2,$$

Where x represents the change in displacement, thus in our case the potential ‘contributed’ by each term is simply:

$$U=frac{1}{2}k(y_{i+1}-y_i^2),$$

because the terms in the bracket represent the ‘stretch’ of each of the springs, i.e. displacement.

I assume the person who gave me the original Lagrangian has given me a spring constant of $T$ rather than $k$, but surely this should return a lagrangian:

$$L=sum_{i=0}^{N}frac{1}{2}mdot{y_i}^2-sum_{i=0}^Nfrac{1}{2}Tleft(y_{i+1}-y_iright)^2,$$

This is clearly not the same as what I was given, the first summation index is out by one place, I can assume this means that we are considering the first oscillator to be fixed, and thus the first velocity term goes to zero leaving:

$$L=sum_{i=1}^{N}frac{1}{2}mdot{y_i}^2-sum_{i=0}^Nfrac{1}{2}Tleft(y_{i+1}-y_iright)^2,$$

Dividing by a would give the ratio of the change w.r.t the original length of the spring, so I don’t understand why this would be done? If anyone can see something that I am missing/done wrong, please let me know!

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