# Hamiltonian of polymer chain

I'm reading up on classical mechanics. In my book there is an example of a simple classical polymer model, which consists of N point particles that are connected by nearest neighbor harmonic interactions. The Hamiltonian of this system is:

$$H = \sum\limits_{i=1}^N \frac{\vec{p}_i^2}{2m} + \frac{1}{2}\sum\limits_{i=1}^{N-1} m \omega^2(|\vec{r}_i - \vec{r}_{i+1}| - b_i)^2,$$

where $p$ stands for momentum, $r$ for position and $m$ for the mass of the respective particle in the chain. Now comes the part that I don't understand: $b_i$ is the equilibrium bond length. I don't understand why one subtracts the bond length from the distance between two neighboring particles in the chain in the above formula. What is the physical meaning of doing this?

I have included the picture below for visualization of the problem. In my case, $k_1 = k_2 = k_3 = k_4 = k$ and also all masses are equal to m. In my understanding, the $b_i$ (equilibrium bond length) would be the black spring (spiral line between two circles) in the picture below.

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Each point of mass has a "default" position where the springs are relaxed, and they don't exert force on the masses. That's when the springs have the length $b$, thus, the distance between $r_i$ and $r_{i+1}$ is $b$. This in turn means
$$|r_i-r_{i+1}|-b_i=0.$$
It's easier to understand if you define $d_i=|r_i-r_{i+1}|$, then the above formula reads $(d_i-b_i)^2$, which means that the potential energy is zero if the actual spring length is equal to the equilibrium spring length.
Furthermore, I would use $r_{i+1}-r_i$ instead of $r_i-r_{i+1}$, because you can get rid of the absolute value $||$ if you define $r_{i+1}>r_i$ (which is the standard case).