Why are steps not visible in the continuum limit of the Frenkel Kontorova model?

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SUMMARY

The discussion centers on the Frenkel-Kontorova model's transition from a discrete to a continuum limit, specifically addressing why steps are not visible in the continuum limit. The model, which consists of many point particles, can be approximated to the sine-Gordon equation in the continuum limit. This approximation illustrates that while the discrete model shows steps, the continuum limit smooths these out, making them unobservable. The sine-Gordon equation describes wave propagation without discrete steps, confirming the transition from a particle-based model to a continuous wave model.

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LagrangeEuler
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http://www.iop.kiev.ua/~obraun/fk-intro.htm#model

Question? Why this model is discreet? As far as I see particles can have any position?

Why then is written in the text then In the continuum-limit approximation we come to the sine-Gordon (SG) equation?
 
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It's discrete in the sense that it's a model of many point particles. You can take a continuum limit. E.g., as the most simple case you can model a string first as a chain of harmonically coupled point particles. The continuum limit leads to the wave equation for the string.
 
Thanks a lot for the answer.

And for example in this paper

http://fmc.unizar.es/people/juanjo/papers/falo93.pdf

steps appear because model is discrete. In continuum limit it would be impossible to see steps in the ##v(F)## (FIG 1). Is it possible to explain why in continuum limit is impossible to see steps?
 
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