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I am reading this paper.
https://www.nature.com/articles/s41467-019-12599-3
The authors derive the equations governing a 1-D mass and spring system with non-reciprocal springs. The final equations, where $u$ is displacement,

$$\frac{1}{c}\frac{d^2u}{dt^2}-\frac{d^2u}{dx^2}+\frac{2\epsilon}{p}\frac{2du}{dx} = 0. \tag{1}$$

The equation is derived in the Methods section. The first steps are just a simple application of Newton's second law to a discrete system, and the following equation is obtained, where $u_j$ is displacement of mass $j$.

$$m\frac{d^2u_j}{dt^2} + k(1+\epsilon)(u_j-u_{j-1})+k(1-\epsilon)(u_j-u_{j+1})=0 .\tag{2}$$

Then the authors consider the continuum limit. $u_j$ becomes $u(x)$, which makes sense. For the displacement of the adjacent masses however, the equation below is presented,
$u_{j\pm1} = u(x) \pm p\frac{du}{dx}+\frac{p^2}{2} \frac{d^2u}{dx^2} \tag{3}$
($p$ is called the rest length)

  1. What is the reasoning behind the transformation above? I get that something has to be substracted/added to $u(x)$, but I don't understand where the second and the third terms come from.

  2. Also the authors state about equation (2),

In such a model, Newton’s action–reaction third law is broken, which means that in practice one needs to add local momentum at each site $j$ to realize such a system.

Why is the third law broken, and what is meant by "adding a local momentum"?

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1 Answer 1

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Your equation $$u_{j\pm1} = u(x) \pm p\frac{du}{dx}+\frac{p^2}{2} \frac{d^2u}{dx^2}$$ just looks like a second order Taylor polynomial. You are using this polynomial to approximate what $u_{j\pm1}$ is from $u_j$, where $p=\Delta x$. i.e. $u_{j\pm1}=u(x\pm p)$

Newton's Third Law is broken here because the system is non-reciprocal. i.e. the springs pull more in one direction than in the other direction. The force of mass $j$ on mass $j+1$ is not equal to the force mass $j+1$ exerts on mass $j$ (through the spring).

Of course, Newton's third law does hold in classical systems like these. So if you actually wanted to make a system like they describe here, you would need to "add momentum" (include additional forces) so that the mass interactions through the springs behave like they do in the model with Newton's third law still bring true overall.

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