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Consider an infinite square grid, where each side of a square is a spring following Hooke's law, with spring constant $k$.

What is the relation between the force and displacement between two points? If they are proportional, what is the equivalent spring constant between the origin and the point $(x,y)$ (integers) ?

Edit 1: I also want to know this: Suppose you make the springs so small that this can be treated as a continuous sheet, at what speed will a wave propagate? Assuming a wave starting as an initial displacement perpendicular to the sheet.

Given some initial state, is there an equation for the time-evolution of the continuous sheet?

Edit 2: Suppose there is a mass at every node, and its $(x,y)$-coordinates is fixed, it only vibrates out of the plane. Consider that we take the continuous limit, such that we get a 2D membrane of mass density $\mu$.

  1. Is the membrane isotropic?
  2. Suppose we use another tiling (like hexagonal) before taking the continuous limit, will this sheet behaves the same way?
  3. If not, but they are both isotropic, how does one characterize their difference, can they be made to behave the same way by changing the spring constant $k$?
  4. What is the equation of motion for the square sheet with spring constant $k$?
  5. What is the equation of motion for the square sheet if the springs obey a generalized Force law, $F=kx^n$, where $n$ is a variable.
  6. What is the equation of motion for a 3D cubic grid?

I am particularly interested in answers to 1., 2. and 3. I dont expect anyone to answer all these and will also accept an answer which does not explain anything but simply provides a good reference.

Qmechanic
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TROLLHUNTER
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  • I'm not sure I understand this. If you just pick two points, you're not getting enough information. You need to know the displacements of every node connected to a given point in order to find the force on that point. – Mark Eichenlaub Dec 21 '10 at 13:52
  • You displace a node(corner) relative to another by applying a force to the two nodes, paralell to the line between them, measure the displacement, and the force applied, are they proportional? – TROLLHUNTER Dec 21 '10 at 14:01
  • Okay. A solution will only be for a static situation, though. – Mark Eichenlaub Dec 21 '10 at 14:05
  • I think this is related. – Raskolnikov Dec 21 '10 at 14:17
  • I think Raskolnikov is right; there is definitely some equivalence... – Noldorin Dec 21 '10 at 14:24
  • @kalle43 All movement/displacements happens on a plane correct? Can you confirm this is a 2D problem? – Sklivvz Dec 21 '10 at 14:36
  • @kalle43 what is the rest length of the springs? is there any tension without displacement? – Sklivvz Dec 21 '10 at 14:39
  • @Sklivz Yes, 2D. Rest length doesnt matter. No tension. – TROLLHUNTER Dec 21 '10 at 14:45
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    This is an infinite dimensional problem. Each connection between springs is roughly one degree of freedom. You can analyse the equilibrium case, by assuming all springs must displace equally (thus reducing the nº of degrees to 1). – Malabarba Dec 21 '10 at 15:06
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    sniping physicists! – Jeremy Dec 21 '10 at 15:28
  • @kalle43: what is the system "attached" to? what is allowing me to stretch/compress the lattice around a point (instead of having the net effect of having the whole lattice simply translate)? Should we assume that the system is fixed at the origin? – Sklivvz Dec 21 '10 at 15:43
  • @sklivvz: I think he means that you apply force to two points equally on opposite directions, so the lattice won't move. Or you could say the lattice is fixed at one point, and force is applied at another point. Either way, won't affect the effective spring constant. – Malabarba Dec 21 '10 at 16:02
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    This looks like a model for a crystal except it can't be stable like this. You can continuously deform the angle between the sides of the square into a rhombus and eventually (letting the angle go to zero) obtain a one-dimensional model. Reasonable crystal models also add diagonal interaction into the square that makes this angle deformation disadvantageous (because you would make the diagonal spring twice longer if you'd collapse the square). – Marek Dec 21 '10 at 18:23
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    @Kalle43 You should simply your question, not complicate it. It is a very complex system you suggested. Reduce it to manageable limits, state assumptions, describe the grid, and you might get your answer. – Malabarba Dec 21 '10 at 19:47
  • @kalle43: either the problem is two dimensional, and therefore there is no vertical displacement, and all waves (if andy) are transversal, or the system is three dimensional. The equations are going to be very different... – Sklivvz Dec 21 '10 at 21:02
  • Minor correction: you'd make the diagonal spring $\sqrt2$ times longer, of course. – Marek Dec 21 '10 at 21:15
  • Let's get some guys from Stackoverflow to create a simulation of this and see what happens! – wrongusername Dec 30 '10 at 21:12

3 Answers3

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I'll answer only the third one (for now at least); the movement with limit to small vertical oscillations will be governed by the drum equation:

$\ddot{s}(x,y)=c^2 \nabla^2 s(x,y)$

where $s(x,y)$ is a vertical displacement in point $(x,y)$ and $c$ is the weave speed; using dimensional analysis I would say that $c\sim\sqrt{\frac{k}{\sigma}}$, where $\sigma$ is the mass density. Of course everything is getting much more complex with larger amplitudes.

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    I don't believe this is really correct. The wave equation is derived from the assumption that nodes oscillate around stable equilibrium positions. This is not the case in the spring model which can be also arbitrarily deformed (i.e. it is not rigid). – Marek Dec 21 '10 at 19:38
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    @marek please find a copy of Goldstein's mechanics. If it is one of the newer editions (with Poole and Safko) then have a look at Ch. 13 "Introduction to Lagrangian and Hamiltonian Formulations for Continuous Systems and Fields." If you find any inconsistencies in that treatment then please post a question because that would be big news. –  Dec 21 '10 at 20:16
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    @Marek This is small deformation approximation. –  Dec 21 '10 at 20:36
  • @space_cadet: I'll try to look it up. – Marek Dec 21 '10 at 21:06
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    @mbq: okay, but then it should be noted that it's not valid (at least not obviously) because you can always deform the square into a rhombus without expending energy in this model. While the small approximations are usually done around stable solutions that you can't deform in this way. – Marek Dec 21 '10 at 21:08
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    @Marek Small vertical deformation than. However, those horizontal motions in linear approximation will also have a speed proportional to $\sqrt{\frac{k}{\sigma}}$. –  Dec 21 '10 at 22:49
  • @mbq: yeah, certain modes will probably still make sense. – Marek Dec 21 '10 at 23:03
  • @mbq Why would it get more complex for larger amplitudes? This is exactly the equation I arrived at for the continuum limit of 2D square grid of springs. – TROLLHUNTER Jan 12 '11 at 12:03
  • @kakemosteret The obvious problem is that the description of the surface as $s(x,y)$ may break -- think of a initial condition when one picks some point up and then horizontally until it wraps over the other part of surface. –  Jan 12 '11 at 12:29
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I stick to the first question.

If you only do small displacements, and the two points are along the same line of springs then the effective spring rate is

$$ k_{eff} = \frac{k}{N} $$

where $N$ is the number of springs between the points. Why? Well split the problem like this

(inf)---[k_out]---(A)---[k_in]---(B)---[k_out]---(inf)

where (A) and (B) are the two points, and the springs are replaces with the effective springs [k_out] between the points and infinity, and [k_in] between the two points. The formula for springs in series is $\frac{1}{k_{eff}} = \frac{1}{k_1}+\frac{1}{k_2}+\ldots$, or $k_{eff}=k/N$ if all the springs have the same rate. So [k_out] is zero because $N=\infty$ and whats left to consider is only the springs in-between the points.

Note that the springs out of the line of the points are un-important for small displacements because they only contribute higher order non-linearities.

Completely different equations are needed for the continious sheet. The wave speed has to do with the mass/density of the sheet also, not just the elasticity and the sitffness.

John Alexiou
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If on every node of the grid you have a small mass, then you have a model for a two dimensional solid. That would behave like a two dimensional membrane. The equation of motion for every disturbance would be a wave equation. In the case of an one-dimensional grid, the wave velocity for such a wave would be

$$c^2=\frac{kl^2}{m}$$

where l is the distance between two neighbouring masses. In the case of the two dimensional grid, you will probably also have a geometric factor.

Update: Regarding the last edit, if you have a tension that characterizes the membrane, then the velocity is the square root of the tension over the mass density. So the geometry of the thing would play some part both to the tension and the mass density. That is because if you change the shape of the cell, then you assign different surface for every mass and you also assign a different number of brunches with springs to every node thus changing the effective spring constant. These are my qualitative guesses.

Vagelford
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    "then you have a model for a two dimensional solid" -> no you don't. See my comment under the question. In short: this model is not stable to perturbation that collapses it to 1D and so it is essentially just a 1D model. – Marek Dec 21 '10 at 19:34
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    How does that happen for an infinite grid? Anyway, that doesn't matter for the wave velocity. – Vagelford Dec 21 '10 at 20:09
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    ""then you have a model for a two dimensional solid" -> no you don't", yes you do @marek. I took a look at your comment and I failed to understand your argument. –  Dec 21 '10 at 20:11
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    @space_cadet: no you don't :-) @Vagelford: the same way it does for square. These things are best illustrated with a picture. All I can say is: think about the same model but with the elementary face being a rhombus instead of a square. You can continuously deform these models into each other. Therefore these can't be good models for a solid because those models need to be... well, solid :-) – Marek Dec 21 '10 at 21:13
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    Yes, but you would have to do it simultaneously for all the infinite grid, so a small perturbation can't do it. Anyway, I didn't say it is a good grid model. It is a good enough for small oscillations. – Vagelford Dec 21 '10 at 21:28
  • @Vagelford: I am not sure what do you consider small perturbation but seeing that you are talking about waves that means a small nudge of every node from its equilibrium position. In this sense you can surely move every node just a little in a compatible manner to make every square into rhombus. This is completely same as what you do to move every node a little here and there for wave modes. And those "deformation modes" are more important because they don't require any energy (so they don't freeze even as you approach $0K$). – Marek Dec 21 '10 at 22:58
  • @Vagelford: well, it's definitely not a good model and I am saying it's not even a model :-) You have to explicitly fix the equilibrium square positions in a lattice to get rid of deformation modes to make any use of this as 2D model. And this feature makes it pretty bad both for analytical and numerical treatment. – Marek Dec 21 '10 at 23:00
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    A small perturbation is a small displacement of a small group of particles (localized). Since we are talking about an infinite grid, it would take infinite energy to move every node and make every square into rhombus. – Vagelford Dec 21 '10 at 23:52
  • @Vagelford: sigh... and how big do you think the energy of the wave modes will be? I just can't understand why you are putting this under the rug. It's as good a mode as any other. If you are allowing waves, you have to allow deformations also, no way around it. – Marek Dec 22 '10 at 10:25
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    As big as the energy you put in to the system. If I pull up (or right or left) one of the masses and let it oscillate, it could not cause a deformation of the type you are describing. The question is, what type of waves do you have in mind? If you have a sin wave in mind, then it is unrealistic since it would have infinite energy. For the type of deformations I have in mind, an infinite grid is analogues to having a rigid support at infinity. – Vagelford Dec 22 '10 at 12:50
  • @Vagelford: okay, I have to agree you have a point if you are talking about realistic perturbations. I was talking about theoretical side. E.g. in EM there is no problem using monochromatic waves even though they carry infinite energy. On the contrary, it is almost always best to work with these unrealistic modes. – Marek Dec 22 '10 at 22:49
  • I am all for unrealistic... :P – Vagelford Dec 23 '10 at 00:09