# Equation satisfied by a rubber sheet

1. May 7, 2013

### Septim

Greetings,

How can I derive the equation satisfied by a rubber sheet? I know that it reduces to Laplace's equation when the sheet is quite planar. I found the expression for it on p.113 in Introduction to Electrodynamics by Griffiths. Thanks in advance.

Edit: I have attached the expression as jpg file.

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Last edited by a moderator: May 9, 2013
2. May 7, 2013

### Simon Bridge

Salutations;
The usual approach would be to consider a free-body diagram for a small element of the sheet.

3. May 8, 2013

### Septim

Thanks for the answer. Can you elaborate more on that plese?

4. May 8, 2013

### Simon Bridge

It involves modelling the it as an array of small masses connected by springs.
The exact details depend on the geometry of your sheet and what you are doing with it - which you haven't said.

eg. a horizontal sheet of uniform thickness $t$ and density $\rho$ can be divided into small volume elements $dV = tdxdy$ each with mass $dm=\rho dV$. The one between (x,y) and (x+dx, y+dy) will be acted on by gravity is the -z direction and a bunch of spring forces along the sheet surface - the spring constants being a material property of the rubber. The directions and magnitudes of the spring forces depends on the nearby distortions of the sheet.

The equilibrium shape of the sheet will be one where all the forces for every volume element sum to zero - given some boundary condition (i.e. what is holding the sheet up?) You'd want to find this by minimizing the action across the sheet (look it up).

The above example will produce an equation z=f(x,y) which has the familiar sag in the middle.

In general, displacements from equilibrium in the sheet will obey the 2D wave equation.

5. May 10, 2013

### Septim

Thanks for the answer. I just saw this equation in Griffiths and wondered how he derived it. I do not know the underlying assumptions or the model.

6. May 10, 2013

### Simon Bridge

By inspection - it looks like Griffith has minimized the action using the general method outlined.
The text suggests that there was an earlier expression derived - probably simpler, using a simplified model.
Examining that should help - but authors will sometimes expose you to stuff like this just so you can get an idea of (a) how more there is to learn and (b) how close such a simple model (the previous one) actually is to a more complete treatment.

7. Apr 23, 2016

### Septim

Dear Simon Bridge,

I am sorry to resurrect this thread, but I am still oblivious on how to proceed to get the equation satisfied by a rubber sheet from scratch. Okay I know the principle of action from classical mechanics, yet how can I apply it to the case of a rubber sheet?

8. Apr 23, 2016

### Simon Bridge

Have you tried looking for other examples?

9. Apr 23, 2016

### Nidum

Look up membrane stress .

10. Apr 23, 2016

### Septim

By other examples you mean 1-D case or something? I could not find a rigorous derivation of the equation satisfied by a rubber sheet.

11. Apr 23, 2016

### Septim

Do you have a particular source in mind?

12. Apr 24, 2016

### Strum

Shouldn't you tell us what $V$ is? At any rate the area of a curved surface can be calculated as $A = \int dx (a) = \int dx \sqrt{1 + (\nabla h(x))^{2}}$ with $h(x)$ being the heigth of the surface. Your equations looks like a ficks type of law with "diffusion" coefficient given by a, ala $\nabla \cdot (\frac{1}{a} \nabla h(x))=0$. I dont know if that helps at all :|

13. Apr 24, 2016

### Nidum

Could you please explain more clearly what you are actually trying to do ?

14. Apr 24, 2016

### Septim

I am sorry for the inconvenience caused, V(x,y) denotes the height of the rubber sheet above the point (x,y) in this case. By the way do you say trying to minimize the area will work?

15. Apr 24, 2016

### Septim

My aim is to drive the equation for the height V(x,y) of the rubber sheet as a function of the position (x,y) on the sheet. This equation is:
$\frac{\partial}{\partial x}(g \frac{\partial V}{\partial x}) + \frac{\partial}{\partial y}(g \frac{\partial V}{\partial y}) = 0$
This equation is a differential equation and when the membrane is more or less planar it reduces to Laplace's equation. The g is defined as follows:
$g = \frac{1}{\sqrt{1 + (\frac{\partial V}{\partial x})^2 + (\frac{\partial V}{\partial y})^2}}$

16. Apr 25, 2016

### Strum

Yes wouldn't that work? If $A[h] = \int dx \sqrt{1 + (\nabla h)^{2} }$ doesn't $\frac{\delta A}{\delta h} = 0$ equal your equation?

17. Apr 25, 2016

### Septim

Okay I should do the math and see I am a bit rusty on variational calculus, yet I will try. Do you think this will really give the answer?

18. Apr 25, 2016

### Nidum

But what forces are causing the deflection ???

19. Apr 25, 2016

### Strum

Lets just do it. Since $A[h]$ is only dependent on the gradient of $h$ the variational derivative reduces to
$\frac{\delta A}{\delta h} =- \nabla \cdot \frac{\partial \sqrt{1 + (\nabla h)^{2} } }{\partial \nabla h} =- \nabla \cdot \left( \frac{1}{\sqrt{1 + (\nabla h)^{2} } } \nabla h \right)$. I might have ignored some important points about vectors and such but the result looks fairly good.

20. Apr 26, 2016

### zinq

"It involves modelling the it as an array of small masses connected by springs."

Suppose we ignored gravity and just considered this infinitely thin sheet as modeled by a square lattice of masses, with masses that are adjacent in the x- and y-directions connected by springs. And then take a limit as the fineness of the lattice approaches 0.

Question: Will this model have the isotropy that a typical rubber sheet has?

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