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After reading a layman's guide to general relativity, I began to wonder what shape a bowling ball on a large rubber sheet would produce. For simplicity, I would like to assume that Hooke's law applies to rubber.

Here's what I've done so far:

The 1-dimensional case is a mass $M$ suspended in the middle of a rubber band with length $L$ and spring constant $k$. In this case, the band would create a V shape, where the angle in the middle (between the band and the horizontal) is constrained by $2Tsin(\theta) = Mg$, and the tension is related to the amount of stretching by Hooke's law, giving $T = k \Delta L = k (\sqrt{L^2 + (L\ tan\ \theta)^2 } - L)$.

However, I'm not sure how to proceed in the 2-dimensional case. Is it valid to consider a rubber sheet as an infinitely fine mesh of rubber bands? Intuitively, at some distance $r$ from the point of mass, it seems that the mass must be supported equally by the material in the circle of radius $r$, so we would have $2\pi r T sin(\theta_r) = Mg$. However, this means that as $r\rightarrow 0$, we have $T\rightarrow\infty$, which is not physical (I think the tension should be constant everywhere in the sheet.)

What is the correct way to model the shape of a rubber sheet with a point mass on it?

tba
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  • Hmmm... there may be an issue with units: $T$ matches $Mg$, so $rT$ cannot also match $Mg$. More importantly, going to more dimensions may very well introduce other stress/strain quantities (Young's modulus $\to$ bulk and shear modulus). –  Dec 27 '12 at 03:27

2 Answers2

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Generalizing Hooke's law to two dimensions requires the introduction of Poisson's ratio: when you stretch a material in one direction, it tends to contract along perpendicular directions. So the radial tension from the weight of the ball, will produce a circumferential contraction that will create circumferential tension to compensate it.

So at any point on your rubber sheet there will be tension in two perpendicular directions, and because the resulting surface will be of revolution, any shape other than a horizontal plane will have at least one non-zero curvature. Curvature plus tension means force perpendicular to the surface, unless your surface main curvatures have opposing signs, and when composed with the corresponding stress they compensate each other.

If instead of a rubber sheet you had a soap film, this leads to the well known minimal surface of revolution, the catenoid. But with a rubber sheet, mean curvature will not be exactly zero, and thus the shape will be different from that, although with the same general aspect: radially it must curve away from the axis of revolution to compensate the circumferential curvature.

I have probably only added more confusion, but the shape in two dimensions will definitely not be a cone.

Jaime
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Gravity does not obey Hooke's law but is an inverse power law (inverse square to be specific) . The rubber sheet analogy is only to provide a means of showing the curved nature of space. The rubber sheet would be as you suggested with a straight line slope.

tpg2114
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Steve Kohl
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  • So you're saying the sheet would form a cone, like the straight-line solution revolved the Z-axis? – tba Dec 26 '12 at 23:38
  • The word you want is not "exponential". Exponential means forms like $e^x$ not forms like $x^n$ which are "polynominal" or "power law" in general. You could use "inverse geometric" if you realy wanted to, but "inverse square" is probably simpler. – dmckee --- ex-moderator kitten Dec 26 '12 at 23:47
  • Agreed dmckee. And tba, the sheet would assume the shape of a cone with straight sides. Gravity is expressed as an inverse square. The exponent gives space its curve. – Steve Kohl Dec 27 '12 at 00:00
  • Thanks for your answer! Can you provide a proof sketch? (Since the other answer claims that a cone is not correct.) – tba Dec 27 '12 at 07:56
  • I'm sorry, looks like I jumped to a conclusion. Here's a web page that predicts a parabola for deflection of thin circular membranes http://www.springer.com/cda/content/document/cda_downloaddocument/9783642194887-c1.pdf?SGWID=0-0-45-1144739-p174105164 – Steve Kohl Dec 27 '12 at 19:33