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Assume I have two convex objects in 3D, $A$ and $B$, and the two are in contact. How can I show that there always exists a vector $v$, such that if I move $A$ along $v$ the contact between $A$ and $B$ is broken?

David G. Stork
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QualsPassed
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  • You can't. Consider two parallels halfspaces. If they're bounded argue any direction works if you move far enough. – CyclotomicField Jan 28 '23 at 22:22
  • Sorry, I probably did not formulate the problem rigorously enough, I am thinking about convex shapes like spheres or polygons. I guess that should be closed convex sets? – QualsPassed Jan 28 '23 at 22:26
  • They don't even have to be compact, just bounded. If they're not convex just consider the convex hull. – CyclotomicField Jan 28 '23 at 22:29
  • I see, but do you know how to show this rigorously, I know this definitely holds if you think about it. – QualsPassed Jan 28 '23 at 22:36
  • I do know how to prove it formally and it's accessible to you. Almost all of the work is in understanding what it means to be bounded. One common definition is that a set $X$ is bounded if there exists some ball $B$ of radius $r>0$ centered at the origin such that $X \subset B$. I would also add that you're allowed to choose a ridiculously large distance to move them. Draw a couple bounded sets in 2D, draw the balls that bound them, and reason it pictoraly in relation to the radius of the balls. It's not calculation intensive so natural language will take you most of the way. – CyclotomicField Jan 28 '23 at 23:29
  • @CyclotomicField: How are your parallel halfspaces a counterexample? For instance, in $\mathbb R^2$, if $A$ and $B$ are the halfplanes above and below the $x$-axis, then you can take $v=(0,1)$. What am I missing here? – TonyK Jan 29 '23 at 00:46
  • @TonyK Choose two half spaces above the $x$-axis rather than one above and below. Now you can't separate them. – CyclotomicField Jan 29 '23 at 18:23
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    @CyclotomicField: Oh yes, of course. For some reason, I erroneously assuming that they were disjoint except at their boundaries. – TonyK Jan 29 '23 at 21:08

2 Answers2

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Not a rigorous proof …

If A and B are in contact at a point P, then they have a common surface normal, N, at that point. Consider the common tangent plane $\pi$ — the plane containing P normal to N. It seems obvious that A must lie one side of $\pi$ and B must lie on the other side, otherwise convexity would be violated. Moving A along the vector N would work, it seems to me.

bubba
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The Hyperplane separation theorem states that two disjoint convex sets in 3D, $A$ and $B$, can be separated by a plane. Let $A$ and $B$ touch at one (or more) points. These must lie on a plane. Now move set $A$ along the normal (perpendicular) to the separating plane. Calculate the new distance to any point on $B$... it has become larger.

David G. Stork
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  • The OP didn't say that $A$ and $B$ were disjoint (or even that they had disjoint interiors). I made the same mistake -- see my comments to the OP's question. – TonyK Jan 29 '23 at 21:10