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I was reading another post in physics.SE about chairs when this question arose. My guess is we would need more than in 3 dimensions as in 2 dimensions you would just need 2 (I think). I've been trying to figure it out but it is imposible for me to imagine 4 orthogonal space directions.

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We could want connect 2 dots , 2 strings , 2 planes or 2 volumes in a 4d space. They must be parallel ; one object is the seat and the other the ground ; gravity or any other force exerts a perpendicular attraction between them.

The container space may have an arbitrarily high dimension. Only accounts the dimension of the objects to connect, obviously lower than the one of the container.

From the point of view of the object seat , the exercise consists now to build an adhesive object to the object ground with the extremities of N legs which are dots in the 1st approximation.

How many legs/dots do we need to define an object of dimension d ? d+1 distinct dots and it must not exist an object of dimension d-1 containing both them. For the common chair problem , a plane where to seat and a plane ground ( the well known solution is 3 ) , this means that the extremities of the 2+1 legs must not be aligned.

Hence, if we assume that a chair in a 4d world must stay on a 3D surface, we will need 4 legs with 4 extremities / dots not belonging to any same plane.

If we assume that a chair in a 123d world must stay on a 122D surface, we will need 122+1 legs with 122+1 extremities / dots not belonging to any same 121D container.

The solution remains the same if the object seat is of dimension lower that the object ground. Finally, Only accounts the dimension of the ground

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  • $\begingroup$ Your answer was the one that convinced me the most, thanks! $\endgroup$ – SaudiBombsYemen Jan 23 '16 at 16:21
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    $\begingroup$ if you want a more mathematical explanation, just ask on math.se , my answer will be among others :) $\endgroup$ – user46925 Jan 23 '16 at 16:24
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I would suggest we need 4 legs to stabilize in 4D space. Consider this: we need at least 3 legs in 3D spacetime to project onto the ground a polygon with non-zero area. If the area of the polygon was zero, the center of mass would be in an unstable equilibrium and would likely tip. So, we draw the analogy to 4D: we project onto a 3D hypersurface which is the ground of our universe, and try to form the 3D polyhedron with fewest vertices (c.f. 2D polygon with fewest vertices is the triangle - each vertex corresponds to the end of a leg in the projection), which is a tetrahedron. But this has 4 vertices! This, we must have 4 legs. And the projection of the legs onto our 3D space would be a tetrahedron.

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I have no idea, but here's a guess. If you think about how many legs a chair would need in 2 spatial dimensions, the answer seems to be 2. If there were only one leg, then a slight push to either the left or right would cause the "chair" to tip over. See the image below for an unstable (left) and stable (right) 2d "chair"

a 2d "chair"

In 3 spatial dimensions (i.e. the real world) we need 4 legs. This makes sense, too. The chair is sitting on a 2d plane, and two of the legs are needed to stabilize the chair in the $x$ direction, and two are needed to stabilize in the $y$ direction. So, with this naive way of thinking, the pattern that emerges is that: in $d$ spatial dimensions the chair sits on a $d-1$ "floor", and for each of these $d-1$ directions two legs are needed. So I would guess that a $d$-dimensional chair would need $2(d-1)$ legs.

Fun fact, in $d=1$ (i.e. a line), apparently no legs are needed. There's no where for the "chair" to fall!

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    $\begingroup$ A chair actualy needs only three legs to stand at a stable state in three dimensions. $\endgroup$ – Al.Ka Jan 22 '16 at 22:40
  • $\begingroup$ I believe that's called a stool ;) $\endgroup$ – Surgical Commander Jan 22 '16 at 22:42
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    $\begingroup$ In d=1 you need no legs unless gravity acts in the direction of your space. Then you need 1 leg. $\endgroup$ – Wolpertinger Jan 22 '16 at 22:48
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In the following I will attempt a slightly ridiculous way of stating that you need as many legs as dimensions in your space. I.e. for 4D: four legs.

I will ignore any peculiarities of n-dimensional Gravity for simplicity. Instead I will assume that there is a uniform directional Newtonian force field acting in the n-dimensional space. I.e. "down" is along that force.

We will chair to be a number of m legs fixed together at one point. The other ends of the legs are then fixed to some other points in the space, but free to rotate about those.

1D is trivial: you need one leg if the force is in the direction of your 1D space (which is the only thing that makes sense really). For higher dimensions the problem becomes an exercise in linear algebra/geometry. As a preliminary note that the concept of rotating around an axis does not make since in dimensions higher than 3, instead one has to think of a rotation in a plane.

We now define a hyperplane that is the subspace spanned by all the legs. We can assume the legs to be linearly independent since linear dependence will increase the number of necessary legs. So this hyperplane is m-dimensional. If you have understood this the problem is essentially solved. I will ignore any peculiarities of 4D Gravity for simplicity. Instead I will assume that there is a uniform directional Newtonian force field acting in the n-dimensional space. I.e. "down" is along that force.

We will chair to be a number of m legs fixed together at one point. The other ends of the legs are then fixed to some other points in the space, but free to rotate about those.

1D is trivial: you need one leg if the force is in the direction of your 1D space (which is the only thing that makes sense really). For higher dimensions the problem becomes an exercise in linear algebra/geometry. As a preliminary note that the concept of rotating around an axis does not make since in dimensions higher than 3, instead one has to think of a rotation in a plane.

We now define a hyperplane that is the subspace spanned by all the legs. We can assume the legs to be linearly independent since linear dependence will increase the number of necessary legs. So this hyperplane is m-dimensional.

Proposition: the construct will then be unstable under gravity if and only if there exists a direction that does not lie in the hyperplane. Proof: if such a direction exists (call is â), then -â is also not in the hyperplane. So unless â is parallel to the axis q that is orthogonal to the hyperplane, the hyperplane will experience an acceleration into a direction it can rotate into => unstable. If â||q then a small perturbation into the same direction will tilt â => same result.

So for stability we need to not have such a direction. Therefore m=n.

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