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Equilibrium Condition Using Tangent Space

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
I am reading a paper on Tensegrity structures by Roth and Whiteley. They have defined equilibrium of forces is a very different-from-normal way and any help on that would be great.
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Write $\underbrace{\mathbb R^3\times \cdots \times \mathbb R^3}_{m\text{ times}}$ as $(\mathbb R^3)^m$.

Let $p=(p_1,\ldots,p_m)\in (\mathbb R^3)^m$.
Let the set of all translations in $\mathbb R^3$ be written as $\mathcal T$ and the set of all special orthogonal transformations in $\mathbb R^3$ be wriiten as $SO(3)$.

Define $M(p)=\{(\tau\circ R(q_1),\ldots,\tau\circ R(p_m)) \in (\mathbb R^3)^m : \tau \in \mathcal T \text{ and } R \in SO(3)\}$.

$M(p)$ happens to be a manifold. (I am not sure about the details of the proof but it's true. See rigid transformation - To Prove That a Certain Set is a Manifold - Mathematics Stack Exchange)

Wrtie $T(p)$ as the tangent space at point $p$ of the manifold $M(p)$.

Let $F=(F_1,\ldots,F_m) \in (\mathbb R^3)^m$.

Imagine a rigid body $\Omega$ in $\mathbb R^3$ which is large enough so that each $p_i$ is on it.

Apply a force $F_i$ at the point $p_i$ of $\Omega$ and write $F=(F_1,\ldots,F_m)$.

Then $\Omega$ is in equilibrium if $F\in T(p)^{\perp}$.
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Can somebody give some insight as to why this should be true? Normally equilibrium is talked about in terms of torques about points etc. How is this characterization equivalent to the usual one?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
I am reading a paper on Tensegrity structures by Roth and Whiteley. They have defined equilibrium of forces is a very different-from-normal way and any help on that would be great.
___

Write $\underbrace{\mathbb R^3\times \cdots \times \mathbb R^3}_{m\text{ times}}$ as $(\mathbb R^3)^m$.

Let $p=(p_1,\ldots,p_m)\in (\mathbb R^3)^m$.
Let the set of all translations in $\mathbb R^3$ be written as $\mathcal T$ and the set of all special orthogonal transformations in $\mathbb R^3$ be wriiten as $SO(3)$.

Define $M(p)=\{(\tau\circ R(q_1),\ldots,\tau\circ R(p_m)) \in (\mathbb R^3)^m : \tau \in \mathcal T \text{ and } R \in SO(3)\}$.

$M(p)$ happens to be a manifold. (I am not sure about the details of the proof but it's true. See rigid transformation - To Prove That a Certain Set is a Manifold - Mathematics Stack Exchange)

Wrtie $T(p)$ as the tangent space at point $p$ of the manifold $M(p)$.

Let $F=(F_1,\ldots,F_m) \in (\mathbb R^3)^m$.

Imagine a rigid body $\Omega$ in $\mathbb R^3$ which is large enough so that each $p_i$ is on it.

Apply a force $F_i$ at the point $p_i$ of $\Omega$ and write $F=(F_1,\ldots,F_m)$.

Then $\Omega$ is in equilibrium if $F\in T(p)^{\perp}$.
___

Can somebody give some insight as to why this should be true? Normally equilibrium is talked about in terms of torques about points etc. How is this characterization equivalent to the usual one?
It doesn't look like it is.

The vector $p$ corresponds to a set of points.
The tangent space $\mathcal T(p)$ corresponds to the velocity of each of those points.
Saying that $F\in \mathcal T(p)^{\perp}$ only means that the forces are perpendicular to the velocities. That is not a sufficient condition for "normal" equilibrium.

My interpretation is that with what they call "equilibrium" they only mean that the forces are perpendicular to the velocities. That would mean that the magnitudes of all velocities are constant.

For "physical static equilibrium", you need that the sum of the forces is zero and that the torques cancel, but that does not appear to be the case here.
Anyway, for physical static equilibrium the forces do not have to be perpendicular to the velocities...
 

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
It doesn't look like it is.

The vector $p$ corresponds to a set of points.
The tangent space $\mathcal T(p)$ corresponds to the velocity of each of those points.
Saying that $F\in \mathcal T(p)^{\perp}$ only means that the forces are perpendicular to the velocities. That is not a sufficient condition for "normal" equilibrium.

My interpretation is that with what they call "equilibrium" they only mean that the forces are perpendicular to the velocities. That would mean that the magnitudes of all velocities are constant.

For "physical static equilibrium", you need that the sum of the forces is zero and that the torques cancel, but that does not appear to be the case here.
Anyway, for physical static equilibrium the forces do not have to be perpendicular to the velocities...
Hey ILS,

Thank you for taking interest in this.

See this:
https://docs.google.com/file/d/0B77QF0wgZJZ7Q1otQnh5SEw2STQ/edit?usp=drive_web

See the last paragraph on pg. 423 and first paragraph on pg. 424. The authors seem to be advocating that the tangent space definition is equivalent to the usual ones.

I have not read much on manifolds so I am finding it quite hard to follow this. Can you also recommend some books for this?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
Hey ILS,

Thank you for taking interest in this.

See this:
https://docs.google.com/file/d/0B77QF0wgZJZ7Q1otQnh5SEw2STQ/edit?usp=drive_web

See the last paragraph on pg. 423 and first paragraph on pg. 424. The authors seem to be advocating that the tangent space definition is equivalent to the usual ones.

I have not read much on manifolds so I am finding it quite hard to follow this. Can you also recommend some books for this?
Ah, I see what you mean.

Let's look at it like this.
Suppose we have a 2-dimensional object that is constrained to be in those 2 dimensions.
Then equilibrium means that all net forces on each point of that object must be zero in both the x-direction and the y-direction.
Torque does not play a role here, since we're only looking at individual points, not the object in its entirety. Put otherwise, if there is a net torque, the net forces on the individual points are not zero since they start accelerating with some angular acceleration.

Suppose we only have a net force in the z-direction at some point. Then its net force in both the x-direction and the y-direction is zero. Ergo, equilibrium. Any conceivable force in the z-direction "does not count" since the object can't move in that direction.

For an object in 3 dimensions the same principle holds. It can't move into an invisible 4th (or higher) dimension. And as long as the projection of the force is zero in the regular 3 dimensions, it is in equilibrium.

Basically, this is how manifolds work. We have a mapping of Rn into some unknown space, which might for instance be 1 dimension higher (although we don't have to make any assumptions here). This is what happens for instance with the unit sphere: we have mappings from 2 dimensions to a sphere in 3 dimensions.
If there is a radial net force on the sphere, any points on the sphere will still stay in place.
 

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
Ah, I see what you mean.

Let's look at it like this.
Suppose we have a 2-dimensional object that is constrained to be in those 2 dimensions.
Then equilibrium means that all net forces on each point of that object must be zero in both the x-direction and the y-direction.
Torque does not play a role here, since we're only looking at individual points, not the object in its entirety. Put otherwise, if there is a net torque, the net forces on the individual points are not zero since they start accelerating with some angular acceleration.

Suppose we only have a net force in the z-direction at some point. Then its net force in both the x-direction and the y-direction is zero. Ergo, equilibrium. Any conceivable force in the z-direction "does not count" since the object can't move in that direction.

For an object in 3 dimensions the same principle holds. It can't move into an invisible 4th (or higher) dimension. And as long as the projection of the force is zero in the regular 3 dimensions, it is in equilibrium.

Basically, this is how manifolds work. We have a mapping of Rn into some unknown space, which might for instance be 1 dimension higher (although we don't have to make any assumptions here). This is what happens for instance with the unit sphere: we have mappings from 2 dimensions to a sphere in 3 dimensions.
If there is a radial net force on the sphere, any points on the sphere will still stay in place.
That's good insight into the manifold definition.

Can we formally establish the equivalence of the manifold definition and the usual thing for $2$ and 3$ dimensions? That would really help me with my thesis.