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My major is mechanical engineering. Recently, I am working on some subject involving three-dimensional finite rotations. More specially, the necessary and sufficient conditions for an applied torque/moment be conservative in the finite rotation range. I have tried to read some math books, but I got more confused. The following is the description of the background.

In mechanics, an externally torque generally exhibits unusual property of configuration-dependent, which means the torque varies from its initial value $\mathbf M_0$ to its current value $\mathbf M$ when moving along a curve lying on SO(3) staring form the identity $\mathbf I$ to the current position $\mathbf R$. In other words, the current counterpart $\mathbf M$ can be viewed as a $\mathbf explicit function$ of the rotation $\mathbf R \in SO(3) $.

Let $\mathbf \delta \omega$ be the spatial spin (an element which belongs to the tangent space of SO(3) at the base point $\mathbf R$, i.e., $\mathbf \delta \omega \in T_{R}SO(3)$). Then the virtual work done by the torque over the spin is given by $$\delta W = \mathbf M \cdot \delta \omega$$

where $\delta W $ is a real number, and "$\cdot$" means dot product. In mathematics, $\mathbf M$ is an element of cotangent space of SO(3) at the base point $\mathbf R$, i.e., $\mathbf M \in T^{*}_{R}SO(3)$.

On the other hand, if the rotation vector (axis-angle representation) $\mathbf \psi = \psi_{i} \mathbf e_{i}$ was used to parameterize the rotation manifold, $\mathbf R = exp(\hat \psi)$, then we can express the torque as $\mathbf M=\mathbf Q \mathbf M_0$ explicitly, where $\mathbf Q=\mathbf Q(\psi)$ is the transformation matrix relating the initial and current values of the torque.

We also can represent the virtual rotation by $\mathbf \delta \psi$, the variation of rotation vector $\mathbf \psi$, $\mathbf \delta \psi \in T_{I}SO(3)$. The relation between $\mathbf \delta \omega$ and $\mathbf \delta \psi$ is given by $ \delta \omega = \mathbf L \delta \psi$, where $\mathbf L= \mathbf L(\psi)$ is the tangential operator, $\mathbf L:T_ISO(3)\to T_RSO(3)$. Thus, the virtual work can be rewritten as $$ \mathbf \delta W = \mathbf L^T \mathbf M \cdot \delta \psi$$

My questions are:

  1. Which expression of the virtual work is a differental 1-form in SO(3) and why?
  2. How to calculate the line integral of the virtual work over a curve lying on SO(3) in terms of a differential 1-form?

Thank you very much!

EDIT 1: In the above description, the spin $\mathbf \delta \omega$ is not a differential, since there does not exist a variable from which the spin can be derived. It comes from the variation of the orthogonality condition of rotation matrix, $\mathbf \delta(\mathbf R \mathbf R^T=\mathbf I)=0$, $\mathbf \delta \mathbf R=\widehat (\delta\omega) \mathbf R$.

However, the variation $\mathbf \delta \psi$ of rotation vector is a differential.

JinsongYang
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  • Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. Also, on this site we use MathJaX to format our maths. Here you can find a basic tutorial. – Nizar Feb 10 '17 at 09:14
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    Ok, thanks a lot. – JinsongYang Feb 10 '17 at 09:32
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    As it stands, neither is a differential form, but rather an output when you apply the desired differential form to a tangent vector $\delta\omega$ as you move along your curve. But you want the first choice, moving along a path in the group. To compute an actual line integral, you'll need either coordinates or an understanding of how to express $\mathbf M$ explicitly as a function of the orthonormal frame $(e_1,e_2,e_3)\in SO(3)$. – Ted Shifrin Feb 10 '17 at 17:53
  • Thank you very much, Ted. I have updated my question. And yes, we can express $\mathbf M$ explicitly as a fucntion of the rotation vector $\mathbf \psi$. You mean we can calculate the line integral on the manifold SO(3) as vector calculus in terms of a set of local coordinates. Does there exist a way of coordinate-free to deal with the line integral? Since if a torque is conservative, then the line integal should be path-independent regardless of the local coordinates (the choice of parameterizations). – JinsongYang Feb 11 '17 at 09:18
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    I'm having some trouble parsing, but adding to Ted's comment, the differential one-form is the collection of all $\mathbf{M}$, viewed as a function on $SO(3)$ whose value at each point $p$ is an element of the cotangent space $T_{p}^{*}SO(3)$. It appears that $\mathbf{R}$ is some representation of a path in $SO(3)$; if that's right, the virtual work done "at time $t$" is $$\delta W = \mathbf{M}(\mathbf{R}(t)) \cdot \mathbf{R}'(t), dt,$$and the work is the integral over the parameter interval of $\mathbf{R}$. – Andrew D. Hwang Feb 12 '17 at 16:56
  • Thank you for your explanation, Andrew. $\mathbf R$ is the rotation matrix. In computational mechanics, several parameterizations of finite rotation were often considered, e.g., rotation matrix, rotation vector (axis-angle), quaternions, Euler angles. – JinsongYang Feb 12 '17 at 17:48
  • Continued. Your equation is definitely right. However, in most of these parameterizations, the "differential" of the rotation parameter is not a true differental, since there does not exist such a variable from which the "differential" of the parameter can be derived. Consequently, the expression of the virtual work in terms of such rotation parameter cannot be integrated. Here comes my question 1, $\mathbf \delta \omega $ is not a differential, then is $\mathbf \delta W= \mathbf M \cdot \delta \omega$ still a differenrial 1-form? – JinsongYang Feb 12 '17 at 17:49
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    I don't have time to write out a more detailed comment/response right now, but regarding your earlier comment, a line integral is always well-defined independent of parametrization of the curve; it's independent of path when the $1$-form is exact (or the force field conservative). But, yes, there's an invariant way to write $1$-forms on any Lie group. (I do not understand how you're thinking of parametrizing the rotation manifold. Care to elaborate?) – Ted Shifrin Feb 12 '17 at 17:54
  • Thank you for your time, Ted. In the context of computational mechanics, paramterizations of rotation manifold play important roles in both theoretical and computational aspects. For instance, if the rotation vector (axis-angle) $\mathbf \psi=\psi_{i}\mathbf e_{i}$ was used, the rotation matrix can be computed with Rodrigues formula $\mathbf R = I + {sin(\psi) \over \psi} \hat \psi+{(1-cos(\psi)) \over \psi^2} \hat \psi ^{2} $. We can deal with finite rotation with only three components of rotation vector and get some other benifits. – JinsongYang Feb 12 '17 at 18:36
  • Continued. The Rodrigues formula is a closed expression of exponential map $exp(\cdot):so(3)\to SO(3)$, $\mathbf R = exp(\hat \psi)$. – JinsongYang Feb 12 '17 at 18:45
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    @JinsongYang You appear to be confusing differential 1-forms with differentials of 0-forms. So, a 0-form f is just a scalar-valued function on a manifold (in this example, SO(3)), and its differential df is indeed a 1-form. But, not all 1-forms arise in this way. For example, $\omega = \frac{-y}{x^2+y^2}\ dx + \frac{x}{x^2+y^2}\ dy$ is a differential 1-form defined on $\mathbb{R}^2 \backslash\ {\vec{0}}$, but it is not the differential of any 0-form on that set (Consider its integral over $S^1$). $\mathbf{M}$ is a 1-form. – Jeffrey Rolland Jun 25 '20 at 02:26
  • Continued. In fact, given any differential 1-form, if its exterior derivative is not 0, then it is not the differential of a 0-form (The one given above does have $d\omega = 0$, it just generates $H^1_{dR}(\mathbb{R}) \backslash\ {\vec{0}}$ and hence is not the exterior derivative of any 0-form). I'll post an answer on how to compute $\int_{[\mathbf{R}]} \mathbf{M}$ later. You'll need a coordinate patch on SO(3), a parameterization of $\mathbf{R}$ in the parameter space (or in SO(3) and you can pull it back to the parameter space), and then the integral is given by @AndrewD.Hwang 's post – Jeffrey Rolland Jun 25 '20 at 02:33
  • Continued. Note $H^1_{dR}(SO(3)) = 0$ (https://topospaces.subwiki.org/wiki/Homology_of_real_projective_space and Universal Coefficients), so $\mathbf{M}$ is a differential of a 0-form ("has a scalar potential function") if and only if $d\mathbf{M} = 0$ as a differential 2-form. – Jeffrey Rolland Jun 25 '20 at 02:40
  • Extra thought. (If $\mathbf{M}$ has a scalar potential function, you could use it to evaluate the line integral (!)) – Jeffrey Rolland Jun 25 '20 at 02:49

2 Answers2

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$\vec{r}: [0,2\pi] \times [0,\pi] \times [0,2\pi] \to \mathbb{R}^4$ given by $$\vec{r}(\psi,\theta,\phi)=\cos(\frac{\phi + \psi}{2})\cos(\frac{\theta}{2})c\hat{t}+\cos(\frac{\phi - \psi}{2})\sin(\frac{\theta}{2})\hat{i}+\sin(\frac{\phi - \psi}{2})\sin(\frac{\theta}{2})\hat{j}+\sin(\frac{\phi + \psi}{2})\cos(\frac{\theta}{2})\hat{k}$$ is a parametrization of $SO(3)$ by the angles yaw ($\psi$), pitch ($\theta$), and roll ($\phi$) https://pasteboard.co/JeFt2eK.png. (Actually, it is a parametrization of half of $S^3$, but, as long as we stay inside the coordinate patch, it will suffice. Note $S^3$ is a double cover, and the universal cover, of $SO(3)$; this is a single "sheet" in the covering space.) (Note $\hat{t}$ is the unit vector in the ''time direction'' and $c$ is the speed of light, converting time dimensions to space dimensions. You do not actually include $c$ in your calculations, $c\hat{t}$ is just heuristic for indicating a fourth spatial dimension.) (I forget which of the standard parametrizations -- rotation matrix, rotation vector (axis-angle), quaternions, Euler angles, etc. -- this is, but you probably know it off the top of your head. I suppose it is the quaternion parametrization.)

We need $\mathbf{R}$ to be oriented to do the integral (a change of orientation only changes the sign of the integral's value); call the oriented path $[\mathbf{R}]$. Parametrize $\mathbf{R}$ in $SO(3)$ by the path $\gamma: [a,b] \to SO(3)$, consistent with the orientation of $[\mathbf{R}]$. Pull $[\mathbf{R}]$ to $[\mathbf{R}^*]$ in $[0,2\pi] \times [0,\pi] \times [0,2\pi]$ and $\gamma$ back to $\gamma^*: [a,b] \to [0,2\pi] \times [0,\pi] \times [0,2\pi]$, with $\gamma(t) = \vec{r}[\gamma^*(t)]$ the desired parametrization of $\mathbf{R}$ in $SO(3)$, consistent with the orientation of $[\mathbf{R}]$. Now, suppose we extend $\mathbf{M}$ to a neighborhood of $S^3$ in $\mathbb{R}^4$ and write $\mathbf{M}(ct, x, y, z) = $ $\mathbf{M}_1(ct, x, y, z)d\{ct\} + \mathbf{M}_2(ct, x, y, z)dx + \mathbf{M}_3(ct, x, y, z)dy + \mathbf{M}_4(ct, x, y, z)dz$, then pull the restriction of this to $S^3$ back via $\vec{r}$ to be $\mathbf{M}^*(\psi,\theta,\phi) \in T^*([0,2\pi] \times [0,\pi] \times [0,2\pi])$ via formulas like $\displaystyle \mathbf{M}_1[\vec{r}(\phi,\theta,\psi)] \ \left(\frac{\partial \{ct\}}{\partial \psi}d\psi + \frac{\partial \{ct\}}{\partial \theta}d\theta+ \frac{\partial \{ct\}}{\partial \phi}d\phi\right)$; alternatively, we could write $\mathbf{M}^*(\psi,\theta,\phi) = \mathbf{M}^*_1(\psi,\theta,\phi)d\psi + \mathbf{M}^*_2(\psi,\theta,\phi)d\theta + \mathbf{M}^*_3(\psi,\theta,\phi)d\phi$ directly. Next, take $\gamma^{*'}(t)$ and $\mathbf{M}^*[\gamma^*(t)]$.

Then $$ \int_{[\mathbf{R}]} \mathbf{M} = \int_a^b \langle \mathbf{M}^*[\gamma^*(t)]\ |\ \gamma^{*'}(t)\rangle\ dt$$, where $\langle\ |\ \rangle$ denotes the action of a differential 1-form on $[0,2\pi] \times [0,\pi] \times [0,2\pi]$ on a tangent vector to $[0,2\pi] \times [0,\pi] \times [0,2\pi]$, is the work done by the torque along the path https://en.wikipedia.org/wiki/Torque#Relationship_between_torque,_power,_and_energy; one could turn the RHS over to a talented Calc II student, in principle.

Here is a Mathematica notebook that does an example; you should only have to change the formulas in the top lines, where $\gamma^*(t)$ and $\mathbf{M}$ are defined, then drag down to "Evaluation $\to$ Evaluate Notebook" to do any reasonable computation you desire. (There could be a teensy problem if your path "leaks out of" or "wraps around" the coordinate patch; you would then just have to break your integral into "sub-integrals" over portions of the oriented path contained entirely in the coordinate patch to do the full integral.)

Jeffrey Rolland
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  • OT: @JinsongYang Where do you study mechanical engineering (if I may ask politely)? This is way beyond anything the standard MSOE student would be expected to understand/do by most of their professors, I should think – Jeffrey Rolland Jun 25 '20 at 05:21
  • That was super-fun, and very much in the spirit of Bullo and Lewis https://www.amazon.com/Geometric-Control.../dp/B07ZKTDN9X/. It was the idea of being able to problems like that that got me into differential topology in the first place. – Jeffrey Rolland Jun 27 '20 at 14:48
  • (I see now why Bullo and Lewis encode generalized forces in Lagrangians as differential forms instead of vector fields: it completely removes the dependence on a Riemannian metric [which is, in a way, exactly what differential forms were designed to do]. It seemed to meaninglessly change the formalism from to what the reader was used to me at first, but now I see it is exactly the correct thing to do. Insert aikido "whetstone analogy" speech here; having a problem to do at exactly the correct level of difficulty/resistance makes concepts clear more than 100 pages of exposition.) – Jeffrey Rolland Jun 27 '20 at 14:48
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(I think I've edited the first answer enough; we'll try another answer for the potential function solution.)

Here is a Mathematica notebook that does the same example using a potential function to compute the integral. Many special thanks to @Michael_E2 on mathematica.stackexchange.com for posting this answer to "help" with the technical details of computing a potential function as a function in Mathematica. It is still caveat emptor for the user to make sure the exterior derivative of $\mathbf{M}$ is 0; recall that part of what makes this work is that $SO(3)$ is a "de Rham cohomology sphere" (see the comments above the first answer).

Here is a Mathematica notebook that uses the DifferentialForms.m package https://library.wolfram.com/infocenter/MathSource/482 to compute the integral directly and using an antiderivative (there's a kludge on the HomotopyOperator I'm not sure why I need).

Jeffrey Rolland
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