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    • CommentRowNumber1.
    • CommentAuthorMike Shulman
    • CommentTimeAug 29th 2011

    This is a bit off the beaten path for what we discuss around here, but I know there are people here who think about physics in nice abstract ways, so maybe someone knows the answer. My understanding of the concept of moment of inertia is that for any rigid body, there’s supposed to be an equation

    τ=Iα\tau = I \alpha

    parallelling F=maF=m a, where τ\tau is torque, α\alpha is angular acceleration, and II is this “moment of inertia” thing. But what type of thing is II?

    In oriented Euclidean 2-space, τ\tau and α\alpha can be identified with scalars, and likewise so can II.

    In oriented Euclidean 3-space, τ\tau and α\alpha can be identified with vectors, and II becomes a symmetric rank-2 tensor.

    (In oriented Euclidean 1-space, there is no room for rotational motion.)

    In oriented Euclidean n-space, my best guess is that τ\tau and α\alpha should both lie in so(n)Λ 2 nso(n) \cong \Lambda^2 \mathbb{R}^n. So what type of beast is II? Is there a good reference for rotational motion in n dimensions?

    • CommentRowNumber2.
    • CommentAuthorUrs
    • CommentTimeAug 29th 2011

    Hi Mike,

    textbooks that have the term “geometric algebra” in their title (or somewhere close to it) will give the general discussion that I think you are looking for.

    I am in a rush right now. Will try to give you a good reference later. A non-optimal one (because it’s more a preview than a lecture) is

    Physical applications of geometric algebra, slide 56

    When they need to write down a number for the dimension they will still write “3” there but their formalism is the general one. On that page 56 you will see them write about “bivectors”, which is precisely the elements in 2 n\wedge^2 \mathbb{R}^n that you mention. These authors will always think of this as embedded canonically into the Clifford algebra Cl nCl_n (and call that the “geometric algebra” of n\mathbb{R}^n.)

    Have to dash off now. I’ll try to come back to you later.

    • CommentRowNumber3.
    • CommentAuthorUrs
    • CommentTimeAug 29th 2011
    • (edited Aug 29th 2011)

    These lectures that I mentioned also exist as a fully-fledged textbook

    Doran, Lasenby, Geometric algebra for physicists (Google Books)

    The moment of inertia in terms of a linear map between bivectors is discussed there from page 74 on.

    I guess that this should be what you are after? (apart, of course, from the math-style of the discussion which hoever you can easily supply yourself) .

    One would think that this general bivector perspective is adopted also in other mathematical physics textbooks. But I am having trouble finding such . For instance also Arnol’d’s otherwise excellent Mathematical methods of classical mechanics has just 3d rigid body dynamics and no bivectors.

    Maybe this means that I haven’t looked in the right places yet. Or maybe it is another piece of evidence that the Geometric Algebra-school is indeed filling a gap.

    • CommentRowNumber4.
    • CommentAuthorzskoda
    • CommentTimeAug 29th 2011

    The group SO(N,R)SO(N,\mathbf{R}) as the configuration space of the rigid body with fixed center of mass in NN-dimensions is obvious. The equations of motion in 3 dimension are called Euler equations and in dimension NN the Euler-Arnol’d equations.

    dMdt=[M,ω] \frac{d M}{d t} = [M,\omega]

    where ω\omega is the vector of angular velocity and MM is the bivector of angular momentum. Now the angular momentum could be written as M=Iω+ωIM = I \omega + \omega I for another symmetric two tensor, II. One can write equivalent system

    dωdt=[I,ωω] \frac{d \omega}{d t} = [I,\omega \wedge \omega]

    with the matrix commutator on the right.

    Where this is coming from ? As the Galilean group is the semidirect group of translations and rotations (and also we can superimpose any motion in rotations and translations) this is still relevant for motion without fixed center of mass. Now the analysis can proceed by considering the angular momentum of a rigid body as an assemblage of a large number of (small) point particles and taking angle variables as the coordinates and then one gets conjugate momenta via Lagrangean formalism to be the corresponding. One takes kinetic energy of each particle, sums up and writes it in terms of these coordinates and momenta and sees that it is still a quadratic form in terms of angular velocities, which are now all the same for a rigid body, so one can sum up the coefficients, which will form a tensor II of second rank, so that one can write kinetic energy as (Iω,ω)/2(I \omega,\omega)/2. This II is the momentum of inertia, no difference in this reasoning, whatever the dimension, from say, classical Landau’s book. One can generalize this to more general real Lie groups,

    • R. Abraham, J. Marsden: The Foundations of Mechanics, Benjamin Press, 1967, Addison-Wesley, 1978; large pdf 86 Mb free at CaltechAuthors

    have in section 4.4 (of the second edition) Hamiltonian systems on Lie groups and rigid body. Even some of the deeper results on integrable systems generalize to such, and some to more general Lie groups. For example,

    • Tudor Ratiu, Euler-Poisson equations on Lie algebras and the N-dimensional heavy rigid body, Proc. Nati. Acad. Sci. USA 78, No. 3, pp. 1327-1328, March 1981, pdf
    • CommentRowNumber5.
    • CommentAuthorzskoda
    • CommentTimeAug 29th 2011

    Remark to Urs: while it may be an advantage over 3d people, the geometric algebra is just an intermediate step towared the more general Lie group case, and other generalizations of dynamics of rigid bodies studied in integrable systems community. I do not think that any of those people could teach, say one Bertram Kostant (who lead some of these discoveries in integrable systems) about Clifford algebras.

    • CommentRowNumber6.
    • CommentAuthorUrs
    • CommentTimeAug 29th 2011
    • (edited Aug 29th 2011)

    I do not think that any of those people could teach, say one Bertram Kostant

    You don’t have to invoke Kostant to make this statement true. These people use the most basic properties of Clifford algebras, only.

    The point here is not the Clifford algebras, but that these people make the good move of not modelling angular variables by vectors – which only works in three dimensions – but by bivectors. Which is what Mike was mentioning.

    I am starting an entry Euler-Arnold equation. Terry Tao has a good survey on his blog.

    • CommentRowNumber7.
    • CommentAuthorUrs
    • CommentTimeAug 29th 2011

    So, as Zoran points out, what the moment of inertia is, abstractly, for nn-dimensional rigid body dynamics, is an (any) left-invariant Riemannian metric on SO(n)SO(n).

    I have started adding some sentences on this to moment of inertia and maybe elsewhere. But I should really be doing something else now.

    • CommentRowNumber8.
    • CommentAuthorMike Shulman
    • CommentTimeAug 29th 2011

    Thanks, both of you! I will have a look at these references.

    • CommentRowNumber9.
    • CommentAuthorzskoda
    • CommentTimeAug 29th 2011

    I wish I had more time to reconcentrate on this. I just came from a mountain camping today and am still in lack of sleep. But good topic to have around…

    • CommentRowNumber10.
    • CommentAuthorMike Shulman
    • CommentTimeAug 29th 2011

    Starting to read the references, especially “Physical applications of geometric algebra”. Maybe I sort of get the point of Geometric Algebra better now. When I first read some philosophical screed about it, I was very put off. But now it sounds like just a different, and perhaps more convenient, way of working with differential forms and their ilk, which are objects I already like a lot.

    It seems like at the heart of it is a canonical (in the presence of a metric) vector-space isomorphism between the Clifford algebra and the exterior algebra, which enables us to identify the exterior product with “part” of the Clifford algebra (“geometric”) product (and, I guess, the inner product with the other part). Plus being willing to treat both of these as graded vector spaces “in the algebraist’s sense” – that is, allowing yourself to add up elements of different degrees. (I guess that perspective is unavoidable for the Clifford algebra, since the Clifford product doesn’t preserve homogeneity, but I’m used to thinking of the exterior algebra as graded “in the topologist’s sense” where you can’t add a scalar to a 2-form.) Is that the (or a) right way to think about it?

    • CommentRowNumber11.
    • CommentAuthorTobyBartels
    • CommentTimeAug 29th 2011

    I think that’s right.

    • CommentRowNumber12.
    • CommentAuthorUrs
    • CommentTimeAug 29th 2011
    • (edited Aug 29th 2011)

    It seems like at the heart of it is a canonical (in the presence of a metric) vector-space isomorphism between the Clifford algebra and the exterior algebra,

    Yes. This is called the symbol map. For instance prop. 28 in Meinrenken’s Clifford algebras and Lie groups .

    There is a deeper mechanism behind this simple idea: the Clifford algebra is the quantization of the exterior algebra (this is a bit off tangent for what we have been discussing here, but is of general fundamental importance):

    where fermions are classically described in supergeometry in terms of Grassmann-valued functions, under quantizaton the canonical anticommutaion relation of differential forms {dx i,dx j}=0\{ d x^i , d x^j\} = 0 becomes the Clifford rule {γ i,γ j}=g ij\{\gamma^i , \gamma^j\} = g^{i j} (where you may imagine an invisible \hbar on the right). I think this is also discussed in the above reference. Also at the beginning of Quantum Fields and Strings.

    it sounds like just a different, and perhaps more convenient, way of working with differential forms and their ilk,

    Yes. I think the whole point of it – which I think is very true – is: Clifford algebras are very usefully introduced in introductory mechanics textbooks right at the beginning, not only much later (if the students hangs on for so long at all) with the Dirac equation. Their use clarifies a lot of rotation-geometry and is a much better teaching tool than the traditional i\mathbf{i}-j\mathbf{j}-k\mathbf{k}-yoga.

    • CommentRowNumber13.
    • CommentAuthorMike Shulman
    • CommentTimeAug 30th 2011

    I definitely believe that the notion of the “cross product” of two vectors being another vector ought to be excised from the collective consciousness. I’ve usually thought that the exterior algebra is the best way to think of this sort of thing, but now that I understand the Clifford approach better I can see that it is quite nice, and fits very well with rotations and so on.