nLab moment of inertia




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In the mechanics of rigid body dynamics in Cartesian space n\mathbb{R}^n, the moment of inertia of a rigid body is the analog of mass for rotational dynamics?. In linear dynamics?, we have the formula

p=mv p = m v

which says that the momentum pp is proportional to the velocity vv. Similarly, in rotational dynamics, we have the analogous formula

L=IΩ L = I \Omega

where LL is the angular momentum, Ω\Omega is the angular velocity, and II is the moment of inertia.

However, the rotational equation is somewhat more complicated than the linear one: firstly because LL and Ω\Omega are not naturally vectors but bivectors; and secondly because they are not necessarily proportional, so that II cannot be a scalar. In general, the moment of inertia is a linear function

I: 2 n 2 n I \colon \wedge^2 \mathbb{R}^n \to \wedge^2 \mathbb{R}^n

so that the above equation becomes simply

L=I(Ω). L = I(\Omega).

This linear function is additionally symmetric with respect to the induced inner product on 2 n\bigwedge^2 \mathbb{R}^n, so it can be represented in coordinates by a symmetric n(n1)2×n(n1)2\frac{n(n-1)}{2} \times \frac{n(n-1)}{2} matrix.

Similarly, differentiating this equation once with respect to time (and assuming that II is constant as it is for a rigid body), we have

τ=Iα, \tau = I \alpha ,

relating the total torque τ\tau to the angular acceleration? α\alpha — this is the rotational analogue of Newton's second law F=maF = m a (where mm must be constant).

In low dimensions

In low dimensions, the situation can be (and usually is) simplified.

In Hamiltonian dynamics

In terms of the discussion at Hamiltonian dynamics on Lie groups, the rigid body dynamics in n\mathbb{R}^n is given by Hamiltonian motion on the special orthogonal group SO(n)SO(n). It is defined by any left invariant? Riemannian metric

,Sym C (G) 2Γ(TG) \langle -,-\rangle \in Sym^2_{C^\infty(G)} \Gamma(T G)

hence a bilinear non-degenarate form on the Lie algebra 𝔰𝔬(n)\mathfrak{so}(n) (not necessarily the Killing form).

This bilinear form is the moment of inertia. (For instance AbrahamMarsden, section 4.6.)

In terms of mass density

If a rigid body has mass density? ρ\rho, then its angular momentum is defined in terms of Ω\Omega by the nn-dimensional integral

L=ρx(xΩ)d nx L = \int \rho \vec{x} \wedge (\vec{x} \cdot \Omega) \,d^n x

over all space, where x\vec{x} is the vector from the origin to the point of integration, \cdot denotes the interior product? of a vector with a bivector (yielding a vector), and \wedge denotes the exterior product of two vectors (yielding a bivector).

When Ω\Omega is the same everywhere (as for a rigid body), then we may view this as a function from Ω\Omega to LL; this function is the moment of inertia.


A classical textbook discussion is for instance section 4.6 of

A pedestrian discussion of moment of inertia in terms of bivectors that applies in any dimension of space(spacetime) is around page 74 of

  • Chris Doran, Anthony Lasenby, Geometric Algebra for Physicists Cambridge University Press

or around page 56 of

  • Chris Doran, Anthony Lasenby, Physical applications of geometric algebra (pdf)

and around slide 6 in

  • Anthony Lasenby, Chris Doran and Robert Lasenby, Rigid Body Dynamics and Conformal Geometric Algebra (pdf)

These authors amplify the canonical embedding of bivectors into the Clifford algebra, which they call “Geometric Algebra”.

Last revised on August 31, 2011 at 16:57:56. See the history of this page for a list of all contributions to it.