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In the mechanics of rigid body dynamics in Cartesian space $\mathbb{R}^n$, the moment of inertia of a rigid body is the analog of mass for rotational dynamics?. In linear dynamics?, we have the formula
which says that the momentum $p$ is proportional to the velocity $v$. Similarly, in rotational dynamics, we have the analogous formula
where $L$ is the angular momentum, $\Omega$ is the angular velocity, and $I$ is the moment of inertia.
However, the rotational equation is somewhat more complicated than the linear one: firstly because $L$ and $\Omega$ are not naturally vectors but bivectors; and secondly because they are not necessarily proportional, so that $I$ cannot be a scalar. In general, the moment of inertia is a linear function
so that the above equation becomes simply
This linear function is additionally symmetric with respect to the induced inner product on $\bigwedge^2 \mathbb{R}^n$, so it can be represented in coordinates by a symmetric $\frac{n(n-1)}{2} \times \frac{n(n-1)}{2}$ matrix.
Similarly, differentiating this equation once with respect to time (and assuming that $I$ is constant as it is for a rigid body), we have
relating the total torque $\tau$ to the angular acceleration? $\alpha$ — this is the rotational analogue of Newton's second law $F = m a$ (where $m$ must be constant).
In low dimensions, the situation can be (and usually is) simplified.
In two dimensions, bivectors form a one-dimensional vector space, so that the moment of inertia is simply a scalar.
In three dimensions, bivectors form a three-dimensional vector space, so that the moment of inertia can be represented by a symmetric $3 \times 3$ matrix. Additionally, in three dimensions, there is an isomorphism between bivectors and vectors (once we choose an orientation to go with our inner product); so that angular velocity and momentum can be (and usually are) identified with vectors, and the moment of inertia with a symmetric rank-2 tensor.
In terms of the discussion at Hamiltonian dynamics on Lie groups, the rigid body dynamics in $\mathbb{R}^n$ is given by Hamiltonian motion on the special orthogonal group $SO(n)$. It is defined by any left invariant? Riemannian metric
hence a bilinear non-degenarate form on the Lie algebra $\mathfrak{so}(n)$ (not necessarily the Killing form).
This bilinear form is the moment of inertia. (For instance AbrahamMarsden, section 4.6.)
If a rigid body has mass density? $\rho$, then its angular momentum is defined in terms of $\Omega$ by the $n$-dimensional integral
over all space, where $\vec{x}$ 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 $\Omega$ is the same everywhere (as for a rigid body), then we may view this as a function from $\Omega$ to $L$; 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
or around page 56 of
and around slide 6 in
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.