Wheeler, Mechanics

This is nine separate equations, one for each value of i and each value of j. However, since the moment of
inertia tensor is symmetric, I ij = I ji , only six of these numbers are independent. Once we have computed
each of them, we can work with the much simpler expression,
N i = d dt I ijω j
The sum on the right is the normal way of taking the product of a matrix on a vector. The product is the
angular momentum vector,
L i = I ij ω j
and we have, simply,
N i = dL i
dt
We can also write the rotational kinetic energy of a rigid body in terms of the moment of inertia tensor,
T = 1 ∫
v 2 dm
2
= 1 ∫
(ω × r) · (ω × r) ρ(x)d 3 x
2
Working out the messy product,
(ω × r) · (ω × r) = ε ijk ε imn ω j r k ω m r n
= (δ jm δ kn − δ jn δ km ) ω j r k ω m r n
= ω i ω j
(
δij r 2 − r i r j
)
Therefore,
T = 1 2
∫
ω · (r × (ω × r)) ρ(x)d 3 x
= 1 2 ω iω j
∫ (δij
r 2 − r i r j
)
ρ(x)d 3 x
= 1 2 I ijω i ω j
The metric tensor We have already used the Pythagorean formulat to specify the length of a curve, but
the usual formula works only in Cartesian coordinates. However, it is not difficult to find an expression valid
in any coordinate system – indeed, we already know the squared separation of points in Cartesian, polar and
spherical coordinates,
ds 2 = dx 2 + dy 2 + dz 2
ds 2 = dρ 2 + ρ 2 dφ 2 + dz 2
ds 2 = dr 2 + r 2 dθ 2 + r 2 sin 2 θ dφ 2
respectively. Notice that all of these forms are quadratic in coordinate differentials. In fact, this must be
the case for any coordinates. For suppose we have coordinates y i given as arbitrary invertible functions of a
set of Cartesian coordinates x i ,
y i = y i (x j )
x i = x i (y j )
Then the differentials are related by
dx i = ∂x i
∂y j
dy j
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