Solution

Let a change of orthonormal coordinates be represented by a rotation matrix

R

L_{i}^{'} = R_{i p} L_{p}, ω_{j}^{'} = R_{j q} ω_{q} .

Using

L_{i} = I_{ij} ω_{j}

in both frames gives

I_{ij}^{'} R_{j q} ω_{q} = R_{i p} I_{pq} ω_{q}

for every vector

ω

. Multiplication by

R^{- 1} = R^{T}

yields

I_{ij}^{'} = R_{i p} R_{j q} I_{pq},

which is exactly the transformation law for a rank-two tensor. Thus

I_{ij}

is a rank-two tensor.

Put

r = x - a

r \times (ω \times r) = r^{2} ω - r (r \cdot ω) .

Therefore

I_{ij} (a) = ρ \int_{B} [(x_{k} - a_{k}) (x_{k} - a_{k}) δ_{ij} - (x_{i} - a_{i}) (x_{j} - a_{j})] d V .

At the centre of mass,

I_{ij} (0) = ρ \int_{B} (x_{k} x_{k} δ_{ij} - x_{i} x_{j}) d V .

Expanding the first formula, the terms linear in

x_{i}

vanish because the centre of mass is the origin, while

ρ \int_{B} d V = M

. Hence the parallel axis theorem in tensor form is

I_{ij} (a) = I_{ij} (0) + M (a_{k} a_{k} δ_{ij} - a_{i} a_{j}) .

Solved by gpt-5.6-sol high.

 Ancestors (11)