Inertia tensor

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 6; Problem 6.11.

In basic physics, the moment of inertia is usually defined by the integral

\displaystyle  I=\int_{V}r^{2}\rho\left(\mathbf{r}\right)d^{3}\mathbf{r} \ \ \ \ \ (1)

where {r} is the perpendicular distance of the volume element from the axis of rotation and {\rho} is the mass density. In fact, this is true only for special cases, where the mass distribution is symmetric with respect to the axis (cases such as a sphere, cylinder, etc). In rotational motion, the angular momentum of a spinning rigid body (that is, a body which does not deform as it moves) is given by {\mathbf{L}=I\boldsymbol{\omega}} where {\boldsymbol{\omega}} is the angular velocity vector.

In the more general case, where the object isn’t symmetric, the moment of inertia becomes a tensor, and the angular momentum equation becomes (in 3-d):

\displaystyle  L^{i}=I^{ij}\omega_{j} \ \ \ \ \ (2)

The derivation of this tensor would take us too far afield here, but basically, the off-diagonal elements of {I^{ij}} measure the amount of asymmetry in various directions. For example, in rectangular coordinates

\displaystyle  I^{xy}=I^{yx}=\int xy\rho\left(\mathbf{r}\right)d^{3}\mathbf{r} \ \ \ \ \ (3)

where the coordinates are measured relative to the centre of mass. For a symmetric object, this integral is always zero, since any mass element at {\mathbf{r}} is balanced by an equal mass element at {-\mathbf{r}}.

If we start with a standard 3-d rectangular system and rotate it by an angle {\theta} about the {z} axis to get the primed system, then the transformation can be found by considering the projection of the unit vectors {\hat{\mathbf{x}}} and {\hat{\mathbf{y}}} onto the {\hat{\mathbf{x}}^{\prime}} and {\hat{\mathbf{y}}^{\prime}} axes.

\displaystyle   \hat{\mathbf{x}} \displaystyle  = \displaystyle  \hat{\mathbf{x}}^{\prime}\cos\theta-\hat{\mathbf{y}}^{\prime}\sin\theta\ \ \ \ \ (4)
\displaystyle  \hat{\mathbf{y}} \displaystyle  = \displaystyle  \hat{\mathbf{x}}^{\prime}\sin\theta+\hat{\mathbf{y}}^{\prime}\cos\theta \ \ \ \ \ (5)

By taking the scalar product of both these equations with {\hat{\mathbf{x}}^{\prime}} we can get the components of {\hat{\mathbf{x}}^{\prime}} along the original axes, and similarly for {\hat{\mathbf{y}}^{\prime}}. For example {\hat{\mathbf{x}}^{\prime}\cdot\hat{\mathbf{x}}=\hat{\mathbf{x}}^{\prime}\cdot\left(\hat{\mathbf{x}}^{\prime}\cos\theta-\hat{\mathbf{y}}^{\prime}\sin\theta\right)=\cos\theta} is the component of {\hat{\mathbf{x}}^{\prime}} in the {\hat{\mathbf{x}}} direction. We get

\displaystyle   \hat{\mathbf{x}}^{\prime} \displaystyle  = \displaystyle  \hat{\mathbf{x}}\cos\theta+\hat{\mathbf{y}}\sin\theta\ \ \ \ \ (6)
\displaystyle  \hat{\mathbf{y}}^{\prime} \displaystyle  = \displaystyle  -\hat{\mathbf{x}}\sin\theta+\hat{\mathbf{y}}\cos\theta \ \ \ \ \ (7)

If we now take an arbitrary vector {\mathbf{r}=x\hat{\mathbf{x}}+y\hat{\mathbf{y}}} and take its scalar product with {\hat{\mathbf{x}}^{\prime}}, we get

\displaystyle   \mathbf{r}\cdot\hat{\mathbf{x}}^{\prime} \displaystyle  = \displaystyle  x\cos\theta+y\sin\theta\ \ \ \ \ (8)
\displaystyle  \displaystyle  = \displaystyle  x^{\prime} \ \ \ \ \ (9)

Similarly for {y^{\prime}}:

\displaystyle  y^{\prime}=-x\sin\theta+y\cos\theta \ \ \ \ \ (10)

The transformation partials are then

\displaystyle  R_{\;\; j}^{i}=\frac{\partial x^{\prime i}}{\partial x^{j}}=\left[\begin{array}{ccc} \cos\theta & \sin\theta & 0\\ -\sin\theta & \cos\theta & 0\\ 0 & 0 & 1 \end{array}\right] \ \ \ \ \ (11)

Now suppose we have an inertia tensor that is diagonal (this is true for symmetric objects, but it is always possible to find some principal axes where this is true for any object) so that

\displaystyle  I^{ij}=\left[\begin{array}{ccc} I_{1} & 0 & 0\\ 0 & I_{2} & 0\\ 0 & 0 & I_{3} \end{array}\right] \ \ \ \ \ (12)

We can transform this to the rotated system using the usual tensor transformation rule:

\displaystyle  I^{\prime ij}=\frac{\partial x^{\prime i}}{\partial x^{k}}\frac{\partial x^{\prime j}}{\partial x^{l}}I^{kl}=\left[\begin{array}{ccc} I_{1}\cos^{2}\theta+I_{2}\sin^{2}\theta & \left(I_{2}-I_{1}\right)\cos\theta\sin\theta & 0\\ \left(I_{2}-I_{1}\right)\cos\theta\sin\theta & I_{2}\cos^{2}\theta+I_{1}\sin^{2}\theta & 0\\ 0 & 0 & I_{3} \end{array}\right] \ \ \ \ \ (13)

This is equivalent to a matrix product as we can see by taking it in stages. First consider the intermediate product {P^{il}}

\displaystyle  P^{il}=\frac{\partial x^{\prime i}}{\partial x^{k}}I^{kl}=R_{\;\; k}^{i}I^{kl} \ \ \ \ \ (14)

This is the sum over the product of elements in the {i}th row of {R} with elements in the {l}th column of {I}. Thus this is a matrix product with the factors in the order {RI}. Now consider

\displaystyle  I^{\prime ij}=P^{il}\frac{\partial x^{\prime j}}{\partial x^{l}} \ \ \ \ \ (15)

This time, we’re summing over the product of elements in the {i}th row of {P} with elements in the {j}th row of {\frac{\partial x^{\prime j}}{\partial x^{l}}}. In order to make this a matrix product, we have to sum over the row elements of one matrix multiplied by the column elements of the second matrix, so we need to take the transpose of {\frac{\partial x^{\prime j}}{\partial x^{l}}} to make this work. That is

\displaystyle  I^{\prime}=PR^{T}=RIR^{T} \ \ \ \ \ (16)

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