# Moment of Inertia Tensor

1. Oct 10, 2013

### dpa

Generally, when we talk about moment of inertia, we talk about rotation and inherently, we talk about moment of inertia about an axis.

But when we talk about inertia tensor, we calculate about a point. Is there a reason for this difference? Am I missing something?

I am new to tensors.

2. Oct 10, 2013

### The_Duck

One way to think about it is that the inertia tensor contains the information that lets you calculate the moment of inertia around *any* axis that passes through a given point.

3. Oct 10, 2013

### SteamKing

Staff Emeritus
The components of the inertia tensor are calculated in the usual fashion. You still do the calculations about the three coordinate axes, if they are convenient to do so. In most instances, you are interested in the moments of inertia transferred from whatever calculation coordinate system to a coordinate system which has its origin at the centroid of the body. This may be the source of your confusion.

4. Oct 10, 2013

### arildno

3-D rotations are nasty. For example, the instantaneous angular velocity vector need not be strictly parallell with the angular momentum vector.

Do remember though, that ESSENTIALLY, all torques are computed relative to a POINT, not relative to an axis.

5. Oct 10, 2013

### D H

Staff Emeritus
Very. One simple example: Perform rotation A followed by rotation B. Then do it again, but this time perform rotation B first. It doesn't make any difference with 2-D rotations. Rotations in 2-D are commutative. That's no longer true in three dimensions (or higher).

Another example is the inescapable fact that angular velocity and angular momentum do not necessarily point in the same direction with 3-D rotations. A rigid body set into rotation in empty space generally will tumble. Angular momentum is a conserved quantity but angular velocity is not.

Rotations in 4-D space or higher are nastier yet. The concept that rotation is about an axis is something that pertains to 3-D space only. On the other hand, the 2-D concept of rotation about a point (or parallel to a plane) does generalize to higher dimensions.

6. Oct 10, 2013

### arildno

However, for FREE rotations in 3-D, a beatiful result is still present, as my new signature shows.

7. Oct 10, 2013

### D H

Staff Emeritus
Yep. Love that expression. It's so nonsensical (or at least counterintuitive), but then again, nonsensical (or at least counterintuitive) is a perfect description of free rotations in 3-D space.

8. Oct 10, 2013

### arildno

H. Goldstein called it a "jabberwockian phrase", and I think that encapsulates the weird elegance of both the phrase and the phenomenon it describes.

9. Oct 10, 2013

### D H

Staff Emeritus
Bringing this discussion back to the original question,
Yes. You are missing something. You were taught in freshman physics that $T=I\alpha$ is the rotational equivalent of $F=ma$. That's only true for those special cases where you can ignore the tensorial nature of the inertia tensor, or where motion is constrained to two dimensions. That was a "lie to children" (http://en.wikipedia.org/wiki/Lie-to-children). It's more complicated in general, and you are now deemed to have adequate mathematical understanding to *start* taking the next step toward understanding rotation.

Filling in that "missing something" starts with understanding Euler's equations. Euler didn't use tensors, so his notation is a bit verbose. It becomes nice and simple with a tensorial notation. To really understand rotations you need to know a bit about group theory (the group SO(3), in particular) and you need to know a bit about Lie groups / Lie algebras.

10. Oct 10, 2013

### arildno

"we talk about rotation and inherently, we talk about moment of inertia about an axis."

As DH points out, freshman courses speedily goes over to rotation about a fixed axis.
This means it is very easy to forget what was breezily derived at the very start of the course, namely that when we compute the torques about a (fixed) point, we gain the formula:
$$\vec{\tau}=\frac{d\vec{L}}{dt}$$
where $\vec{\tau}$ are the externally applied torque, and $\vec{L}$ the quantity called "angular momentum".

THAT equation is (almost*) perfectly general for torques about a fixed point in classical mechanics, but it is a wolf in sheep's clothing.

In freshman courses, most of the wolf's teeth are pulled out to begin with (DH has given you the names of a few of those teeth), by saying we limit ourselves to cases of rotation about a fixed axis.
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*It holds perfectly for the ideal, perfectly rigid body. For other types of objects, the left hand side of the equation might get a lot nastier, not just RHS (which is, in 3-D, already exceedingly nasty for the perfectly rigid body as well).

Last edited: Oct 10, 2013