Definition of inertia tensor from a differential geometry viewpoint

[cianfa72]

TL;DR

About the definition of moment of inertia tensor (inertia tensor) from a differential geometry point of view

Hello,
I'd like to better understand the definition of inertia tensor from a mathematical viewpoint.

As discussed here, one defines the (0,2)-rank system's moment of inertia tensor (inertia tensor) $\mathbf I$ w.r.t. the system's CoM. Of course such a tensor $\mathbf I$ depends on the system's orientation in space, i.e. it depends on the "point" representing it in configuration space.

Just to fix ideas, consider a rigid dumbbell. To describe its orientation in 3D space 2 parameters are required (e.g. spherical coordinates $\theta, \varphi$ w.r.t. its CoM), hence the dumbbell configuration space is two-dimensional. Now, at each point in such configuration space, one can assign the (0,2)-rank inertial tensor.

My question: does this assignment actually define a tensor field? AFAIK, a tensor field requires the underlying notion of "tensor bundle" built from a differentiable manifold $M$. Here $M$ is the system's configuration space, however the dimensions do not match.

Namely $M$ as configuration space is two-dimensional (like the dimension of the tangent space at each point) while the inertia tensor "eats" three-dimensional vectors (i.e. angular velocities).

What is actually going on? Thanks.

 

[Orodruin]

The configuration space is 3-dimensional.

You need three Euler angles to uniquely specify the orientation of a frame.

 

[Orodruin]

Yes, however if one restricts to inertia tensors w.r.t. the system's CoM, then in dumbbell case the configuration space is just 2-dimensional, I believe.

No.

https://en.wikipedia.org/wiki/Euler_angles

 

[cianfa72]

Ah yes, my bad (of course we are talking of rigid bodies otherwise the number of dimensions increase).

Therefore the configuration space $M$ as manifold is 3-dimensional. Then, since dimensions match, given a rigid body the inertia tensor w.r.t. its CoM can be understood as tensor field.

Btw, I'm not sure whether the product tensor bundle over the configuration space $M$ is built as product tensor of cotangent spaces as fiber or just 3-dimensional covector spaces as fiber.

P.s. my definition of tensor field is a smooth section of a tensor bundle over a smooth manifold $M$, i.e. a smooth section of $(T^*M)^{\otimes ^p}\otimes (TM)^{\otimes q}$.

In the specific case of inertia tensor field $p=2, q=0$.

 

[cianfa72]

Btw some source claims that inertia tensor is actually a pseudo-tensor. What does it mean ?

My definition of (0,2)-rank tensor $T$: a bilinear map $$T: V \times V \to \mathbb R$$ where $V$ is a vector space over $\mathbb R$.

 

[robphy]

Btw some source claims that inertia tensor is actually a pseudo-tensor. What does it mean ?

Possibly related...
see my answer to
"Inertia tensor as two times contravariant pseudotensor" on physics.stackexchange.com
which quotes the section about the "fourth-order moment of inertia tensor in N-dimensions"
from Synge and Schild's Tensor Calculus (1949, 1969, Dover:1978, ISBN 9780486636122 ).

 

[Orodruin]

It is incorrect that it is a pseudotensor.

Just looking at $T = I_{ab}\omega^a \omega^b/2$ with $T$ a scalar and $\omega$ a pseudovector, $I$ must be a tensor and not a pseudotensor. The LHS being a scalar and the RHS containing two insertions of a pseudovector, there would be a parity mismatch if $I$ was a pseudotensor.

 

[cianfa72]

Just looking at $T = I_{ab}\omega^a \omega^b/2$ with $T$ a scalar and $\omega$ a pseudovector, $I$ must be a tensor and not a pseudotensor.

So the question is what a pseudo-vector like $\omega$ is.

A vector is an element of a vector space, and its components transform accordingly as vector basis changes. What about a pseudo-vector?

 

[Orodruin]

So the question is what a pseudo-vector like $\omega$ is.

A vector is an element of a vector space, and its components transform accordingly as vector basis changes. What about a pseudo-vector?

A pseudo tensor has the same transformation properties as a tensor under rotations, but transforms with an additional sign under reflections.

 

[cianfa72]

A pseudo tensor has the same transformation properties as a tensor under rotations, but transforms with an additional sign under reflections.

Ok, consider the case of pseudo-vector like $\omega$. As far as I can tell, a pseudo-vector is an element of a set with a structure of exterior algebra over a vector space (see PSE).

Said $\mathbb R^3$ the underlying vector space then, for instance, $\omega$ is an element of $\bigwedge(\mathbb R^3)$. The latter has itself a structure of vector space since an algebra is a vector space.

Now let's do a change of basis in $\mathbb R^3$. How are the $\omega$'s components affected by a change of basis in $\mathbb R^3$ (i.e. by the matrix transformation $T$) ? From what you said if the $\mathbb R^3$ basis's transformation is a reflection then the $\omega$'s components in the "new" $\bigwedge(\mathbb R^3)$ associated basis will pick a negative sign w.r.t. its components in the "old" associated basis.

 

[Orodruin]

You really do not need to complicate things. Using the regular matrix representation of $A \in O(3)$, a vector $x$ transforms as $x \to Ax$ whereas a pseudo vector $y$ transforms according as $y \to \det(A) A y$.

 

[cianfa72]

You really do not need to complicate things. Using the regular matrix representation of $A \in O(3)$, a vector $x$ transforms as $x \to Ax$ whereas a pseudo vector $y$ transforms according as $y \to \det(A) A y$.

Yes, however the point is: said $A$ the regular matrix representation of transformation $T$ in $\mathbb R^3$, the transformation law for components $x$ is always the same, namely $x \to Ax$.

The confusing point here is that the associated matrix representation in $\bigwedge(\mathbb R^3)$, say $C$, is related to $A$ by $$C=\det(A)A$$ For the pseudo-vector components $y$ in fact we get: $y \to Cy$

 

[Orodruin]

Why is this confusing? There are several examples of objects that transform like this, such as angular momentum and the magnetic field.

 

[cianfa72]

Why is this confusing? There are several examples of objects that transform like this, such as angular momentum and the magnetic field.

Yes, however the discussion is about the formal viewpoint. There exists only one transformation law for the components of a vector when changing the vector space's basis. Full stop.

Since pseudo-vector $w$ is element of a space with a vector space structure (i.e. it is a vector by very definition), its components transform according that unique law. The point is that a pseudo-vector $w$ is an element of a vector space built "on top" of the underlying vector space $V$.

Pick $V=\mathbb R^3$ as vector space and take the exterior algebra $\bigwedge(\mathbb R^3)$ built on top of it. Then angular momentum or magnetic field are actually pseudo-vectors, that is elements of $\bigwedge(\mathbb R^3)$.

The main feature is that, from the regular matrix representation $A$ of a change of basis in $V=\mathbb R^3$, the corresponding/associated pseudo-vector's components transformation is $y \to \det(A)Ay$. Note that $y$ is supposed to be the tuple of the pseudovector's components in a basis of $\bigwedge(\mathbb R^3)$).

 

[Orodruin]

I still don’t see what is complicated about this.

 

[cianfa72]

Following the same line of reasoning, I believe a rank 2 pseudo-tensor is an element of the exterior algebra built over the relevant tensor product space, e.g. $$\bigwedge({\mathbb R^3}^{\otimes ^2})$$ where the underlying vector space $V$ is taken as $\mathbb R^3$.