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I am confused about translating the definition of the inertia tensor I know into the language of differential geometry. Part of this confusion arises because in every physics textbook I read, the term "coordinates" is used in different places with different meaning (even within the same book).

Anyway, let me try to illustrate what I am familiar with first, and then try to explain my doubts. For example, the definitions I am familiar with are:

Given a (say real) vector space $V$, an $(r,s)$ tensor over $V$ is a multi-linear map $T: \underbrace{V^* \times \dots \times V^*}_{\text{$r$ times}} \times \underbrace{V \times \dots \times V}_{\text{$s$ times}} \to \Bbb{R}$.

and

Given a smooth manifold $M$, a smooth $(r,s)$ tensor field on $M$ is a smooth section $\xi : M \to T^r_s(M)$ of the $(r,s)$ tensor bundle over $M$. (i.e at each point $p$ of the manifold, we have an $(r,s)$ tensor $\xi(p)$ over the tangent space $T_pM$, such that the association $p \mapsto \xi(p)$ is smooth).

Of course, I know a few examples of tensors and tensor fields from basic linear algebra. For example, if $V$ is a finite-dimensional vector space, we can always equip it with an inner product (a $(0,2)$ tensor over $V$). Also, a typical example of a tensor field is a metric tensor field $g$ on a smooth manifold (a $(0,2)$ tensor field). I'm also "acquainted"(i.e I've seen them but haven't had any practice with them) with other examples of tensor fields from physics, such as the stress-energy tensor field in the context of electromagnetism.

Now, the reason I'm not so confused in these cases is because I know precisely what the spaces $V$ and $M$ are, and I know the exact definition (i.e the rule for the map). When it comes to the inertia tensor however, I'm not so sure.


Now, given a rigid body, here's the definition I know (from Landau and Lifshitz, Volume $1$, $\S 32$).

We take a "moving system of coordinates $x_1, x_2, x_3$, which is supposed to be rigidly fixed in the body, and to participate in its motion", and in these coordinates, we define \begin{align} I_{ij} &= \int_{\text{Body}}(\delta_{ij} \lVert x\rVert^2 - x_i x_j) \cdot \rho \, dV. \end{align} (this discussion occurs over a few pages, so I'm summarizing the essentials).

I understood their motivation for such a definition: namely by defining $I_{ij}$ like this, the rotational kinetic energy can be expressed as $T_{\text{rot}} = \dfrac{1}{2}I_{ij} \Omega^i \Omega^j$ ($\vec{\Omega}$ being the angular velocity of the rigid body). However, here are some things I am unclear about:

$1$. What does it mean in a precise and technical manner to have a system of coordinates attached to a point in the body? I understand the intuitive notion that I'm supposed to think of myself as being anchored to a point in the rigid body and "describe how I see things". But I'm having trouble precisely formulating this "simple" idea as a precise mathematical definition. I'm hoping someone can fill in something along the lines of "A coordinate system (____ on a certain space ____ )is a (_____ type of object ____). And a coordinate system attached to a point is (____ something_____)"

$2$. My next doubt is whether the inertia tensor is actually a tensor over a fixed vector space $V$ (if so which vector space? is it $\Bbb{R}^3$? the tangent space at a point of the rigid body? ) or whether it is actually a tensor field over a certain manifold $M$ (if so which one?). The reason I ask this is because I'm aware that in the physics literature, it's not uncommon to leave out the term "field", because it's usually clear from context... but unfortunately it isn't clear to me $\ddot{\frown}$.

$3.$ Following up with ($2$), what is the type/rank of the inertia tensor(field?) (i.e what are $r$ and $s$)? My guess is that based purely on the way it is written, it is a $(0,2)$ tensor (field?) based on the index structure. But I'm not sure because this is only Volume $1$ of Landau and Lifshitz, and from my understanding, at this point they do not make any distinction between upper vs lower placement of indices. Another reason I ask this is because people usually identify $(2,0), (1,1), (0,2)$ tensor (fields) all together using an inner product/ Riemannian metric tensor field. So, I'm wondering, which "type" is the most natural to begin with.

$4.$ Is it possible to define the inertia tensor in a manner which manifestly makes it clear that it is actually a tensor (field?). For example, if we consider $M = \Bbb{R}^4$ as a smooth manifold, then with the identity chart $(\Bbb{R}^4, \text{id}_{\Bbb{R}^4})$, where we denote its four component functions as $\text{id}_{\Bbb{R}^4}(\cdot) = \left( t(\cdot), x^1(\cdot), x^2(\cdot), x^3(\cdot)\right)$, we can define $g := -dt \otimes dt + \delta_{ij}dx^i \otimes dx^j $. Written in this manner, although we have used the component functions of the identity chart, all the operations used (exterior derivative, tensor product) etc are all clearly chart-independent and purely geometric operations. So, the result is very clearly chart-independent and is indeed easily seen to be a (symmetric) $(0,2)$-tensor field on the manifold $M$. So, my question is if we can describe the inertia tensor (field?) in a similar terminology.


So, really, my issue is one of translating between terminology in the math books I'm familiar with and the physics texts which I also read simultaneously, and one about the geomtric way of defining such objects.

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Nice question you got here! Let's go through this step by step:

1) Embedding of the body

To understand the coordinates "rigidly fixed" on the rotating body, think of the object as a manifold $B$ embedded in space $M=\mathbb{R}^3$, i.e. an injective immersion $$\imath:B\rightarrow M.$$ Since we want to discribe a motion of this body, we let this mapping be dependent on time: $$\imath:T\times B\rightarrow M,$$ where $T\subseteq \mathbb{R}$ is a time interval. Now we can introduce those "rigidly fixed" coordinates as coordinates on the manifold $B$. Since the dynamics of this object are fully containt in the embedding map $\imath$, these coordinates are fixed on the body.

2) Moment of inertia

The moment of inertia is a tensor, but of which vector space? We will see that it is a tensor on the tangent space $T_p M$ at the point $p\in M$ around that the body rotates. More specifically, it is a mapping $$I:T_p M \otimes T_p M \rightarrow \mathbb{R}.$$ At this level, this is not a tensor field, but rather a tensor on the tangent space of a specific point in space. However, it can be made one by mapping each point $p\in M$ to the inertia tensor $I_p$ of a rotation around this point. Nevertheless, I'm not sure wheter it is of some use to consider this as a tensor field.

Now we want to connect this to the stuff above. Choose a submanifold $B$ of $\mathbb{R}^3$, then the motion of $B$ around the point $p\in\mathbb{R}^3$ and an angular velocity vector $\omega\in T_p\mathbb{R}^3 \cong \mathbb{R}^3$ is described by the embedding \begin{align} \imath: T\times B &\rightarrow \mathbb{R}^3 \\ (t,x) &\mapsto R[t\omega](x-p) + p. \end{align} Here, $R[v]:\mathbb{R}^3\rightarrow\mathbb{R}^3$ is the rotation map by angle $|v|$ around the axis $v$ through the origin of $\mathbb{R}^3$.

To motivate the definition of the moment of inertia, we will take a look at the kinetic energy $$E=\frac{1}{2} \int_B dV \rho(x) \langle \partial_t \imath(x),\partial_t \imath(x)\rangle$$ of the system, where $\rho:B\rightarrow \mathbb{R^+}$ is the mass density of the body and $\partial_t \imath(t,x)$ its "local velocity". Note that $\partial_t \imath(t,x) = \omega \times (x-p)$ with $\bullet\times\bullet$ the vector product on $\mathbb{R}^3$. Using the identity $$\langle a\times b,a\times b\rangle = |a|^2|b|^2 - \langle a,b\rangle^2$$ then gives rise to $$E=\int_B dV \rho(x) (|\omega|^2 |x-p|^2 - \langle \omega, x-p \rangle^2).$$ Motivated by this expression, we define the moment of inertia tensor as $$I_p(v,w):=\int_B dV \rho(x) (|x-p|^2 \langle v,w\rangle - \langle v,x-p \rangle \langle w,x-p \rangle)$$ and, thus, $E=\frac{1}{2} I_p(\omega,\omega)$. As we see, the moment of inertia is a type $(2,0)$ tensor on the tangent space $T_p M$ and, thus, depends on the point $p$ in space, the body is rotated around. As mentioned above, you can consider the map $I: p\in M \mapsto I_p$ as a tensor field, but I am not sure whether this brings any advantages. I hope this answers your question.

Cheers!

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  • $\begingroup$ This was very helpful, but I'm afraid I still don't fully understand point 1. So I understand that the map $\iota: T \times B \to M$ allows us to think of for each $t \in T$, $\iota[\{t\} \times B] \subset M$ as "how the rigid body $B$ is situated in the ambient space $M$ at time $t$". But I don't understand how the map $\iota$ allows us to introduce coordinates (charts?) on the manifold $B$. Also, I recently had this question on regularity: is it really appropriate to model rigid bodies as smooth manifolds? Because there are several examples where they're clearly not, like pointy cones $\endgroup$ – peek-a-boo Apr 3 at 18:04
  • $\begingroup$ and also cylinders (their edges are not smooth), and I'm sure these are not the only examples. Also, for the most part of your section ($2$), it seems that the only key properties of $B$ that are needed is that it is a subset of an ambient manifold $M$ (in our case $\Bbb{R}^3$), and that we can integrate over it. So it seems more fitting to model a rigid body as a Lebesgue-measurable subset of $M = \Bbb{R}^3$ (perhaps you considered the case $B$ is a smooth submanifold to simplify the discussion?). I appreciate your input on this. $\endgroup$ – peek-a-boo Apr 3 at 18:09
  • $\begingroup$ You don't have to restrict yourself to $B$ being a manifold, however, if you want to talk about coordinates it is an assumption making live easy. You don't need the map $\imath$ to introduce coordinates on $B$. If it is a manifold (not necessary smooth), it is possible to introduce coordinates. The point is that you don't need coordinates to define the inertia tensor. If you want to calculate $I$, you choose appropriate coordinates on $B$ and calculate the components $I_{ij}$ of its coordinate representation. Landau and Lifshitz defines the tensor by this components in a certain coordinate rep $\endgroup$ – Johnny Longsom Apr 4 at 11:41
  • $\begingroup$ I think using a Lebesgue-messurable subset of $\mathbb{R}^n$ instead is possible, however, from a physical perspective describing a body by a topological manifold seems more reasonably to me. $\endgroup$ – Johnny Longsom Apr 4 at 12:03

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