7
$\begingroup$

The tensors used in physics, 2 rank tensors, have to be thinked as linear operators which send a vector into another vector or as multilinear maps which send two different vectors into a scalar?

I'm getting confused as long as I read texts about inertia tensor or polarizability tensor, in "The Raman effect, a unified treatment of the theory of Raman Scattering by Molecules-Derek A. Long, A10 paragraph pag. 417" is said that a rank two tensor is to be understood as a direct product $AB$ of two vectors and the tensor $T^{2}(V)=AB(V)=A(B\cdot V)$ so that it takes a vector ($V$) and gives another vector since $B\cdot V$ is the dot product. According to this definition one can associate a $3X3$ matrix to this tensor, in cartesian coordinates, with components given by the result of the direct product of two vectors $A$ and $B$ so $A_xB_x$, $A_xB_y$ ecc.

On the other hand, like is said in "Manifolds, tensor analysis and applications-Marsden, Ratiu and Abraham" and in this question Understanding and Expressing the Definition of Inertia Tensor in the Language of Differential Geometry, a tensor can be intended as a multilinear map which takes vectors and covectors and gives a scalar, with this definition the inertia tensor has components given by $I(v,w)=\int_B(r^2\langle v,w\rangle - \langle \vec{r},v\rangle \langle \vec{r},w\rangle)\,dm$ and of course it takes two vectors $(v,w)$ and gives a scalar.

My questions are:

1) What is the truth? In this physical cases how may I intend a tensor? Are these equivalent definitions? For example for inertia tensor we have $\vec{P}=I\vec{\omega}$, is $I$ to be intended as a tensor that takes the vector $\omega$ and gives another vector $\vec{P}$ or as the integral I wrote previously?

2) If one can intend tensors as a direct product of two vectors $A,B$ what is the physical meaning of these two vectors?

$\endgroup$
1
  • 3
    $\begingroup$ At least computationally, these can be equivalent notions because of currying, i.e., the 1-1 correspondence between a function $A\times B\to C$ and a function $A \to (B \to C)$. $\endgroup$
    – march
    yesterday

2 Answers 2

4
$\begingroup$

tldr.

For Q1, there is no one correct way of writing things. Describing inertia as a $(0,2)$ tensor as in the link, and as an endomorphism are both valid approaches.

Next, the term "a rank 2 tensor" is not specific enough, so if you're just starting to learn about tensors I suggest you be much more precise (as Abraham, Marsden, Ratiu definitely are); you should speak of tensors of type/valence $(r,s)$. Also, the definition of a rank-2 tensor you quoted, as a direct(tensor) product of two vectors is not general enough.


Different Types of Tensors.

Fix an $n$-dimensional vector space $V$ over $\Bbb{R}$. In the linked answer, there is a precise definition for an $(r,s)$ tensor over $V$ in terms of multilinear maps, and the symbol $T^r_s(V)$ denotes the set of all such tensors. If you know some more abstract algebra, you can equivalently define $T^r_s(V):= \underbrace{V\otimes\cdots\otimes V}_{\text{$r$ times}}\otimes \underbrace{V^*\otimes\cdots \otimes V^*}_{\text{$s$ times}}$, with the understanding that $T^0_0(V):=\Bbb{R}$. Now, the definition of rank-2 tensor you quoted, as a direct (tensor) product of two vectors, is not general enough. Elements of the form $v_1\otimes\cdots\otimes v_r\otimes\omega^1\otimes\cdots\otimes\omega^s$, where each $v_i\in V,\omega^i\in V^*$ are referred to as decomposable tensors in $T^r_s(V)$ or as pure tensors in $T^r_s(V)$. The set of pure tensors is not equal to $T^r_s(V)$, rather we have $T^r_s(V)$ equals the linear span of the set of pure tensors. So, the tensor space $T^r_s(V)$ is much larger than that of just pure tensors.

In the link you quoted, the moment of inertia tensor about the point $p\in\Bbb{R}^3$ is denoted as $I_p$, and it is defined as a $(0,2)$ tensor on $\Bbb{R}^3$ (actually on $T_p\Bbb{R}^3$, but there's no real need here for the distinction), so $I_p\in T^0_2(\Bbb{R}^3)$. As a fun fact, note that if $p$ is the center of mass of the body $B$ in question, $M$ the total mass, and $\xi\in\Bbb{R}^3$ denotes a 'separation vector', then the moment of inertia about the different points $I_{p+\xi}$ and $I_{p}$ are related as follows: for all $v,w\in\Bbb{R}^3$, \begin{align} I_{p+\xi}(v,w)&=I_p(v,w)+M\left(\|\xi\|^2\langle v,w\rangle-\langle \xi,v\rangle\langle \xi,w\rangle\right). \end{align} In components, $(I_{p+\xi})_{ij}=(I_p)_{ij}+M(\|\xi\|^2\delta_{ij}-\xi_i\xi_j)$, or in even more common notation, the points are suppressed in the notation, and we instead use primes: \begin{align} I'_{ij}&=I_{ij}+M\left(\|\xi\|^2\delta_{ij}-\xi_i\xi_j\right). \end{align} This is the standard transformation formula for the inertia tensor about two points, where one is the center of mass (I know there's a more colloquial term for this, and it's used often in introductory mechanics courses to calculate off-center moment of inertia, but I forget what it's called).


Relating different types of Tensor Spaces.

Typically in physics when one uses the term rank 2 tensor or second rank tensor (on a given vector space $V$), one means an element belonging to one of the following three spaces $T^2_0(V)$ or $T^1_1(V)$ or $T^0_2(V)$. Now, you can easily check that these vector spaces are all of dimension $(\dim V)^2=n^2$, so by basic linear algebra, they're all isomorphic (they have the same dimension as the space of matrices $M_{n\times n}(\Bbb{R})$ as well). However, two spaces being isomorphic by itself is not a good enough reason to arbitrarily identify them. On the other hand, if we have an inner product (as is the case in $\Bbb{R}^3$) then there is a standard isomorphism $V\cong V^*$ given to us by the inner product. Namely, the mapping $g^{\flat}:V\to V^*$ defined as $g^{\flat}(v):= \langle v, \cdot\rangle = \left(w\mapsto \langle v, w\rangle\right)$. This is injective because if $g^{\flat}(v)=0$, then $\langle v,\cdot\rangle=0$, and so in particular $\langle v,v\rangle=0$, so by positive-definiteness, $v=0$. Now, by finite-dimensionality, the rank-nullity theorem tells us $g^{\flat}$ is in fact an isomorphism; it is comon to denote the inverse map as $g^{\sharp}:V^*\to V$. These maps are called the musical isomorphisms; $g^{\flat}$ is called the index-lowering map and $g^{\sharp}$ is called the index raising map. Summarizing, we've just established the basic fact that:

A choice of inner product provides us with an isomorphism of a finite-dimensional vector space and its dual.

If you're reading Abraham, Marsden, Ratiu then you'll definitely see this.

As a result of this isomorphism, we get induced isomorphisms $T^2_0(V)\cong T^1_1(V)\cong T^0_2(V)$ is a natural way (this is a stronger statement than just having the same finite dimension, because in that case, the isomorphism can be pretty arbitrary, but here there is a very specific one which utilizes the inner product structure). As a final note, there is also a natural isomorphism $T^1_1(V)\cong \text{End}(V)$ (independent of any inner products).

For the particular case of the inertial tensor, the link describes it as $I_p\in T^0_2(\Bbb{R}^3)$. So, you can either use this general theory to say there is a corresponding linear transformation $J_p:\Bbb{R}^3\to\Bbb{R}^3$ (i.e $J_p\in\text{End}(\Bbb{R}^3)$). For the sake of clarity, I'll unwind the above definitions in this special case. So, the integral formula gives a bilinear map $I_p:\Bbb{R}^3\times\Bbb{R}^3\to\Bbb{R}$. As a first step, consider the mapping $\Bbb{R}^3\to (\Bbb{R}^3)^*$ defined as $v\mapsto I_p(v,\cdot)$. The empty slot can still be fed a vector to produce a number overall, that is why $I_p(v,\cdot)$ lives in the dual space $(\Bbb{R}^3)^*$, by definition. Now, one can use the inner product-induced isomorphism to map a covector in $(\Bbb{R}^3)^*$ to a vector in $\Bbb{R}^3$. If you carry this out in full, you'll see that the linear transformation $J_p:\Bbb{R}^3\to\Bbb{R}^3$, very explicitly written out has the following rule: \begin{align} J_p(v)&:=g^{\sharp}(I_p(v,\cdot)) = \begin{pmatrix} I_p(v,e_1)\\ I_p(v,e_2)\\ I_p(v,e_3) \end{pmatrix}\in\Bbb{R}^3, \end{align} where $e_i\in\Bbb{R}^3$ is the vector with $1$ in the $i^{th}$ spot and $0$ elsewhere. Conversely, if you're given the inertia endomorphism $J_p:\Bbb{R}^3\to\Bbb{R}^3$, then you can recover $I_p:\Bbb{R}^3\times\Bbb{R}^3\to\Bbb{R}$ by the formula \begin{align} I_p(v,w)= \langle J_p(v),w\rangle. \end{align}


Final Remarks.

  • In $\Bbb{R}^3$ people are so accustomed to these isomorphisms that they don't care about distinguishing these spaces. In more physicsy language: raising and lowering indices by the Kronecker delta is trivial, so in $\Bbb{R}^3$, we avoid distinguishing the index placement $T^{ij}$ vs $T^i_{\,j}$ vs $T_{ij}$.

  • Strictly speaking, $I_p$ and $J_p$ as defined above are different mathematical objects; the former is a $(0,2)$-tensor on $\Bbb{R}^3$, while the latter is an endomorphism on $\Bbb{R}^3$. Due to this difference, I have used different letters for them. However, once you understand how the isomorphisms work, it's very common to drop the strict notational distinction, and refer to both guys as "the inertia tensor of the body about the point $p$", and use the same symbol $I_p$ for them. The placement of indices indicates the intended meaning $I_{ij}$ vs $I^{i}_j$, but again as mentioned above, in $\Bbb{R}^3$ people usually don't pay so much attention to these details.

  • In more matrix terms, we can say that different matrix representations of the endomorphism $J_p$ are related by matrix similarity (i.e $A=PA'P^{-1}$, where $P$ is an invertible matrix) whereas different matrix representations of the $(0,2)$-tensor $I_p$ are related by a congurence (i.e $A= Q A' Q^t$, where $Q$ is an invertible matrix).

  • A final curious fact is that $I_p$ is a symmetric $(0,2)$-tensor, so the corresponding inertia endomorphism $J_p$ is self-adjoint (with respect to the standard inner product on $\Bbb{R}^3$). The converse is also true: if $J_p$ is self-adjoint, then the corresponding $I_p$ is symmetric.

  • So far I've addressed your first question. The answer to your second question is that in general there no physical significance, because $A\otimes B=\left(cA\right)\otimes \left(\frac{1}{c}B\right)$ for all non-zero scalars $c$. So, it makes no sense to attach any real significance to these guys. On the other hand, for the inertia tensor, we can try to be more specific. For instance, the symmetry of $I_p$ implies that the endomorphism $J_p$ is self-adjoint as I mentioned above. The real spectral theorem tells us that there is an orthonormal basis $\{e_1,e_2,e_3\}$ of $\Bbb{R}^3$ consisting of eigenvectors of $J_p$, say with corresponding eigenvalues $\lambda_1,\lambda_2,\lambda_3$ (be warned, although I'm using the $e_i$ notation here, I do not necessarily mean $(1,0,0),(0,1,0),(0,0,1)$... the point is that after rotating the coordinate system, this is what the vectors look like). Note that unless the mass-distribution is degenerate in some sense (eg concentrated on a line/plane), these eigenvalues are strictly positive. Now, let $\{\epsilon^1,\epsilon^2,\epsilon^3\}$ denote the dual basis for $(\Bbb{R}^3)^*$. Then, we can write \begin{align} I_p&=\sum_{i=1}^3\lambda_i\,\epsilon^i\otimes\epsilon^i. \end{align} This is the eigen-decomposition for $I_p$ (induced by that of $J_p$). So, in this case, we can assign a specific interpretation for each term: $\lambda_i$ is the eigenvalue/principal moment of inertia, and $e_i$'s are an orthonormal basis of eigenvectors (so the frame $(e_1,e_2,e_3)$ is said to constitute the principal axes of inertia), and the $\epsilon^i$ are the duals.

$\endgroup$
1
$\begingroup$
  1. They are not equivalent definitions because they express two different cases of what an object a rank-2 tensor is. First, let's assume you have a tensor $\mathbf{T}$ in a $3 \times 3$ matrix representation. If this tensor acts upon a 3-vector $v$, then:

\begin{equation} \mathbf{T} v = v' \, , \quad v , v' \in \mathbb{R}^{3} \end{equation}

It's a simple fact of linear algebra, since a square matrix $n \times n$ acting on a $n$-vector (essentially a $n \times 1$ matrix) will give back another $n$-vector. This is precisely your first definition.

Now, if you take the covector $w = u^{T}$ (where $u \in \mathbb{R}^{3}$) and the vector $v$, then the quantity $w v$ is a scalar (it's essentially the inner product $\vec{u} \cdot \vec{v}$). We can however also find a scalar had we inserted a $3 \times 3$ matrix in the middle. For our analysis, we'd like that matrix to be a tensor:

\begin{equation} w \mathbf{T} v = u^{T} \mathbf{T} v \end{equation}

So the tensor $\mathbf{T}$ is "sandwiched" between a vector and a covector and gives a scalar. In more formal mathematical language (and as your second definition suggests), the tensor acts as a multilinear map that takes a vector and a covector and outputs a scalar.

Now comes the crucial question: are these two definitions of what $\mathbf{T}$ is the same? The answer is no. They may be reconcilable, less often expressed by the same matrix, but they are mathematically distinct. To truly distinguish between these two definitions, you have to think about how they transform.

Let's go back to the first case. I shall name this tensor $\mathbf{T_{1}}$ for clarity. If we transform a vector $v$ by some transformation matrix $S$ so that we switch basis, then every vector must change accordingly:

\begin{equation} \begin{split} &\tilde{v} = Sv \\ &\tilde{v}' = Sv' \end{split} \end{equation}

However, by our first equation, what this means is that in the new basis:

\begin{equation} \mathbf{\tilde{T}_{1}} \tilde{v} = \tilde{v}' \end{equation}

which implies that the tensor transformed as:

\begin{equation} \mathbf{\tilde{T}_{1}} = S\mathbf{T_{1}}S^{-1} \end{equation}

Let's try the same idea for the second definition (where we shall name the tensor $\mathbf{T_{2}}$). The transformed vectors under a change of basis shall now be:

\begin{equation} \begin{split} &\tilde{v} = Sv \\ &\tilde{w} = (Su)^{T} = u^{T}S^{T} \end{split} \end{equation}

Thus the new scalar product shall be:

\begin{equation} \tilde{w} \mathbf{\tilde{T}_{2}} \tilde{v} = (u^{T}S^{T}) \mathbf{\tilde{T}_{2}} (Sv) = u^{T} (S^{T}\mathbf{\tilde{T}_{2}}S)v \end{equation}

Of course a scalar quantity shouldn't change after a simple change of basis, hence the tensor has to transform like:

\begin{equation} \mathbf{\tilde{T}_{2}} = (S^{-1})^{T} \mathbf{T_{2}} S^{-1} \end{equation}

In general, these are mathematically inequivalent statements which means $\mathbf{T_{1}}$ and $\mathbf{T_{2}}$ - even though they are both technically rank-2 tensors - are mathematically different objects. The coincidence between them occurs if the change of basis one imposes is a rotation i.e. $S \in O(3)$. Then the transformation matrices are orthogonal which means $S^{-1} = S^{T}$, meaning an object that transforms as the tensor $\mathbf{T_{1}}$ behaves just like $\mathbf{T_{2}}$. This is precisely what happens with the inertia tensor and which is why you find two conflicting definitions of it.

  1. As per the definition in the first case, a tensor could potentially be written as the tensor product of two other objects:

\begin{equation} \mathbf{T} = \mathbf{v} \otimes \mathbf{u} \end{equation}

$\mathbf{v}$ and $\mathbf{u}$ in the matrix representation can be any element of any dimensionality. So this statement about tensors is less so a definition about them as it is a definition of the tensor product itself. There can also be tensors that cannot even decompose in this manner (thanks @fqq for the correction). Assuming $\mathbf{T}$ is once again a $3 \times 3$ matrix that may be decomposed as such, then $\mathbf{v}$ must be a vector (i.e. a $3 \times 1$ matrix) and $\mathbf{u}$ a covector (i.e. a $1 \times 3$ matrix). Then the tensor is expressed as:

\begin{equation} \mathbf{T} = \bf{v} \otimes \bf{u} = \begin{pmatrix} v_{1}u_{1} & v_{1}u_{2} & v_{1}u_{3} \\ v_{2}u_{1} & v_{2}u_{2} & v_{3}u_{2} \\ v_{3}u_{1} & v_{3}u_{2} & v_{3}u_{3} \\ \end{pmatrix} \end{equation}

For the vectors $\mathbf{v}$ and $\mathbf{u}$ to acquire any physical meaning, one has to specify what kind of tensor you are constructing. For example, the electromagnetic tensor is given as:

\begin{equation} \mathbf{F} = (\partial \otimes \mathbf{A}) - (\partial \otimes \mathbf{A})^{T} \end{equation}

or in a more familiar form:

\begin{equation} F_{\mu \nu} = \partial _{\mu} A_{\nu} - \partial _{\nu} A_{\mu} \end{equation}

where $\partial _{\mu}$ the 4-gradient vector and $A_{\mu}$ the EM potential 4-vector.

$\endgroup$
2
  • $\begingroup$ It's not true that any tensor can be decomposed like that. $\endgroup$
    – fqq
    yesterday
  • $\begingroup$ True, I was thinking within the confines of the definition given in the OP. I shall edit to correct it. Thanks for pointing it out. $\endgroup$
    – rhomaios
    yesterday

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.