# Is the inertia tensor a tensor field?

The inertia tensor seems like it cannot depend in any way on position, but every other tensor in physics is a tensor field (stress tensor, electromagnetic tensor...) so, which is it?

• Do you know what a field is? Maybe that is the confusion here. – Aaron Stevens Jul 25 at 21:34
• In some sense every tensor is a tensor field because the basis vectors necessary to define this tensor, arise from the tangent space of the spacetime manifold. Sometimes there are no implications due to this, but sometimes the implications are significant, for example $\mathbf{\hat{x}}$ is a position independent-vector in Cartesian basis, but it has to be treated as position dependent in most other coordinates, e.g. in spherical coordinates. – Cryo Jul 26 at 0:38
• @Cryo The position vector is weird, too, because it doesn't live in the tangent space of the point it points to, yet we use the basis on the tangent space to that point to describe it. I'd love to know how that can be defined with, like, more mathematical rigor. – Pablo T. Sep 30 at 9:15
• @ Pablo T. How about defining the position vector it via the gradient, i.e $\mathbf{r}=\boldsymbol{\nabla}\left(\frac{r^2}{2}\right)$, where $r^2$ is the square of the distance from the origin? Or, in index notation, $r^\alpha=g^{\alpha\beta}\partial_\beta\left(\frac{r^2}{2}\right)$, where $g^{\alpha\beta}$ is the inverse metric? – Cryo Sep 30 at 10:44
• @Pablo T. IMHO the tangent space, at any point, is spanned by the partial derivatives, which form a vector space isomorphic to the normal vector space you are used to. i.e. vector $\mathbf{V}=V^x \mathbf{\hat{x}}+\dots$ can be thought of as $V=V^x \partial_x + \dots$ – Cryo Sep 30 at 10:48

Tensors do not have to be tensor fields. Similarly, scalars do not have to be scalar fields, and vectors do not have to be vector fields.

• I know, I'm just asking if the inertia tensor is a tensor field or just a tensor of a vector space (which vector space?) – Pablo T. Jul 25 at 21:42
• What is "a tensor of a vector field"? – ggcg Jul 25 at 21:43
• @ggcg a typo (sorry!) I edited the comment just now. – Pablo T. Jul 25 at 21:44
• "tensor of a vector space"? It has transformation properties inherited from the "space" in which the body resides (x, y, z). What else are you thinking of? Are you referring to it as a vector function (tensor(vector space))? Well then, I guess yes it is a function of the configuration space of all the particles that make up the large object. A mapping from R^(3*N) --> SM(3, 3, R) (3N dim space --> the set of symmetric real valued matrices in 3d). – ggcg Jul 25 at 21:48

The inertia tensor is defined very specifically and is related to the distribution of matter about a set of axes in space (which can be chosen to move with the object's CoM). This definition produces a matrix (or tensor) by integrating over the entire body, the entire mass distribution. As such, once the evaluation is complete it cannot possibly depend on position within the matter distribution (or outside it). The situation is different for the strain tensor for example. Since small elements of matter can be compressed by different amounts that will be a field in general. However, at any instant the MoI will be a matrix. The same could be said for the CoM of a soft object. As it moves the mass elements will continuously redistribute in space and this may cause the CoM and MoI to require being calculated at each instant but the CoM will always just be a single vector (not a vector field).

Now, based on the above, it is pretty clear that these quantities can depend on time.

The inertia tensor is just a tensor, not a tensor field.

It is also not the only example in physics of a tensor that isn’t a tensor field. A slightly more exotic example of such a tensor is the polarizability tensor, that describes the tendency of a material to become electrically polarized in the presence of an electric field. For materials that are anisotropic, the direction of the polarizability vector isn’t necessarily the same as that of the electric field vector, because the material has a different tendency to become polarized along different axes. The polarizability vector encodes the relevant information about how the material will behave. A good discussion of this can be found in the Feynman lectures.

In general, tensors over a three-dimensional space are useful for describing the properties of anisotropic materials. For example, there is also the Cauchy stress tensor in continuum mechanics. But that’s a tensor field.