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If I have a rank-two tensor that I want to analyze ─ say, an electric quadrupole moment, or a moment of inertia ─ it can often be very easy to analyze by moving to its principal-axes frame: one rotates to a reference frame where the tensor is diagonal, and this simplifies all sorts of understandings about it, whether one sees the tensor as a linear transformation or as a two-form or what have you.

When one is faced with a tensor of rank three and higher, though, it's a lot less obvious how to proceed. Given a general tensor with arbitrary entries (though probably asking that it be symmetric in all pairs of dimensions, just to keep things simple), presumably there exists a frame of reference where the tensor is much easier to understand (so, as an example, a tilted octupole that behaves as $Y_{30}$ has axial symmetry, and it'll look way simpler if you put your $z$ coordinate along that axis), but what is that frame, what properties does it have, and how does one find it?

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A rank-two tensor is going to have 3 principle axes that can be visualized. You will end up with 3 axes and the best way to visualize these at a point is with an oblate spheroid at each point. Effectively you just rotate the ellipse created by 2 of the principle axes about the third. Various software packages can do this, I prefer ParaView but to each their own.

Higher order than that though and you will not have a very good way to visualize it. I would prefer to move to a different type of analysis and instead begin to look at the topology of the field instead and study the tensor invariants. These invariants are the coefficients of the characteristic equation, which for rank-two is:

$$ \lambda^3 + P \lambda^2 + Q \lambda + R = 0 $$

Similar expressions exist of course for higher dimensions. This paper and this paper provide algorithms for computing the invariants, and provide the expressions for a rank-four tensor. In the context of my work, fluid dynamics, this is done with the velocity gradient tensor, or the rate-of-strain tensor or the rate-of-rotation tensor. The $(P,Q,R)$ space is divided by the discriminant surfaces, and these can be assigned to topological features. In the rank-two example, there are 8 sectors that correspond to focuses, saddles, nodes etc that may be stable or unstable (see for example this paper for applications in fluids). Again, drawing back to my work, these can be assigned physical properties. For example, unstable focus/compressing corresponds to vortex compression. Unstable node/saddle/saddle is a vortex sheet while stable node/saddle/saddle is a vortex tube. I'm sure other descriptions could be attributed in your case, and for higher order invariants. The invariants themselves also may have physical meaning. For fluids, $P$ is the volumetric expansion/compression and $Q$ is related to the rotation.

The last topological technique I am familiar with is the Morese-Small Complex. In this, you take a field and identify the critical points -- local minima, maxima, saddles, and nodes. These points are then connected together through the field and the boundaries around each critical point identify the flow of information along the topology. It is useful for creating topological maps of high dimensional datasets.

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  • $\begingroup$ I'll try to add references/citations soon... defending my PhD thesis tomorrow, but I have a chapter on topology so I have these references handy. I just need to pull them out. $\endgroup$ – tpg2114 Jun 1 '17 at 20:29
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    $\begingroup$ defending your PhD tomorrow and on SE today... you are a brave soul! $\endgroup$ – enderland Jun 1 '17 at 20:37
  • $\begingroup$ I don't understand your first comment. A symmetric rank-two tensor has the principal axes in three dimensions and $n$ principal address in $n$ dimensions. $\endgroup$ – Emilio Pisanty Jun 1 '17 at 20:48
  • $\begingroup$ @EmilioPisanty Sorry, fixed it. Brain fry. I got ahead of myself. I am unaware of a visual tool for the vectors in higher dimensions. I would think the better choice is to move to other techniques that reduce the dimensionality. But hopefully somebody else comes along with some interesting techniques. This is a great question. $\endgroup$ – tpg2114 Jun 1 '17 at 20:52
  • $\begingroup$ I'm primarily interested in two and three dimensions, but I suspect that if you know how to visualise a rank-$k$ tensor in $3$ dimensions, then visualising over in $n$ dimensions is not much harder than visualising $n$ dimensions to begin with. $\endgroup$ – Emilio Pisanty Jun 1 '17 at 20:59

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