Three dimensional tracefree tensor?

[Pencilvester]

TL;DR Summary

I’m not sure how that works.

Hi PF, I’m working through “A Relativist’s Toolkit” by Poisson, and I’m in the section on geodesic congruances, subsection: kinematics of a deformable medium. I got through the section on the 2-dimensional example that introduced expansion, shear, and rotation just fine, but I’m having trouble with the generalization to three dimensions, specifically with shear.

I’ll have to start by defining tensor B as he does:$$\frac{d\xi^a}{dt}=B^a_b (t)\xi^b +O(\xi^2)$$where $\xi$ is the displacement vector. In the 2-dimensional example, shear was introduced as the case where B is symmetric and tracefree, so the top left and bottom right components were equal but with opposite signs. That’s fine. But how does that generalize to three dimensions? The only way I can think to make it tracefree is by making the diagonal components equal without varying signs, but scaling at least one of them by some coefficient in order to make them cancel out. Is that right?

 

[Pencilvester]

Or I guess one component could equal 0?

 

[Orodruin]

The sum of the diagonal elements is zero. There is nothing else to it.

 

[Pencilvester]

Ha. So the relationship between the diagonal components is irrelevant if one even exists at all (other than they must all add to zero). I think the fact that in two dimensions one component equals the other (with opposite sign of course) threw me. I didn’t even think about the fact that that’s the only possibility when you’re only adding two things together. Well thanks!

 

[pervect]

Suppose you have a cube with sides (1+a), (1+b), (1+c). a,b, and c can be regarded as displacements of the faces. Then the volume of the cube is $$(1+a)(1+b)(1+c) \approx 1+a+b+c$$ to the first order.

So when a+b+c=0, the trace is zero, and the volume of the element doesn't change, which is consistent with the idea that the expansion scalar is zero.