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- 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

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?