Gutierrez and Raabe have reported amazingly low dislocation density in Fe-3Si (wt.%) alloy after deforming to 500 MPa, using electron channeling contrast imaging in SEM-EBSD. In their paper “Dislocation density measurement by electron channeling contrast imaging in a scanning electron microscope“, published in Scripta Materialia, Vol 66, Issue 6, 2012. They report dislocation density of 10 ± 4 × 10^{-13} and 17 ± 6 × 10^{-13} m^{-2}.

Boo typos 😦

Just for fun… lets imagine shortest dislocation to be 3 unit cells long and calculate the volume it would be found in (any shorter and we might be considered out dislocation to be some sort of point defect, but I wonder if there is criteria for shortest length?). 3 unit cells, so 2.8 * 3 = 9 nm. Volume = length/ density. 9 × 10^{-9} / 10 × 10^{-13} = 9000 m^{3}. So one of our short dislocations would be found in a block 20 m × 20 m × 20 m.

A more usual dislocation density would be 10 × 10^{13}. That means a length of 10^{14} m of dislocations in 1 cubed meter. That is equivalent to 100,000 km of dislocations in a 1 cm cube (a block of 1 cm × 1 cm × 1 cm). Or 100 km of dislocations in 1 mm cube. — A distance that would take more than 1 hour to ride on a motorbike, although it would be balance on the dislocations travelling at that speed.

The distance to the moon is 385,000 km, so we can only go 1/4 of the way there using the dislocation density I started with, also dislocations are space, and space is space too, so there are already dislocations there. It makes more sense to dig a tunnel with them…

Impression of what earth would look like if we introduced dislocations from surface (Burgers Vector not to scale).

The radius of Earth varies from 6,353 km to 6,384 km (Wikipedia), so there is a long enough length of dislocations in 1 cm cube of steel (if we take dislocation density to be 1^{14} m/m^{3} still) to stretch to the centre of the earth 16 times, or to go all the way through the earth and come out the other side 8 times.

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