Why do we not consider three polarisations in one- and two-dimensional solids?

Physics Q&A
1

When we consider the total amount of states in a Debye solid, there is a term corresponding to the ‘dofs’ of polarisation: 1 for 1D, 2 for 2D, 3 for 3D.
However, if we, for example, consider a two-dimensional solid, wouldn’t the atoms still be able to have a displacement perpendicular to the plane?
When you make a guitar string vibrate (though I don’t know how far you could stretch the macroscopic analogy), you can make it vibrate perpendicular to the propagation direction (but we aren’t allowed to do so in a 1D solid?).

A classmate explained this as ‘when we’re dealing with 2D solids, we don’t consider the third dimension, and so there isn’t a third possible polarisation to be defined’, but this seems hardly satisfactory :frowning:

2

It’s a fair point: we could be considering atoms that can move in any direction, but forming a 2D sheet or a 1D chain. In those cases we’d have 3 polarizations regardless of the dimensionality of the system. If, on the other hand, we consider a true 2D plane with atoms forced to move in that plane, then there’s only 2 polarizations. So both interpretations make sense.

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