1. The Hall Effect LL8 Section 22

2. A conductor in an external magnetic field H Onsager’s principle doesn’t hold Instead v.5 section 120, and v.2: Time- reversal symmetry only if H -H

3. Separate conductivity tensor into symmetric and antisymmetric parts. This is always possible for a rank 2 tensor. But

5. a ik has only 3 components, like a vector. Any antisymmetric a ik is dual to an axial vector, which has no sigh change under inversion

6. Joule heat Determined for given E only by the symmetrical part of  ik.

7. External H-fields are usually weak. Expand  (H) in powers of H. a(H) is odd, so it contains only odd powers of H. axial Ordinary polar tensor, product of components that change sign under inversion, x -> -x, etc. axial

8. s ik (H) is even. Expansion of s ik has only even powers Zero-field conductivity tensor Symmetrical in (i,k) and in (l,m)

9. First order effect of H-field is linear in H. This term might also have a component perpendicular to E.

10. Inverse formula Symmetric part Resistivity tensor Antisymmetric part

11. Math arguments repeat First term is ordinary Ohm’s law Second term gives Hall effect – The axial vector b – dual to b ik – Linear in H for small H

13. For isotropic conductor, including cubic semiconductors Axial vectors a and b must be parallel to H More generally,

14. Symmetry: All tensors that characterize an isotropic medium must be invariant under all rotations about H

15. Likewise, symmetric parts of conductivity and resistivity tensors must be invariant under rotations about H

16. Let j lie in the xz plane x

17. In an isotropic conductor, the Hall field is the only E-field that is perpendicular to both j and H.

19. Next terms in expansion of  ik j k must be – quadratic in H, – linear in j, – And be a vector Only possible combinations of H & J are