Physics History (Maxwell) Rotary Vectors?

[Storm3371]

I am reading the text 'Innovations in Maxwell's Electromagnetic Theory'. on page 44 there is a discussion on Ampere's circuital law .
The passage is below. I don't understand the final statement. "In general represent a kind of relationship that obtains between certain pairs of phenomena , of which one has a linear and the other a rotary character. if a,b,y is linear, then p q r is rotary, and if a b y is rotary then p q r is linear."

I sort of understand why p q r would be 'rotary' since these equations are essentially the curl. But I don't really get how you would get a linear vector out if a b y are rotary - or even what it means by them being rotary.

thanks for any clarification that you can offer!

 

[vanhees71]

I'm not sure. I've always had great difficulties reading old papers not using our modern vector notation. So just take my guess with a grain of salt.

I think what's meant is the difference between a polar and an axial vector, which refers to the behavior of the vectors under space reflections. Both types of vectors transform the same under rotations SO(3) but not under the rotation group including space reflections O(3).

An axial vector can be interpreted as the Hodge dual of an antisymmetric tensor, for which you need the Levi-Civita symbol which forms tensor components under SO(3) but not under O(3).

If I understand the text right, it's stated that for an polar vector field $\vec{A}$ the curl $\vec{B}=\vec{\nabla} \times \vec{A}$ is axial. Indeed under space reflections by assumption $\vec{A} \rightarrow -\vec{A}$ and by definition of a space reflection $\vec{\nabla} \rightarrow -\vec{\nabla}$. Thus $\vec{B} \rightarrow (-\vec{\nabla}) \times (-\vec{A}=\vec{\nabla} \times \vec{A}=\vec{B}$, i.e., $\vec{B}$ behaves indeed as an axial vector. In the same way it's clear that when $\vec{A}$ is an axial then $\vec{B}$ is a polar vector field.