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In his book Differential Forms with Applications to the Physical Sciences, on pages 19–20, Harley Flanders writes:

"a one-form $$ \omega = P\,dx+Q\,dy+R\,dz$$ may be identified with an ordinary vector field $(P,Q,R)$ in $\mathbf E^3$, a two-form $$ \alpha = A\,dy\,dz+B\,dz\,dx+C\,dx\,dy $$ may be identified with a polar vector field in $\mathbf E^3$."

What do "ordinary" and "polar" mean here?

http://books.google.com/books?id=pG0PllIO08kC&pg=PA20&lpg=PA20&dq=harley+flanders+differential+forms+ordinary+polar+vector+field&source=bl&ots=P3-grA4gZv&sig=7P3lI4AWzhhPj9MunTAz_vRaIE8&hl=en&sa=X&ei=THbpUZG2JfGEygGn0YHoCw&ved=0CCoQ6AEwAA#v=onepage&q=harley%20flanders%20differential%20forms%20ordinary%20polar%20vector%20field&f=false

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    $\begingroup$ Very strange. Usually I see "polar" to be identified with "true" or "ordinary" mathworld.wolfram.com/PolarVector.html but in the book you linked to it appears that the two are opposites. $\endgroup$ – Willie Wong Jul 23 '13 at 11:13
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    $\begingroup$ Ah, here's a better link for the physics terminology. $\endgroup$ – Willie Wong Jul 23 '13 at 11:16
  • $\begingroup$ I think normal and polar refer to cartesian coordinates and polar coordinates. $\endgroup$ – The Great Duck Nov 15 '16 at 1:14
  • $\begingroup$ @TheGreatDuck : I think your remark is plainly mistaken. How can polar coordinates be applied in this three-dimensional setting where the three variables $x$, $y$, and $z$ play symmetrical roles, and why would $2$-forms have anything to do with polar coordinates, where $1$-forms do not? $\qquad$ $\endgroup$ – Michael Hardy Nov 15 '16 at 17:57
  • $\begingroup$ @MichaelHardy It was just a thought. I don't understand the question. I just know what vector fields are and that vectors are commonly written as polar coordinate vectors in certain circumstances. $\endgroup$ – The Great Duck Nov 16 '16 at 2:15
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Physicists tend to define geometrical entities by how they change under a coordinate transformation. Those transformations are almost always rotations and parity, that is, reversing all directions (in 3 dimensions, this reverses the orientation too). If something "rotates like a vector", physicists call it a vector, no matter what.

Now in a Euclidean setting, if you only look at rotations, you can't really distinguish a vector from a 1-form, because they change with the same transformation. If a vector transforms with the linear map $A$, a 1-form transforms with the map $(A^t)^{-1}$, which for rotations is equal to $A$. The very same can be said about 2-forms in 3 dimensions.

Under parity, 1-forms and vectors transform the same way: they simply reverse. 2-forms, instead, are invariant, they don't change sign.

Now, before the discovery of differential forms (and bivectors), physicists already had to use something that rotates like a vector but is invariant under parity. This was needed, for example, for angular momentum, or for any vector that was the vector product of two usual vectors (can you see why?). Such a "vector" is called pseudo-vector, or axial vector, or polar vector.

The magnetic field is the nicest example of a "polar vector" that can be modelled better by a 2-form.

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  • $\begingroup$ The thing is, when I was taught physics, I was taught polar vectors are conventional vectors and axial vectors are pseudovectors. But I think the terminology is kinda regionalized and not universal. $\endgroup$ – Muphrid Jul 27 '13 at 1:47
  • $\begingroup$ That is what I was told too! Until I read Flanders' book. Maybe the author was confused. I don't know. Anyway, differential forms clear most of such confusion, especially in space-time. Or Clifford algebras, they work wonders when the metric is fixed. $\endgroup$ – geodude Jul 27 '13 at 11:16
  • $\begingroup$ There has been work on Clifford algebra for non-fixed metrics. Look for "vector manifolds" for an extrinsic view. Gauge theory gravity uses formalism that is more intrinsic, but it hasn't been explored in as much detail. One of the things I want to do is recast differential geometry in that latter language because it is powerful to have Clifford algebra at one's disposal. $\endgroup$ – Muphrid Jul 27 '13 at 14:32
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I'm actually not familiar with that language, even though I've perused that book. In the former case, by integrating $\omega$ you compute the work of the vector field, and in the latter case, by integrating $\alpha$ you compute the flux of the vector field.

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