Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Why this form of the vector product?

  1. May 17, 2017 #1
    Hi. I was investigating through this week why there are the differential forms, why are they anti-symmetric, why do we have the Jacobian when expressing the volume in a different coordinate system. This was just fantastic! I found all the connections between these topics. And I found that all of these things are due to the fact that the volume element (the same works for an area element) is defined through what is called the triple product, namely ##(A \times B) \cdot C## for vectors ##A, B## and ##C##.

    But now I wonder: why does the vector product is defined in terms of a determinant in first place? I thought at first that it had to do with the way basis vectors transform between coordinate systems, but I noticed that it doesn't matter, because covectors are defined in such a way that the inner product is left invariant anyway. So why the vector product is defined that way?
     
  2. jcsd
  3. May 17, 2017 #2

    jedishrfu

    Staff: Mentor

  4. May 17, 2017 #3
    I think I see what this article says. I actually considered this before:

    Suppose we want to form a vector out of two other vectors. We also want that vector to be orthogonal to the plane containing the other two vectors. We would discover that the determinant of their components mnemonically give us the third one we wish, by doing this for the most trivial (and perhaps, fundamental) case:

    ##(1,0,0) \times (0,1,0) = (0,0,1)##.

    But how to be sure that this procedure is correct for any pair of vectors (in any dimension)?
     
  5. May 17, 2017 #4

    jedishrfu

    Staff: Mentor

    The key is that the basis unit vectors are orthogonal for the determinant method to apply.
     
  6. May 17, 2017 #5
    I don't understand. The spherical polar basis vectors aren't orthogonal to each other and we still have a determinant.
     
  7. May 17, 2017 #6

    jedishrfu

    Staff: Mentor

    I thought they were an orthonormal set.
     
  8. May 17, 2017 #7

    jedishrfu

    Staff: Mentor

  9. May 18, 2017 #8

    Mark44

    Staff: Mentor

    This isn't something I've ever lost much sleep over, wondering why this definition is as it is.

    However, technically speaking, the cross product (or vector product) isn't really a determinant -- it's called a pseudodeterminant, because the top row consists of unit vectors, while the second and third rows consist of the coordinates of the two vectors making up the cross product.

    Another definition of the cross product is ##\vec u \times \vec v = \left( |\vec u| |\vec v| \sin(\theta) \right) \vec w##, where ##\vec w## is a vector that is orthogonal to both ##\vec u## and ##\vec v##. I don't know who came up with the idea of a pseudodeterminant that produces the coordinates of the vector that is normal to both of the other vectors, but whatever, the formula does what it's supposed to.
    I'm pretty sure the spherical basis vectors are all orthogonal. In fact, there are an orthonormal set. See the definitions of the basis vectors here -- https://en.wikipedia.org/wiki/Spherical_basis#basis_definition.

    It's not applicable in just any dimension. For the usual cross product, the vectors involved are three-dimensional. I've heard there are extensions to some higher dimensional spaces, such as quaternions and octonions, but this isn't something I've ever studied.
     
  10. May 18, 2017 #9

    fresh_42

    Staff: Mentor

    As it is always the case with "why" questions: it depends on what you will accept as an answer.

    From a historical point of view we have to mention Graßmann who was indeed driven by geometric considerations like generalizing the one-dimensional length of a vector.
    From a geometric point of view you can find an interesting treatise about it on the first pages here: https://arxiv.org/pdf/1205.5935.pdf.
    From an algebraic point of view, one might refer to the cross product as being the multiplication in the three dimensional simple Lie algebra.
    From a topological point of view, one might consider simplicial complexes and corresponding boundary operators.
    From a physicist's point of view, we will probably have to mention the right hand rule.
    For a differential geometer, you already mentioned the connection to differential forms which brings us back to the geometrical treatment I quoted and Hermann Graßmann, which closes in the circle.

    So as to why is it as it is depends on which approach you would prefer and consider "natural".
     
  11. May 18, 2017 #10
    Indeed. I'm sorry, but I messed things up earlier.
    yea
    by "in any dimension" in post #1, I meant one carries over any dimension the determinant (Jacobian determinant).
    never the less, I think one can prove by induction that for any dimension, using the canonical basis components in the determinant ##(1,0,...,0), (0,1,...,0),...,(0,0,...,1)## one can obtain the next one basis vector, i.e. it's as if we were displacing the ##1## one slot to the right.

    Now, if we accept that the canonical basis I mentioned above is the most fundamental one, because it arises naturally from Euclidean Spaces and all curvilinear coordinates can be written as a function of them, e.g. ##x = x(r, \theta), y = y(r, \theta)## (derivatives with respect to the coordinate directions give the basis vectors), then it's obvious why we should carry over the vector product defined that way to other coordinate systems.
    This raises another question: why does the determinant is defined in that way we know? :biggrin:
    Note that I'm not getting into a loop of why things are the way they are, but I'm trying to understand what I said in post #1.
     
  12. May 18, 2017 #11
    Thanks. This is a helpful reply.
     
  13. May 18, 2017 #12

    jedishrfu

    Staff: Mentor

    The why is perhaps rooted in Cramer and the rule he discovered for solving a system of equations:

    https://en.wikipedia.org/wiki/Cramer's_rule

    However, there are earlier uses found in the math of other cultures as described in the HISTORY section of the wikipedia article and quoted below:

     
  14. May 18, 2017 #13

    jedishrfu

    Staff: Mentor

  15. May 18, 2017 #14
    I will take a look at these links. Thanks!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Why this form of the vector product?
  1. Vector product (Replies: 5)

Loading...