1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Question on the History of the Cross Product

  1. Jan 20, 2012 #1
    I am trying to get an intuition behind the cross product but i seem to get stuck with understanding why we make the vector perpendicular to the other two? I understand the need to define an orientation for physical systems like torque (ccw or cw) but then why make it a vector and why choose perpendicular? Is this just by definition or does it arise from the geometry? What if we re defined the cross product to be at a 30 degree angle from both of the vectors, would it still work in physical analysis (albiet more difficult becuase of the skewed angle?)
     
  2. jcsd
  3. Jan 20, 2012 #2

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    The cross product comes from vector geometry, for instance, the direction perpendicular to a plane is not a turning vector like torque but is still determined by the cross-product of two vectors in the plane.

    We've just adopted the notation for physics as a kind of notation to represent twisty things. Turning still have directional qualities and the cross product is a handy way to describe them. This is why the torque is called a pseudo-vector.

    In physics, math is used as a language.
     
  4. Jan 21, 2012 #3
    One way to see that A×B is orthogonal to both A and B is to invoke the geometric fact that two vectors are orthogonal if their dot product is 0. You can verify this result for yourself by using the matrix definition of the cross product in ℝ3 (3 dimensions).

    The cross product in ℝ3 can be shown to be equivalent to the determinant of a matrix with the vector (i, j, k) in the first row, the entries of A in the second row, and the entries of B in the third row. It's very easy to dot a vector into a cross product. Simply replace (i, j, k) with the vector being dotted into the cross product. (You should verify for yourself that this is true.)

    By observe that the determinant of a matrix with two identical rows is 0. You'll realize that A.(A×B)=B.(A×B)=0. Therefore, both A and B are orthogonal to A×B. Nobody "defined" it to be this way; that's just how it turned out.

    On the note of definitions, you can certainly "redefine the cross product to be at a 30 degree angle from both of the vectors." Nobody is stopping you from defining the cross product this way. However, the definition would not be meaningful, as it would contradict the results every time. To compare it to a simpler example (often used in Mathematical Analysis), you can redefine a/b+c/d as (a+c)/(b+d). Nobody is stopping you from defining the addition of rational numbers this way. However, if a/b or c/d aren't in lowest terms, then the formula does not yield unique results. Therefore, the formula a/b+c/d=(a+c)/(b+d) is not meaningful.

    In physics, we define torque in this way because the vector that arises is unique to the situation. You can verify this for yourself.
     
    Last edited: Jan 21, 2012
  5. Jan 21, 2012 #4
    Harrisonized, looking at the cross product as a matrix may explain why we consider the new vector to be perpendicular. However, it just replaces the problem with trying to understand why you would then define the product to be the determinate of a matrix. This seems just as arbitrary.
    Is it true that re defining it would result in physical non sense? Or would everything just shift equally so that the physics would still work out (albiet with more work involved becuase of the 30 degree angle)?
     
  6. Jan 21, 2012 #5

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi Storm Butler! :smile:
    Because it's the dual of a second-order tensor, so its order is n-2, so …

    in 3D it's a vector (technically, a pseudovector)

    in 4D or higher, it wouldn't be any sort of vector

    in 2D, it's a scalar, as you've probably noticed! (technically, a pseudoscalar). :wink:

    Why is it defined in terms of tensors (ie mutli-linear functionals)?

    Because we want to be able to do things with it (for example, to sum it over all points in a solid), and frankly if it's not (multi-)linear, it's not much use! :biggrin:
     
  7. Jan 23, 2012 #6

    lavinia

    User Avatar
    Science Advisor
    Gold Member

    I don't know the history but certainly in EM the cross product represents physical quantities
    that have both magnitude and direction. I just found this example on Wikipedia.

    http://en.wikipedia.org/wiki/Poynting_vector
     
  8. Jan 23, 2012 #7
    Hi Storm Butler,

    I don't know if this can be useful for your purposes, but I can tell you that I personally abandoned long ago the commonly taught conception of cross product.
    In the domain of Clifford (Geometric) Algebras, you can introduce the algebraic operation of wedge [itex]\wedge[/itex] between two vectors. For example [itex]\mathbf{a} \wedge \mathbf{b}[/itex] would represent a geometrical object called 2-blade that roughly speaking represents the subspace spanned by a and b, with a certain orientation and a certain magnitude given by the area of the parallelogram spanned by a and b. You dont need anything more than this.

    For "backward compatibility", if you consider the space [itex]\mathbb{R}^3[/itex], and you calculate the dual of [itex]\mathbf{a} \wedge \mathbf{b}[/itex], you obtain exactly the classical cross product [itex]\mathbf{a} \times \mathbf{b}[/itex]. Keep in mind that the cross-product is something that is usable only in 3-D, while the concept of n-blade (and its dual) is very intuitive, more natural, and much easier to grasp and visualize.

    I read somewhere, some time ago, that actually the foundations of this powerful algebraic language to "do geometry" were laid thanks to the work of Hamilton, Grassmann, and Clifford, more or less at the same time when vector analysis was being promoted by Gibbs and Heaviside.
    Partly because Clifford died very young, and partly because vector analysis was easier to adopt by the physicists of the time, Clifford algebras were forgotten, and rediscovered only recently.
     
    Last edited: Jan 23, 2012
  9. Jan 23, 2012 #8

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    The wedge-product is actually how I do things now too.
    It is often not taught before grad-school in physics courses since the vector method is easier to grasp since it requires less math education.

    In fact - keeping pace with the math education side is what governs the timing of a lot of physics education. I've often found that teaching the subset of the math needed for the physics ahead of the math courses is quicker and simpler.

    notice that in 3D, the resulting 2-blade magnitude is just ab.sinA: A is the angle between a and b ... eg: it is the magnitude of the vector cross product. The vector for the plane surface of the 2-blade is this magnitude perpendicular to the surface just the same as any plane surface.

    The main downside is that it appears more abstract: it is quite easy to imagine vectors as having some physical meaning. When you start using math script languages you come to realize that a vector is just a way of organizing numbers and it's physical meaning depends on the numbers, and rules you use to organize them.

    So when you were learning about things with a direction quality - you learned how to represent things describable as arrows on paper as a list of numbers in a mathematical object called a vector. Any list of numbers can be represented as a vector though ... the mathematical vector is the map of the thing, not the thing itself. It is important not to confuse the map with the territory.

    in 3D we end up just writing:
    [tex]\newcommand{bv}[1]{\mathbf{#1}}


    \bv{u}\wedge\bv{v}= \det \left [
    \begin{array}{ccc}
    \bv{i} & \bv{j} & \bv{k} \\
    u_x & u_y & u_z \\
    v_x & v_y & v_z
    \end{array} \right ]
    \qquad \text{where} \qquad
    \bv{u}=\left ( \begin{array}{c}
    u_x \\ u_y \\ u_z
    \end{array} \right )
    \qquad
    \bv{v}=\left ( \begin{array}{c}
    v_x \\ v_y \\ v_z
    \end{array} \right )

    [/tex]... which has been known to make some mathematicians cringe.

    I think the takeaway lesson here is that it is just a notation. Don't sweat it.
     
    Last edited: Jan 23, 2012
  10. Jan 24, 2012 #9

    chiro

    User Avatar
    Science Advisor

    Hey Storm Butler.

    This kind of thing has history in geometric calculus. To understand this properly you need to know the history of geometric calculus.

    People like Hermann Grassmann did research into this kind of thing. This work was unfortunately largely ignored but people like Clifford revived a lot of the ideas and added new insights into the area.

    Also it might help to understand the contributions of Sir Rowan Hamilton on the work of quaternions. I don't know if the book is still available but at some point in the past I downloaded the entire book written by Hamilton that was a treatise on quaternions that went through all of his results, ideas and thoughts on the matter of quaternions. I downloaded it from Google (Google books) and hopefully if you want to check it out it should be there.

    One way to think of geometric calculus is to consider what multiplication and division of vectors actually 'mean'. It turns out that a good way to interpret this is to use the idea found in the complex numbers: you use imaginary units to represent rotations and multiplication/division is an inverse process that is related to rotation.

    As a result of research into this idea, a thing called the "geometric product" was formed that combined the ideas of the "exterior" and "interior" (or "outer" and "inner") products. You can think of the exterior in terms of your "cross product" and your interior as your "dot product" for three dimensional geometry.

    Of course there is a whole area of multi-linear mathematics that deals with this kind of thing, but if you view it (at least in one way), but trying to interpret how you would multiply and divide vectors in some sense then this can give one perspective. There are of course others, but this is one way and it will also help you if you ever look at highly complex structures involving imaginary quantities like those found in theoretical physics.
     
  11. Jan 24, 2012 #10

    chiro

    User Avatar
    Science Advisor

    Also forgot to add this:

    It would be handy if you want to understand this, to read about what Heaviside did to EM. In the process the vector algebra that includes the cross-product was defined as a simplification of existing results.

    This will give you a bit of understanding as to where it come from in a physical sense.
     
  12. Jan 24, 2012 #11
    ... That is incorrect. You have basically rewritten the cross-product.
    The outer product does not work that way. If you still want to work with coordinates in R3 then you would have:

    [tex]\mathbf{u} \wedge \mathbf{v} = (u_x \mathbf{i} + u_y \mathbf{j} + u_z \mathbf{k}) \wedge (v_x \mathbf{i} + v_y \mathbf{j} + v_z \mathbf{k})[/tex]

    and using the following properties:

    *) [itex]\mathbf{x} \wedge \mathbf{x} = 0[/itex]
    **) [itex]\mathbf{x} \wedge \mathbf{y} = - \mathbf{y} \wedge \mathbf{x}[/itex] (anticommutativity)
    ***) distributivity of the outer-product with respect to addition

    you would obtain:

    [tex]\mathbf{u} \wedge \mathbf{v} = \lambda \mathbf{B}[/tex]

    where [itex]\lambda \in \mathbb{R}[/itex] is a real scalar (the magnitude) that turns out to be exactly the determinant that you mentioned to compute the cross product. Instead [itex]\mathbf{B} \in \bigwedge^2 \mathbb{R}^3[/itex], is a unit bivector (actually a 2-blade in this case), that directly represents the 2 dimensional subspace spanned by u and v.
    The cross product (as a vector perpendicular to u and v), is obtained by the dual of the 2-blade mentioned above:

    [tex](\mathbf{u} \wedge \mathbf{v})^* = (\mathbf{u} \wedge \mathbf{v}) \rfloor \mathbf{I}^{-1} = \lambda \mathbf{n}[/tex]

    where [itex]\rfloor[/itex] is the contraction operator, I is the pseudoscalar for R3 given by [itex]\mathbf{I}=\mathbf{i} \wedge \mathbf{j} \wedge \mathbf{k}[/itex], and n is a unit vector normal to u and v.
     
    Last edited: Jan 24, 2012
  13. Jan 25, 2012 #12
    ...I also have to disagree on the statement that Geometric Algebra is "more difficult and abstract" than ordinary vector analysis and linear algebra. For me it was exactly the opposite, as I found it made relatively "advanced" topics, completely accessible.

    Also, what's so difficult about replacing the cross-product of two vectors, with the concept of a 2-blade: an entity that in textbooks is simply depicted as a parallelogram spanned by two vectors.

    Due to its simplicity, I have heard of a bunch of experimental programs in some schools around the world aimed at teaching high-school students the basics of geometric algebra. One of the first proposed exercises, was to familiarize with the concept of 2-blade and use it to replace the cross-product.

    The end of the story, is that I would suggest the original poster of this thread to have a look at the history of geometric calculus, as chiro suggested, and if interested in making sense of the cross-product (and all the sloppy notations adopted for it), read some introduction to geometric algebra. There are many free sources available, but a very accessible, and not overly rigorous introduction is found in Leo Dorst's book "Geometric Algebra for Computer Science".
     
  14. Jan 29, 2012 #13
    Thank you all for the recommendations. I am currently reading a book on the history of Vector Analysis which goes through the highlights of Gibbs, Heavyside, Grassman, Clifford, Tait, and Hamilton. I'll look at that book on geometric algebra alongside...hopefully I will find it as revealing as you did mnb96.
     
  15. Jan 29, 2012 #14
    yes, go ahead.
    I am sure you will find the effort very rewarding.
    If you learn by making sense of geometrical concepts by visualizing them, you will probably find geometric algebra quite revealing, as it better exposes the "true" nature of such concepts in a coordinate-free manner. You might also be surprised at how apparently involved artificial constructions like the determinant of a matrix have an incredibly simple and visual interpretation.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Question on the History of the Cross Product
  1. The Cross Product (Replies: 4)

Loading...