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Units and the Cross Product

  1. Jun 12, 2009 #1
    I've been reading a bit on vector calculus trying to get a solid grasp on it. I am wondering how it can be stated that given C=AXB (all vectors), that C is equivalent to the area of the rectangle formed by A and B in the A,B plane. How can you equate length and area. One is in say...mm and the other would be in mm2. So are they saying that if A is 1 mm and B is 1 mm 90 degrees apart, the C is 1 mm3 perp to A and B? I thought vectors had only length and direction. Do you see my quandry here?
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  3. Jun 12, 2009 #2

    Doc Al

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    They have magnitude and direction. They can be represented by "arrows" with length, but that length is merely proportional to the magnitude--the units of the magnitude depend on the particular vector being described. For example, force and velocity are both vectors, but neither is a length.
  4. Jun 12, 2009 #3
    Thanks. So a cross product vector can represent velocity, acceleration, curl and volume or area among other things? I quess I don't know why I'd ever want a vector that is proportional to area in the first place.
  5. Jun 12, 2009 #4
    Oh, I know what I was thinking. I was thinking that a vector can represent any one dimenional quantity but it strains my thinking to see multi-dimensional quantities displayed by a vector.
  6. Jun 12, 2009 #5

    Doc Al

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    If the vectors A and B have units of length, then C = A X B will literally be a vector representing an area (units of length*length). But in general, the units of the cross product will be the product of the units of A and B. For example: Torque = r X F. r is a position vector with units of length and F is a force with units of force, thus torque has units of length*force. It's still useful to think of the cross-product as an "area" in a more general sense.
  7. Jun 13, 2009 #6


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    The "length" or "magnitude" of a vector is a number and numbers can represent many different things.

    One reason we would want "a vector that is proportional to area" is to be able to define the "vector differential of surface area", [itex]d\vec{S}[/itex], normal to the surface and whose length is the differential of surface area.
  8. Jun 13, 2009 #7


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    Welcome to PF!

    Hi Rocky! Welcome to PF! :smile:

    AxB is technically the dual of the 2-form (exterior product) A∧B …

    we can also make 3-forms and so on (A∧B∧C∧…), and a k-form is k-dimensional (as one would expect!) …

    the dual of a k-form is an (n-k)-form (where n is the dimension of the space)

    and there's an isomorphism between all the k-forms and all their duals (the (n-k)-forms) …

    in this case, n = 3, so the vector AxB is the dual of something that is two-dimensional. :smile:

    (and just to make you even more confused … and I admit this has virtually nothing to do with your question :rolleyes:

    AxB technically is a pseudovector, not a vector (and (AxB).D is a pseudoscalar), since if you reflect in the origin, a vector comes out negative, but a pseudovector stays the same …

    for that reason, we can never add vectors and pseudovectors. :wink:)
  9. Jun 13, 2009 #8
    Thanks everyone. The torque example reallly cleared it up for me, I think. Here's my summary of what everyone said.

    C=AXB (equation should be bold)
    C is a pseudovector since you can't have a negative area
    The units of C are the combination of the units of A and B as in dist and force in torque

    So basically I am thinking that C is a kind of graph of whats going in the the plane that contains the two vectors A and B and creates the parallogram formed by them.

    Just a crazy question: Since C is a sort of graph of the changes of the area of the parallelogram, is C a kind of integral of what is no longer an equation but a pair of vectors??
  10. Jun 13, 2009 #9
    Why is the Cross Product needed in the first place? As a simplification, I would think that if one is presented with two vectors and asked to find the parallelogram formed, it's much simpler to just multiply the two magnitudes and have done with it. Just like in non-vector mathematics.
    And why would the vector have to be normal to the plane that contains the two vectors? Is that arbitrary?
    Can I now create a new vector calculus rule called the ROCKY PRODUCT that defines a paralellogram fromed by A and B in a plane, but the pseudovector is at 45 degrees to the plane holding the vectors. Or an infinite amount of new rules at differing angles?
    Anyone see what I'm driving at?
  11. Jun 13, 2009 #10


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    the rocky product

    Hi Rocky! :smile:

    (are you the rocky product of your parents? :wink:)
    You certainly can :smile:

    but it wouldn't satisfy certain very useful rules, such as:

    (A+B)xC = AxC + BxC

    and (AxB).C wouldn't be the volume of the paralellepiped (the wedgy thingy) :wink:
  12. Jun 13, 2009 #11
    You may be right but it may satisfy other yet unknown equations.
    I think there is a world of as of yet untapped inbetween rules that will in the future all of a sudden become markedly useful. Call me the Nostrodamus of Vector Calculus but I believe that there is significance in currently insignificant correlations.

    Here's a quick question. If AXB is the perp pseudovector that implies the area of the two vectors A and B in a certain plane for torque let's say. What is the significance of the normal vector C to torque? Cannot the torque magnitude plus the combined units of A and B, convey the same exact information about torque not matter what angle the pseudovector has in relation to the plane of A and B? for example. A=1 meter, B=1 Newton, so C=1 Newton/meter.
    I would say that C's relative angle to A and B is insignificant.
  13. Jun 13, 2009 #12


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    Hi Nostrodamus of Vector Calculus! :biggrin:
    (btw, i wouldn't go around calling it a pseudovector, or people will think you're odd :redface:)

    The significance is that torque is a vector (pseudovector) …

    torque isn't just a number, you have to specify the direction (axis) of it …

    the basic torque equation is torque equals rate of change of angular momentum, and both torque and angular momentum are vectors (pseudovectors). :smile:
  14. Jun 13, 2009 #13
    I will buy that but the cross product seems to imply that torque would neccessarily be perpendicular to the plane of A and B by definition. That's highly unlikely that all torque is applied in that direction so what about torque that is applied at an angle NOT perp to the plane of A and B?
    The Karnac of Vector Calculus (downgrade lol)
  15. Jun 13, 2009 #14


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    Nostradamus would have seen that downgrade coming! :smile:
    Nope, you've lost me :redface:

    the A and B of torque are the force and the distance to the axis …

    with that A and B, torque has to be perpendicular to them both.
  16. Jun 13, 2009 #15


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    "Have done with it" and just ignore the fact that it gives the wrong answer? The area of a parallelogram is the product of the two magnitudes only if the parallelogram happens to be a rectangle. The area of parallelogram is [itex]|u||v| sin(\theta)[/itex] where [itex]\theta[/itex] is the angle between the sides, exactly the formula for magnitude of the cross product of two vectors.

    I answered that before. Distinguishing the normal to a surface is an important ability in applications and that is an important use of cross product. If a surface is given in terms of two parameters, u and v, with "position vector", [itex]\vec{r}(u,v)[/itex], then the derivatives, [itex]\vec{r}_u[/itex] and [itex]\vec{r}_v[/itex], are two vectors in the tangent plane to the surface. The "fundamental vector product", [itex]\vec{r}_u\times\vec{r}_v[/itex] is a vector normal to the surface. It's magnitude, [itex]\left|\vec{r}_u\times\vec{r}_v\right|[/itex] times "dudv", [itex]\left|\vec{r}_u\times\vec{r}_v\right|dudv[/itex] is the "differential of surface area". To integrate a real-valued function over the surface, you write it as [itex]\int\int f(u,v)\left|\vec{r}_u\times\vec{r}_v\right| dudv[/itex]. Even simpler is integrating an vector valued function over a surface (as for example, finding the flow through a surface or the flux of a magnetic field through the surface). That is just [itex]\int\int \vec{f}(u,v)\cdot\vec{r}_u\times\vec{r}_v dudv[/itex]

    Of course you can! Is it useful? I can think of many situations in which it is useful to separate a vector into components along and perpendicular to a surface: only the component of force perpendicular to a surface causes pressure on the surface, only the component of force along a surface will move an object over the surface. Can you give examples where a component at 45 degrees to a surface or at some other specific angle would be useful?

    Apropos your post on "torque", you don't see to understand that torque is always measured relative to some axis. And is always perpendicular to that axis. Perhaps instead of railing against things you don't understand, you should learn them.
    Last edited: Jun 13, 2009
  17. Jun 14, 2009 #16
    Railing against things in a gentile way is the best method for me to understand things I do not understand. It is rather like an exclamation point at the end of a sentence for executing emphasis. :)
    As far as the torque analogy to the cross product, I'm still not grasping it properly. Could the entire thing be put into a non-mathematical statement for easier digestion? ie

    The cross product vector (pseudovector), which is always normal to A and B, relates to the the axis of rotation around which the torque is applied and the magnitude of the cross product vector (pseudovector) is equal to the magnitude of the torque applied in whatever units the vector A and B are set to.

    If that's pretty close to being accurate, I would still ask this really crazy question.

    I would argue that the area of a parallogram is not A time B sin theta but simply A times B.

    Here's my rational.

    Vector A has a magnitude and B has a magnitude. Imagine a parallogram. Hold the bottom two points in place and rotate the top two points such that they line up with the bottom two points in the vertical axis. Now you have the same magnitudes just at different angles. And the area is not the actual magnitude's products. So you are saying that by shifting the top two points, you are actually changing the area inside the geometric object? It seems counter-intuitive. If you can explain how it does change the area, then I will believe it.

  18. Jun 14, 2009 #17


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    Go the other way. Rotate the top two points so that they are in the same line as the bottom two points. Now the area is zero. I conclude that the area of any parallelogram is zero? Or I conclude that the area is not AB

    Calculating the area of a parallelogram is fairly easy and straightforward; why don't you try to do it?
  19. Jun 14, 2009 #18


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    As opposed to a Jewish way?:tongue:

    You can "argue" that 2+ 2= 5 if you like.

    I don't know how to "explain" this to you since you don't seem to want to accept the evidence of your eyes. Instead of rotating so that the top two points "line up with the bottom two points in the vertical axis", rotate so that the lie along the same horizontal line. Do you not see that the area is decreasing toward 0? Yes, of course, changing that angle changes the area.

    I don't see any way you can "understand" things like the cross product if you do not know basic geometry.
  20. Jun 14, 2009 #19
    By shifting the top two points the magnitudes do not change if you image they are metal rods rotating around the two bottom pivot points. Then you have a rectangle and the magnitudes are exactly the same. So in that case is the area the same as before it was rotated?
  21. Jun 14, 2009 #20
    So you would have maximum volume at 90 degree angles from what you say.
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