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Cross product torque

  1. May 24, 2005 #1
    Why does torque, magnetic fields and others, work with something like the vectorial cross product? A plane director vector of intensity equal to the area of the paralelogram formed by the other two?

    Anyone?
    Basic Statistics Physics?
    Thank you all in advance.
     
  2. jcsd
  3. May 24, 2005 #2

    dextercioby

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    Symmetry & behavior under space inversion (parity transformation) is the key & the reason for using peudovectors.

    Daniel.
     
  4. May 24, 2005 #3

    OlderDan

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    Some physical phenomena are observed to depend on the extent to which directional quantities are parallel. For example, work done by a force depends on the motion of an object in the direction of the force. Motion in other directions is irrelevant when it comes to work. There are infinitely many directions an object could be moving when a force is applied, but for any given force there is only one direction for that force. If the component of the displacement in the direction of the force can be determined, the work can be calculated. This component is always the product of the magnitude of the displacement times the cosine of the angle between the displacement and the force. Work has no direction associated with it.

    There are other phenomena that are observed to depend on the extent to which directional quantities are perpendicular, such as the ones you have mentioned. Torque depends on the position at which a force is applied relative to an axis of rotation, but only the component of the force perpendicular to the position vector tends to produce rotation. Furthermore, the tendency to produce rotation is proportional to the distance of the point of application from the point of rotation. The perpendicular component of the force times the distance of application is the magnitude of the force times the distance times the sine of the angle between the direction of the force and the position vector. It happens that this is numerically equal to the area of a parallelogram formed by the force and position vectors, but the important thing is that the product is the product of two perpendicular vector components.

    In cases where parallel components are the important consideration, there is a unique direction for the component of one vector quantity in the direction of the other quantity. This is not the case when the perpendicular component of one vector relative to another vector is involved. There are infinitely many directions the perpendicular component can take, and each of them tends to produce a different result. In the case of torque, the perpendicular component of force tends to produce motion in a plane that contains the position vector and the force vector. It is convenient to describe the orientation of that plane by giving a single direction perpendicular to that plane, in other words by specifying the axis of rotation.

    All the physical quantities that involve the vector cross product involve the product of the perpendicular components of the vectors, and have direction that depends on the orientation of the plane containing the original vectors. The vector cross product is a convenient representation of both the perpendicular component interaction, and the direction of the plane of those components. That the numerical value happens to be the area of the parallelogram is a coincidence of no real importance.
     
  5. May 26, 2005 #4
    Dextercioby, could you be more specific?

    OlderDan, could you elaborate on the 'coincidence' of the paralelogram area equal to the vector intensity?

    Thank you both in advance
     
  6. May 26, 2005 #5

    OlderDan

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    In any vector cross product, the magnitude of the product depends on the component of one vector that is perpendicular to the other vector. It does not matter which of the two vectors you break into components. The component parallel to the other vector contributes nothing, and the perpendicular component contributes everything. The magnitude of the resulting product is the magnitude of the one vector times the magnitude of the perpendicular component of the second vector. When resolving the second vector into two components, the parallel component is the magnitude of the vector times the cosine of the angle bewteeen the two vectors, and the perpendicular component is the magnitude of the vector times the sine of that angle. The result is that the magnitude of the cross product is the magnitude of each vector times the sine of the angle between them, which also happens to be the area of the parallelogram constructed by placing the two vectors tail to tail and tail to head. The geometry is inescapable. The magnitude of the product will always equal the area, but the physically important thing is that only the vertical component of the vector is contributing.
     
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