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Tensors in Gemotric Algebra

  1. Jun 1, 2009 #1
    Hello,
    can anyone explain simply what is a tensor, using the language of Geometric Algebra?
    Thanks!
     
  2. jcsd
  3. Jun 1, 2009 #2

    CompuChip

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    A (rank-n) tensor is just any homogeneous multivector (n-vector).
    For example, a bivector is a rank 2 tensor.
     
  4. Jun 1, 2009 #3
    Wow, that sounds very simple.

    However I still have some troubles. If you open any book on tensor analysis, and look for the general definition of a (mixed) tensor of order (m+n), you'll find something pretty obscure (for the beginner) which involves Jacobians, weights, partial derivatives, transformation laws, covariant/contravariant components.

    I am really missing how all those "ingredients" can be absorbed into such a simple definition in Geometric Algebra.
     
  5. Jun 1, 2009 #4
    You can start with understanding rank 1 tensors in terms of vectors (a special case of multivectors). Try:

    'Gradient and tensor notes'

    in:

    http://sites.google.com/site/peeterjoot/math2009/gabook.pdf

    I have a lot of other worked examples here:

    http://sites.google.com/site/peeterjoot/electrodynamics

    that translate to and from tensors and GA (as I am learning both simultaneously). In particular, try taking somerthing like the Lorentz force equation in GA form:

    [tex]
    \dot{p} = q F \cdot v/c
    [/tex]

    and translate this to index form. That is a good exercise to get some comfort with the index manipulation, and to see how the vector and bivector objects are related to their tensor equivalents.

    Also note that GA doesn't neccessarily have a natural representation for any arbitrary tensor. Any completely antisymetric tensor has a blade representation. I'm not so sure that you'd neccessarily find natural representations for symmetric tensors, or more general tensors. The stress energy tensor which is symmetric does happen to have a slick GA representation, but I don't currently have a clue how one would figure out that out without knowing it beforehand (I can't currently follow the derivations I've seen).
     
  6. Jun 1, 2009 #5

    tiny-tim

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    Hi mnb96! :smile:
    A tensor is a formula for converting one vector to another vector.

    For example, the https://www.physicsforums.com/library.php?do=view_item&itemid=31" tensor converts the angular velocity vector of a rigid body into the angular momentum vector: Iω = L.

    (surprisingly, angular momentum is not generally aligned with rotation. :wink:)
     
    Last edited by a moderator: Apr 24, 2017
  7. Jun 10, 2009 #6
    That's an interesting statement. Angular momentum is MLT^-1 where velocity is measured in radians per second. This implies rotation. I can see how a particle moving along a curving path (not a circle) has an angular velocity at every point but is not in a rotary path around some point. Is this what you mean? (Strictly speaking the particle is in a rotary path around some point, but only instantaneously depending on the trajectory.)
     
    Last edited: Jun 10, 2009
  8. Jun 10, 2009 #7

    tiny-tim

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    ?? :confused:

    I mean that the angular momentum vector of a rigid body is not generally in the same direction as its angular velocity vector.
     
  9. Jun 10, 2009 #8
    I don't think that's true in a gravitational field where a massive body is following a geodesic.
     
    Last edited: Jun 10, 2009
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