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Question about tensors

  1. Jul 31, 2012 #1
    Hi, I've seen in some texts where a tensor is only supplied with one(or two) of it's arguments when it has more than that, and produce a tensor with a lower order than the original.
    Is this a formal operation?
    For example, the moment of inertia tensor has 2 arguments, supplying it with an angular velocity vector gives an angular momentum vector.
  2. jcsd
  3. Jul 31, 2012 #2


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    Science Advisor

    Hey GarageDweller.

    The term for reducing the number of indices (both upper and lower) of a tensor is known as contraction and the idea is that things eventually 'sum out' to remove that index from being one that can vary in the summation as opposed to something that is constant.

    You'd probably be better off reading about contraction in tensor algebra than me telling to you because I don't think I'd do as much justice.
  4. Jul 31, 2012 #3
    Yes, it's a general property of [2nd order] tensors that when summed over one index with a vector the operation produces a new vector that in general is not in the same direction as the original vector. So with the moment of inertia tensor you would have [itex]L_{i} = I_{ij} \omega_{j}[/itex]. Similarly, if you sum 2nd order tensor over both indices with two vector you will get a scalar invariant value. Again, in the case of the moment of inertia tensor this would be the rotational kinetic energy [itex]2T = I_{ij} \omega_{i} \omega_{j}[/itex].

    And yes it is a general property of tensor no matter the rank because it's simply the inner product that you're seeing. The inner product always reduces the rank of the tensor that is performing the operation by 1.
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