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Components of a tensor

  1. Nov 16, 2008 #1
    hello forum,

    a tensor, say a 3X3, has nine components. Upon change of coordinate system, those 9 numbers change accordingly. If the change is a special rotation, then the nondiagonal elements vanish and only the diagonal ones remain. (diagonalization).

    That sounds all good, but then I think about the stress tensor. The non-diagonal elements represent shear forces, which are physically real. Diagonalizing makes them disappear.

    For a surface subject to both compressive and shearing forces, after diagonalization, we only see the compressive (or stretching) forces.

    what am i missing? it seems that simplifying the mathematics through diagonalization hides the physical reality. I know it cannot be, but..

  2. jcsd
  3. Nov 16, 2008 #2


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    It does not hide reality, just a changes perception. Take a step down to a vector. Say you have some force vector with components 5# up and 5# right. Those are real forces you could measure or use to perform work. Now instead think of the vector as sqrt(50)# at a 45 degree angle and 0# perpendicular to it. A force has vanished! If we measure the force we get the same measurement. We can perform the same work.. So your shearing forces have not vanished they can still be calculated and measured as before. There are shearing forces in an infinite number of directions. Diagnalization is simply chosing the direction where they are zero. Would you be equally upset if we chose a coordinate system with different nonzero components?
  4. Nov 16, 2008 #3
    Thanks lurflur,

    that is very clear. I need to focus on the vector itself and not on the components.
    My doubts derived from the fact that, in the stress tensor case, we find the stress force components on the surface by multiplying the stress tensor by the surface area vector.

    It would seems, naively, that the only way to have zero shearing (tangential) force components is to have the net stress force exactly parallel to the area vector.
    In that case we only have compressive or tensile effects. One of the axes of the coordinate system would then have to lie along the same direction as the normal surface area.

    However, I was overlooking the fact that a component of the force in a certain direction is a linear combination of the area vector components....
    Need to do a few exercises to prove that to myself.

    Also, this idea of diagonalization is purely local. The orientation of the cood. system that gives a diagonal matrix at one point, does not give a diagonal matrix at another one. That is in the case the material is inhomogeneous.
    If the material is homogeneous then choosing a local coord. system that works, diagonalizes the stress tensor everywhere.

    So diagonalization is useful only locally....

  5. Nov 16, 2008 #4
    actually, let me correct what I just said and mumble a little more:

    Say the reality is that there is a net stress force exactly along the surface normal. That would cause only tensile effects on the surface. No shearing is occurring physically.
    BUT the stress tensor, even in that case, could have shearing components in it. What derives from those components eventually has to cancel out....

    My naivety was that if you only have compressive action, the stress tensor at that point should not contain elements called shearing components. But what happens it that those shearing components are there, exist in the chosen frame of reference, and give rise to tangential forces that eventually cancel each other out.

    So two forces of equal magnitude but opposite direction=zero. The same way that a zero=zero.
    What a discover I made!

    Thanks for listening.
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