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Parity of stress tensor versus stress-energy tensor

  1. Jun 15, 2014 #1

    bcrowell

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    The stress-energy tensor is an actual tensor, i.e., under a spacetime parity transformation it stays the same, which is what a tensor with two indices is supposed to do according to the tensor transformation law. This also makes sense because in the Einstein field equations, the stress-energy tensor is related to the Einstein tensor, which is tensorial.

    However, in three-dimensional continuum mechanics, the stress tensor takes a normal vector as an input and gives a stress vector as an output. In three dimensions, a normal vector is an axial vector (even under parity), while a stress vector is a true vector (odd under parity). Therefore it seems that the stress 3-tensor must be odd under parity, which makes it not a real tensor.

    Is this analysis correct? I'm used to thinking of the stress 3-tensor as a block of elements in the stress-energy tensor, when they're expressed in Minkowski coordinates. Doesn't that imply that they should have the same parity properties?
     
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  3. Jun 15, 2014 #2

    Bill_K

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    I'd say that the quantities are all defined relative to a small element of the matter, with the convention that "outward" is the positive sense. The normal vector ni is the outward normal, and when you do the parity reflection, volume element and all, the reflected normal vector points in the other direction, but is again outward.

    Likewise for the stress vector, Ti = τij nj the positive sense is taken to be outward, and remains so under reflection.
     
  4. Jun 15, 2014 #3

    bcrowell

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    Last edited: Jun 15, 2014
  5. Jun 15, 2014 #4

    Bill_K

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    Um, no, quite the opposite! :smile: I'm saying that if τij is positive, it will remain positive under reflection, so it's a real tensor.
     
  6. Jun 15, 2014 #5

    bcrowell

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    Hmm...OK, I see what you mean. Let's say we have a cube with sides of length 2, centered on the origin, with edges parallel to the Cartesian axes. Let's take a normal vector at point P=(1,0,0). This normal vector equals (1,0,0), pointing in the outward direction.

    Now define new coordinates (x',y',z')=(-x,-y,-z). In these coordinates, the point P is (-1,0,0). The normal vector points in the outward direction, so it's (-1,0,0).

    So under a parity inversion, the normal vector has flipped signs. That makes it a vector, not an axial vector.

    The reason I was thinking of it as an axial vector was that if it's defined by a vector cross product, then it doesn't change under a parity flip. I was influenced by this argument: http://mathoverflow.net/a/171888/21349

    I think what's going on is that you can define the normal vector as a cross product, in which case it does not necessarily point outward, it's even under parity, and the stress tensor needs to be odd under parity. Or if you have a closed surface, you can define the normal vector as pointing outward, in which case it's odd under parity, the stress tensor is even under parity, and everything is tensorial.
     
    Last edited: Jun 15, 2014
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