- #1
MadRocketSci2
- 48
- 1
I have been trying to learn and visualize a bit of tensor algebra recently, and have been confused by the transformation properties of the stress tensor:
Background:
*The transformation properties of other tensors have been fairly straightforward for me to grasp so far - one example is the angular momentum tensor (which, if you want to compose it properly, should not be represented by a vector created by the cross product - instead substitute the wedge product and get a 2-vector). Under transformation, the 2-vector (asymmetric rank-2 tensor) transforms properly - the vector does not. (Polar vectors are only a useful concept for the group of proper and improper rotations).
The stress tensor is commonly given in terms of a rank two tensor - the tensor appears to be composed with the components of the force density vector over a given differential area, and *the normal vector of that differential area*. This is where my difficulties lie.
The differential area is also properly formulated as a two-vector. The transformation properties of the differential area map to the normal tensor transformation rules of a rank-2 tensor, not anything having to do with the normal vector. If you have a differential area oriented y-z, and scale the x axis, the differential area should not scale!
In this case then, the stress tensor should be a rank three tensor, with one rank for the force vector, and the other two for the differential area two-vector. When I set it up this way, I can transform the stress arbitrarily, and get results that make physical sense
S(i;j,k) (y) = S(l;m,n) (x) * dy(i;1)/dx(l;1)*dx(m;1)/dy(j;1)*dx(n;1)/dy(k;1)
vs the original.
S(i,j;1) (y) = S(m,n;1) (x) * dy(i;1)/dx(m;1) * dy(j;1)/dx(n;1)
Suppose you have a rod under a load, aligned with axis 1, with a given cross sectional area. Suppose the transformation from coordinate x to coordinate y involves scaling down the 2,3 axes by a factor of two. Under the traditional representation of the stress tensor, I get no change in stress.
If the cross sectional area is four times larger in terms of the new coordinates, I should see 1/4 the stress. This appears to be represented in my setup, but not the tradiational setup.
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However, I know that in relativity, they represent the stress energy tensor the tradional way, one direction for the force vector, the other direction for the *normal* to the bounding differential 3-form. (I would want to set it up as rank 4 with one for force, and three for the bounding differential 3 form)
Clearly, this posed no difficulty for 100 years of reletavistic physics, so they must have some way of working non-metric-preserving transforms that I'm not getting with only the normal vector.
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I'm not sure what I'm doing wrong with the traditional stress tensor representation, can anyone help?
Background:
*The transformation properties of other tensors have been fairly straightforward for me to grasp so far - one example is the angular momentum tensor (which, if you want to compose it properly, should not be represented by a vector created by the cross product - instead substitute the wedge product and get a 2-vector). Under transformation, the 2-vector (asymmetric rank-2 tensor) transforms properly - the vector does not. (Polar vectors are only a useful concept for the group of proper and improper rotations).
The stress tensor is commonly given in terms of a rank two tensor - the tensor appears to be composed with the components of the force density vector over a given differential area, and *the normal vector of that differential area*. This is where my difficulties lie.
The differential area is also properly formulated as a two-vector. The transformation properties of the differential area map to the normal tensor transformation rules of a rank-2 tensor, not anything having to do with the normal vector. If you have a differential area oriented y-z, and scale the x axis, the differential area should not scale!
In this case then, the stress tensor should be a rank three tensor, with one rank for the force vector, and the other two for the differential area two-vector. When I set it up this way, I can transform the stress arbitrarily, and get results that make physical sense
S(i;j,k) (y) = S(l;m,n) (x) * dy(i;1)/dx(l;1)*dx(m;1)/dy(j;1)*dx(n;1)/dy(k;1)
vs the original.
S(i,j;1) (y) = S(m,n;1) (x) * dy(i;1)/dx(m;1) * dy(j;1)/dx(n;1)
Suppose you have a rod under a load, aligned with axis 1, with a given cross sectional area. Suppose the transformation from coordinate x to coordinate y involves scaling down the 2,3 axes by a factor of two. Under the traditional representation of the stress tensor, I get no change in stress.
If the cross sectional area is four times larger in terms of the new coordinates, I should see 1/4 the stress. This appears to be represented in my setup, but not the tradiational setup.
----
However, I know that in relativity, they represent the stress energy tensor the tradional way, one direction for the force vector, the other direction for the *normal* to the bounding differential 3-form. (I would want to set it up as rank 4 with one for force, and three for the bounding differential 3 form)
Clearly, this posed no difficulty for 100 years of reletavistic physics, so they must have some way of working non-metric-preserving transforms that I'm not getting with only the normal vector.
-----
I'm not sure what I'm doing wrong with the traditional stress tensor representation, can anyone help?