Tensor Rank of Stress Tensors

Good luck!In summary, the stress tensor is a rank-2 tensor composed of the force density vector and the normal vector of the differential area. The normal vector is not a separate component, but is included in the force density vector. When it comes to transformations, tensors are invariant and the components may change, but the physical quantity they represent remains the same. This is why the traditional representation of the stress tensor may not seem to change under transformations.
  • #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.

----
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?
 
Physics news on Phys.org
  • #2




I understand your confusion with the transformation properties of the stress tensor. It can be a tricky concept to grasp, but I will try my best to explain it in a clear and simple manner.

Firstly, it is important to note that the stress tensor is a rank-2 tensor, meaning it has two indices. This is because it represents the stress at a point in a material, which has both a magnitude and a direction. As you mentioned, the stress tensor is composed of the force density vector and the normal vector of the differential area. This is where the confusion lies, as the normal vector is not a separate component of the stress tensor, but rather it is included in the force density vector.

Let's take your example of a rod under a load. The force density vector at a point on the rod would have three components - one in the direction of the force, and two in the direction of the differential area. The magnitude of the force density vector would represent the magnitude of the stress at that point, while the direction would represent the direction of the stress. This is why the stress tensor is a rank-2 tensor, as it has two indices to represent the two components of the force density vector.

Now, when it comes to transformations, it is important to remember that tensors are invariant under coordinate transformations. This means that the components of the tensor may change, but the physical quantity it represents remains the same. In your example, when you transform the coordinates from x to y, the components of the force density vector may change, but the stress at that point remains the same. This is why you do not see a change in stress when using the traditional representation of the stress tensor.

I hope this helps to clarify your confusion. If you have any further questions, please don't hesitate to ask. Tensor algebra can be a complex topic, but with practice and a solid understanding of its concepts, it can become more intuitive. Keep up the good work in learning and visualizing tensor algebra.
 

What is the tensor rank of a stress tensor?

The tensor rank of a stress tensor is the number of independent components required to fully describe the tensor. For a 3D stress tensor, the tensor rank is typically 6, as there are 6 independent components (3 normal stresses and 3 shear stresses).

How is the tensor rank of a stress tensor determined?

The tensor rank of a stress tensor can be determined by counting the number of independent components in the tensor. In general, for a n-dimensional stress tensor, the tensor rank will be n(n+1)/2.

What is the significance of the tensor rank of a stress tensor?

The tensor rank of a stress tensor is important as it determines the complexity of the tensor and the number of equations needed to fully describe it. It also affects the computational cost and accuracy of simulations that use stress tensors.

Can the tensor rank of a stress tensor change?

No, the tensor rank of a stress tensor is a fundamental property and cannot change. However, it may appear to change when transforming the tensor to a different coordinate system, but the underlying rank remains the same.

How does the tensor rank of a stress tensor relate to its physical properties?

The tensor rank of a stress tensor is directly related to the physical properties of the material it describes. For example, for an isotropic material, the tensor rank will be 6, while for a transversely isotropic material, the tensor rank will be 5.

Similar threads

  • Differential Geometry
Replies
2
Views
442
  • Differential Geometry
Replies
9
Views
332
  • Differential Geometry
Replies
8
Views
2K
  • Differential Geometry
Replies
7
Views
2K
  • Differential Geometry
Replies
6
Views
2K
Replies
3
Views
1K
  • Special and General Relativity
Replies
3
Views
1K
  • General Math
Replies
1
Views
997
  • Differential Geometry
Replies
3
Views
2K
  • Differential Geometry
Replies
3
Views
1K
Back
Top