Is the Cauchy stress tensor really a tensor?

[Philip Wood]

I've just started learning about tensors from Jeevanjee's highly praised 'An Introduction to Tensors and Group Theory for Physicists'. He defines a tensor as a function, linear in each of its arguments, that takes some vectors (maybe only 2) and produces a number. [The components of the tensor are the values of this number when base vectors are the arguments.] The Cauchy stress tensor, $\boldmath {\sigma}$, is, I read, defined by $\textbf{T}_e=\textbf{\sigma}~\textbf{e}$ in which the right hand side can be written as a multiplication a column vector of the components of the unit vector $\mathbf e$ normal to a surface by a matrix of components representing $\textbf{\sigma}$. $\textbf T$ is another vector, the traction vector. So we seem to be inputting one vector and outputting another. This doesn't seem to fit Jeevanjee's definition of a tensor. Can someone explain?

Sorry about the Latex failure; I've no idea why I can't make it work on this site.

 

[renormalize]

I've just started learning about tensors from Jeevanjee's highly praised 'An Introduction to Tensors and Group Theory for Physicists'. He defines a tensor as a function, linear in each of its arguments, that takes some vectors (maybe only 2) and produces a number. [The components of the tensor are the values of this number when base vectors are the arguments.] The Cauchy stress tensor, $\textbf \sigma$, is, I read, defined by $\textbf{T}_e=\textbf{\sigma}~\textbf{e}$ in which the right hand side can be written as a multiplication a column vector of the components of the unit vector $\mathbf e$ normal to a surface by a matrix of components representing $\textbf{\sigma}$. $\textbf T$ is another vector, the traction vector. So we seem to be inputting one vector and outputting another. This doesn't seem to fit Jeevanjee's definition of a tensor. Can someone explain?

A rank-2 tensor like $\sigma_{ij}$ can indeed "take" two vectors and make a number (i.e., a scalar), but it can also "take" just one vector and turn it into another vector.

 

[Philip Wood]

Many thanks, renormalize. So Jeevanjee's being too restrictive? No wonder he doesn't use the Cauchy stress tensor as an example.

 

[jbergman]

Many thanks, renormalize. So Jeevanjee's being too restrictive? No wonder he doesn't use the Cauchy stress tensor as an example.

It's a long journey to learn about tensors. His definition is correct. As, @renormalize stated if you input a vector in one of the slots, you are left with a tensor that takes one less vector. This is similar to currying, or partial application if you are familiar with programming.

 

[WWGD]

In a Mathematical sense, Tensors are generalizations of matrices for dimensions beyond 2, i.e., multilinear maps.

 

[Chestermiller]

My understanding is that a 2nd order tension (like the stress tensor) maps a given vector into another vector, the case of the stress tensor, the given vector is a unit normal vector to a surface, and the output vector is the traction vector on the surface.

 

[Steve4Physics]

Sorry about the Latex failure; I've no idea why I can't make it work on this site.

Looks like the \textbf command is treating its argument as literal text - i.e. not processing commands embedded in the text.

Using \mathbf doesn't seem to resolve this, - possibly because bold Greek letters aren't included in its font.

However \boldsymbol seems to work:
- a plain sigma looks like this: $\sigma$.
- using \boldsymbol gives: $\boldsymbol \sigma$

There may be other ways of course.

 

[WWGD]

IIRC, one of the reasons is that it requires more than $n ; n^2$ entries to fully described in n-dimensions,
while a vector in n dimensions can be fully described with n components. The force, the normal, their interaction along each dimension.

 

[Philip Wood]

My understanding is that a 2nd order tension (like the stress tensor) maps a given vector into another vector, the case of the stress tensor, the given vector is a unit normal vector to a surface, and the output vector is the traction vector on the surface.

Many thanks. I'd better look into the concept of a second order tensor.