I am currently studying Reliability engineering and hence need to deal with material properties like elasticity modulus and poisson's ratio. I am basically an electrical engineer and hence never had to deal with Tensors previously. What I figured out about the difference between a tensor and a vector is that, a vector when broken down to it's most basic components, is nothing but an n-dimensional array where every element in the array represents a scalar, but on the other hand if I try to represent a Tensor in an "array" and not a matrix then every element in the array is a vector too. Kindly help me out, because I am a zero in Tensor mathematics because I never had to deal with it. The knowledgeable man or woman answering my question, if they do have an idea, can also help my understanding by referring to the "stress tensor in material" and the "conductivity tensor in the Current density and Electric field intensity" examples, since I will need to deal with these.
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$\begingroup$ I would interpret them in these cases to be $3 \times 3$ matrices. The current density and electric field are related by $J = \sigma E$ (where $\sigma$ is a $3 \times 3$ matrix), in an isotropic material we have $\sigma = c I$ for some scalar $c$, so this reduces to the familiar case. However, in general, the current is not necessarily parallel to the field. $\endgroup$– copper.hatJan 8, 2015 at 6:47
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$\begingroup$ yes the current is not necessarily parallel and that is exactly the case that I was trying to figure out, since I am overtly familiar with the dry and drab case of σ=cI since I am an electrical engineer and have only worked with circuits, I am more interested in the non-particularization because I am unable to understand the meaning of Tensors. $\endgroup$– ubuntu_noobJan 8, 2015 at 6:52
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1$\begingroup$ Some materials, such as cadmium or graphite have a crystalline structure that results in different conductivities in different directions. $\endgroup$– copper.hatJan 8, 2015 at 7:03
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3$\begingroup$ Note, it would have been a little more precise to characterise the $\sigma$ tensor as a linear operator, rather than a $3 \times 3$ matrix, partly because the latter implies some basis has been chosen. $\endgroup$– copper.hatJan 8, 2015 at 7:13
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$\begingroup$ Yes, I can understand what you are talking about. It is basically hence a linear operator on a vector which results in a vector in a different direction. But, if we want to operate on a vector to get another vector, why don't we just use the cross product instead, why do we have to define a new kind of mathematical operator for that? $\endgroup$– ubuntu_noobJan 8, 2015 at 7:42
1 Answer
This is really something that you should learn from a textbook, but I'll try to answer the second part of your question.
When you apply forces on any deformable body in equilibrium, it must develop internal resisting forces to keep every piece of it in equilibrium, by Newton's 2nd law. This is usually demonstrated by cutting the body into two by an imaginary plane and replacing material on one side of the body by a system of internal forces.
These internal forces vary from point to point in the body (think of the body as a well-behaved subset of $R^3$), but also depend on the direction in which you make the cut.
Since you can slice a 3D body in two through a point in an infinite number of ways, it would seem hopeless to specify these internal forces.
Cauchy proved that this need not be the case. All one needs is an entity called the stress tensor that, at a given point, maps normal vectors to these cutting planes to the vector force per unit area at that point (this is called the traction vector).
Mathematically, the stress tensor is a linear map $\mathbb{\sigma}$ such that (in a euclidean background space)
$$ \mathbb{t} = \mathbb{\hat{n} \cdot \sigma = \sigma^T \cdot \hat{n}} $$ where $\hat{n}$ is the unit normal to the plane and $\mathbb{t}$ is the traction vector.
Due to symmetry, $\sigma = \sigma^T$.
So, in a Cartesian basis, you can think of $\sigma$ as a 3 x 3 symmetric matrix that premultiplies a 3 x 1 column vector containing the components of the unit normal $\hat{n}$ to produce a 3 x 1 column vector of traction components.
$$ t_i = \sum_{j=1}^{3} \sigma_{ji} n_j \quad i=1,2,3 $$
(The summation sign is normally omitted)
In matrix form, $$ \{t\}_{3 \times 1} = [\sigma]_{3 \times 3} \{n\}_{3 \times 1} $$
The stress tensor $\sigma_{ij}$ is a function of space and time - it is a tensor field - so you can think of it as a system of functions $\sigma_{ij}(x,y,z,t)$.
