Tensor quantities [duplicate]

Question

What is a tensor quantity , how is it different from vector quantity or scalar quantity and why is pressure a tensor. It would be very helpful if you could give me an intuitive idea of tensor quantities.

Answer 1

So first, pressure itself is not a tensor quantity, though stress is.

The best, most intuitive understanding of tensors I've come across, is to imagine a tensor as some machine. A function. However instead of being $f(x)$, a function of a single variable, it is instead a function of a vector, or another tensor. (A vector can be seen as a tensor of rank 1).

Let's consider stress. Let's say I have a small box, and I want to know the force on each side of the box. For one side (say constant $x$ side), you could imagine a force acting on that side, say $\vec{F}_x$. Now we move to another side, we have a force $\vec{F}_y$. Another side, and it's $\vec{F}_z$. Each of these forces will have 3 components themselves, so you can see we need 9 pieces of information to specify the stress. Now let's consider how I could represent this as a function, a machine, a tensor. I want to be able to specify a side of the box, maybe with a vector $\hat{n}$, hand this to my tensor $\textbf{T}$, and have it spit out the force on that side of the box, $\vec{F}$.

This machine sounds complicated! However, we are lucky in knowing that the function it is representing should be linear. That is, if I know I give my tensor $\textbf{T}$ a vector $\vec{v}$, and I get the output vector $\vec{w}$, then I know if I give it a vector $2\vec{v}$, then the output must be $2\vec{w}$. We also know that $\textbf{T}(\alpha\vec{a}+\beta\vec{b}) =\alpha\textbf{T}(\vec{a})+\beta\textbf{T}(\vec{b})$, where the greek letters are just scalars. This is very useful. In short, it means we can specify our tensor entirely, just by looking at how it acts on a basis vector.

For instance, consider feeding our machine the basis vector $\hat{e}_1$. Then we know this will give a vector $\vec{V}=\textbf{T}(\hat{e}_1)$. But we could look at the components of the vector $\vec{V_1}$, as $\sum_i{V_{1i}\hat{e}_i}$, where $i$ goes from 1 to 3, (e.g. one for each $x, y, z$).

Now if we have any scalar multiple of $\hat{e}_1$, we know that $\textbf{T}(\alpha \hat{e}_1) = \alpha\sum_i{V_{i1}\hat{e}_i}$. But now we can repeat the same process for each of the three basis vectors. Now we can write the relationship: $\textbf{T}(\hat{e}_j)=\sum_i V_{ij}\hat{e}_i$. This quantity $V_{ij}$ now completely specifies our tensor, and so I rename it $T_{ij}$.

It is a 3x3 matrix with values. The $i$ value is the $i$th component of the vector out of the machine, upon feeding the machine the basis vector $\hat{e}_j$, and using this matrix, we can calculate the output from $\textbf{T}$ for any arbitrary vector. For completeness, that is: $$ \textbf{T}\bigg(\sum{A_k \hat{e}_k}\bigg)=\sum_{ij}T_{ij}A_j\hat{e}_i. $$

The actual values of $T_{ij}$ will have some functional form which is given by the physics of the situation. There are several other mathematical properties of tensors, which I have mainly left out in this intuitive post. One of the most useful is that a tensor equality is frame-independent. If you can equate two tensors in one frame, you will know they are equal in all frames.

So, in conclusion, intuitively a tensor is a tensor valued function, of tensors. (Remember a rank 0 tensor is a scalar, a rank 1 tensor is a vector). You can get far by remembering you want to feed your tensor a vector (or tensor) and get out a physically meaningful tensor (or vector, or scalar) back. There are just several mathematical niceties that make them incredibly useful in physics ( nearly everything can be approximated as being linear if you look at it closely enough!). I think finally it's important to remember that $\textbf{T}(~)$ is the more fundamental thing here, the matrix of values only has meaning in one specific basis, and can be seen as a way of extracting the information the tensor, as a function, holds.