# Why is stress a tensor quantity?

1. Why is stress a tensor quantity?

2. Why is pressure not a tensor?

3. According to what I know pressure is an internal force whereas stress is external so how are both quantities not tensors?

I am basically having a confusion between stress pressure and tensor.

I am still in school so please give a very basic answer.

1. Stress is a tensor1 because it describes things happening in two directions simultaneously. You can have an $x$-directed force pushing along an interface of constant $y$; this would be $\sigma_{xy}$. If we assemble all such combinations $\sigma_{ij}$, the collection of them is the stress tensor.

2. Pressure is part of the stress tensor. The diagonal elements form the pressure. For example, $\sigma_{xx}$ measures how much $x$-force pushes in the $x$-direction. Think of your hand pressing against the wall, i.e. applying pressure.

3. Given that pressure is one type of stress, we should have a name for the other type (the off-diagonal elements of the tensor), and we do: shear. Both pressure and shear can be internal or external -- actually, I'm not sure I can think of a real distinction between internal and external.

A gas in a box has a pressure (and in fact $\sigma_{xx} = \sigma_{yy} = \sigma_{zz}$, as is often the case), and I suppose this could be called "internal." But you could squeeze the box, applying more pressure from an external source.

Perhaps when people say "pressure is internal" they mean the following. $\sigma$ has some nice properties, including being symmetric and diagonalizable. Diagonalizability means we can transform our coordinates such that all shear vanishes, at least at a point. But we cannot get rid of all pressure by coordinate transformations. In fact, the trace $\sigma_{xx} + \sigma_{yy} + \sigma_{zz}$ is invariant under such transformations, and so we often define the scalar $p$ as $1/3$ this sum, even when the three components are different.

1Now the word "tensor" has a very precise meaning in linear algebra and differential geometry and tensors are very beautiful things when fully understood. But here I'll just use it as a synonym for "matrix."

• thanks for the answer. just one confusion what is the meaning of tensor in physics? are they the quantities showing scalar and vector geometry both? how? Dec 12, 2014 at 16:09
• @user166748: the introduction to the Wikipedia article on tensors does a fair job of summarising what a tensor is. Dec 12, 2014 at 16:25
• If you have some geometric basics, I recommend you to take a look at this clip, beautifully explained: youtube.com/watch?v=uO_bW2zzrNU Oct 18, 2018 at 6:08

For each surface on a unit cube (see below), the stress on that surface can point in each of the three directions. (source)

Since it is not necessarily the case that $\sigma_{11}=\sigma_{31}=\sigma_{21}$ (all pointing the in the same $\mathbf{e}_1$ direction)--or any of the other $\sigma_{ij}$ combinations, we need to have 9 components describing it, hence the tensor being the natural choice: $$\boldsymbol\sigma=\sigma_{ij}=\left(\begin{array}{ccc}\sigma_{11}&\sigma_{12}&\sigma_{13}\\ \sigma_{21}&\sigma_{22}&\sigma_{23}\\ \sigma_{31}&\sigma_{32}&\sigma_{33}\end{array}\right)$$

Since pressure acts on all sides equally, it can be naturally described as a scalar quantity and is related to the stress tensor with a Kronecker delta, $\delta_{ij}$, under hydrostatic equilibrium conditions: $$\sigma_{ij}=-p\delta_{ij}$$

$\newcommand{\vect}{{\bf #1}}$

Imagine you have a body $\Omega$ and you want to understand how it responds to some external load. Do do this, imagine you cut the body with a plane, say the plane whose normal vector is the $\hat{x}$ vector. Now select an element of area $\delta A$ and measure how the force on the element, I will call it ${\rm d}\vect{F}$.

Define the numbers

$$\tau_{xx} = \lim_{\Delta A \to 0}\frac{\Delta F_x}{\Delta A}, ~~~ \tau_{yx} = \lim_{\Delta A \to 0}\frac{\Delta F_y}{\Delta A}, ~~~ \tau_{zx} = \lim_{\Delta A \to 0}\frac{\Delta F_z}{\Delta A}$$

You can see that

• It is important to label this element with out choice of the orientation for the cutting plane. In this case we select $\hat{x}$, but in general it could be any normal vector $\vect{n}$

• The definition of $\tau$ resembles that of pressure. Indeed, if you consider the element $\tau_{xx}$ you'd get the force per unit area perpendicular to the are element along the $x$ direction.

The stress tensor is just the collection

$$\tau = \left(\begin{array}{ccc} \tau_{xx} & \tau_{yx} & \tau_{zx} \\ \tau_{xy} & \tau_{yy} & \tau_{zy} \\ \tau_{xz} & \tau_{yz} & \tau_{zz} \\ \end{array}\right)$$

The average pressure could then be defined as

$$\pi = \frac{1}{3}(\tau_{xx} + \tau_{yy} + \tau_{zz})$$