Tensors from first principles

Cut a balloon open and lay the rubber flat, or take a wide rubber band. Draw a small circle on it, and two lines through the circle that cross at a right angle, both slanting at 45 degrees. Add a horizontal and a vertical line too. Now pull the rubber in one direction only.

Before you do, make three guesses. Does the circle stay a circle? Does the rubber get narrower across the pull? And do the two slanted lines, which you did not pull on, still cross at a right angle? Commit to your answers; Figure 1 asks for them before it lets you pull.

This article starts from that experiment and ends with a mathematical object, the tensor, that predicts every one of its results. You need to know what an arrow with a length and a direction is, and to be comfortable with a small table of numbers. Everything else is explained on the way, and the figures are live.

A circle on a stretched sheet#

To compute anything we need a model of the rubber. Take an idealized sheet: the same everywhere, the same in every direction, and incompressible, so its volume does not change when it is pulled. Pull it to $\lambda$ times its length. To keep its volume, the sheet has to get thinner and narrower, and if it is free at its long edges it does both by the same factor, $1/\sqrt{\lambda}$. A real rubber band is close to this away from your fingers, but it is not exactly this, so read the numbers below as those of the model.

Make your three guesses, then pull.

Does the circle stay a circle?

Stays a circleBecomes an oval

Does the rubber get narrower across the pull?

NarrowerSame width

Do the two slanted lines still cross at a right angle?

Still 90°Not anymore

Skip the guesses

The circle becomes an ellipse, long along the pull. The sheet gets narrower: at a pull of 1.5 it is 0.82 times as wide. And the slanted lines no longer cross at a right angle. At a pull of 1.5 they cross at 57 degrees on the side of the pull, and at 123 degrees on the other side. Nobody pulled on those lines, and nothing pushed them sideways; still, the angle between them changed by 33 degrees.

The lengths tell the same story. The horizontal line is now 1.5 times as long, the vertical line 0.82 times, and each slanted line 1.21 times. Every line changed by its own amount, and the slanted ones also turned toward the pull.

One number is not enough#

So how much did the sheet stretch? The honest answer is "it depends which line you ask about": 1.5 along the pull, 0.82 across it, 1.21 on the slant, and something else for every other direction. No single number can say this. An arrow can't either. An arrow along the pull, 1.5 long, says nothing about the shrinking across the pull, and nothing about lines turning.

What we need is a rule that takes any line drawn on the sheet and tells us where it ends up. Here is one. Mark a tiny arrow $d\mathbf{X}$ on the sheet before the pull. After the pull it has become another tiny arrow, $d\mathbf{x}$. If the deformation is smooth, then over a small enough region it is nearly uniform, and the new arrow depends on the old one linearly (Holzapfel, 2000):

$$d\mathbf{x} = \mathbf{F}\, d\mathbf{X}.$$

$\mathbf{F}$ is the deformation gradient. Once we pick a horizontal axis $x$ and a vertical axis $y$, it can be written as four numbers. For the pull, a line along $x$ grows by $\lambda$ and a line along $y$ shrinks by $1/\sqrt{\lambda}$, and neither turns, so

$$\mathbf{F} = \begin{bmatrix} \lambda & 0 \\ 0 & 1/\sqrt{\lambda} \end{bmatrix}.$$

Now apply it to a slanted line, from $(0, 0)$ to $(1, 1)$. It goes to $(\lambda, 1/\sqrt{\lambda})$, which at a pull of 1.5 is $(1.5, 0.82)$: longer, and tilted toward the $x$ axis. Its partner, from $(0, 0)$ to $(1, -1)$, goes to $(1.5, -0.82)$, tilted toward the axis from the other side. The two have closed in on each other, which is the 57 degrees of Figure 1.

Stretch of every line, by the direction it started in

Direction of the line, before the deformation

30°

Change the deformationStart again

Show the lines that stretch most and least

Stretch of the orange line: 1.362

$F = \begin{bmatrix} 1.50 & 0.00 \\ 0.00 & 0.82 \end{bmatrix}$

$C = F^{\mathsf T} F = \begin{bmatrix} 2.25 & 0.00 \\ 0.00 & 0.67 \end{bmatrix}$

Largest and smallest stretch: 1.500 and 0.816, the square roots of C's eigenvalues 2.250 and 0.667

Turn the line's direction in Figure 2 and watch its stretch on the right. Before you tick Show the lines that stretch most and least, decide: does the most-stretched line lie exactly along the pull, or somewhere else? For a straight pull it is along the pull. Now press Change the deformation and add some shear. The longest line moves away from both axes, and the four numbers of $\mathbf{F}$ no longer show where it is.

A turn that stretches nothing#

One more experiment before any more math. Leave the rubber unstretched, and turn it 30 degrees on the table. In Figure 2, that is the Pure rotation button. Before you press it, decide: will any line on the sheet stretch?

None does. The polar plot sits on the dashed circle, every stretch is 1, and yet $\mathbf{F}$ has changed: it is now a rotation, with four numbers that are not those of "no deformation". So $\mathbf{F}$ is not a measure of how much the sheet stretched. It mixes stretching with turning, and a turn by itself does nothing to the material.

To keep the stretching and drop the turning, look at lengths. The squared length of the new arrow is

$$|d\mathbf{x}|^2 = d\mathbf{x} \cdot d\mathbf{x} = d\mathbf{X} \cdot \mathbf{F}^{\mathsf T}\mathbf{F}\, d\mathbf{X}.$$

The combination $\mathbf{C} = \mathbf{F}^{\mathsf T}\mathbf{F}$ is called the right Cauchy-Green tensor. Write the old arrow as a length times a unit direction $\mathbf{N}$, and the stretch of the line in that direction is

$$\lambda(\mathbf{N})^2 = \mathbf{N} \cdot \mathbf{C}\,\mathbf{N}.$$

For a pure rotation, $\mathbf{C}$ is exactly the identity, so every stretch is 1, which is what the figure showed. The usual measure of strain is built from $\mathbf{C}$ for the same reason: the Green-Lagrange strain, $\mathbf{E} = (\mathbf{C} - \mathbf{I})/2$, is zero for any rigid turn (Holzapfel, 2000).

$\mathbf{C}$ also answers the question from Figure 2: which lines stretch most and least? They are always perpendicular to each other, and their stretches are the square roots of the two eigenvalues of $\mathbf{C}$; the lines are its eigenvectors. After the deformation they are still perpendicular, and they become the long and short axes of the ellipse. That is why a circle always becomes an ellipse under a uniform deformation, never a lopsided oval.

The diagonal numbers of $\mathbf{C}$ are not the answer, except when the axes happen to line up with those two lines. For $\mathbf{F}$ with rows $(1.5, 0.4)$ and $(0, 0.8)$, the square root of the first diagonal number is 1.500, but the largest stretch is 1.570, along a line about 20 degrees from the $x$ axis.

Stretch, then turn

Any deformation gradient with a positive determinant can be split into a stretch followed by a turn, $\mathbf{F} = \mathbf{R}\mathbf{U}$, where $\mathbf{R}$ is a rotation and $\mathbf{U}$ is symmetric, with positive eigenvalues equal to the principal stretches. Then $\mathbf{C} = \mathbf{U}^{\mathsf T}\mathbf{R}^{\mathsf T}\mathbf{R}\mathbf{U} = \mathbf{U}^2$, because a rotation undoes itself when transposed: $\mathbf{C}$ sees the stretch $\mathbf{U}$ and not the turn $\mathbf{R}$. This is the polar decomposition (Holzapfel, 2000).

Turn the axes, not the sheet#

The four numbers of $\mathbf{C}$ were written in our $x$ and $y$ axes, and we chose those axes. Nothing about the rubber made us draw $x$ along the bottom edge of the page. So are those numbers the physics?

Figure 3 freezes one deformation and lets you turn only the axes. The sheet, the ellipse and the orange line do not move. Before you move the slider, decide: when the axes turn by 30 degrees, will the numbers of $\mathbf{C}$ change? And will the stretch of the orange line?

Angle of the axes

0°

$C' = \begin{bmatrix} 1.96 & 0.42 \\ 0.42 & 0.73 \end{bmatrix}$

Did not change

Stretch of the orange line

1.444

Largest and smallest stretch

1.446 and 0.775

Sum of the diagonal of C′

2.690

Start again

The numbers change; nothing physical does. The stretch of the orange line stays the same, and so do the largest and smallest stretches. At one angle, and again 90 degrees from it, the two off-diagonal numbers become zero: the axes then lie along the two perpendicular lines that stretch most and least, the principal directions, and the diagonal holds the squares of the principal stretches.

The numbers change by a fixed rule. If the new axes are the columns of a rotation $\mathbf{P}$, written in the old axes, then the components of a direction become $\mathbf{N}' = \mathbf{P}^{\mathsf T}\mathbf{N}$ and those of $\mathbf{C}$ become

$$\mathbf{C}' = \mathbf{P}^{\mathsf T}\mathbf{C}\,\mathbf{P},$$

so that $\mathbf{N}' \cdot \mathbf{C}'\mathbf{N}' = \mathbf{N} \cdot \mathbf{C}\,\mathbf{N}$: every prediction comes out the same.

This is a different thing from the pure rotation in Figure 2. There the sheet turned and the axes stayed: $\mathbf{F}$ changed, and $\mathbf{C}$ stayed the identity. Here the axes turn and the sheet stays: nothing happened to the material, and the numbers change anyway. Keeping those two apart is most of what it takes to read a tensor correctly.

What makes it a tensor#

If $\mathbf{C}$ were its four numbers, Figure 3 would show the physics changing as we turned the axes, and it didn't. So $\mathbf{C}$ is something else, and its four numbers are a description of it in one choice of axes.

What $\mathbf{C}$ is, is the rule itself. Give it two directions on the sheet, and it returns a number, $\mathbf{a} \cdot \mathbf{C}\,\mathbf{b}$; give it the same direction twice, and the number is that line's squared stretch. The rule is linear in each direction, and it does not refer to any axes at all. Its components are what we get once we choose axes: the four numbers. When we change the axes, the components change by the law above, and they change in exactly the way that keeps everything the rule says the same. A table of numbers that changes by that law is the set of components of a tensor; a table that doesn't, isn't.

Counting is not what matters. A list of four numbers is not a tensor just because there are four of them, and $\mathbf{F}$ and $\mathbf{C}$ both have four components in two dimensions while being different kinds of rule. $\mathbf{F}$ takes a direction on the sheet before the deformation and returns an arrow after it, so it connects two different settings; $\mathbf{C}$ works entirely on the sheet before. Both are tensors of order two: they act on directions twice over, and their components carry two indices and change with two factors of $\mathbf{P}$.

Now the familiar ladder makes sense, as a summary rather than a starting point. A temperature at a point has no direction to change with when the axes turn: it is a tensor of order zero, a scalar. A velocity has one direction; its two components change with one factor of $\mathbf{P}$: order one, a vector. The stretch rule of the rubber needs two: order two. In three dimensions the same rule has nine components, and the transformation law has the same form.

Index notation, and the general definition

With components $C_{kl}$ in the old axes and the rotation's components $P_{ki}$, the law $\mathbf{C}' = \mathbf{P}^{\mathsf T}\mathbf{C}\,\mathbf{P}$ reads $C'_{ij} = P_{ki}\, P_{lj}\, C_{kl}$, summed over the repeated indices $k$ and $l$: one factor of $\mathbf{P}$ for each index.

That form assumes axes that are perpendicular and of unit length. With skewed or stretched axes, two kinds of component appear for the same vector: one kind changes like the axes themselves (covariant components) and the other changes the opposite way (contravariant components), and indices are written low or high to keep them apart. For perpendicular unit axes the two kinds coincide, which is why this article never needed the distinction. In general, a tensor of type $(p, q)$ is a rule that takes $p$ covectors (linear functions of vectors) and $q$ vectors and returns a number, linear in each argument; the bilinear rule $\mathbf{a} \cdot \mathbf{C}\,\mathbf{b}$ is of type $(0, 2)$.

Three lines decide every line#

Nobody draws circles on a bridge. Engineers read the same rule with strain gauges: thin metal foil grids bonded to the surface, whose electrical resistance changes as they stretch. A gauge reports one number: how much the surface stretched along the gauge's own direction.

The stretches in a steel structure are tiny. A steel with a yield strength of 355 MPa and a stiffness of 210 GPa starts to yield at a strain of about 0.0017, so gauges work in millionths, microstrain (µε). At that size the strain measure simplifies. Write the displacement of each point as $\mathbf{u}$; when its gradient is small, the Green-Lagrange strain is almost the small strain $\boldsymbol\varepsilon$, the symmetric part of that gradient (Holzapfel, 2000). The normal strain along a direction at angle $\theta$ is then

$$\varepsilon(\theta) = \varepsilon_{xx} \cos^2\theta + \varepsilon_{yy} \sin^2\theta + 2\varepsilon_{xy} \sin\theta \cos\theta,$$

the small-strain version of $\lambda(\mathbf{N})^2 = \mathbf{N} \cdot \mathbf{C}\,\mathbf{N}$. Three numbers describe the strain at a point, so three gauges in three different directions are enough to find them. The usual arrangement is a rosette, three gauges at 0, 45 and 90 degrees. The first and last give $\varepsilon_{xx}$ and $\varepsilon_{yy}$ directly, and the middle one gives the third:

$$\varepsilon_{xy} = \varepsilon_{45} - \frac{\varepsilon_{0} + \varepsilon_{90}}{2}.$$

The manufacturers' guidance says the same thing from the other side: when the principal directions are not known in advance, three independent strain measurements are needed (Micro-Measurements, 2014). Many texts and data sheets use the engineering shear strain $\gamma_{xy} = 2\varepsilon_{xy}$ instead; check which one a formula expects before you use it.

Direction of the fourth gauge

120°

Your estimate of its reading, in microstrain

Show the readingNew loadStart again

Gauges at 0°, 45°, 90°

420, 315, −130 µε

Fourth gauge, 120°

hidden

Choose a direction for the fourth gauge in Figure 4 and estimate its reading before you reveal it. The three dots are all the information there is. Once revealed, the curve passes through them and through the fourth reading, and it is always the same shape, a wave that repeats every 180 degrees, because the strain along a line and along the same line reversed are the same. Its highest and lowest points are the principal strains, 90 degrees apart, exactly like the longest and shortest lines on the rubber.

Stress is a tensor too#

Strain describes how lines on a material stretch and turn. Stress describes the forces inside it, and it is a different tensor, with the same structure.

Imagine a steel bar pulled along its axis, and cut it in your mind along a plane. The material on one side of the cut pulls on the other with some force per unit area, the traction $\mathbf{t}$. Which way it points, and how large it is, depends on how the cut is oriented. Cauchy's result is that a single tensor at the point, the stress $\boldsymbol\sigma$, gives the traction on every cut from that cut's unit normal $\mathbf{n}$ (Gurtin, Fried & Anand, 2010):

$$\mathbf{t} = \boldsymbol\sigma\,\mathbf{n}.$$

A prediction to try: the bar is pulled straight, along its length, and nothing pushes it sideways. On a cut tilted at 45 degrees, is there any force sliding one side past the other? With the bar's axis as $x$ and a stress $\sigma$ along it, the stress tensor has components $\sigma$ in the first diagonal place and zeros elsewhere. A cut whose normal is at 45 degrees has $\mathbf{n} = (1, 1)/\sqrt{2}$, so the traction is $(\sigma/\sqrt{2}, 0)$. Split it along the normal and along the cut, and both parts have the size $\sigma/2$: the cut carries a sliding force as large as its pulling force. This is the slanted lines of the rubber again, in forces instead of shapes: a straight pull, and something happening in directions nobody pulled along.

Stress and strain are linked only through the material. For small strains in a material that behaves the same in every direction, Hooke's law with two constants does it; in general the link is itself a tensor.

Where this picture stops#

Everything above treated the deformation as uniform. A real deformation is uniform only over small regions: $\mathbf{F}$ is a field that changes from point to point, near your fingers on the rubber most of all, and the analysis applies point by point.

The rubber model assumed a material that is incompressible and the same in every direction. Woven fabric is neither: it stretches much more along the bias, at 45 degrees to the threads, than along the threads. Its deformation still has an $\mathbf{F}$ and a $\mathbf{C}$, but the relation between strain and stress depends on direction, and describing that takes a stiffness tensor of order four.

The figures stay in the plane. In three dimensions $\mathbf{F}$ and $\mathbf{C}$ have nine components each, $\mathbf{C}$ is still symmetric and so has six independent ones, and every idea here carries over. And the rosette formula is a small-strain formula: it assumes strains of the order of a thousandth, not the 50 percent of a rubber band, and a strain that is uniform under the three gauges.

Back to the circle#

Why did the slanted lines close? Because the two lines that stretched without turning were the ones along and across the pull: for a straight pull they are the principal directions, and they became the axes of the ellipse. Every other line turned toward the pull, and the two slanted lines turned toward it from opposite sides, so they closed in on each other. Two lines that started perpendicular and are no longer perpendicular show a shear strain, and a straight pull produced it without any sideways force.

The rule predicts the number too. The angle between two deformed lines follows from $\mathbf{C}$, since for unit directions $\mathbf{a}$ and $\mathbf{b}$, $\mathbf{a} \cdot \mathbf{C}\,\mathbf{b}$ is the product of the lines' new lengths and the cosine of their new angle. For the two slanted lines under a pull $\lambda$, that cosine is $(\lambda^2 - 1/\lambda)/(\lambda^2 + 1/\lambda)$, which at $\lambda = 1.5$ gives 57 degrees: the number Figure 1 showed.

In one sentence#

A tensor is a rule that acts on directions, here the rule that says how every line drawn on a material stretches and turns; its numbers depend on the axes we choose, and change with them by a fixed law, so that what the rule predicts does not.

References#