# Why is the stress on a body not a vector?

In my textbook, Physics, Part II—Textbook for Class XI, there's a line which talks about why stress is not a vector:

Stress is not a vector quantity since, unlike a force, stress cannot be assigned a specific direction. Force acting on the portion of a body on a specified side of a section has a definite direction.

It does not elaborate why stress cannot be assigned a specific direction. I know that the stress on a body is the restoring force (applicable when the body is deformed) per unit area. Has it got something to do with the fact that area itself is a vector? Moreover, we often say that tensile (or compressive) stress is applied perpendicularly to the surface. That's specifying direction, isn't it?

An intuitive explanation (instead of a rigorous mathematical one) is highly appreciated.

• Essentially a duplicate of physics.stackexchange.com/q/152927/25301 Commented Dec 18, 2022 at 15:33
• @KyleKanos I don't consider closing this question though; it has received a lot of good answers, I am unsure though. Commented Dec 18, 2022 at 16:11
• Would it be possible for a moderator to merge these two questions into one? If ever there were a candidate for merging questions, this is it. Commented Dec 19, 2022 at 2:13
• Yep, here are a few good answers, but they're eager to jump in to tensor definitions without even considering that a school textbook might not even introduce them yet! Seems like a curse of knowledge to me. One vector is not enough, but combine them and you'll get from the regular Hooke's law to the generalized version and invent tensors in the process, win/win! Philip got the closest to actually acknowledging this possible gap, IMHO. And the linked question surely helps with additional explanations. Commented Dec 19, 2022 at 3:13
• @KyleKanos I had no idea it was a duplicate, as I had no idea what tensors were (which Lodinn talks about). It really begs the question: do you close a question in this site if the author doesn't know about the main concept involved in another question, which it's essentially a duplicate of? I'm genuinely interested in knowing this. Like, if my question were closed, I'd have never got a satisfactory answer to my question, and I'd have also kept scratching my head about tensors. Commented Jan 14, 2023 at 17:25

Draw a square on an elastomer strip and stretch it:

"OK, I get this:"

The lengthwise load (comprising two force vectors, to the left and to the right) applies a stress state on the shape. What kind of stress?

"I'll assume the stress state can be expressed as a vector. I guess the vector corresponds to normal stresses to the left and to the right (i.e., forces acting perpendicular to the left and right sides). This is consistent with normal stresses changing side lengths of infinitesimal elements. I guess I'll call the vector [1 0 0], where I've normalized by the load magnitude."

Now consider drawing not a square but a diamond, for the same load.

"OK, now I get this:"

What is the stress state?

"It now includes some shear stress, since interior angles are now changing. Effectively, some forces on the sides are now parallel instead of solely perpendicular. The vector [1 0 0] doesn't capture this change, nor can I transform it rotationally to scale with [1 1 0] or [1 -1 0], say, because the diamond doesn't deform that way either; it stretches more to the left and right than it shrinks up and down. Hmm.

"Nature doesn't care which way we draw our coordinate systems, so we need a mathematical representation that transforms correctly. I have to conclude that a vector is incapable of representing the stress. However, a tensor would work:

$$\left[\begin{array}{ccc} 1 &0 &0\\0& 0& 0\\0& 0 &0\end{array}\right]$$

would transform upon a 45° rotation into

$$\left[\begin{array}{ccc} 1/2 &1/2 &0\\1/2& 1/2& 0\\0& 0 &0\end{array}\right],$$

which is consistent with the observed deformation of the diamond. Specifically, the side lengths stretch equally from an equibiaxial stress—from the diagonal elements—of 1/2, and this is superimposed on a shape change from a shear stress—from the off-diagonal elements—of 1/2.

"Furthermore, the tensor satisfies the standard requirements, such as invariance of the trace (here, 1) and two other invariants. These invariants capture the true essence of the stress state, which must be coordinate independent."

Why not just list those indices as, say, [½ ½ 0 ½ ½ 0 0 0 0] to make a vector?

"That not a true Cartesian vector, which has three elements and a well-defined direction. It's just a list."

One more question. When we apply a load on a surface, the resulting stress state has a well-defined direction that corresponds to the load. Why isn't a tensor needed here?

"A tensor is still needed to describe the stress state because of the above reasoning, but neither the surface nor the load are free to rotate, so any infinitesimal element aligned with the surface is constrained. Although this appears to suggest that stress has a single direction, its a particular result of the constraint and doesn't hold in general."

(Images from my site, adapted from a photograph by Nelson Fitness.)

• That matrix is wrong. The matrix $\pmatrix{1&0\cr 0&0}$ transforms to $\pmatrix{t&t\cr -t&t}$ under a 45 degree rotation, where $t=1/\sqrt 2$. The sign error is more important than the scale. Commented Dec 20, 2022 at 11:55
• The transformation is correct for a rank-2 tensor. A rotation matrix is premultiplied and postmultiplied. It looks like you’ve performed only one of these multiplication steps. Commented Dec 20, 2022 at 13:20

We can put a wire under tensile stress by pulling each end with a force of equal magnitude. If the wire has an East-West alignment we need to pull its eastern end to the East and its western end to the West (even though one of these forces may be inconspicuously applied by a fixed anchorage to which one end of the wire is attached). You need both these forces to put the wire under uniform stress. So the stress doesn't point to the East or to the West. It has an alignment rather than a direction. That doesn't stop us from calculating the tensile stress by dividing the magnitude of either force by the area over which it acts.

Stress, then, is not a vector quantity. [Instead it can be represented by a mathematical object called a second rank tensor. Tensors enable us to represent not just tensile stress but other sorts of stress (such as shear stress), or combinations of different sorts of stress.]

• Thanks a lot! Most of the answers here talk a lot about tensors, but unfortunately, as a high-school student, I heard about them for the first time here! You've explained it intuitively, and then went on to explain what tensors are and how it all connects together—which was really helpful! Commented Jan 14, 2023 at 17:18

Suppose a ball so full of air that is almost exploding. Due to the symmetry of the situation, it is clear that each small area of its surface is being stretched equally to all tangential directions. It is not possible to represent such a stress state to an entity with a modulus and a given direction as it is the case of a vector.

However it is possible to define a $$x$$ and $$y$$ axis for a small squared region along $$2$$ orthogonal tangential directions. There is an equal force $$F$$ for each of this directions. As forces are vectors, when we project and add them to a direction $$45^{\circ}$$, the result is $$F\sqrt{2}$$.

But if we divide $$F$$ by the side of the square, and divide $$F\sqrt{2}$$ by its diagonal (because it is the length associated with an angle of $$45^{\circ}$$), the result is the same, as should be due to the symmetry.

It is possible to see in this example that the state of stress can not be adequately represented by one vector, but by a stress tensor formed with $$2$$ tensile stresses in mutually orthogonal directions.

Stress is not a vector because it needs more information than what a vector can provide.

Stress can be represented as a $$2^{nd}$$ order tensor quantity $$\mathbb{T}$$ that relates two vectors, the unit normal vector, $$\mathbf{\hat{n}}$$, of a surface passing through a chosen point and the force per unit surface acting on it, $$\mathbf{t_n}$$,

$$\mathbf{t_n} = \mathbf{\hat{n}} \cdot \mathbb{T}$$.

If you change the orientation of the surface, the force per unit surface changes as well. This can be interpreted with the equilibrium of the Cauchy tetrahedron, see wiki page https://en.m.wikipedia.org/wiki/Cauchy_stress_tensor