I know that a vector is a tool to help with quantities that have both a magnitude a direction. At a given point in space, a vector has a particular magnitude and direction and if we take any other direction at the same point we can get a projection of this vector in that direction.

Tensor is a tool to help with quantities that have a magnitude and 2 directions. But what are its magnitude and directions? Like if a tensor is defined at a point does it have a magnitude and 2 directions? For example, the stress tensor gives the stress at a point but as far as I have understood, at a given point as we change the plane (ie one of the direction) we get different stress value in a different direction. So at that point, the stress has different magnitude and direction as we keep changing the direction of the plane and in a given plane it's not like the projection of some 1 stress quantity.

When we say 'this' is a vector at 'this' point, we know what its magnitude and its direction are. Similarly, if we say 'this' is the tensor at 'this' point does it mean a particular magnitude and particular 2 directions? Can stress be defined by having a magnitude and 2 directions at a point like how a force is given by having a magnitude and 1 direction at a point?

EDIT: By defining a Coordinate system to space(consider a 3D space), we can define the vector by 3 coordinates and these 3 coordinates would give both magnitude and direction of the vector. Do the 9 coordinates of the tensor give a particular magnitude and 2 particular directions as well?

  • $\begingroup$ No, a tensor does not have a magnitude and directions. Instead, one way to think about a (second-order) tensor is as a function that takes in a vector and gives you another vector. For example, you give the stress tensor a vector which is the normal of the plane, and it gives you another vector which is the stress across that plane. $\endgroup$
    – user856
    Jun 1, 2018 at 13:45
  • $\begingroup$ @Rahul Tensor as linear mapping is fine, but I was wondering how to picture a tensor, just like how a vector can be given by an arrow. $\endgroup$
    Jun 2, 2018 at 7:28

2 Answers 2


This is a fantastic question that I have wrestled with myself. A second rank tensor is a linear map, but a tensor is also an object in itself. It can be rotated and scaled like other objects (it obeys certain laws under coordinate transformation).

Imagine an ellipsoid oriented arbitrarily in space. If you were to take the eigen values and vectors of your tensor, the eigen vectors correspond to a rotation that will align the coordinate system with the major axes of the ellipsoid, while the eigenvalues represent the major radii. The visualization breaks down if you have negative eigenvalues and perhaps other conditions as well.

Ellipsoid in space


I am going to go on a limb here using my memory of stress tensor and stress-traction-vector, and say that the "second direction" you're talking about is actually the orientation of the volume's surface areas (note edit) upon which the stress-traction-vector sits. I think of the Pascals Law for fluids, where you 'cut' the cube of material (either fluid or solid) and its orientations/normal-vectors on the two faces and the also the orientation/normal-vector on the cut face is minimally sufficient to define the orientation of that volume.

I would say that the stress tensor requires the magnitude and direction, i.e. the stress-traction-vectors, and then also the volume's orientations / normal-vectors. This is a great question!

edit: there 1st paragraph above, I said that the stress traction vector sits on the volume, the edit now says the stress traction vector sits on the volume's surface areas

edit2: see this answer from 2015 Significance of 'faces' in Stress tensor components?

  • $\begingroup$ let me clarify that those normal-vectors/orientations of the volume are the really the orientations of the volume's surface areas (two sides and cut face) $\endgroup$
    – L92MD14
    Jul 10, 2021 at 1:43

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