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Is there any easy way to visualize tensors (such as stress) using paper and pencil with the help of a matrix? If it is possible, kindly post a picture of that tensor.

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  • $\begingroup$ This video might help: youtu.be/…. I recommend watching the video from the beginning after you watch the part I timestamped and all of this person's videos as they help you understand tensors in a nice way (assuming you have learned and understand the math). $\endgroup$
    – Tachyon
    Oct 22 at 14:39
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    $\begingroup$ @Terrific I would strongly suggest that you watch this video by Eugene Khutoryansky he has an almost 'supernatural ability' to explain everything with animations and hardly ever makes reference to equations. The guy has a whole host of Youtube videos under his channel - "Physics Videos by Eugene Khutoryansky". Definitely worth a look in my opinion, the best i've seen. $\endgroup$ Oct 23 at 1:21
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The following answer is entirely my opinion: the best way to visualize tensors in Euclidean space is by considering them in their spherical basis. The spherical basis is standard in quantum mechanics, where tensor operators couple states of known initial and final angular momenta quantum numbers. (Note that this does not help visualization, but it's mathematically powerful, e.g. the Wigner-Eckart theorem).

If the tensors are rank-1, then it decreases visual understand by replacing our orthogonal triplet of easily visualized arrows:

$$({\hat x},{\hat y},{\hat z})$$

with a triplet of complex vectors:

$$ ({\hat e}^0=\hat z, {\hat e}^{\pm q}=(\mp \hat x -i\hat y)/\sqrt 2)$$

It seems like a step back, but helps explain what a vector is beyond the high-school "arrow" description. The $\hat e^q$ are the 3 eigenvectors of $z$-rotations (by $\theta$), with eigenvalue $\exp{(iq\theta)}$, where $q=(-1,0,1)$.

Spherical vectors are the things that can be rotated into themselves. The same is true of rank 2 spherical tensors, except the allowed eigenvalues are derived from $q=(-2,1,0,1,2)$. ($|q|=2$ describes time averaged linear polarization, where a 180-degree rotation is equal to the identity).

In general, for rank-$l$, there are $2l+1$ "natural" tensors with eigenvalues $q=(-l,-l+1, \ldots, 0, \ldots, l-1,l)$. Of course $2l+1 < l^2$, so you have to get rid of some degrees of freedom. That is done by symmetrizing and subtracting all traces.

For rank-2, we take our Cartesian tensor, $T_{ij}$, and symmetrize it:

$$ S_{ij}= \frac 1 2 \big(T_{ij}+T_{ji})$$

and then subtract the trace:

$$ N_{ij} = S_{ij}-\frac 1 3 {\rm Tr}({\bf T}) $$

For the remaining degrees of freedom, the antisymmetric part:

$$ A_{ij}= \frac 1 2 \big(T_{ij}-T_{ji})$$

looks like a vector:

$$v_i = \epsilon_{ijk}A_{jk}$$

and a scalar:

$$ s = {\rm Tr}({\bf T}) = A_{ii}$$

In the case of strain, the tensor is symmetric, and the scalar part becomes the pressure (and pressure is a scalar). The antisymmetric part would be a deformation that is equivalent to a rotation, so it's not used.

The five pure rank-2 tensors are:

$$ N^{(2,0)}= \sqrt{\frac 3 2}N_{zz}$$ $$ N^{(2,\pm 1)}= \frac 1 2[N_{zx}\pm iN_{zy}]$$ $$ N^{(2,\pm 2)}= \frac 1 2[N_{xx}-N_{yy}\pm 2iN_{xy}]$$

from that, the real combinations will describe fundamental shapes of strain deformation, which look like the real combinations of the spherical harmonics of order two:

enter image description here

This method of visualizing tensors holds for any rank $N$, where the shapes are combinations of $Y_N^{\pm m}(\theta, \phi)$.

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This may be a controversial opinion. But I've actually found it unhelpful to try to directly visualize tensors (other than vectors, or special cases with a lot of symmetry).

I'd like to make an analogy. In relativity, we often deal with 4-dimensional spacetimes. But I rarely, if ever, try to actually visualize what is happening in 4 dimensions. This is too much for my human brain. Instead, I have learned to trust the math. And, I have also learned a series of tricks that apply in different situations that can reduce the problem to something in 2 or 3 dimensions, where I can more easily visualize what's going on. In fact, I rarely directly try to visualize things directly in 3d! Often, I am really thinking about 2d projections (sometimes multiple 2d projections of one object that each give a different point of view).

Similarly, a general tensor has a lot of components and structure. I don't find it particularly useful, or easy, to try and simultaneously visualize the effect of all these components. Instead, I think of a rank $(M, N)$ tensor as a machine which takes as input $M$ "1-forms" and $N$ "vectors" and returns a number (the machine also has a lot of other nice properties, like linearity, which are useful for combining tensors and determining the behavior under linear transformations -- but I won't focus on that).

Then, when faced with something like a stress tensor, which is a $(0, 2)$ tensor, I think "in order to turn this into a number, I need to specify 2 vectors." In the case of the stress tensor, we can think of the two vectors as (1) the direction of an applied force, and (b) the orientation of the normal vector of a surface to which that force is applied. So, the component $T_{xx}$ represents a force in the $x$ direction applied to a face normal to the $x$ direction. This represents pressure. Meanwhile, the component $T_{xy}$ represents a force in the $x$ direction applied to a surface normal to the $y$ direction. This represents shear.

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The most common ways to think of a vector are a little arrow or an ordered tuple of coordinates such as

$$X = (x, y, z)$$

But this isn't what a vector really means.

To a mathematician, vectors are things that behave reasonably when added together or multiplied by a number. Little arrows are an easy example of this. You add them tip to tail to get a resultant vector. They stretch or shrink when multiplied by a number.

Mathematicians have a list of $8$ rules that a set of objects must satisfy to be considered a vector space. The rules encapsulate what addition and multiplication by a number should be. They are simple things like the associative law:

$$a + (b + c) = (a + b) + c$$

Any set of objects that satisfy the $8$ rules are a vector space. For example, the set of all quadratic polynomials

$$ax^2 + bx + c$$

is a vector space. You can add them and multiply by a number using the laws of algebra.


Physicists have slightly bigger idea of what a vector is. In addition to the $8$ rules, they require that vectors behave as expected under coordinate transforms.

Going back to little arrows on the x,y plane, suppose you have a vector of length $1$ in the x direction. Now leave the arrow fixed, but rotate the coordinates so the vector points in the y direction. In a sense the vector is the same. But it now has new coordinates.

You can use the coordinates to calculate the length of the arrow. You will still get $1$ after the coordinate transformation. The coordinate transformation must not change things like the length of an arrow, or the angle between two arrows.

Note that rotating the vector $90^o$ to the left and leaving the coordinate system alone is the same transformation as rotating the coordinate system $90^o$ to the right and leaving the vector alone. Either way, a vector in the x direction becomes a vector in the y direction.


Physicists also think about more kinds of things than vectors. For example scalars (numbers) are simpler than vectors. You can add numbers together and multiply them by numbers. They follow the $8$ rules. But they transform differently.

Suppose you have $3$ rocks sitting on the x,y plane. If you rotate the coordinates, the number of rocks does not change at all. Not changing at all makes this a scalar.


Physicists also think about more complex objects than vectors. Sometimes this requires more complex coordinate transformations, and more complex behavior under the transformations.

The electromagnetic field is an example. This is a field that one charge produces and exerts forces on another charge.

When first introduced, it is presented as two different E and B vectors. But it turns out the vectors are related to each other. They actually behave like a single more complex object under the right coordinate transformation.

A stationary charge produces an electric field. If you run by it, it is a moving charge that produces both an electric field and a magnetic field. The charge did not change because you ran by it. The effect on a nearby charge does not change. What changed was the coordinate system. Now it is moving. In this coordinate system the combined effect of the electric and magnetic fields will have the same effect on a charge as the electric field did in the stationary frame. The electromagnetic field is an example of a rank $2$ tensor.

To understand the electromagnetic field as a tensor requires special relativity. Instead of coordinates in $3$-D space, it requires $4$-D spacetime.

Instead of $4$ coordinates like a rank $1$ tensor (a vector), it requires $16$ coordinates, a $4x4$ matrix. In this case, the matrix is anti-symmetric for physics reasons. This means the elements on the diagonal of the matrix are $0$. Each element in the upper right is the negative of the corresponding element in the lower left. So you really just need the $6$ coordinates from the E and B fields.

$$F = \begin{pmatrix} 0 & -E_x & -E_y & -E_z \\ E_x & 0 & -B_z & B_y \\ E_y & B_z & 0 & -B_x \\ E_z & -B_y & B_x & 0 \\ \end{pmatrix} $$

Matrices can be added together and multiplied by a number using coordinates much like a vector can.

When you apply a change-of-velocity coordinate transformation to this matrix, you get another matrix with different coordinates. But it represents an electromagnetic field that has the same strength and effect as the untransformed field.

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