Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I have a pretty good knowledge of physics but couldn't understand what a tensor is. I just couldn't understand it, and the wiki page is very hard to understand as well. Can someone refer me to a good tutorial about it?

share|improve this question
If you like this question you may also enjoy reading this Phys.SE post. –  Qmechanic Jul 14 '12 at 7:45
I wrote a short one. The main problem is that the mathematical definitions are absurdly abstract, and don't match the domain of discourse well. The abstract definition is that a tensor is a linear map from vectors and covectors to numbers. The concrete definition is below. –  Ron Maimon Jul 14 '12 at 8:52
Really short answer: A $[p, q]$ valent tensor is a thing that has $p$ upper and $q$ lower indices. Or it is something, that has to be multiplied by $q$ vectors and $p$ covectors to become a scalar. –  queueoverflow Nov 2 '12 at 17:12
A covector is a row vector ($1 \times n$ matrix), or transposed (i. e. rotated) vector as opposed to a vector, which is a column vector ($n \times 1$ matrix). So a regular matrix $M$ is a $[1, 1]$ valent tensor because you have to multiply it with a vector $v$ and a covector $c$ to get a scalar $s$: $s = c M v$. –  queueoverflow Nov 3 '12 at 9:04
a tensor is a 3D matrix(cube/box) written in 2D (screen/plane/paper for ease). –  Waqar Ahmad Apr 25 '14 at 3:59

7 Answers 7

up vote 27 down vote accepted

A (rank 2 contravariant) tensor is a vector of vectors. If you have a vector, it's 3 numbers which point in a certain direction. What that means is that they rotate into each other when you do a rotation of coordinates. So that the 3 vector components $V^i$ transform into

$$V'^i = A^i_j V^j$$

under a linear transformation of coordinates.

A tensor is a vector of 3 vectors that rotate into each other under rotation (and also rotate as vectors--- the order of the two rotation operations is irrelevant). If a vector is $V^i$ where i runs from 1-3 (or 1-4, or from whatever to whatever), the tensor is $T^{ij}$, where the first index labels the vector, and the second index labels the vector component (or vice versa). When you rotate coordinates T transforms as

$$ T'^{ij} = A^i_k A^j_l T^{kl} = \sum_{kl} A^i_k A^j_l T^{kl} $$

Where I use the Einstein summation convention that a repeated index is summed over, so that the middle expression really means the sum on the far right.

A rank 3 tensor is a vector of rank 2 tensors, a rank four tensor is a vector of rank 3 tensors, so on to arbitrary rank. The notation is $T^{ijkl}$ and so on with as many upper indices as you have a rank. The transformation law is one A for each index, meaning each index transforms separately as a vector.

A covariant vector, or covector, is a linear function from vectors to numbers. This is described completely by the coefficients, $U_i$, and the linear function is

$$ U_i V^i = \sum_i U_i V^i = U_1 V^1 + U_2 V^2 + U_3 V^3 $$

where the Einstein convention is employed in the first expression, which just means that if the same index name occurs twice, once lower and once upper, you understand that you are supposed to sum over the index, and you say the index is contracted. The most general linear function is some linear combination of the three components with some coefficients, so this is the general covector.

The transformation law for a covector must be by the inverse matrix

$$ U'_i = \bar{A}_i^j U_j $$

Matrix multiplication is simple in the Einstein convention:

$$ M^i_j N^j_k = (MN)^i_k $$

And the definition of $\bar{A}$ (the inverse matrix) makes it that the inner product $U_i V^i$ stays the same under a coordinate transformation (you should check this).

A rank-2 covariant tensor is a covector of covectors, and so on to arbitrarily high rank.

You can also make a rank m,n tensor $T^{i_1 i_2 ... i_m}_{j_1j_2 ... j_n}$, with m upper and n lower indices. Each index transforms separately as a vector or covector according to whether it is up or down. Any lower index may be contracted with any upper index in a tensor product, since this is an invariant operation. This means that the rank m,n tensors can be viewed in many ways:

  • As the most general linear function from m covectors and n vectors into numbers
  • As the most general linear function from a rank m covariant tensor into a rank n contravariant tensor
  • As the most general linear function from a rank n contravariant tensor into a rank m covariant tensor.

And so on for a number of interpretations that grows exponentially with the rank. This is the mathemtician's preferred definition, which does not emphasize the transformation properties, rather it emphasizes the linear maps involved. The two definitions are identical, but I am happy I learned the physicist definition first.

In ordinary Euclidean space in rectangular coordinates, you don't need to distinguish between vectors and covectors, because rotation matrices have an inverse which is their transpose, which means that covectors and vectors transform the same under rotations. This means that you can have only up indices, or only down, it doesn't matter. You can replace an upper index with a lower index keeping the components unchanged.

In a more general situation, the map between vectors and covectors is called a metric tensor $g_{ij}$. This tensor takes a vector V and produces a covector (traditionally written with the same name but with a lower index)

$$ V_i = g_{ij} V^i$$

And this allows you to define a notion of length

$$ |V|^2 = V_i V^i = g_{ij}V^i V^j $$

this is also a notion of dot-product, which can be extracted from the notion of length as follows:

$$ 2 V\cdot U = |V+U|^2 - |V|^2 - |U|^2 = 2 g_{\mu\nu} V^\mu U^\nu $$

In Euclidean space, the metric tensor $g_{ij}= \delta_{ij}$ which is the Kronecker delta. It's like the identity matrix, except it's a tensor, not a matrix (a matrix takes vectors to vectors, so it has one upper and one lower index--- note that this means it automatically takes covectors to covectors, this is multiplication of the covector by the transpose matrix in matrix notation, but Einstein notation subsumes and extends matrix notation, so it is best to think of all matrix operations as shorthand for some index contractions).

The calculus of tensors is important, because many quantities are naturally vectors of vectors.

  • The stress tensor: If you have a scalar conserved quantity, the current density of the charge is a vector. If you have a vector conserved quantity (like momentum), the current density of momentum is a tensor, called the stress tensor
  • The tensor of inertia: For rotational motion of rigid object, the angular velocity is a vector and the angular momentum is a vector which is a linear function of the angular velocity. The linear map between them is called the tensor of inertia. Only for highly symmetric bodies is the tensor proportional to $\delta^i_j$, so that the two always point in the same direction. This is omitted from elementary mechanics courses, because tensors are considered too abstract.
  • Axial vectors: every axial vector in a parity preserving theory can be thought of as a rank 2 antisymmetric tensor, by mapping with the tensor $\epsilon_{ijk}$
  • High spin represnetations: The theory of group representations is incomprehensible without tensors, and is relatively intuitive if you use them.
  • Curvature: the curvature of a manifold is the linear change in a vector when you take it around a closed loop formed by two vectors. It is a linear function of three vectors which produces a vector, and is naturally a rank 1,3 tensor.
  • metric tensor: this was discussed before. This is the main ingredient of general relativity
  • Differential forms: these are antisymmetric tensors of rank n, meaning tensors which have the property that $A_{ij} =-A_{ji}$ and the analogous thing for higher rank, where you get a minus sign for each transposition.

In general, tensors are the founding tool for group representations, and you need them for all aspects of physics, since symmetry is so central to physics.

share|improve this answer
"A (rank 2 contravariant) tensor is a vector of vectors": I like this picture! From the three numbers of a vector, we construct its length which is an invariant to rotations. Does something analogous exist for a rank 2 contravariant tensor wrt rotations? –  Physiks lover Nov 15 '14 at 1:18
@Physikslover: This is just the classical index definition. The process of finding invariants is by raising and lowering indices using the metric tensor (which is just the identity for Euclidean space, so you can ignore up and down position) and taking arbitrary traces by contracting indices. With a two-tensor, there are lots of invariants T^a_a, the trace, T_ab T_ab (sum of squares of all components--- the "length"), T_abT_bcT_ca (a cubic invariant that doesn't have a name) etc. There is one other invariant, the epsilon tensor, which is used to construct chiral invariants like the cross prod. –  Ron Maimon Nov 15 '14 at 19:39
Why do so many physicists insist on defining tensors by how their representations transform under coordinate transformations? This is such a backasswards way to think about the underlying object. –  DanielSank Jan 16 at 20:20
@DanielSank: Because this is how you represent the tensor on a computer. It is not only NOT "backasswards", it is the preferred way to do things. The mathematician's way is only tolerable because it is equivalent to the correct business. The issue here is whether you have a fetish against coordinate systems. Physicists take coordinate systems, but do not distinguish between two different systems--- they make gauge equivalence classes. Mathematicians strain very hard to make definitions which do not mention a coordinate system. The latter approach is NOT preferrable, as many examples show. –  Ron Maimon Jan 17 at 14:44
... The best example of a quantity you can define in a coordinate system which changes extremely nontrivially from one coordinate system choice to another is the Einstein stress-energy pseudotensor. Even conceiving of such a quantity would be impossible for a person indoctrinated by coordinate free methods--- the location of the stress-energy depends on coordinate choice! Coordinate systems are not something to sweep into the closet, you just have to make sure you understand the transformation law. It is not always possible or desirable to make coordinate free definitions. –  Ron Maimon Jan 17 at 14:47

A tensor is a generalization of the notion of scalars and vectors. A tensor of rank 0 is a scalar (it has $3^0$ compenent), while a tensor of rank 1 is a vector (which has $3^1$ components). In general, a tensor of rank $n$ has $3^n$ components.

See http://www.grc.nasa.gov/WWW/k-12/Numbers/Math/documents/Tensors_TM2002211716.pdf for a nice introduction.

share|improve this answer
The above link sometimes doesn't work. If so, google "tensor" and "nasa", which will bring up the article by JC Kolecki. You can then use Google QuickView. –  Joebevo Jul 14 '12 at 8:04
I've edited the link so that it points directly to the pdf, rather than going through a google search result. I hope that's ok... –  Nathaniel Jul 14 '12 at 9:34
This was an excellent tutorial! The example on page 10 on why not all scalars are rank-0 tensors was very clear. –  recipriversexclusion Jul 27 '12 at 22:27

Tensors are objects with usually multiple indices, a generalization of vectors and matrices, with definite transformation properties under a change of basis. They are introduced differently in different traditions, with different notations.

You may find the entry ''How are matrices and tensors related?'' from Chapter B8 of my theoretical physics FAQ relevant to disentangle some of the associated problems.

share|improve this answer

There are a lot of answers already hope I can make it even more clear.

Tensors are the generalization of the linear transformations.

Tensor is something that takes $m$ vectors and makes $n$ vectors from it.

The $n+m$ is the order (or rank) of the tensor.

Their type is denoted by $(n,m)$ (n: output vectors, m: input vectors)

When a tensor takes 0 vectors it means it calculates something from a scalar (or is a constant), if a tensor makes 0 vectors, it produces a scalar.

Some examples of tensors by type:

  • (0,0): scalar, just a number.
  • (1,0): single vector.
  • (2,0): a bivector
  • (1,1): Linear transformation.
  • (0,2): dot product of two vectors.
  • (1,2): cross product of two vectors in 3D.
  • (1,3): Riemann curcature tensor (if you are interested in general relativity, you will need this.)

Tensors can be described using an $n+m$ dimensional array of numbers. So the tensor's elements can be accessed using $n+m$ indexes.

For example, linear transformation is a 2nd order tensor.

The elements of the multidimensional tensor can be accessed by index, a matrix has obviously 2 indexes.

Now something about the notation. Tensor elements usually has multiple indexes, some upper indexes and some lower ones. Lower indexes going for the input vectors, upper indexes are for the output vectors. Note: upper indexes has nothing to do with exponents!

So a linear transformation tensor would look like this: $L_j^i$.

You do a linear transformation (aka calculating the elements of the resulting vector) like this:

$b^i = \displaystyle\sum_j L_j^i a^j $

So assume you are in 3D and multiply a 3×3 matrix with a column vector. In this case the upper index is for the lines, and the lower is for the columns of the matrix. $i$ and $j$ runs from 1 to the dimension you are in (usually 3).

You can chain these linear transformations like this:

$c^k = \displaystyle\sum_i M_i^k \displaystyle\sum_j L_j^i a^j $

Einstein noted, that in these summation formulas the index below the summation sign appears exactly twice. So it can be removed. So the previous two expressions will look like this:

$b^i = L_j^i a^j $

$c^k = M_i^k L_j^i a^j $

Which is very analogous with the matrix formulas you use in linear algebra. The upper index kills the lower index during calculation, while the lone indexes remain intact.

So you can multiply the two matrixes as tensors like this:

$T_j^k = M_i^k L_j^i = \displaystyle\sum_i M_i^k L_j^i $

And finally a cross-product with tensors would look like this:

$r^k = C_{ij}^k a^i b^j$

The $C$ is a 3×3×3 array of numbers multiplied by a vector will produce and ordinary matrix, which multiplied by another vector will produce the final vector.

A dot product in the language of tensors would look like this:

$r = D_{ij} a^i b^j = \displaystyle\sum_{i,j} D_{ij} a^i b^j$

Where $D_{ij}$ is an identity matrix.

Now the wiki article on Tensors should be more comprehensible.

Hope this will give an aha moment someone.

share|improve this answer

This is what A.Zee says about a tensor from his book Einstein Gravity in a Nutshell (Hardcover)

A tensor is something that transforms like a tensor

Long ago, an undergrad who later became a distinguished condensed matter physicist came to me after a class on group theory and asked me, 'What exactly is a tensor?' I told him that a tensor is something that transforms like a tensor. When I ran into him many years later, he regaled me with the following story. At his graduation, his father, perhaps still smarting from the hefty sum he had paid to the prestigious private university his son attended, asked him what was the most memorable piece of knowledge he acquired during his four years in college. He replied, "A tensor is something that transforms like a tensor."

share|improve this answer

My very simple answer is really just one of many situations where a tensor is handy when describing the forces on a body...they are used almost everywhere in physics however...this just one SIMPLE example.

A cubic body is moving through air and is feeling the resistance to motion orthogonal to it's trajectory. This normal force could occur on any SIDE of of the cube. OR, If the cube sits still, it experiences pressure from the atmosphere, the pressure can be decomposed into normal forces on each side.

Now there's the SHEAR force, of the viscose air that clings to the top of the cube and the drag deforms the top of the cube. This shear force is on the sides parallel to the motion of the moving cube. This can occur FOR EVERY parallel surface.

Tensors are handy when ALL the possibilities are actually possible and do occur. Then there are tricks for summing the forces. That is what all the fancy tenor math above is about.

I was told by a great fluid mechanics professor, that tensors should only be used when we understand the forces and/or system well. Typically, when learning something new, we start with each dimension separately and tediously work out all the math....then when we know what's going on, tensors can be used.

share|improve this answer

There are some good answers already, and they show how varied the equivalent definitions can be. For a succinct answer, the Wikipedia page has a good one, depending on how familiar you are with the notion of a dual space.

For application to physics, though, know that many different subfields use tensors, and all in different ways. We might be able to provide more pertinent references if you specify the context you're interested in. For example, Carroll's Spacetime and Geometry gives a physically intuitive introduction to tensors in general relativity without any loss of rigor.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.