# What is a tensor?

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.

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.

In the context of physics, the most illuminating description I have found is that a tensor is a generalized quantity **whose algebraic/analytical properties don't depend on the coordinate system being used***

Now, the traditional way to represent a generalized quantity is as a linear combination of basis vectors, or as a scalar. For example, momentum can be represented by $p_x\mathbf{\hat{i}} + p_y\mathbf{\hat{j}} + p_z\mathbf{\hat{k}}$. If you change coordinates, say by a passive rotation, the component $p_\alpha$'s might change, and, of course, the basis vectors will change, but the momentum **won't,** precisely because **both** the basis vectors and the components change. You can imagine how important it is for a physical quantity to have this property. Thus, tensors serve as a natural mathematical object to do theoretical physics with.

Really, they **are just a mathematical formalization of the almost all the physical quantities you should have studied by now**. The utility of this formalization comes to the forefront once you start studying things like Relativity, which is **all about** the fact that physical laws are independent of a very general class of linear coordinate transformations.

This behavior is perhaps best captured by a (the?) fundamental theorem of Tensors, where any tensor whose components are all $0$ in one system of coordinates has its components as $0$ in all others, as well.

This implies that if an equation involving tensors is true in one coordinate system, it is true in all others.

This theorem, as best as I can tell, follows from one of the many axiomatic frameworks for defining tensors. Some frameworks start by introducing tensors as being multilinear maps. Many start by defining covariant/contravariant tensors as multi-indexed sets of components that follow certain transformation rules.

The end result is, however, the same. You get something that can be represented by a bunch of components, and whose algebraic/analytic properties don't change no matter what system of coordinates you use.

It is important to note that Tensors are **not** simply collections of components. In fact, some treatments of tensors are entirely component free. For example, Geometric Algebra represents tensor (think generalize geometric) operations in terms of something called the Geometric Product. And yet, the things being studied are still tensors precisely because their properties don't depend on "how you look at them".

*By coordinate system, I mean a "typical" one that can be gotten to by invertible linear transformations.