I have a pretty good knowledge of physics, but couldn't deeply understand what a tensor is and why it is so fundamental.


12 Answers 12


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.

  • $\begingroup$ Maybe I'm misunderstanding "takes" and "makes". In my mind , the dot product takes two vector and makes a scalar, so I would think it would be (2,0) based on your description of n and m, but it seems to be the opposite. $\endgroup$
    – Tyberius
    Commented Jun 26, 2017 at 21:07
  • 1
    $\begingroup$ @Calmarius Your explanation is such a great, thanks! $\endgroup$
    – Amalaric
    Commented May 28, 2020 at 12:21

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.

  • 1
    $\begingroup$ OP was asking why tensors are fundamental and the answer is basically that all physical laws must be independent of our choice of coordinate system - which gives rise to tensors. This is the only answer that addresses to a good degree what the OP was truly asking. +1 $\endgroup$
    – joseph h
    Commented Jun 22, 2021 at 0:27

There are several equivalent ways to define and understand tensors, and it is worthwhile to understand all of the different perspectives and the relationships between them. The perspective that I find most intuitive and well-motivated is the perspective of tensors as multilinear functions.

A tensor is a multilinear function that takes as input a collection of vectors, and outputs a scalar. By multilinear, it is meant that the function is linear in each input independently.

You can imagine multilinear functions as a local approximations to nonlinear functions that depend on several variables, where the approximation structurally takes into account the fact that there are several inputs from several spaces. Just like when you zoom in on a nonlinear function of one vector it looks approximately linear, if you zoom in on a function of many vectors it looks approximately multilinear.

If collections of basis vectors are chosen that span each input vector space, then a multilinear function is completely defined by its action on all possible combinations of basis vectors. The results of applying the multilinear function to all combinations of basis vectors can be arranged into a multidimensional array of numbers, and this array can be considered as a representation of the multilinear function, with respect to the given bases.

If one changes the bases, obviously the entries in the multidimensional array representation will change, but in a predictable way. The exact way in which the entries of the array change when you change the bases are known as "transformation rules". In many physics classes, boxes of numbers obeying these transformation rules are presented as the definition of a tensor, which is a perfectly legitimate definition, but can be jarring and unmotivated if the multilinear context that these rules come from is not explained.

Holding some of the inputs to a multilinear function fixed (in programming language terminology, "closing over" those inputs) yields a multilinear function in the remaining inputs. The array representation of the induced multilinear function in the remaining inputs can be computed by performing a certain sum involving the original array, and the coordinate vectors for the fixed input vectors. This process of forming a new multilinear function by holding certain inputs fixed is known as a tensor contraction.

Since fixing all but one input to a tensor yields a linear function in the remaining input, and since linear functions on a vector space can be identified with elements in the dual to that vector space, a tensor can be equally viewed as a multilinear function that takes one less than the original amount of vectors as input, and produces a (dual-)vector as output. This output can then be used as one of the inputs to another tensor that has an input spot for a vector in the dual space that was output. More generally, complicated networks can be constructed where various inputs to a tensor are reinterpreted as outputs, and then those outputs are used as inputs into other tensors in the network (see Penrose graphical notation).

If one attaches a different tensor to every point on a surface (or manifold), the collection of all of these tensors is a tensor field (just like if one attaches vectors to all points on a manifold one gets a vector field). A common special case is where the input vector spaces that form the domain of the tensor at a point are copies of the tangent space and cotangent space of the manifold at that point. In this case, convenient bases to use can be formed from the tangent (or co-tangent) vectors at every point associated with some pre-existing coordinate charts for the manifold. If one changes the parameterization of the manifold, the coordinate charts will change, so the bases for the tensor field at every point will change, so the array representations of the tensors at each point will change (but in a predictable way..).

Since tensor fields arise in physics much more often than single tensors, often the term "tensor" is used to refer to a tensor field. This terminology works because most of the terms for operations on tensors can also used for tensor fields, with the understanding that that the operation is simultaneously done to all the tensors in the field, pointwise.

Hope this helps with the intuition. This perspective was forged slowly through years of fighting to understand tensors myself from first principles; going from utterly confused at the beginning, to having tensors feel natural and clear now. This post is what I wish someone had told me at the start.

  • $\begingroup$ "A tensor is a multilinear function that takes as input a collection of vectors, and outputs a scalar." <-- This seems to contradict another highly-voted answer here, which states that: "Tensor is something that takes 𝑚 vectors and makes 𝑛 vectors from it." $\endgroup$
    – kennysong
    Commented May 21, 2020 at 6:07
  • 2
    $\begingroup$ @kennysong These are the same because of the multilinear isomorphism: $\mathcal{L}^{n}\left(X_1 \times \dots \times X_n, \mathcal{L}^m\left(X_{n+1} \times \dots \times X_{n+m}\right)\right) \simeq\mathcal{L}^{n+m}\left(X_1 \times X_2, \dots \times X_{n+m}, \mathbb{F}\right)$, where $\mathbb{F}$ is the base field like $\mathbb{R}$ or $\mathbb{C}$. Here $\mathcal{L}^k(A,B)$ is the space of $k$-multilinear maps from $A$ to $B$. The isomorphism is by currying (treating some inputs as fixed constants, which yields a multlinear function in the remaining inputs) $\endgroup$
    – Nick Alger
    Commented May 21, 2020 at 20:22
  • $\begingroup$ @kennysong The other answer is not strictly correct (Taken literally, "something that takes 0 vectors and outputs 2 vectors" should be just a pair of vectors, but a 2-tensor is something else) while this one is. $\endgroup$
    – Will Sawin
    Commented Mar 30, 2021 at 20:44

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.


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."


(This answer was originally posted for a newer question, asked on 2018 Oct 26, that was later marked as a duplicate question. That newer question was specifically about the context of Rotational Dynamics. I relocated my answer here for the benefit of other visitors looking for an answer that is specific to that context.)

In the context of Rotational Dynamics, a vector $v$ is something whose components $v_i$ transform under rotations like $$ v_i\mapsto\sum_j R_{ij}v_j $$ where $R$ is a rotation matrix. A tensor (such as the moment-of-inertia tensor) is something whose components $I_{ij}$ transform under rotations like $$ I_{ij}\mapsto\sum_{k,\ell} R_{ik}R_{j\ell}I_{k\ell}. $$ More specifically, this is a $2$-index tensor. A vector is a $1$-index tensor. In general, an $N$-index tensor is a quantity with $N$ indices (of course) that transforms under rotations according to the pattern illistrated above, with one $R$ per index.

An easy example of a tensor with $3$ indices is $T_{ijk}=a_i b_j c_k$, where $a,b,c$ are vectors. This automatically transforms correctly under rotations because of the way $a,b,c$ transform.

Tensors often have special symmetries. For example, the moment of inertia tensor $I_{ij}$ is symmetric: $I_{ij}=I_{ji}$.

Such symmetries do not affect the general rule for how the tensor transforms under rotations, but they can lead to interesting coincidences. For example, the appropriate generalization of angular momentum to $D$-dimensional space is represented by an antisymmetric $2$-index tensor, $L_{ij}=-L_{ji}$. Because of antisymmetry, this has $D(D-1)/2$ independent components, namely those with $i<j$. (Those with $j>i$ are determined by antisymmetry, and those with $i=j$ are zero by antisymmetry.) In the physically-relevant case $D=3$, $L_{ij}$ happens to have $3$ components, like a vector. Even more interestingly (and less trivially), those three components turn out to transform like the three components of a vector under rotations, even though the general rule for transforming a two-index tensor involves two rotation matrices instead of just one! This is why most formulations of rotational dynamics represent things like angular momentum, angular velocity, and torque as though they were vectors, even though they would more properly be represented as two-index antisymmetric tensors.

Actually, even in $D=3$, there is one clear symptom that quantities like angular velocity (etc) are not really vectors: they transform like vectors when $R$ is an ordinary rotation, but not when $R$ is a reflection. Terms like "axial vector" or "pseudovector" refer to this situation. The direction of angular velocity cannot be reversed by a mirror reflection (because a two-index tensor transforms with two factors of $R$, so the minus signs cancel), but the direction of a legitimate vector (one-index tensor) can be reversed by a mirror reflection.

A more subtle symptom that angular momentum should be represented as a two-index antisymmetric tensor (instead of a vector = one-index tensor) is the way it is constructed. For example, an object with momentum $p$ (a vector) on the end of a massless rotating rod of length $r$ (a vector) is usually written as $L=r\times p$, or something like that (the sign is a matter of convention). The cross-product only makes sense in three-dimensional space. It really should be written $L_{ij}=r_i p_j-r_j p_i$ instead. In three-dimensional space, this has the same list of independent components as the cross-product, but the two-index tensor representation makes sense in any number of dimensions.

In other contexts, such as relativity, the definition is analogous but with the group of Lorentz transformations (in special relativity) or all coordinate transformations (in general relativity) in place of the group of rotations.


I think groups representations are the most natural way to understand tensors. The following parts can be read essentially independently. The 2nd is the most relevant.

Why tensors in physics can be confusing

I think the primary source of confusion is that $X$-tensors always, always, always include structure that physics literature isn't too worried about making explicit. It's the transformation law that I mean by structure, under the transformations as implied by $X$ (this could be $X\in\{\text{"Lorentz" or 4},\text{"Euclidean" or 3},\dots\}$).

The infamous question "Is my matrix here a rank 2 $X$-tensor?" can only be answered by "well, it depends". Because how are we to test whether your matrix is transforming like a tensor if you don't tell us how your matrix transforms.

Some element $A$ of a vector space (note that any $X$-tensor lives in a vector space: adding $X$-tensors makes an $X$-tensor and multiplying by $\lambda\in\mathbb{R}$ as well) can be confirmed to be a tensor only if you specify the following data: $$Data:=\{A \text{ how it looks in frame }1,A \text{ how it looks in frame } 2,\dots,A \text{ how it looks in last frame}\}.$$ Then we can compare with an $X$-tensor and can decide if this is exactly how our tensor transforms or not. If yes, A is by def. an $X$-tensor, otherwise not.

Tensor as an element of the representation of a group

Fix a group $X$ (this could be $X\in\{\text{lorentz trafos},\text{rotations},\dots\}$ in direct correspondence to the first paragraph). Recall that an $n$-dim. representation of $X$ is by def. a group homomorphism ($\text{GL}$ are all invertible matrices) $$\rho:X\rightarrow\text{GL}(\mathbb{R},n),\quad g\mapsto\underbrace{(T\mapsto g.T)}_{\text{matrix is a function}}\quad (T\in\mathbb{R}^n\text{ for "tensor"}).$$ If this is new, think of the action "$g.$" of $g$ as a matrix. Recall that homomorphism ("equal structure") means only $$\rho(g\cdot h)=\rho(g)\rho(h)\quad\text{i.e.}\quad (g\cdot h).T=g.(h.T)\;\forall T\in\mathbb{R}^n\tag{*}$$ where in the first $=$ on the l.h.s. it's $X$'s group operation and on the r.h.s. it's $\text{GL}_n$'s, which is matrix multiplication. I'll write three equivalent sentences in different notations explaining what an $X$-tensor is:

  1. It's $T\in\mathbb{R}^n$ on which $g\in X$ acts like $g.T$
  2. It's $T\in\mathbb{R}^n$ that transforms under $g\in X$ as $g.T$
  3. It's an object $T$ that transforms under $g\in X$ like $T\rightarrow g.T$

Note that you can find many such $\rho$ (for one $X$ and one $n$), so which $\rho$ does a physicist mean? This is hidden in some nomenclature and notation, including: the rank, whether the index is up- or downstairs, saying spinor instead of tensor. Examples:

  1. $X=\text{SO}(3)=\text{rotations in }3\text{ dim.}$, rank 1: A rank $1$ 3-tensor $\vec{v}\in\mathbb{R}^3$ is an element of the fundamental rep. of $\text{SO}(3)$. That's the easiest rep. as $R\in \text{SO}(3)$ acts like (read $\equiv$ as "was defined as" and $:=$ as "is now defined as") $$\rho(R)(v)\equiv R.\vec{v}:=R\vec{v}$$ i.e. by matrix multiplication. (This seemingly tautological definition is worth dwelling on! Is (*) fulfilled? It has to be.) This is what people in school call vector, it's the arrow with the 3 numbers "that rotate into each other,...". All that makes a vector is that he rotates in the fundamental way. All 3-tensors build on this representation, like...
  2. $X=\text{SO}(3)=\text{rotations in }3\text{ dim.}$, rank $k>1$: A rank $k>1$ 3-tensor extends this behaviour to $k$ indices where every index has to be multiplied by $g=(R_{ij})$, e.g. $T$ is a $k=2$ 3-tensor iff $$\rho(g)(T)\equiv g.T = g.(T_{ij}):=(R_{ik}R_{jl}T_{kl})\text{ (repeated indices summed from 1..3)}.$$ In this case, $n=3^2=9$ is the dim. of the 2-tensor representation. If you have read this far, you should definitely check that the single rep. requirement (*) is satisfied for this action! Note that it's purely convenience to arrange $3^2$ numbers in a square matrix. Writing it like $(T_{11},T_{23},T_{13},T_{31},..)$ stays a rank 2 3-tensor if you modify the action of $g$ accordingly. (Why does it stay one? Because ( *) is still satisfied!)
  3. $X=\text{some Lorentz group}$: different choices of rep. have very different, in general super important, applications in physics. The fundamental rep. (acting in analogy to exp. 1 but in this case $n=4$) consists of contravariant 4-vectors $(x^\mu)$ with $\mu=0,..3$ summed in the following (without any signs from metric or so). Another 4d rep. is obtained for a Lorentz trafo $g=\Lambda=(\Lambda^\mu_\nu)$ by $$\rho(g)(x)\equiv g.(x_\mu):=((\Lambda^{-1})_\mu^\nu x_\nu)=\Lambda^{-1T}x$$ (matrix multiplication in last term) dubbed dual representation that contains all covariant 4-vectors. Some comments on this:
  • Check again (*) if you want or at least check that just $\Lambda^{-1}$ instead of additionally transposing like $\Lambda^{-1T}$ in the last term would not furnish a rep. Conversely, just transposing without inverting gives no rep. either!
  • The need for transposing $\Lambda$ is why sometimes people refer to covariant vectors as row vectors. It's probably not as complex as you think: A column vector's entries are summed over with one row of the multiplying matrix $\Lambda^{-1}$ and a row vector's entries with one column of the matrix. Here $\Lambda^{-1}$ is transposed and then multiplied with $x$ by matrix multiplication. So $x$'s entries are summed with each of $\Lambda^{-1}$'s columns. Hence the lower index in $(x_\mu)$. The place of the index tells you which rep. the author wants to consider!
  • In $\text{SO}(3)$ the dual rep. is literally the same as the fundamental rep. because $\text{SO}(3)\ni R=R^{-1T}$ by def. of $\text{SO}(3)$. So no need to distinguish between upstairs and downstairs indices! There's simply no difference between covariant and contravariant 3-tensors of any rank. In the course of writing this, I wondered why not take $$R.\vec{v}:=\vec{v}^TR^T$$ ($R\in \text{SO}(3)$ and matrix multiplication on the right) as the "downstairs index" rep. Fortunately this doesn't work at all - try to write out $S.(R.\vec{v})$ to see this.
  1. $X=\text{some Lorentz group}$: Lorentz rep. theory is vast! Have a look at this subsubsubsection of common reps. for example.

Upshot: Whenever the t-word drops, please know that, along the lines, this needs to include two things: a group $X$ and a rep. of it $\rho$.

Let me now answer OP's question,

What is a tensor?

A tensor is anything that lies in some representation of some group.

Deducing the transformation law from notation

In real-world physics language, one will hardly ever be given the precise transformation behaviour, i.e. rep., under which a physical quantity transforms. But it is very often understood. Let's exemplify this in special relativity:

We start off with the rank 1 4-tensor, $x^\mu$. We know by notation that it's what it is, namely $(t,x,y,z)$, so we deduce its transformation behaviour under a boost $\Lambda$, namely $x^\mu\rightarrow\Lambda^\mu_\nu x^\nu$. The covariant version like $x_\mu\rightarrow(\Lambda^{-1})^\sigma_\mu x_\sigma$. Now a standard question would be "How does $x^\mu x_\mu$ (indices summed 0,..3 without signs) transform under $\Lambda$?". And from the first part of my answer you could be inclined to say "You have to tell me how it transforms, don't fool me!".

But physicists mean something very natural when they write composites of tensors: The known transformation law of the pieces carries over. Every object you write has to have a definite trafo law, to naturally define the trafo of the composite as transforming each component separately. The answer to the question would hence be $$x^\mu x_\mu \rightarrow (\Lambda^\mu_\nu x^\nu)\; ((\Lambda^{-1})^\sigma_\mu x_\sigma) = x^\nu \delta^\sigma_\nu x_\sigma = x^\mu x_\mu.$$ So combining fundamental and dual rep. of the Lorentz group like $x^2:=x^\mu x_\mu$ produces an object transforming in the trivial representation (and is hence defined to be a 4-scalar).

Soft equivalence to tensors in differential geometry

I'll be sloppy about the difference of tensor fields and tensors. Let me also keep this short and only elaborate on where in a rank $(p,q)$ tensor $T$ as defined by $$T=T^{\mu_1\dots\mu_p}_{\nu_1\dots\nu_q}\frac{\partial}{\partial x^{\mu_1}}\dots\frac{\partial}{\partial x^{\mu_p}}dx^{\nu_1}\dots dx^{\nu_q}\in TM^{\otimes p}\otimes T^*M^{\otimes q}$$ one can read off the transformation law. In the above decomposition into coordinates of the implicitly specified chart/coordinate system $x$ on a patch $U\subset M$, the transformation is read off from the definition of the (co)vector fields $\frac{\partial}{\partial x^{\mu}}$ and $dx^{\nu}$. A $(1,0)$ tensor (i.e. a tangent vector) $V$ is defined independently of a chart as the derivative along an equivalence class of curves (I won't write out the definition here). So for two different charts $x,y$ we get two different sets of coordinates because $$V^\nu \frac{\partial}{\partial x^{\nu}}=V=\bar{V}^\mu\frac{\partial}{\partial y^{\mu}}$$ which is really a beautiful equation showing why differential geometry is the language in which general relativity has to be formulated. The ingenious way of defining a tangent vector as a directional derivative along a curve gives rise to the object $V$ which is well defined without specifying a coordinate system! In GR, we rarely deal with these abstract objects, we always pick a frame and write the coordinates. But keeping in mind that there is an abstract thing from which we derive the coordinates already tells us that it has fulfill the tensor transformation properties because we calculate the transformation behaviour and don't check it. We calculate it from the above as $$\bar{V}^\mu=\frac{\partial y^\mu}{\partial x^{\nu}}V^\nu.$$ Analogously, a covector field transforms with the transpose inverse, $$\omega_\nu dx^{\nu}=\omega=\bar{\omega}_\mu dy^{\mu}\quad\Rightarrow\quad\bar{\omega}_\mu=\frac{\partial x^\nu}{\partial y^{\mu}}\omega_\nu.$$ If we put only Lorentz frames in our atlas of $M$, i.e. $\exists\Lambda:\Lambda=(\frac{\partial y^\mu}{\partial x^{\nu}})\;\forall\text{ charts }x,y$ then $V$ lies in the fundamental and $\omega$ in the dual representation of the Lorentz group.

Hence, every tensor in the differential geometry sense indeed lies in some representation of some group (the group of coordinate transformations with the fundamental and dual representation) but I am honestly not sure whether the converse also holds. E.g. the spinor rep. is surely not realisable if $M$ is taken as our spacetime but you might find other $M$ (such as the spin bundle) to realise a spinor as a tangent (co)vector but this is leaving my area of expertise.

Thank you for reading this monstrous answer! :)


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.

  • 3
    $\begingroup$ Why must a vector have 3 components? $\endgroup$
    – Mark C
    Commented Dec 4, 2016 at 20:13

Tensor is a multi dimensional vector in colloquial language.

Where the variations in one direction effects the other.

In Newtonian mechanics we assume all forces, velocities etc that are mutually orthogonal $\Rightarrow$ mutually independent.

$$F=F_x\vec i + F_y\vec j +F_z\vec k $$

$F_x\vec i . F_y\vec j =0$ since $\vec i . \vec j =0$ Whenever we apply some force, we resolve into components and calculate the network done to be zero if the components contribute zero along given direction.

Ie, force applied in one direction will not have any effect in a direction perpendicular to it.

Whereas some physical quantities like pressure applied in one direction can produce effect in other directions also. The corresponding directional quantity is the stress tensor.

If we press a baloon in one direction, we can see expansion in other directions which are mutually perpendicular also.

If we push a solid cube onto a wall, it won't move up whereas a baloon rises up. This is a simple analogy that could distinguish a vector and tensor.

Any variation in x component of stress tensor has its effects reflecting in y z directions as well.

$$ σ = \begin{pmatrix} σ_{11} & σ_{12} & σ_{13}\\ σ_{21} & σ_{22} & σ_{23} \\ σ_{31} & σ_{32} & σ_{33}\end{pmatrix}$$

Similar is the case of moment of inertia.

Here still we deal the effects of cause on other directions independently.

In other words, we deal the effects in y, z directions independently.

Hence stress, moment of inertia are rank 2 tensors.

Ie, at a time we could connect only 2 spatial dimensions.

Levi-Civita is a rank 3 tensor that we use in angular momentum

$$[L_i, L_j]=i\hbar \epsilon_{ijk} L_k$$

Here order of all the three i,j,k decide the value/sign of the function $\epsilon_{ijk}$.

In electrodynamics and relativistic mechanics we may come across rank 4 tensors as well.


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.


Old question, offering some new insight. The description below presents an aspect of tensors that may help in understanding them intuitionally. For formal definition and other explanations, please do look at other answers.

Tensors in physics and mathematics have two different but related interpretations - as physical entities and as transformation mapping.

From a physical entity point of view, a tensor can be interpreted as something that brings together different components of the same entity together without adding them together in a scalar or vector sense of addition. E.g.

  1. If I have 2gm of Calcium and 3gm of Calcium together, I immediately have 5gm of Calcium - this is scalar addition, and we can perceive the resulting substance.
  2. If I am moving at 5i m/s and 6j m/s at the same time, I'm moving at (5i+6j) m/s. This is vector addition, and once again, we can make sense of the resulting entity.
  3. If I have monochromatic pixels embedded in a cube that emit light at different angles, we can define pixels per unit area ($\chi$) in the cube as $\begin{bmatrix} \chi_x&\chi_y&\chi_z \end{bmatrix}$ where $\chi_x$ is the number of pixels emitting light perpendicular to the area in yz plane, and so on.
    This entity, $\chi$, has three components, and by writing $\chi$, we are writing the three components together. Apart from that, the three components cannot be added like a scalar or vector, and we cannot visualize $\chi$ as a single entity.

$\chi$ above is an example of a tensor. Though we may not be able to see $\chi$ as a single perceivable thing, it can be used to fetch or understand perfectly comprehensible entities, e.g. for a given area $\vec{s}$, we can get the total number of pixels emitting light perpendicular to it by the equation: $$ \begin{bmatrix}\chi_x&\chi_y&\chi_z \end{bmatrix}\cdot \begin{bmatrix}s_x\\s_y\\s_z \end{bmatrix}$$

Change the monochromatic pixels in this example to RGB ones, and we get something very similar to the stress tensor (a tensor of rank 2), and we can get the traction vector (force per unit area for a given unit area n) by the equation:

$$\textbf{T}^{(\textbf{n})} = \begin{bmatrix} T_x\\T_y\\T_z \end{bmatrix}^{(n)} = \textbf{n} \cdot \boldsymbol{\sigma} = \begin{bmatrix}\sigma_{xx}&\sigma_{xy}&\sigma_{xz}\\ \sigma_{yx}&\sigma_{yy}&\sigma_{yz}\\ \sigma_{zx}&\sigma_{zy}&\sigma_{zz}\\ \end{bmatrix} \begin{bmatrix}n_x\\n_y\\n_z \end{bmatrix} $$

Though it's difficult to visualize the stress tensor in totality, each of its components tells us something very discrete, e.g. $\sigma_{xx}$ tells us how much force in x-direction is being experienced by a unit surface area that is perpendicular to the x-direction (at a given point in a solid). The complete stress tensor, $\sigma$, tells us the total force a surface with unit area facing any direction will experience. Once we fix the direction, we get the traction vector from the stress tensor, or, I do not mean literally though, the stress tensor collapses to the traction vector.

Note that the possibility of interpreting tensors as a single physical entity or something that makes sense visually is not zero. E.g., vectors are tensors and we can visualize most of them (e.g. velocity, electromagnetic field).

I needed to keep it succinct here, but more explanation on similar lines can be found here.

  • 3
    $\begingroup$ A tensor isn’t just a collection of components, there’s a lot more to it than that. Also, please do not leave answers that are just advertisements for your blog. $\endgroup$
    – knzhou
    Commented Jul 6, 2021 at 21:28
  • $\begingroup$ Thanks for the advice, I'll modify the answer. As for the blog, it's basically for the formatting, video embeds and better control of math equations actually that I left a link for the blog. I'll see how much detail I can possibly add here. $\endgroup$
    – manisar
    Commented Jul 6, 2021 at 21:34
  • $\begingroup$ "..it's basically for the formatting, video embeds and better control of math equations..." then why you post image in place of equation written with $\LaTeX$ (MathJax herein) ??? $\endgroup$
    – Frobenius
    Commented Jul 6, 2021 at 23:42
  • 1
    $\begingroup$ I had added the image at one place for saving me some time... I changed it to Latex now! Thanks for pushing. I'm a new contributor here, and it seems having external links is not appreciated here. But I do feel there is quite more in the link than I can or would want to add here. $\endgroup$
    – manisar
    Commented Jul 6, 2021 at 23:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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