I am wondering why do we really need the concept of tensor. I think it is like vectors, just as a notation of a set of related parameters. I could write the Navier–Stokes equations with scalars, or vectors, or tensors. If this is the case, we struggle to learn the algebra and calculus rules of tensors only to simplify the notation of a complex equation.

My question is:

Are there some examples to show the power of tensors instead of just simplifying the notation?

I know that when I use tensors of rank 3 or higher, it is often hard to use a matrix (equivalent to a rank-2 tensor). Tensors do make life easier in this kind of situation.

Here is a quote from the preface to Wilhelm Flügge's Tensor Analysis and Continuum Mechanics:

Many of the recent books on continuum mechanics are only "tensorized" to the extent that they use cartesian tensor notation as a convenient shorthand for writing equations. This is a rather harmless use of tensors. The general, noncartesian tensor is a much sharper thinking tool and, like other sharp tools, can be very beneficial and very dangerous, depending on how it is used. Much nonsense can be hidden behind a cloud of tensor symbols and much light can be shed upon a difficult subject. The more thoroughly the new generation of engineers learns to understand and to use tensors, the more useful they will be.

  • 11
    $\begingroup$ One point is that it goes beyond "just notation". But still, I think you may appreciate this: profoundphysics.com/… $\endgroup$
    – Gold
    Jan 31 at 20:20
  • 8
    $\begingroup$ That tensors simplify notations is a secondary thing. Most importantly, different physical laws have the same form in different coordinate system provided they are written in tensor form. Otherwise, one needs to derive (formulate) separate equations for each coordinate system of interest. What a waste of time! $\endgroup$
    – yarchik
    Feb 1 at 15:18

5 Answers 5


It is well known that the exercise of logic never adds to our knowledge: its role is to make a certain aspect of that knowledge clearer or more explicit, while keeping all the rest conveniently out of our sight.

This is a quote by Tommaso Toffoli, in "Entropy? Honest!". Entropy 18, 247 (2016). doi: 10.3390/e18070247. Reading your question reminded me of it because it is pretty much the point: we don't need to write our equations in tensorial form, we could indeed just write them in terms of each individual component (important note: those would not be scalars, for they wouldn't transform as scalars). However, this often will hide interesting properties of what we are dealing with that might make our life considerably easier (see, e.g., this brilliant answer by Terence Tao on a similar question in Math Overflow).

In the case of tensors, our main interest is that their components have well defined transformation properties under changes of coordinates and are, deep down, geometrically invariant, and that allows you to see and comprehend much more than you otherwise would.

For example, take the expression $\mathbf{F} = m \mathbf{a}$ in Classical Mechanics. This is written in terms of rank-1 tensors (vectors), but we could write it as three equations in components. It would read $$ \left\lbrace \begin{aligned} F_x &= m a_x, \\ F_y &= m a_y, \\ F_z &= m a_z, \end{aligned} \right. $$ where I employed Cartesian coordinates. Let us suppose you already measured $F_x$, $F_y$, and $F_z$ experimentally and now wants to determine the acceleration in each direction. However, you notice you messed up your setup and actually you wanted the components of the acceleration in a slightly different angle, rotated in $45º$ around the $z$ axis with respect to the system you chose. Then you compute the transformation and finds that the previous equations read now, after we transform them, $$ \left\lbrace \begin{aligned} \frac{\sqrt{2}}{2} F_x - \frac{\sqrt{2}}{2} F_y &= m a_{x'}, \\ \frac{\sqrt{2}}{2} F_x + \frac{\sqrt{2}}{2} F_y &= m a_{y'}, \\ F_z &= m a_{z'}, \end{aligned} \right. $$ assuming I didn't get anything wrong. Had you chosen to work with vectors, the equation would read $$\mathbf{F} = m \mathbf{a},$$ because while vector components transform under a change of basis, vectors themselves are geometrical objects which do not depend on the basis.

Of course, this is just an example, you'd have to compute the components to get to your final computation either way. However, notice how writing in vector notation employs automatically the invariance properties of vectors. You can choose to write component-wise, but that won't destroy the vectors that lie beneath your computations, it just hides them. The components still transform as vectors and the vectors are still there, you are just not looking at them, and, as a consequence, you are missing some information that could make things way simpler. The point is not that they are a way of organizing calculations (you'll often have to work with the components sooner or later), but that there are "hidden" symmetries and properties that become explicit once you cast the formulae into an appropriate notation.

At this point, it is worth pointing out the comment made by Dvij D.C. (which I'm adding here in case it is deleted in the future)

Tensors not only make the underlying symmetries more manifest but they are the only basis-independent/coordinate-independent way of expressing physics. The tensor-components $T_{\mu\nu}$ transform under a change of basis but the tensor itself $\textbf{T} = T_{\mu\nu} \textbf{e}^{\mu} \otimes \textbf{e}^{\nu}$ remains invariant.

Notice how this is similar to us writing Newton's second law in components and in vector form. Vectors are simply rank-1 tensors, and when employing them our equations automatically are expressed in an invariant way under rotations.

As a second example, consider the expression $\mathbf{p \cdot q}$ written in two other different notations. The first one, in Einstein notation, which would read $$\mathbf{p \cdot q} = p_i q^i,$$ and the second being the expression in Cartesian coordinates, $$\mathbf{p \cdot q} = p_x q_x + p_y q_y + p_z q_z.$$

Both expressions are tautological. Neither of them adds any intrinsic knowledge to us. However, the expression in Einstein notation makes it clear that the object is invariant under rotations, while the second one has a lot of objects that transform in a sort of complicated way, and you are not sure whether the expression would change if you chose to work with different coordinates.

In principle, one can component-wise everything we do with tensors. It is similar to searching for a needle in a haystack: you can do it by looking carefully and eventually you'll find it, but you can also use the extra knowledge that the needle is made of iron and use a magnet.

Extra Examples

Quantum Mechanics

One other example of usage of tensors in Physics, although in a particularly specific notation, is within Quantum Mechanics. For simplicity, I'll stick to the rotation example. Suppose you made a measurement, noticed you should have rotated your coordinates before, and so on. If you denote the states in terms of wavefunctions, you'll figure out that the probability a particle prepared in the state $\psi(x)$ to be measured in the state $\phi(x)$ is $$P(\phi|\psi) = \left\vert \int \phi^*(x) \psi(x) \mathrm{d}^3{x} \right\vert^2$$ before the rotation and, after the rotation, $$P'(\phi|\psi) = \left\vert \int \phi'^*(x) \psi'(x) \mathrm{d}^3{x} \right\vert^2,$$ where $\psi'$ is the wavefunction $\psi$ after the rotation of the system of coordinates. Opening up the appropriate expression for $\psi'$ and $\phi'$, one will eventually find out that the probability is, of course, the same. In Dirac notation, however, which employs the fact that the objects we're dealing with are vectors, one would write $$P(\phi|\psi) = \vert\langle \phi \vert \psi \rangle\vert^2$$ before the rotation and $$P'(\phi|\psi) = \vert\langle \phi' \vert \psi' \rangle\vert^2.$$ Since rotations are implemented in Quantum Mechanics by means of unitary operators, one has then that $\vert\psi'\rangle = U \vert\psi\rangle$ for some unitary operator $U$. Hence, the rotated version is $$P'(\phi|\psi) = \vert\langle \phi' \vert \psi' \rangle\vert^2 = \vert\langle \phi \vert U^\dagger U \vert \psi \rangle\vert^2 = \vert\langle \phi \vert \psi \rangle\vert^2 = P(\phi|\psi),$$ and the result is immediate using properties of linear algebra, without the need to manipulate anything with calculus.

Another example is how one can solve the quantum harmonic oscillator in terms of ladder operators instead of solving differential equations. It exploits the linear structure hidden in the wavefunctions and allows one to get a solution with much less labor.


Expressions written in terms of tensors in Relativity are assured to work in all reference frames, since tensors are geometrical objects. This is similar to my previous example of how working in vector notation made taking the rotations into account much easier.

Take, for example, the famous expression $$E^2 = p^2 c^2 + m^2 c^4.$$ In this form, it is not obvious that this formula holds in any inertial frame (unless, of course, you are already well acquainted with Relativity and know the fact by heart). However, the same formula can be written as $$p^\mu p_\mu = m^2 c^2,$$ where $p^\mu = (\frac{E}{c}, \mathbf{p})^\intercal$, which explicitly shows the expression is invariant.

Another example is the fact that the relativistic Doppler effect for a source in uniform linear motion boils down to noticing that $k^\mu u_\mu$ is an invariant, $k^\mu = (\frac{\omega}{c}, \mathbf{k})^\intercal$ being the wave's four-momentum and $u^\mu$ being the source's four-velocity. By computing the invariant in two different frames, one of which has the source at rest, one arrives at the expression for the frequency shift in an incredibly straightforward way. Indeed, suppose that in the frame of rest of the source the wave has $k^\mu = (\frac{\omega_0}{c}, \mathbf{k}_0)^\intercal$, while in some other frame the source has $u^\mu = (\gamma c, \gamma \mathbf{v}$ and the wave has $k^\mu = (\frac{\omega}{c}, \mathbf{k})^\intercal$. Then $$ \begin{align} k^\mu u_\mu\vert_{\text{rest}} &= k^\mu u_\mu\vert_{\mathbf{v}}, \\ - \omega_0 &= \gamma(-\omega + \mathbf{k \cdot v}). \end{align} $$

Let's write $\mathbf{k} = \frac{\omega}{c} \mathbf{\hat{n}}$, where $\mathbf{\hat{n}}$ is a unit vector (the fact that this is possible comes from usual wave mechanics). Then $$ \omega_0 = \omega\gamma\left(1 - \frac{\mathbf{n \cdot v}}{c}\right), $$ and we conclude that $$\frac{\omega}{\omega_0} = \frac{\sqrt{1 - \frac{v^2}{c^2}}}{1 - \frac{\mathbf{n \cdot v}}{c}}.$$

This derivation just comes in such a direct way because we noticed from the start that $k^\mu u_\mu$ is a relativistic invariant, something that is favored by tensor notation: the absence of "free" indices tells us that the quantity is an invariant.

  • 2
    $\begingroup$ Remark: I chose to try to work with examples from standard Newtonian mechanics, since I am not sure how much I could assume of other theories. If you prefer, I can also provide other examples coming from Eletromagnetism, Quantum Mechanics, Relativity, etc. Tensors are often hidden, but they are pretty much everywhere in Physics $\endgroup$ Jan 31 at 21:01
  • $\begingroup$ I know the basics of Quantum mechanics and Relativity. Could you share some examples of them? $\endgroup$
    – Jack
    Jan 31 at 21:18
  • 3
    $\begingroup$ +1: I'd just add that tensors not only make the underlying symmetries more manifest but they are the only basis-independent/coordinate-independent way of expressing physics. The tensor-components $T^{\mu\nu}$ transform under a change of basis but the tensor itself $\mathbf{T}=T^{\mu\nu}\mathbf{e}_\mu\otimes\mathbf{e}_\nu$ remains invariant. $\endgroup$
    – ACat
    Jan 31 at 22:15
  • 1
    $\begingroup$ @DavidConrad Thanks for pointing it out! I edited the phrase $\endgroup$ Feb 3 at 19:59
  • 1
    $\begingroup$ @DvijD.C. Thank you very much for the remark! I added it to the post mentioning you in case the comment ends up being deleted after a while. $\endgroup$ Feb 3 at 20:01

I guess others will give longer answers, but one obvious point is that tensor notation makes it easy to form and to spot scalar invariants (i.e. quantities which are independent of rotations of the coordinate system).

More generally, if a phenomenon has certain symmetries with respect to changes of coordinates, such as rotations, then it must be possible to express it in a equations where every term is a tensor (of some appropriate rank). This observation sometimes makes it easy to deduce the appropriate expression. One investigates the phenomenon in a choice of coordinates where it happens to come out in a simple way, or at least in a way you know how to calculate, and one expresses its form in that coordinate frame, but using suitably chosen tensors. One thus has immediately an expression which can in principle be valid in all frames. One just has to check that there are no further terms which happened to be zero in the first frame.

  • 1
    $\begingroup$ I heard about that invariants are often important, but I am not sure how to find them in an equation expressed with tensors. Could you give me a clue of how to do it? $\endgroup$
    – Jack
    Jan 31 at 20:25

Here's an answer from a mathematician's perspective:

Tensors are necessary for differential geometry. Differential geometry is, by necessity, the language of physics.

I'll explain that in reverse. Physicists need to study space, which was found to be curved, from within that space.

Differential geometry is the study of particular structure on particular objects.

The objects in question are "manifolds," space-like objects that "look like" flat space if you zoom in enough. Our world behaves like this, because the classical limit works. The structures on them are called "metrics," and encode information that allows one to construct something "like" standard geometry, defining lengths of curves in the manifold, angles between vectors tangent to the manifold, and so on. We would like to have angles and lengths to do physics, so these structures are necessary as well.

Critically, mathematicians choose to describe these objects in an intrinsic way, only using objects defined in terms of points of the manifold, and not appealing to the manifold being "inside" some larger space (this is called an embedded surface, and interestingly there exist a lot of manifolds that aren't able to be embedded in any surface, hence why mathematicians chose to study manifolds intrinsically). This is very helpful for physics, as we can't observe the structure of space from outside.

To explain tensors in differential geometry, one must understand dual vector spaces: a dual vector is a function that takes in a vector, and outputs a scalar. A $(r, k)$ tensor is then a function of multiple variables, taking in $r$ normal vectors and $k$ dual vectors and outputting a scalar.

Why this is useful: in order to describe the curvature of a manifold intrinsically, without appealing to circles tangent to the space and other things one might do in calculus 3, mathematicians use an object called a curvature tensor. It takes in two tangent vectors (once again defined intrinsically; it's really counterintuitive and not necessary here) to the manifold, and outputs a number corresponding to how tightly the manifold is bent.

Another example is the metric tensor with which one is familiar in general relativity. On the 4-dimensional pseudo-Riemannian manifolds studied in GR, this is the Minkowski metric represented as $$\eta = \begin{pmatrix}-1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \\ \end{pmatrix}$$

Something that's important to note is that the matrix representing a tensor and the tensor itself are distinct. This confusion is due to natural scientists not teaching proper finite-dimensional vector spaces to their undergraduates, but I digress. A matrix is a way to store information about a linear map or tensor, it is not the tensor itself in either the physics notion (which is compatible to the one here, but that's complicated and requires appealing to changing the so-called atlas of the manifold to have different coordinate functions) or the mathematical. For example, the Minkowski metric measures length. So it doesn't act like a linear map, which produces a vector and is what you're likely used to seeing matrices represent. The metric actually acts like $$\eta(v,w)=v^T\eta w=\begin{pmatrix} v_1 & v_2 & v_3 &v_4 \end{pmatrix} \begin{pmatrix}-1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \\ \end{pmatrix} \begin{pmatrix} w_1 \\ w_2 \\ w_3 \\ w_4\end{pmatrix}$$ So the linear map represented by $\eta_{ij}$ and the tensor represented by $\eta_{ij}$ are two extremely different objects. One is a linear map between two four-dimensional vector spaces; the other is a map from pairs of elements of a four-dimensional vector space to the field of scalars.

Interestingly, there was a work I saw by Moon and someone else defining something called "holors;" they were interested in studying the properties of these multi-dimensional arrays we use to store information on their own merits. From what I could gather, there wasn't a tremendous amount of shatteringly novel work there, but very good exposition. I seem to remember either it or the authors' biographies gave me a bit of a crankish impression somewhere, but it's very hard to be a crank in math, so I'll still suggest the book.

EDIT: As a general example of how easy this makes physics, I'm currently in a single-semester mathematical QFT class with differential geometry as a prerequisite. The class is teaching several people who may remember snippets from 8-years-gone freshman general physics, nothing more. Yet, the professor intends to explain all of classical mechanics, all of traditional quantum mechanics, gauge theory, and the Batalin-Vilkovsky formalism of quantum field theory meeting 3 hours a week for a semester. By treating everything in this very familiar (to math graduate students) way, even classical mechanics, gaining a respectably technical understanding of physical theories is very quick and easy.

  • 1
    $\begingroup$ The someone else was Moon's (second) wife, Domina Spencer. The book they wrote is Theory of Holors. More on Moon: en.wikipedia.org/wiki/Parry_Moon. $\endgroup$
    – KCd
    Feb 3 at 3:51

Tensors are higher dimensional vectors. Here it is not the ambient dimension that is higher, but the dimension of the arrow itself.

This means that a 2-vector has two ways to add, whilst a 5-vector has 5 ways to add and so on.

They are useful in continuum mechanics, quantum mechanics and general relativity. In the latter, what are called tensors are actually tensor fields.

Even in classical mechanics we need tensors. For example, angular momentum is usually given by the cross product of displacement and momentum and is an axial vector. This turns out only to work in 3d, in higher dimensions there is no cross product (apart from 7d) and it is replaced by a 2d anti-symmetric tensor. This shows tensors naturally describe this concept rather than axial vectors. In fact, anywhere where a cross product is used should be replaced by such a tensor.

  • $\begingroup$ Take classic mechanics as an example, is there something we can’t do or can’t easily do without tensors? $\endgroup$
    – Jack
    Jan 31 at 20:22
  • 2
    $\begingroup$ @Jack: Sure, angular momentum is actually a tensor and not a vector. This is usually hidden from us because in 3d, 2d anti-symmetric tensors can alternatively be represented by an axial vector. This is the cross product. Generally, the cross product ought to be replaced by a 2d antisymmetric tensor. $\endgroup$ Jan 31 at 20:50
  • 1
    $\begingroup$ Related to the (excellent) angular momentum example: physics.stackexchange.com/q/682505/168783 $\endgroup$ Jan 31 at 21:02
  • 1
    $\begingroup$ In classical mechanics we need the inertia tensor to evaluate general three dimensional rotation of a rigid body about a non-principal axis, where the products of inertia are not zero. $\endgroup$
    – John Darby
    Jan 31 at 22:16
  • $\begingroup$ It's a bit shortsighted to say tensors are higher-dimensional vectors. Some tensors can be represented as a list of numbers, sure, and tensors themselves form a vector space. But higher-dimensional vectors exist, like (1,2,3,4,5), that only correspond to the most trivial of tensors (in this case, the tensor $(v_1,v_2,v_3,v_4,v_5)\mapsto v_1+2v_2+3v_3+4v_4+5v_5$). It's better to think about what a tensor does rather than how it looks; doing the opposite can only be confusing when one comes to 2-tensors. $\endgroup$
    – Duncan W
    Feb 2 at 6:18

For many applications, the use of tensor notation or non-tensorial notation is just a matter of convenience or personal preference. For instance, if you're an engineer calculating something about the motion of a car going fast around a curve, it may be helpful to work with a moment of inertia tensor in Cartesian 3-space, and write it in expressions using notation such as the Einstein summation convention. But this is all optional.

The area where it gets to be non-optional is general relativity. GR tells us that time and space simply don't work the way we think they do, and in particular it tells us that there is not even any such thing as a global frame of reference or a global set of space-time coordinates that is uniquely well suited to a particular observer in a particular state of motion.

Furthermore, GR tells us that parallel transport of vectors is not well defined unless we specify the path along which the vector is transported. As a concrete example, the Gravity Probe B satellite transported a gyroscope along a certain path through spacetime, and it measured how much that gyroscope's axis had wobbled at the end of that process, relative to a gyroscope axis such as the one defined by the earth's rotational axis (which amounts to the same thing as a wobble relative to the "fixed stars").

So in the context where GR effects are significant, you could tell me about some quantity you've defined that is notated like a "vector," but if it isn't really a vector in the sense of being a rank-1 tensor, then you haven't really told me how you're defining this thing. Specifically, suppose we point this "vector" toward Arcturus, seal it in a box, and transport it along some path through spacetime, then open the box back up at the end. We want to know whether it will still be pointing toward Arcturus or not, and if not, then in what direction. If you can't guarantee me that it's an actual rank-1 tensor, then we don't know the answer to this question, and I would submit that you haven't really defined what this thing is at all.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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