It seems, at times, that physicists and mathematicians mean different things when they say the word "tensor." From my perspective, when I say tensor, I mean "an element of a tensor product of vector spaces."

For instance, here is a segment about tensors from Zee's book Einstein Gravity in a Nutshell:

We already saw in the preceding chapter that a vector is defined by how it transforms: $V^{'i} = R^{ij}V^j$ . Consider a collection of “mathematical entities” $T^{ij}$ with $i , j = 1, 2, . . . , D$ in $D$-dimensional space. If they transform under rotations according to $T^{ij} \to T^{'ij} = R^{ik}R^{jl}T^{kl}$ then we say that $T$ transforms like a tensor.

This does not really make any sense to me. Even for "vectors," and before we get to "tensors," it seems like we'd have to be given a sense of what it means for an object to "transform." How do they divine these transformation rules?

I am not completely formalism bound, but I have no idea how they would infer these transformation rules without a notion of what the object is first. For me, if I am given, say, $v \in \mathbb{R}^3$ endowed with whatever basis, I can derive that any linear map is given by matrix multiplication as it seems the physicists mean. But, I am having trouble even interpreting their statement.

How do you derive how something "transforms" without having a notion of what it is? If you want to convince me that the moon is made of green cheese, I need to at least have a notion of what the moon is first. The same is true of tensors.

My questions are:

  • What exactly are the physicists saying, and can someone translate what they're saying into something more intelligible? How can they get these "transformation rules" without having a notion of what the thing is that they are transforming?
  • What is the relationship between what physicists are expressing versus mathematicians?
  • How can I talk about this with physicists without being accused of being a stickler for formalism and some kind of plague?
Ellie Kesselman
  • 453
  • 7
  • 24
  • 2,574
  • 1
  • 10
  • 21
  • 14
    Take this with a grain of salt, since I know very little about how physicists think, but I've done a course in electrodynamics in the past using a physicist book (of course) which made similar definitions. From what I understood, I think they usually have something which is expressable as coordinates in mind (something 'real', not mathematical). If when they measure this something in another "system of coordinates", the change that happens is according to *this and that*, then it is "tensorial/vectorial". – Aloizio Macedo Jan 28 '18 at 02:44
  • 2
    I just have to say that if you think physicists lack formalism, try theoretical chemistry. They confuse themselves. –  Jan 28 '18 at 02:45
  • 11
    So, answering your first question: they *know*, a priori, what it is that they are transforming. It is something they are measuring. To the second question: when you formalize that structure mathematically, you get the tensors we know. To the third: Try to make him understand that this sort of definition is not smth you are comfortable with. And also try to make yourself more willing to understand different points of views and needs for definitions ("can someone translate what they're saying into some language intelligible by humans" seems like you don't want to try to understand how they do). – Aloizio Macedo Jan 28 '18 at 02:52
  • It seems like you should be reading better authors: https://www.amazon.com/Geometry-Physics-Introduction-Theodore-Frankel/dp/1107602602 – user14717 Jan 28 '18 at 03:01
  • 14
    As a final note, you seem to understand mathematical definitions of tensors. Since you already understand the mathematical definition, and you want to understand "what the hell the *physicists* are saying", why are you asking *mathematicians*? Why not ask this in physics stackexchange? (I don't think this question is off-topic here, but if you genuinely want your question answered then maybe hearing them out is the best option). – Aloizio Macedo Jan 28 '18 at 03:58
  • 3
    Obligatory: [Fields arranged by likelihood of practitioner being accused of being some kind of plague for being a stickler for formalism](https://xkcd.com/435/) (_Most likely_ to be accused to the left...) – davidbak Jan 28 '18 at 05:16
  • Related: https://math.stackexchange.com/questions/789956/whats-the-term-for-a-physical-vector-space – David Z Jan 28 '18 at 05:33
  • 26
    "What the hell are the physicists saying, and can someone translate what they're saying into some language intelligible by humans?" — physicists are humans, too. – celtschk Jan 28 '18 at 08:08
  • 16
    Just a minor curiosity: that is pretty much the historical definition of tensor given by Ricci Curbastro in his works. While studying differential forms he recognized that was the peculiar pattern characterized "good " changes of coordinates and other "good" formal phenomena, so he started elaborating an abstract calculus (later called tensor calculus by Einstein and Levi-Civita) based on such formal rules: "things with multiindexes which behave well". So yeah, that's a tensor in the old way, even before abstract algebra was really born – TheMadcapLaughs Jan 28 '18 at 08:45
  • 5
    I am slowly gaining the opinion that 50 years ago, people were thinking in terms of what we would now call "representation theory"; the emphasis on "how things transform" was *essential* to the mode of thought. Tensor algebra gradually edged its way in to take over the mathematical foundations, leading to the representation theory POV tinting the lenses through which tensors are viewed. –  Jan 29 '18 at 04:11
  • One further thing to keep in mind, especially in GR, is that most of the time (but not always), physicists take partial derivatives $\partial_i$ as the basis for the tangent space at a given point (of spacetime). Then, a change of coordinates induces a change of basis. Of course, you can use an anholonomic basis, and the vectors won't notice a change of coordinates at all. – Toffomat Jan 30 '18 at 09:36

7 Answers7


What a physicist probably means when they say "tensor" is "a global section of a tensor bundle." I'll try and break it down to show the connection to what mathematicians mean by tensor.

Physicists always have a manifold $M$ lying around. In classical mechanics or quantum mechanics, this manifold $M$ is usually flat spacetime, mathematically $\mathbb{R}^4$. In general relativity, $M$ is the spacetime manifold whose geometry is governed by Einstein's equations.

Now, with this underlying manifold $M$ we can discuss what it means to have a vector field on $M$. Manifolds are locally euclidean, so we know what tangent vector means locally on $M$. The question is, how do you make sense of a vector field globally? The answer is, you specify an open cover of $M$ by coordinate patches, say $\{U_\alpha\}$, and you specify vector fields $V_\alpha=(V_\alpha)^i\frac{\partial}{\partial x^i}$ defined locally on each $U_\alpha$. Finally, you need to ensure that on the overlaps $U_\alpha \cap U_\beta$ that $V_\alpha$ "agrees" with $V_\beta$. When you take a course in differential geometry, you study vector fields and you show that the proper way to patch them together is via the following relation on their components: $$ (V_\alpha)^i = \frac{\partial x^i}{\partial y^j} (V_\beta)^j $$ (here, Einstein summation notation is used, and $y^j$ are coordinates on $U_\beta$). With this definition, one can define a vector bundle $TM$ over $M$, which should be thought of as the union of tangent spaces at each point. The compatibility relation above translates to saying that there is a well-defined global section $V$ of $TM$. So, when a physicist says "this transforms like this" they're implicitly saying "there is some well-defined global section of this bundle, and I'm making use of its compatibility with respect to different choices of coordinate charts for the manifold."

So what does this have to do with mathematical tensors? Well, given vector bundles $E$ and $F$ over $M$, one can form their tensor product bundle $E\otimes F$, which essentially is defined by $$ (E\otimes F)_p = \bigcup_{p\in M} E_p\otimes F_p $$ where the subscript $p$ indicates "take the fiber at $p$." Physicists in particular are interested in iterated tensor powers of $TM$ and its dual, $T^*M$. Whenever they write "the tensor $T^{ij...}_{k\ell...}$ transforms like so and so" they are talking about a global section $T$ of a tensor bundle $(TM)^{\otimes n} \otimes (T^*M)^{\otimes m}$ (where $n$ is the number of upper indices and $m$ is the number of lower indices) and they're making use of the well-definedness of the global section, just like for the vector field.

Edit: to directly answer your question about how they get their transformation rules, when studying differential geometry one learns how to take compatibility conditions from $TM$ and $T^*M$ and turn them into compatibility relations for tensor powers of these bundles, thus eliminating any guesswork as to how some tensor should "transform."

For more on this point of view, Lee's book on Smooth Manifolds would be a good place to start.

  • 3,790
  • 1
  • 11
  • 23

Being a physicist by training maybe I can help.

The "physicist" definition of a vector you allude to, in more mathematicians-friendly terms would become something like

Let $V$ be a vector space and fix a reference frame $\mathcal{F}$ (mathematicians lingo: a basis.) A collection $\{v^1, \ldots, v^n\}$ of real numbers is called a vector if upon a change of reference frame $\mathcal{F}^\prime = R ^{-1} \mathcal{F}$ it becomes the collection $\{ v^{\prime 1}, \dots, v^{\prime n}\}$ where $v^{\prime i} =R^i_{\ j} v^j$.

If you like, you are defining a vector as an equivalence class of $n$-tuples of real numbers.

Yes, in many physics books most of what I wrote is tacitly implied/shrugged off as non-important. Anyway, the definition of tensors as collections of numbers transforming according to certain rules is not so esoteric/rare as far as I am aware, and as others have pointed out it's also how mathematicians thought about them back in the past.

Physicists often prefer to describe objects in what they find to be more intuitive and less abstract terms, and one of their strength is the ability to work with vaguely defined objects! (Yes, I consider it a strength and yes, it has its drawback and pitfalls, no need to start arguing about that).

The case of tensors is similar, just think of the collection of numbers with indices as the components with respect to some basis. Be warned that sometimes what a physicist calls a tensor is actually a tensor field.

As to why one would use the definition in terms of components rather than more elegant invariant ones: it takes less technology and is more down-to-earth then introducing a free module over a set and quotienting by an ideal.

Finally, regarding how to communicate with physicists: this has always been a struggle on both sides but

  1. Many physicists, at least in the general relativity area, are familiar with the definition of a tensor in terms of multilinear maps. In fact, that is how they are defined in all GR books I have looked at (Carroll, Misner-Thorne-Wheeler, Hawking-Ellis, Wald).

  2. It wouldn't hurt for you to get acquainted, if not proficient, with the index notation. It has its own strengths and is still intrinsic. See Wald or the first volume of Penrose-Rindler "Spinors and space-time" under abstract index notation for more on that.

  • 5,016
  • 15
  • 28
  • 12
    Just adding to the last point: all physicists know the formal definition of a tensor. It's just that the component definitions are much, much more useful in almost every computation. Nobody actually defines a tensor as "something that transforms like a tensor" anymore; that is a joke physicists occasionally make to rile up mathematicians. Zee does this several times in his books. – knzhou Jan 28 '18 at 18:22
  • Even the first ~25 pages of Wald would be enough to be acquainted. – gabe Jan 29 '18 at 04:28
  • 5
    I don't really understand your quoted definition at the top, or understand how it is the same as defining a vector to be an equivalence class of tuples of reals. Your condition seems to be true for all tuples of reals; in any case, your definition defines a *subset* of $\mathbb R^n$, not an equivalence relation on it. – John Gowers Jan 29 '18 at 09:10
  • 3
    Re: why one would work in terms of components. I have seen some of the lectures on tensor calculus by Pavel Grinfeld on youtube, and he use them because he is an applied mathematician. He wants to be able to put these things into a computer. Also, as one work with expressions like $\nabla_iT^j$ one does start more and more to see that as a single symbol at the same time as one sees it as the nine components it has in a given basis. Kindof like that famous rabbit / duck optic illusion. – Arthur Jan 29 '18 at 14:14
  • All this gets nasty with Chtisoffel symbols of the third kind. – richard1941 Jan 31 '18 at 13:16

I had this exact discussion a few months ago in the comments of this answer on Academia.SE. Let me report here my argument, with a few additions.

Let's start with your first point:

It seems, at times, that physicists and mathematicians mean different things when they say the word "Tensor."

There's really no difference in the modern definitions of vectors and tensors in physics and mathematics. There was such a difference 50 or more years ago, but not really today. However, many old-school physicists still insist with the old point of view.

This does not really make any sense to me. Even for "Vectors" and before we get to "Tensors," it seems like we'd have to be given a sense of what it means for an object to "transform." How do they divine these transformation rules?

In the olden times physicists defined vectors and tensors as arrays of numbers which transformed in two possible ways under changes of the coordinate system. An example of such a definition can be found in E. Persico, Introduzione alla fisica matematica, p. 27 (pdf) from 1943. It's in Italian, but I think that the different mathematical approach with respect to modern books is evident (I chose this Italian book as example because I studied tensors on it some thirty-odd years ago, when I was 17, and I'm still fond of it). For a physicist, coordinate systems (which are pure mathematical objects) are associated to reference frames, that is, to physical operations that allow to associate coordinates with events. The transformation rules mentioned in your post are then those associated to a change of basis, in a vector space or its dual.

Of course, with such a definition, you don't go very far. Modern physics has a different approach: for instance, in Newtonian mechanics, the space-time is assumed to be an affine space with a certain structure (in relativity, one assumes a different structure). Vectors, then, are just the vectors of the vector space associated to this affine space, exactly as you would do in mathematics. A German physics book which adopts this approach is N. Straumann, Theoretische Mechanik. One about special relativity that adopts this modern view is E. Gourgoulhon, Special Relativity in General Frames.

In modern physics, tensors are defined as multilinear maps, showing that those old arrays of numbers were just the components of the tensor with respect to a certain base. And the transformation rules follow directly from this definition. There is probably still a difference with respect to the mathematical definition: in fact, tensors in algebra are defined as element of a tensor space which possesses a certain universal property, and multilinear maps are used in a constructive proof of the existence of a tensor space, but this is just one possible construction out of many others which are isomorphic. Physicists usually skip on this and define tensors as multilinear maps. I studied the modern approach already in 1997 during a relativity course, and the professor stressed several times that the old approach was outdated since long.

Massimo Ortolano
  • 696
  • 6
  • 17

You are bothered by tensors defined as tuples of numbers that "transform" in some way because you wish to have specified what a tensor is first. But what would you say to someone who is confused about the meaning of quotient groups or quotient rings, and wishes to know what those quotient constructions are first? Quotient constructions are something in math that baffles many students, because there often is no first description in terms of something simpler. They just are what they are: you lump together stuff by some equivalence relation.

Do the same thing to define the physicists tensors: on a $D$-dimensional space $V$ consider pairs $(L,B)$ where $L$ is an ordered list of $D$ real numbers and $B$ is a basis of $V$, with a pair $(L,B)$ and a pair $(L',B')$ considered equivalent if $L$ and $L'$ are related by the tensor transformation rule you described. A tensor is then an equivalence class of such pairs. Does that satisfy you?

See section 7 of http://www.math.uconn.edu/~kconrad/blurbs/linmultialg/tensorprod.pdf for a discussion of how the physicist's description of tensors in terms of transformation rules is related to the mathematician's description of tensors.

  • 30,396
  • 2
  • 67
  • 110

Before giving my answer to this question, I'd like to make a comment about an assumption implicit in the question. There actually is no one way in which physicists think about tensors: different (equally good) physicists would answer this question in different ways. Some prefer to place mathematical well-definedness first and foremost, others to focus on experimental measurements and are happy to jettison the mathematical formalism as soon as it gets confusing. (A physicist once told me, semi-jokingly, "It's always nice when a group of definitions is circular - that way you know they're self-consistent.")

That having been said, here's the way I (a physicist) personally like to think about tensors: you can't "divine the transformation rules" theoretically. The actual definition of a "physics tensor" is always phenomenological (e.g. the electromagnetic tensor $F_{\mu\nu}$ is defined in terms of the force experienced by a charged particle at that point in spacetime). To check whether it's an actual tensor, you need to measure the electric and magnetic fields in one frame, then build an experimental apparatus that moves through the same system at a decent fraction of the speed of light and measure the electric and magnetic fields at the same point in spacetime, and check whether the components are the appropriate $\mathrm{SO}(3,1)$ conjugation of each other. (In practice, of course, we can almost never actually do this, so instead we need to logically work out what it would look like if we could do this, and then derive the phenomenology of that system in the frame that we can access. For spatial $\mathrm{SO}(3)$ rotations, we really can actually rotate the apparatus, and then check that our measured numbers change appropriately.) If we were to ever measure any (precise, reproducible, etc.) deviations from the predicted tranformation behavior, then we would conclude that the the electromagnetic "tensor" is not actually tensor.

So when we speculate about the behavior of exotic physical fields like a (hypothetical) axion field, for which we have no experimental data at all, why do we always assume that they transform as tensors? Because we assume, based on experimental experience, that all the laws of physics transform covariantly under Lorentz boosts (when gravitational effects are negligible), including those describing exotic new physical quantities. This is an extremely useful assumption, because it hugely cuts down the number of possible physical theories to consider. It's always just a provisional guess though; if we ever measured any violations of the Lorentz-invariance of physics, then the question of how a given exotic physical quantity transforms would be entirely empirical.

  • 5,421
  • 1
  • 17
  • 51
  • "*Lorentz boots*" much surely be an auto-correct error, but I have no idea what it was meant to say. – Peter Taylor Jan 29 '18 at 11:04
  • 3
    @PeterTaylor I'm assuming they meant to type *[Lorentz boo**st**](https://physics.stackexchange.com/questions/30166/what-is-a-lorentz-boost-and-how-to-calculate-it)*. *"The Lorentz transformation is a linear transformation. It may include a rotation of space; **a rotation-free Lorentz transformation is called a Lorentz boost**."* –  Jan 29 '18 at 13:13

The components approach make sense when you are working in crystal systems or anistropic media (condensed matter, material science, solid state chemistry, or even MechE or geophysics). Also, I think it is easier to have a picture of tensors and why we need them, in that context.



If you are seriously wondering what vectors are, they are magnitude and direction. I think of them as forces (push, pull, punch in the nose) but of course we can have other vectors. Tensors are when the the vectors aren't good enough because the media is anisotropic. (Easier to break a mica rock by sliding layers past each other than perpendicular to plane.)

P.s. I don't know GR but that is a tough area. Maybe if you looked at tensors first in something more tangible, it would be easier for you.

  • 11

There are two types of tensors, general tensors and Cartesian tensors. I think that is the distinction you're looking for. Both mathematicians and physicists use general tensors, engineers use Cartesian tensors. Most tensors are rank 2 tensors and can be represented by a square matrix. That's true if it's a general tensor tensor or a Cartesian tensor. General tensors start with vectors and covectors, with a covector being a map of a vector to real number. Cartesian tensors only use vectors. When the basis for a tensor's underlying vector space changes, a vector and covector transform differently if non-orthonormal bases are allowed (because the dot product is not preserved transforming to a non-orthonormal base). General tensors account for this difference in their basis transformation formulas. If only orthonormal bases are allowed vectors and covectors transform the same way so there is no distinction, and the basis transformation formulas are simpler. General tensors are mandatory in general relativity, but not elsewhere (I think.).

  • 61
  • 6