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In many books on Physics, I read this:

(i) Some quantities are represented in terms of numbers; they are called scalars.

(ii) Some quantities are expressed in terms of numbers and directions; they are vectors.

(iii) Some quantities, which are neither scalars nor vectors, are called tensors.

Q. Does this mean that any quantity in Physics is one of these three only?

[I am a mathematics person, not an expert in Physics. The question arose naturally in mind since vectors are defined/introduced as in (ii), and then the question comes - what about quantities which are neither scalars nor vectors? I saw in some books that they are tensors. So, I just want to confirm from experts in this subject, whether all the things around us coming in Physics have to be one among these only? ]

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    $\begingroup$ "Does this mean that any quantity in Physics is one of these three only?" No. $\endgroup$ Commented Sep 17 at 10:32
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    $\begingroup$ Basically, everything is a tensor. A scalar is a tensor of rank $(0,0)$. A vector is a tensor of rank $(1,0)$. A co-vector is a tensor of rank $(0,1)$. $\endgroup$ Commented Sep 17 at 10:45
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    $\begingroup$ No, for starters, there are also densities and spinors. $\endgroup$ Commented Sep 17 at 10:50
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    $\begingroup$ @kangermu, density is not simply a scalar. It will transform in a non-trivial way under change of coordinates. This could be important. $\endgroup$ Commented Sep 17 at 12:45
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    $\begingroup$ @MathsRahul, there are also forms (e.g. differential forms). Also there can be a lot of groups, e.g. in crystallography or particle physics. Also functionals etc... $\endgroup$ Commented Sep 17 at 12:48

5 Answers 5

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(i) Some quantities are represented in terms of numbers; they are called scalars.

(ii) Some quantities are expressed in terms of numbers and directions; they are vectors.

(iii) Some quantities, which are neither scalars nor vectors, are called tensors.

I would say that this example represents the typical attempt to classify everything too early, without a good understanding of the mathematical definitions.

Let me list a few weaknesses of such a classification, and where to find a direct answer to the question of whether only scalars, vectors, and tensors have a role in Physics.

Definition (i) could be accepted at the introductory level (but only with the additional specification that each quantity is represented by a single number), even though in mathematics, the name "scalar" is usually reserved for the elements of a field (or a module) that is a member of a vector space algebraic structure. The real meaning of "such a quantity is a scalar" in Physics is more like "this quantity is not a vector", although such a definition is not enough to characterize it fully.

Definition (ii) is poor: it conveys the idea that a vector is characterized by magnitude and a direction, but it misses two crucial points. The first is that even directions (not explicitly defined but left as intuitive concepts) can be expressed as numbers (so, what's the real difference between numbers and directions?). The second is the well-known (?) fact that there are quantities represented by magnitude and directions that are not vectors. For a vector, one needs to add some information about the sum of such quantities. More on that at the end of this answer.

Definition (iii) is probably the most critical. It assumes that everything that is not classifiable as scalar or vector should be a tensor. That is false. From a mathematical point of view, tensor spaces can be introduced from vector spaces, adding a tensor product operation, and it is immediate to show that tensor spaces have again the structure of a vector space, thus tensors can be considered a special kind of vectors. At the elementary level in Physics, quite often vectors and tensors are not defined intrinsically, in terms of algebraic structures, but in terms of their components on a basis. Therefore, the distinction between vectors and tensors is related to the different transformation laws of their components.

Let me go back to the definition of vector and on to the fact that there are quantities represented by magnitude and directions that are not vectors. A well-known example is the representation of a rotation in three dimensions. It can be characterized by magnitude and direction, but it fails to be a vector if we define the sum of two rotations as the compound final rotation. Such a sum/composition is not commutative; therefore, the resulting algebraic structure fails to satisfy the definition of a vector space.

3D rotations form a group of linear transformations of vectors, but are not vectors themselves. They cannot be tensors, because tensors are (a special kind of) vectors. Therefore, we have an elementary example of something that is not a scalar, nor a vector, nor a tensor.

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  • $\begingroup$ Thanks for an interesting answer, which would benefit from some more examples though I think : ) $\endgroup$ Commented Sep 19 at 8:25
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    $\begingroup$ The example of a non-tensor (of any rank) is good, but I note that it is an operation rather than a physical property. Another example could be the Levi-Civita symbol but again that's a kind of operation. Can you think of an example which is a physical property? With General Relativity in mind, I am guessing that there, at least, everything can be done in terms of tensors, although one is not required to employ them. $\endgroup$ Commented Sep 19 at 17:52
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    $\begingroup$ @AndrewSteane At a certain level of sophistication, one could ask if quantities that belong to some vector space are not also part of additional algebraic structures, thus adding more possibilities to the choice between scalars, vectors, or tensors. For example, elements of an exterior algebra are certainly elements of a vector space, but not all vectors belong to an exterior algebra. $\endgroup$ Commented Sep 19 at 22:09
  • $\begingroup$ There is a difference between the concepts of a vector in mathematics and physics. In mathematics it is an element of a vector space and in physics it is a quantity with a magnitude and a direction. These two concepts don't always overlap. That does not mean the use of the term in physics is incorrect. $\endgroup$ Commented Sep 23 at 3:40
  • $\begingroup$ @flippiefanus There is no difference, provided one remembers that in addition to being a quantity with a magnitude and a direction, the physical definition of vector includes that they add according to the parallelogram rule. This point is often forgotten/omitted by physicists, but it is a mistake. $\endgroup$ Commented Sep 23 at 6:53
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No. Most notably, we have spinors that are not of this "classification". The exact notion of a "quantity" as of physics is very broad, usually we consider them as objects in mathematics. Here, scalars and vectors are also tensors, being rank-$0$ and rank-$1$ tensors respectively. Considering physical representations of symmetry groups, we have typically tensor representations and spinor representation as two distinct classes of quantities.

The general construction of this usually on manifolds with $\operatorname{Spin(n)}$-structure, a $G$-structure by restriction of the structure group of the frame bundle. From this, we have a principal $\operatorname{Spin}(n)$-bundle, and hence fields as associated bundles, for chosen representation. For physics, we have by the short exact sequence $$1 \rightarrow \mathbb{Z}_2 \rightarrow \operatorname{Spin}(n) \rightarrow \operatorname{SO}(n) \rightarrow 1$$ the two distinct representations:

  • Tensor Representation: a representation of $\operatorname{Spin}(n)$ that factors through $\operatorname{SO}(n)$, $$\rho : \operatorname{Spin}(n) \rightarrow \operatorname{SO}(n) \rightarrow \operatorname{GL}(V)$$ which is not faithful, as we have the element $-1 \in \operatorname{Spin}(n)$ (the non-trivial element of the kernel $\mathbb{Z}_2$) acting as identity $$\rho(-1) = \operatorname{Id}_V$$ The sections of $\rho$-associated bundles correspond to scalars, vectors, and higher rank tensor fields.
  • Spinor Representation: a representation of $\operatorname{Spin}(n)$ that does not factor through $\operatorname{SO}(n)$, $$\rho : \operatorname{Spin}(n) \rightarrow \operatorname{GL}(S)$$ which is faithful, as we have $-1\in \operatorname{Spin}(n)$ acting as $$\rho(-1) = -\operatorname{Id}_S$$ The sections of $\rho$-associated bundles correspond to spinor fields.

Note: there is a slight difference in how we define tensors in mathematics and in physics. In mathematics, we have in $\operatorname{Vect}_k^\otimes$ a $(p,q)$-tensor as an element in the tensor product space $$T^{(p,q)}(V) = \underbrace{V \otimes \dots \otimes V}_{p\text{ times}} \otimes \underbrace{V^* \otimes \dots \otimes V^*}_{q\text{ times}}$$ While, in physics, we define tensors by their behavior under change of coordinates.

If you regard "quantities" specifically as units, see the discussion at What kind of mathematical objects are units of measurement?.

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    $\begingroup$ I think this depends on your definition of tensors. Spinors as representations of Spin can be classified as linear maps on the representation space and as such are tensors in the mathematical sense (I guess this depends if you want to view them as linear reps on the universal cover or as projectives on the quotient). However of course this does not fit into the classification mostly used in GR where tensors are expected to act on the tangent/cotangent and tensor products of these. Maybe tensorial densities coming from rational reps. are a more clear example $\endgroup$ Commented Sep 17 at 11:42
  • $\begingroup$ @ThomasTappeiner Well, yes, in the mathematical sense of tensors being elements of tensor products, this is indeed the case. However, it doesn't fit our purpose of having a physical classification here, we still define these quantities by how they transform $\endgroup$ Commented Sep 17 at 11:52
  • $\begingroup$ I agree was mostly pointing this out since the original question mentioned a mathematics background $\endgroup$ Commented Sep 17 at 12:55
  • $\begingroup$ @ThomasTappeiner Yes, indeed, I've now clarified that $\endgroup$ Commented Sep 17 at 13:03
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I apologize in advance for being more of a mathematician than a physicist. I hope this answer is to your tastes.

You are basically looking for different ways that a quantity can transform under change of coordinates. The most common non-tensorial, non-scalar quantity is a connection. Let me illustrate this.

Let $M$ be a smooth manifold of dimension $n$, and let $\nabla$ be a linear connection on $TM$. On a coordinate chart $(U, x)$ with coordinates $x^1,\dots,x^n$, write the coordinate vector fields as $\partial_i := \partial/\partial x^i$.
We represent connections in a coordinate chart via their Christoffel symbols. They are defined by \begin{equation} \nabla_{\partial_i}\partial_j \;=\; \Gamma^k{}_{ij}\,\partial_k \quad\text{on } U. \end{equation} If $(U, x')$ is another coordinate chart on the same domain with coordinates $x'^1,\dots,x'^n$ and coordinate vector fields $\partial'_i := \partial/\partial x'^i$, then \begin{equation} \partial'_i \;=\; \frac{\partial x^p}{\partial x'^i}\,\partial_p \quad\text{and}\quad \partial_k \;=\; \frac{\partial x'^r}{\partial x^k}\,\partial'_r. \end{equation}

Now we get to the transformation law. The Christoffel symbols $\Gamma'^k{}_{ij}$ of $\nabla$ in the primed chart are related to the unprimed symbols $\Gamma^\ell{}_{pq}$ by \begin{equation} \Gamma'^k{}_{ij} \;=\; \frac{\partial x'^k}{\partial x^\ell}\, \frac{\partial x^p}{\partial x'^i}\, \frac{\partial x^q}{\partial x'^j}\, \Gamma^\ell{}_{pq} \;+\; \frac{\partial x'^k}{\partial x^m}\, \frac{\partial^2 x^m}{\partial x'^i\,\partial x'^j} \end{equation} where $x \mapsto x'(x)$ is the coordinate change and $x = x(x')$ denotes its local inverse. As you can see the connection transforms in an affine way, instead of the linear way that tensors transform.

There are also more quantities beyond connections. A connection is a special case of a differential operator. Differential operators (of higher order) are even more general.

As a final comment, maybe for when you come back to this interesting question some years from now, it is in fact known what all the possible natural transformation laws, and quantities obeying them, under coordinate changes are. This is detailed in my favourite book, Natural operations in differential geometry, by I. Kolář, J. Slovák, and P. Michor. It turns out that they are all (sections of) associated bundles given by representations of the jet groups.

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An oversimplified classification as scalar/vector/tensor/spinor could fail since a physics quantity could be one thing in one bundle, but it could be another thing in a different bundle.

An example: for the Weak gauge field strength tensor $F^i_{\mu\nu}$, index $i = 1, 2, 3$ indicates that it's a vector in the adjoint representation of the isospin $SU(2)$ gauge bundle, whereas the space-time index $\mu\nu$ indicates that it's a 2-form/antisymmetric tensor in tangent bundle.

So is Weak gauge field strength $F^i_{\mu\nu}$ a vector, a 2-form, a rank-2 antisymmetric tensor, or a rank-3 tensor including all three indices $i$ and $\mu\nu$?

The following is a comprehensive categorization of all physics quantities emphasizing the differentiation between various fiber bundles.

Tangent Bundle

For the tangent bundle of the coordinate basis, the concept of metric is not needed at all. All we need is that there is a smooth basis manifold. Example physics quantities pertaining to this bundle are differential forms of different order:

  • 0-forms, usually called scalar fields, such as Sine-Gordon scalar field $\Phi$.
  • 1-forms, usually called vectors, such as electromagnetic gauge connection field $A_\mu$.
  • 2-forms, usually called tensors, such as electromagnetic curvature field $F_{\mu\nu}$. Note that the Lagrangian of electromagnetism DOES need the contracting of "upstairs" $F^{\mu\nu}$ with "downstairs" $F_{\mu\nu}$, which DOES require the metric to do the trick of moving ${\mu\nu}$ upstairs. However, the official definition of the 2-form $F = F_{\mu\nu}dx^\mu \wedge dx^\nu$ per se does NOT need the metric.

Frame Bundle

It is sometimes called tangent bundle as well. But it is actually something extra to the tangent bundle. The frame bundle lives locally at each space-time point and it is detached from the tangent bundle which is constrained by the the coordinate basis. This is where the concepts of metric and Euclidian/Minkowski/Riemannian space-time come into place. Some example physics quantities:

  • Vectors, such as Tetrad/Frame/Vierbein 1-form $e^a_\mu$, where Roman index $a$ indicates that it's a vector in the frame bundle, and the Greek index $\mu$ indicates that it's a 1-form/vector in tangent bundle. In our (almost) flat universe, Tetrad/Frame/Vierbein 1-form $e^a_\mu$ gains the vacuum VEV $e^a_\mu=\delta^a_\mu$, resulting in (almost) one-to-one correspondence between the Tangent Bundle and Frame Bundle. Therefore, Tetrad/Frame/Vierbein 1-form $e^a_\mu$ is also sometimes called soldering form, soldering together the Tangent Bundle and Frame Bundle.
  • Connections, such as the spin-connection 1-form $\omega^{ab}_\mu$, where Roman index $ab$ indicates that it's related to the rotation in the $ab$-bivector direction of the frame bundle, and the Greek index $\mu$ indicates that it's a 1-form/vector in tangent bundle.
  • Tensors, such as spin-connection curvature 2-form $R^{ab}_{\mu\nu}$, where Roman index $ab$ indicates that it's related to the $ab$-bivector direction of the frame bundle, and the Greek index $\mu\nu$ indicates that it's a 2-form in tangent bundle.
  • Derived quantities using the above 3 listed items (plus Minkowskian $\eta_{ab}$), such as metric tensor $g_{\mu\nu}$, Christoffel connection $\Gamma^{\eta}_{\mu\nu}$, Riemann curvature tensor $R^{\eta}_{\rho\mu\nu}$, and density (density is actually a 4-form pseudo-scalar instead of a scalar). For instance, metric tensor $g_{\mu\nu}$ can be derived from Tetrad/Frame/Vierbein 1-form $e^a_\mu$ as $g_{\mu\nu}= e^a_\mu e^b_\nu\eta_{ab}$.
  • Generic spinors, such as the Dirac spinor $\psi$. The absence of Greek index indicates that it's a 0-form/scalar in tangent bundle (invariant under diffeomorphism, any coordinate-related transformation of spinor is a secondary effect resulting from soldering between Tangent Bundle and Frame Bundle via the VEV of Tetrad/Frame/Vierbein 1-form $e^a_\mu$, as noted earlier). Note that the definition of the spinor (such as spin-up vs. spin-down) DOES need the Frame bundle, since its non-trivial transformation properties in the Frame bundle (which is $Spin(1,3)$, the double cover of Lorentz $SO(1,3)$) are tied to the rotations in the Frame bundle and its covariant derivative requires the presence of spin-connection 1-form $\omega^{ab}_\mu$.

Gauge Bundle

Specifically we are talking about the principle bundle of Yang-Mills gauge group. This is where the gauge groups of the Standard Model $U(1)*SU(2)*SU(3)$ coming into picture. Some example physics quantities:

  • Yang-Mills gauge connection vector fields, such as Weak gauge field connection 1-form $W^i_\mu$, where index $i$ indicates that it's a vector in the adjoint representation ($i= 1, 2, 3$) of the isospin $SU(2)$ gauge bundle, and the Greek index $\mu$ indicates that it's a 1-form/vector in tangent bundle.
  • Yang-Mills curvature tensor fields, such as Weak gauge field curvature 2-form $F^i_{\mu\nu}$, where index $i$ indicates that it's a vector in the adjoint representation of the isospin $SU(2)$ gauge bundle, and the Greek index $\mu\nu$ indicates that it's a 2-form/tensor in tangent bundle.
  • Standard model spinors, such as the electron spinor $\psi^e$. The absence of Greek index indicates that it's a 0-form in tangent bundle. The index $e$ indicates that it's an electron spinor (rather than a neutrino spinor) in the fundamental representation of the isospin $SU(2)$ gauge bundle. Similar to the generic spinor field, the definition of Standard model spinors also needs the Frame bundle, as stated earlier.
  • Higgs field, such as the Standard Model Higgs field $\phi$. The absence of Greek/Roman index indicates that it's a 0-form/scalar in tangent/frame bundle. Note that the definition of the Higgs field DOES need the Yang-Mills gauge bundle, since its transformation properties are tied to the gauge transformations in the Yang-Mills gauge bundle and its covariant derivative requires the the presence of Yang-Mills gauge fields such as Weak gauge field connection 1-form $W^i_\mu$. We usually regard the Standard model Higgs field $\phi$ as a scalar given that it's a 0-form/scalar in tangent/frame bundle. However, Higgs field $\phi$ transforms more in line with a pair (right-handed plus left-handed) of Standard Model spinors in the fundamental representation of the Yang-Mills gauge symmetry. In this sense, we could have called Higgs field $\phi$ a bi-spinor as well.
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A great question, and there are already some good answers. One way to look at this is through "Coordinate Free Physics" (https://faculty.washington.edu/seattle/physics544/2011-lectures/thorne-GR.pdf).

The goal is to represent physical law as relations between geometric objects (that do not depend on, nor require, coordinates). As it turns out, these geometric objects are tensors.

I'll be addressing tensors in 3D euclidean space, $\mathbb{E}^3$, in which case tensors are the things that can be rotated "nicely" under the the rotation group ($SO(3)$).

By "nicely" I mean a set of $N$ objects are Eigen-tensors of rotations about an arbitrary axis ($z$), and they are closed under rotations. (That is, they transform as an $N$-dimensional representation $SO(3)$, but I will avoid technical stuff like that, and only mention it parenthetically).

Step one is to unlearn two things we all have learned.

  1. Numbers are scalars: While a scalar can be represented by a number, it doesn't mean it is a number. A scalar is a geometric object that is invariant under rotations, so it's an eigenstate of any rotation with eigenvalue $1$. Take, for instance, the pressure at some point, in SI units: $$ p = \sigma_{ii} = 101325. $$ where $\sigma_{ii}$ is the Cauchy stress tensor field (at the point). The pressure, $p$, is manifestly a scalar, as it is the trace of a rank-2 tensor. It's represented by $101325\,$Pa. If we rotate coordinates, the Cauchy tensor's components change, but the scalar trace is the same, and its representation as a number is unchanged. The number 101325 is just a number. It can't be rotated, it doesn't live in $\mathbb{E}^3$. It is not a scalar. (Note: unit scalars transform as the trivial representation of $SO(3)$).

  2. Vectors are arrows, they have a magnitude and direction: That's not wrong, you have 3 basis vectors $(\hat x, \hat y, \hat z)$ that span the space of all possible vectors. Intuition goes a long way here, but when moving to higher rank tensors, ones is likely to get lost: what is a dyad, physically? This is because the standard coordinate unit vectors are not fundamental, they transform as the adjoint representation of $SO(3)$.

Now consider so-called spherical vectors (the fundamental representation os $SO(3)$):

$$ \hat e_0 = \hat z $$ $$ \hat e_{\pm 1} = -(\hat x \mp i\hat y)/\sqrt 2 $$

Each basis vector is an eigenvector of rotation about the $z$ axis:

$$ R_z(\theta)[\hat e_m] = e^{im\theta}\hat e_m $$

with eigenvalue:

$$ \lambda = e^{im\theta} $$

For higher ranks $N$, the so-called natural form tensors are symmetric in all indices, and traceless in all indices, so for $N=3$:

$$ T_{ijk} = T_{jik} $$ $$ T_{ijk} = T_{ikj} $$ $$ T_{ijk} = T_{kji} $$

and

$$ T_{iik} = T_{ijj} = T_{kjk} = 0 $$

In general, there are $2N+1$ in each $N$-dimensional irreducible representation of $SO(3)$. The eigenvalue equation is valid for $m\in[-N, -N+1, \ldots, N-1, N]$.

The astute reader will note that that paragraph describes the spherical harmonics, too.

The natural form rank $N$ tensors are the things that can be rotated nicely: they're closed, irreducible, and are eigentensors for $z$-axis rotations.

Physics-wise, we're used to dipole moments being a vector (rank 1). At rank-2, we have quadrupole moments or inertia tensors. We don't see a lot of rank-3 tensors coming up, except the isotropic Levi-Civitta symbol, $\epsilon_{ijk}$.

At rank-4 is the generalized Hooke's law. Here $k$, the spring constant, is generalized to the elasticity tensor, $c_{ijkl}$, which relates the stress and the strain:

$$ \sigma_{ij} = c_{ijkl}\epsilon_{kl} $$

Final note:

At rank-$N$, there are $3^N$ cartesian tensors, but only $2N+1$ natural from tensors (say, for spherical operators in quantum mechanics) and only $N+1$ for classical real tensors (see: solid-harmonics).

The cartesian tensors are reducible, this is the domain of Schur-Wyle duality and the Robertson Schensted correspondence. The integer partitions of $N$ are made into representations of $Sym(N)$, the group of permutations on $N$ letters. From there, the standard Young tableaux each describe the index permutations that form an irreducible representation, and its symmetry. Through the remarkable hook-length formula, the dimensions of the representations can be calculated.

At rank-2, you get:

$${\bf 3}\otimes{\bf 3}={\bf 5}_S\oplus{\bf 3}_A\oplus{1} $$

The ${\bf 5}_S$ is the natural form symmetric tensor, while ${\bf 3}_A$ is the cross-product, and ${\bf 1}$ is the isotropic trace.

tl;dr: rank-$N$ tensors are $N$-nomial products of $(\hat x, \hat y, \hat z)$ that form an $N^3$ reducible representations of the rotation group. That is then a sum of irreducible representations of dimension $2l +1$ for $l \in [1, \cdots, N] a la spherical harmonics. They are the Eigen-shapes of rotations.

Addendum: One of the answers mention spinors. We've all struggled with an intuitive understanding of spinors...the vector with a flag on top that rotates? idk. I think of a spin 1/2 object as the "geometric thing" that can be rotated nicely (under $SU(2)$), where "nicely" again means "a set of things that is closed under rotations" and "are eigenvectors of rotations about the $z$-axis".

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