0-rank tensor vs vector in 1D

What is the difference between zero-rank tensor $$x$$ (scalar) and vector $$[x]$$ in 1D?

As far as I understand tensor is anything which can be measured and different measures can be transformed into each other. That is, there are different basises for looking at one object.

Is length a scalar (zero rank tensor) or is it a 1D vector (rank 1 tensor)?

In books it is said that temperature, pressure and other "numbers" are 0rank tensors, they are invariant under transformations and posess no "direction" (that is there is no basis for them). But I'm cofused about units... I thinks of units as somekind of basis.

ex.:

• physical parameter: writing pen's length
• tensor: $$l$$
• length in "inches basis": $$[5.511811023622]$$
• length in "centimeters basis": $$$$
• transformation law: 1cm = 2.54inch

so $$l$$ is a scalar (0rank), but on the other hand it's a vector (1rank).

The same logic can be applied to mutate classical examples of 0rank tensors: pressure, temperature (which is used as a primer of zero rank tensor most in any book) in C and K units. I'm confused.

• Tensors are maps, you should look up the formal definitions. Vectors are not (1,0) tensors (as very often mistaken by physicists), they are isomorphic to (1,0) tensors, to be precise. Mar 31 '19 at 13:08

“Scalar”, “vector”, and “tensor” have no meaning without specifying the group of transformations. In physics we focus on groups such as rotations, Galilean transformations, Lorentz transformations, Poincaire transformations, and gauge transformations because these are symmetries of various physical theories, built in to reflect symmetries of the natural world.

The length of a writing pen is a scalar under rotations and Galilean transformations. This is a significant physical fact about our world.

But the fact that you can measure its length in various units is not significant, because units are inventions of humans, not of Nature. Physicists never say that the length of a writing pen “transforms” because you can choose to measure it in different length units. Different units such as inches and centimeters for a particular physical quantity like length do not have any physical significance at all.

Going back to your original question, the difference between a scalar and a vector under rotations should now be obvious: a scalar is a single number that stays the same under a rotation, while a vector is a directed quantity that requires three numbers to describe it, and under rotations these numbers transform into linear combinations of each other, as specified by the relevant rotation matrix.

Under any other transformation group, the distinction between scalars and vectors is similar.

• I'm sorry if it might sound dumb, but ... Is 1D vector invariant under rotation? I mean is there rotation in 1D space? If so how it's different from scalar? Mar 30 '19 at 21:29
• Whoops, I completely overlooked the fact that you were asking about 1D. (Since you had referred to scalars as rank 0, I was thinking "rank 1" , not "1D", when you said "vector".) There are no proper rotations in 1D. As @ChiralAnomaly explains, you can consider 1D reflections, and scalar and vectors transform differently under these, even though both are only a single number. Mar 30 '19 at 22:05
• @coobit there's a difference in what a mathematician and a physicist mean when they say "vector". A mathematician would just use "vector" to mean "an element of a vector space", where the vector space in question could be anything depending on context. In physics, "vector" almost always means "an element of the tangent space of some manifold", where the manifold in question depends on context (often Euclidean or Minkowski space). See also this question on math stack exchange Mar 31 '19 at 1:52
• I'm sorry but nothing makes sence to me. Let me purify the question: "In books they say that 1)"temperature" is an example of 0rank tensor 2)0rank invariant "number" but how is that true for temperature if I can transform it's number from C to Kelvin??? Looks like temperature is 1D vector (1rank tensor). Is it? Mar 31 '19 at 7:31
• Changing a unit system is not a valid transformation for judging a vector from a scalar, clearly the components of a vector and the numerical value of a scalar both can change with a different set of units. So this is not a criterion for separating them. Mar 31 '19 at 9:32

First of all, I'll constrain the discussion assuming:

1) Finite-dimensional vector spaces

2) Real Vector spaces

3) Talking just about contravariant tensors

4) Physics which use the standard notion of Spacetime

$$* * *$$

I) The tensor object and pure mathematics:

The precise answer to the question "What is a tensor?" is, by far:

A tensor is a object of a vector space called Tensor Product.

In order to this general statement become something that have some value to you, I would like you to think a little bit about vectors and their algebra : the linear algebra.

I.1) What truly is a Vector?

First of all, if you look on linear algebra texts, you'll rapidly realize that the answer to the question "What are vectors after all? Matrices? Arrows? Functions?" is:

A vector is a element of a algebraic structure called vector space.

So after the study of the definition of a vector space you can talk with all rigour in the world that a vector isn't a arrow or a matrix, but a element of a vector space.

Consider then a vector formed by a linear combination of basis vectors:

$$\mathbf{v} = \sum_{j = 1}^{n} v^{j}\mathbf{e}_{j} \tag{1}$$

This is well-known fact about vectors. So, there's another key point about basis vectors: the vector space is spanned by these basis vectors. You can create a "constrain machinery" to verify if a set of vectors spans a entire vector space (i.e. forms a basis):

A set $$\mathcal{S}$$ is a basis for a vector space $$\mathfrak{V}$$ if:

1) the vectors of the set $$\mathcal{S}$$ are linear independent

2) the vectors of the set $$\mathcal{S}$$ spanned the vector space $$\mathfrak{V}$$, i.e. $$\mathfrak{V} \equiv span(\mathcal{S})$$

So another point of view to "form" an entire vector space is from basis vectors. The intuitive idea is that, more or less, like if the basis vectors "constructs" (they span) and entire vector space.

Another fact is that you can change the basis $$\mathbf{e}_{j}$$ to another set basis of basis $$\mathbf{e'}_{j}$$. Well, when you do this the vector components suffer a change too. And then the components transforms like:

$$v'^{k} = \sum^{n}_{j=1}M^{k}_{j}v^{j}\tag{2}$$

but, of course, the vector object, remains the same:

$$\mathbf{v} = \sum_{j = 1}^{n} v^{j}\mathbf{e}_{j} = \sum_{j = 1}^{n} v'^{j}\mathbf{e'}_{j}$$

So, a vector, truly, is a object of a vector space, which have the form of $$(1)$$ and their components transforms like $$(2)$$.

I.1.2) The "physicist way" of definition of a Tensor

When you're searching about tensors on physics/engeneering texts you certainly will encounter the following definition of a tensor:

A Tensor is defined as the kind of object which transforms, under a coordinate transformation, like:

$$T'^{ij} = \sum^{n}_{k=1}\sum^{n}_{l=1} M^{i}_{k}M^{j}_{l} T^{kl} \tag{3}$$

This definition serves to encode the notion that a valid physical law must be independent of coordinate systems (or all that G.Smith said). Well, there's some interesting happening here. A vector, is a object which have a precise formulation in terms of a algebraic structure, have a precise form (that of $$(1)$$, which the basis vectors spans the entire $$\mathfrak{V}$$) and their components have "transformation behaviour" like $$(2)$$. If you compare what I exposed about vector and $$(3)$$, you may reach the conclusion that, concerning about tensors, some information about their nature is missing. The fact is, the definition $$(3)$$ isn't a tensor, but the "transformation behaviour" of the components of a tensor $$\mathbf{T}$$.

I.2) What truly is a Tensor?

Well, you have the transformation of components of a tensor, i.e. $$(3)$$, well defined. But what about their "space" and "form" (something like $$(1)$$)?

So, the space is called tensor product of two vector spaces:

$$V\otimes W \tag{4}$$

The construction of tensor product is something beyond the scope of this answer [*]. But the mathematical considerations about tensor products are that they generalize the concept of products of vectors (remember that, in linear algebra and analytic geometry you're able to "multiply" vector just using the inner product and vector product), they construct a concept of products of vector spaces (remember that a Direct sum of vector space gives you a notion of Sum of vector spaces), and they construct the precise notion of a tensor. Also, by the technology of the construction of the tensor product we can identify (i.e. stablish a isomorphism between vector spaces) the vector space $$V\otimes W$$ and $$\mathfrak{Lin}^{2}(V^{*} \times W^{*}; \mathbb{K})$$:

$$V\otimes W \cong \mathfrak{Lin}^{2}(V^{*} \times W^{*}; \mathbb{K}) \tag{5}$$

where $$\mathfrak{Lin}^{2}(V^{*} \times W^{*}; \mathbb{K})$$ is the dual vector space of all bilinear functionals.

So a tensor have the form:

$$\mathbf{T} = \sum^{n}_{i=1}\sum^{n}_{j=1} T^{ij} (\mathbf{e}_{i}\otimes\mathbf{e}_{j}) \tag{6}$$ And $$\mathbf{T} \in V\otimes W$$.

Well, given the transformation rule $$(3)$$ the space, $$(5)$$, and form, $$(6)$$, you can talk precisely about what tensor really is. It's clear that the "object tensor" isn't just a transformation of coordinates. Also, in $$(6)$$ the tensor basis $$(\mathbf{e}_{i}\otimes\mathbf{e}_{j})$$ spans $$V\otimes W$$.

By virtue of the general construction of tensor product and the identification given by $$(5)$$, you'll also encounter the definition of a tensor as a multilinear object which spits scalars:

$$\begin{array}{rl} \mathbf{T} :V^{*}\times W^{*} &\to \mathbb{K} \\ (\mathbf{v},\mathbf{w})&\mapsto \mathbf{T}(\mathbf{v},\mathbf{w})=: v^{i}\cdot_{\mathbb{K}}w^{j} \end{array}$$

Where the operation $$\cdot_{\mathbb{K}}$$ is the product defined in the field.

With this picture we say that a tensor like $$(6)$$ is a tensor of rank 2. And a vector a tensor of rank 1. Furthermore a scalar a tensor of rank 0.

II) The tensor object and physics

The well stablish physics, in general, deals with spacetime (like Newtonian physics), and the theory of spacetime is geometry. So, in order to really apply the tensor theory in physics first we have to give the geometry of physics. The geometry is basically classical Manifold Theory (which, again, is beyond the scope). And by Manifold Theory we can apply tensors on Manifolds introducing the concept of a tangent vector space. In parallel, we can construct another algebraic structure called Fibre Bundle of tangent spaces and then create the precise notion of Vector Field and Tensor Field.

Tensor Fields are the real objects defined in physics books as tensors and we use the word of a tensor and tensor field as synonyms (IN FACT THEY ARE NOT THE SAME CONCEPT!). A tensor field is a section of the tensor bundle and a vector field, a section of vector bundle. But the intuitive definition (by far, general to physics) of a tensor field is then:

$$[\mathbf{T}(x^{k})] = \sum^{n}_{i=1}\sum^{n}_{j=1} [T^{ij}(x^{k})] ([\mathbf{e}_{i}(x^{k})]\otimes[\mathbf{e}_{j}(x^{k})]) \tag{5}$$ A tensor field is the object which attaches a tensor to every point p of the Manifold.

With the manifold theory, the transformation rule becomes:

$$[T'^{ij}(x^{m})] = \sum^{n}_{k=1}\sum^{n}_{l=1} \frac{\partial x'^{i}}{\partial x^{k}}\frac{\partial x'^{j}}{\partial x^{l}} [T^{kl}(x^{m})] \equiv T'^{ij} = \sum^{n}_{k=1}\sum^{n}_{l=1} \frac{\partial x'^{i}}{\partial x^{k}}\frac{\partial x'^{j}}{\partial x^{l}} T^{kl} \tag{7}$$

Notice that the partials are simply the transformation matrices $$M$$. The matrices $$M$$ are called the Jacobians transformation matrices and the matrices $$M$$ became these jacobians by virtue of Manifold theory.

In a restric way, these Jacobians are rotations,lorentz transformations,galilean transformation, and so on.

III) What is the difference between zero-rank tensor x (scalar) and 1D vector [x]?

So, in order to talk about lengths we have to realize that we are talking about a scalar field, or a tensor of rank 0 (i.e. we've stablish an isomorphism and then we can say that a scalar field is isomorphic to a rank zero tensor). Then the difference between a scalar and a 1D vector (which is isomorphic to a tensor of rank 1) is that one is a scalar field and the other is a vector field. From a "Pure" mathematical point of view, (section I) if this answer) one is a member of the field $$\mathbb{K}$$ and the other is a member of a vector space. Also, you're quite right, a scalar (or a scalar field) is a rank 0 tensor or "a object which do not have "matrices of change"; an object which do not suffer a change under a transformation of coordinates (we say that a scalar quantity is a invariant quantity).

$$\phi'= \phi$$

$$* * *$$

[*] ROMAN.S. Advanced Linear Algebra. Springer. chapter 14. 1 ed. 1992.

• Thanks for the commitment and such a big post, but I'm afraid it was in vein.... :( I undestand all this linear algebra stuff and basis ect... I DON"T understand whether "length" is 0rank or 1rank tensor (1Dim vector, that is). Books say it's 0rank, but my example shows it's 1rank.... What i'm doing wrong? Mar 31 '19 at 7:35
• Even as a pure mathematician who has done some differential geometry already, I found this answer is incredibly insightful. Fantastic post! Mar 31 '19 at 12:00
• Thank you so much for taking the time to explain this. Jan 3 '21 at 8:38

I can understand why you would think they are the same, but even in the 1D case I think the answer is: no they are not.

You might think that a scalar function is that that outputs a single number, whilst a vector function outputs a single vector - but a vector in 1D can be described by only one number and so they must be the same! In fact, no this is not true. You might have heard that the difference is that vectors transform in such a way that "their components mix together", but again it's 1D so maybe that can't happen. In fact, the 'mixing' terminology comes from rotations, but if we focus on scaling our coordinates then the difference becomes clearer. The rest of this answer is an explicit example, along with what we mean by a scalar and vector field.

Scalar field

We want to consider a space $$M$$ consisting of points $$p$$ that we give coordinates $$x(p)$$ to. Note that it's very important that we don't confuse the coordinates we give points, and the points themselves! The coordinates are numbers, whilst points are physical points in space. This distinction is often lost, but it really is key here. We can always pick a different set of coordinates $$x'(p)$$ but the underlying point $$p$$ is the same. A scalar function is any function that assigns to each $$p\in M$$ real number:

$$f:p\mapsto f(p)$$.

However, we often abuse notation and write the function as a function of the coordinates:

$$f:x\mapsto f(x)$$ and $$f':x'\mapsto f'(x')$$

the issue that whilst these are easier functions to work with, you have different functions for each coordinate system. The consistency requirement is that they at least have the same value at the same value of $$p$$:

$$f'(x'(p)) = f(x(p)) = f( \Lambda^{-1}(x'(p))$$

where $$\Lambda$$ is the function that changes coordinates i.e. $$x\mapsto x' = \Lambda (x)$$.

Vector Field

There are two definitions of these that are equivalent - the physicist's and the differential geometer's. We'll only give the physicists as the other is a little abstract (it has the advantage of being shorter though). A physicist's vector field picks out a vector space $$V_p$$ at each point $$p\in M$$ of the same dimension as $$M$$ (so 1 in our case). If then maps each point $$p$$ in $$M$$ to a vector in the vector space at that point. Finally, it also needs to have a transformation rule $$v^i\mapsto v^{i'} = J^{i'}_{i} v^i$$ on the individual components of and vector $$v\in V_p$$. Here, $$J$$ is the Jacobian matrix of the transformation and changing the components of the vector can be thought of as adjusting for the fact that changing coordinates also changes the basis in your vector space.

Example

Consider the scalar function $$f(x)=e^x$$ and the vector function $$\vec{v}(x)=e^x\hat{x}$$*. OPs question is whether there is any meaningful distinction - after all, both encode the same information. Let us look at the change of coordinates $$x\mapsto x' = ax$$. Then to give the same values $$f(x)$$ must change to become $$f'(x')=e^{x'/a}$$.

The vector function transforms to become $$\vec{v}'(x') = e^{x'/a} \hat{x}$$. However as noted above we must also transform the basis vector and the rule amounts to $$\hat{x} = a \hat{x}'$$ in this case. So the new vector function is $$\vec{v}'(x') = a e^{x'/a} \hat{x'}$$.

The case of length

For your specific example of length, we can just consider the effect of a scaling of the coordinates. If we scale our coordinates by a factor of $$2$$ (i.e. measure in units corresponding to twice as large as previously) then the number representing the length halves. This is consistent with the scalar transformation law for the quantity $$L(x)=x$$ rather than the vector transformation law for $$\vec{L}=x\hat{x}$$ which would be different.

*The notations $$\hat{x}$$ and $$\hat{x}'$$ mean the basis vector appropriate for the $$x$$ and $$x'$$ coordinate systems respectively. I'm not sure how to simply explain why scaling affects the natural basis vector to use, but it does.

• Thanks for the answer. It does not help. I will try to help my self and reformulate the question. Is Length of a pen (as given in my example) a 1 rank tensor? If "yes" how to reformulate the example with length to make it 0rank? Mar 31 '19 at 18:09
• @coobit I have updated the answer. If it is still unclear tell me. Mar 31 '19 at 20:03
• Still no clue :( Sorry. Let me give you more of my thinking, and maybe you would be able to pin-point the error. I know that physically Length is invariant, no metter what ruler you will use. That reasoning hints to me that Length might be a some kind 1D vector (with pressure, temperature ect.) in mathematical sence. Of coure length is not a vector in physical sence. Why not? No idea. On the other hand scalars (rank 0 tensors) must be a number which does not change from basis to basis (according to some books). Such numbers (for me) are unitless: 7, or $\pi$, $e$ or 0, -1 ect. Mar 31 '19 at 21:25
• But Length is not some unitless number, there is always a ruler to measure it. So length "transforms" between rulers, unlike $\pi$ which looks like true 0rank tensor... I know, I've got a mess in my mind...... Mar 31 '19 at 21:27
• @coobit one resolution to this is to consider how you actually measure anything. First, you take a metre stick / ruler etc and you tick out regular markings along a line. Then you align your pen with the tick marks and count how many 'ticks' long the pen is. This does ultimately give you a pure number for a length. This is the same notion of 'coordinate' system I use in my answer: to every physical point $p$ you assign a number $x(p)$ based on ticks from an arbitrary origin. But different choices of ruler give different tick spacings.... Mar 31 '19 at 21:41

You seem to confuse transformations with conversion of units. A transformation is a purely mathematical concept and as such has no notion of (physical) units. Physics gives rise to the concept of units which are used to describe physical quantities.

Coming back to your example, let's consider a one-dimensional coordinate system and a pen therein. The pen's head is located at $$\vec{x}_{head}=0$$ and its tail is located at $$\vec{x}_{tail} = t$$. These two positions are (one-dimensional) vector quantities while the pen's length is a scalar quantity, defined by $$L = ||\vec{x}_{tail} - \vec{x}_{head}||$$ where $$||\cdot||$$ is an appropriate metric with which the vector space is equipped (usually Euclidean metric in physics context but Minkowski metric is used as well). Now consider a reflection of the coordinate system about $$\vec{x}=0$$. The tail position is going to change to $$\vec{x}'_{tail} = -t$$ but the pen's length is still $$L$$, i.e. it is invariant under such transformations.

Now you brought in the concept of (physical) units but, as mentioned above, this is distinct from (mathematical) transformations. While you wrote "transformation law" the appropriate term would be "conversion law" and this also emphasizes the distinction between these two concepts. As you can observe from the previous paragraph, no notion of a unit was necessary to describe the transformation (reflection) of the coordinate system. Units on the other hand determine the value of $$t$$ (the pen's tail position) that we are going to obtain by our measurement. They ensure that you and I, if we measure the same pen, arrive at the same value for that position. A conversion of unit means that we change to a different ruler and hence obtain a different number (magnitude) for the position but this is all valid and compatible as long as we report this number together with its unit of measurement and we know the relevant conversion rules.