12
$\begingroup$

Physics texts like to define vector as something that transform like a vector and tensor as something that transform like a tensor, which is different from the definition in math books. I am having difficulty understanding this kind of definition. To me, it goes like this:

  1. First we have a collection of bases or coordinate systems (do these represent reference frames?) and the transformations between them.

  2. A vector/tensor is an assignment of an array of numbers to each basis, and these arrays are related to each other by coordinate transformations.

I am wondering how coordinate systems and transformations are specified.

  1. Do we need all possible coordinate systems and transformations to define a vector space or just a few of them?

  2. How to define the concept of basis and coordinate transformation without completing the very notion of vector?

$\endgroup$
2
13
$\begingroup$

There are two mathematical concepts which are both called vector. The first one, the vector from linear vector space is the basic "multicomponent object" which you seem to mainly talk about. The second notion of a vector is of a member of the so-called "tangent bundle" of a manifold. The second notion is the one which is defined equivalently with the physical vector.


The mathematical linear vector space

The mathematical linear vector space just refers to any object which can be added, multiplied,... but may be more than a mere number (i.e. have more components). Your mathematical vector might be for example "fruit collection" with the basis being one apple and one orange. You can have two apples and 3 and a half oranges represented by a vector $(2,3.5)$. You can double your fruit collection $2\cdot(2,3.5)=(4,7)$, or you can add your fruit collection to a friends fruit collection of $A$ apples and $O$ oranges $(2,3.5)+(A,O) = (2+A,3.5+O)$. You can even get into "fruit debt" by borrowing and eating fruits and have negative fruit collections. This is a perfect mathematical linear vector space.

You can find out that for example certain kinds of functions span an infinite-dimensional linear vector space with the values of the function acting as components $V_i \to V_x = V(x)$. Once again you can even define a "dot product" by summing over the infinite components $\vec{V}\cdot \vec{W} = \sum_i V_i W_i \to \int V(x) W(x)$. And so on. This was just to show that the mathematical linear vector space is a label for very general objects far beyond the extent of a physical vector in 3D space.


The physical vector

We talk about a vector in Euclidean space without distinguishing where it "lives". When we draw a picture of the arrow of say the velocity vector in 3D space, do we mean the tip of the arrow actually does touch the point in space where it ends? Surely not, this would be only true for the distance vector.

We must now draw a line between objects we call vectors in physics. One kind is the distance vectors, which do not transform as a vector unless they point from the origin, and "polar" vectors, the real physical vectors. These polar vectors include: velocity, force, acceleration and electric intensity. (Vector products of "axial" and polar vectors are also polar vectors.)

We know how points in space transform under coordinate transformations and the velocity vector is actually a tangent to two infinitely close points in space. From this fact, we can deduce how velocity transforms - the points transform according to the coordinate transformation, and the velocity vector transforms according to the difference of the transformation in infinitely close points - the Jacobian matrix. This can be shown to be true also for the acceleration vector.

Now we want to formulate equations for velocities and accelerations - it is necessary that these equations give us the same results no matter how we choose to describe the situation. Thus, we require that all the other terms in the equation transform the same way as velocity/acceleration under coordinate transforms. You probably already know this as the principle of covariance. This is the only motivation of the definition of "the true physical vector".

However, when considering a new physical quantity, we cannot just define it to be covariant with velocity - we must explicitly show that an object is consistently transforming this way due to physical arguments. This is the reason why the physicists usually give a down-to-earth technical definition of a vector by transformation, because in many cases this is the most practical formulation for actual computation.


The physical tensor

Physical tensors arise mainly in the context of continuum mechanics. Say we don't want to trace the evolution of one point, but an infinitely small deformation of an infinitely small cube. For this, we need three vectors showing the deformation of every edge of the cube (9 components in total) making it a "vector of vectors".

It is intuitive that every of these 3 deformation vectors showing an infinitely small deformation will transform with the Jacobian matrix. These vectors are however not independent - say the cube rotates by an angle - then the 3 deformation vectors mix and for an infinitely small cube this is again executed by the Jacobian matrix. That is, we multiply every vector of the "vector of vectors" by the Jacobian and then also mix them together by multiplying the whole "vector of vectors" by the Jacobian.

In general, the concept of vectors and tensors is linked to the linearization or differential localization of a given physical fact always leading to transformations via the Jacobian matrix. It is a non-trivial statement of classical physics which tells us that through the description of these linearizations, we can describe the behaviour of the whole.


Tangent spaces and vectors living in them

The mentioned physical ideas can be straightforwardly reconciled with some mathematical ones. The most intuitive way to see why vectors aren't living in the same space as the physical points, is to picture a curved surface with a trajectory on it. The velocity vector of the trajectory points generally "outside" the surface. Nevertheless, all the possible velocity vectors at a given point span only a two-dimensional surface which is tangent to the curved surface at that point.

Mathematicians take this notion and define a tangent manifold at every point of a space as the space of tangent vectors to trajectories at a given point (they do it with a clever trick which makes the term "tangent to a trajectory" well defined). The tangents to trajectories can than be shown to transform once again with the Jacobian matrix and span a mathematical linear vector space in every point. When we take the whole ensemble of "tangent surfaces" or tangent manifolds, we get something we call a tangent bundle.

So, physical vectors actually live in these tangent manifolds attached to every point in space, not in the space itself. Tensors can be also very easily generalised as "vectors of vectors" living in "tangent manifolds times tangent manifolds".

The coincidental structure of the flat space and the flat tangent manifold leads to more or less harmless confusion, but this has to be resolved once we are moving around in a curved space(-time).

$\endgroup$
0
2
$\begingroup$

Let me start with a tautology: Vectors are geometrical objects living on a vector space XD So far it says nothing, but we always have had the mental image of a vector as an arrow.

A bit further into abstraction (still with or idea of an arrow representing a vector), one can find a set of transformations of vectors which preserves the properties of vectors, e.g., in $\mathbb{R}^3$ the rotations keep the properties of vectors${}^*$, including their norm. Mathematically expressed${}^\dagger$, $$\vec{V}\to \vec{V}^\prime = \mathbf{R}\vec{V}.$$

So, a vector transforms homogeneously, i.e.,there is a transformation for each vector, but no additional terms.

Now, imagine that you have two copies vector spaces... and you kind of put them together, e.g., start with $\mathbf{V}$ and construct the put together operation to obtain this new thing (let me call it) $\mathbf{V}\otimes\mathbf{V}$.

This construction is defined such that if I want to rotate the vectors... I necessarily rotate both spaces (because they are replicas of the first one). So, if I call $\mathbf{T}\in\mathbf{V}\otimes\mathbf{V}$ an element in this new vector space, it will transform (under the same transformations of the vector) as $$\mathbf{T}\to \mathbf{T}^\prime = \left(\mathbf{R}\otimes\mathbf{R}\right)\mathbf{T},$$where the first rotation acts on the first copy of $\mathbf{V}$ and same for the second. Moral: $\mathbf{T}$ transform homogeneously, but with two of the expected transformations of a vector.

One can spend the life constructing bigger and bigger new vectors spaces by considering more and more copies of $\mathbf{V}$.

Tensors... How are they defined?

Ingredients:

  • A set of (geometrical) objects.
  • A field (real or complex numbers are Ok)
  • A "scaling" (multiplication) operation between elements of the field (numbers) and our geometrical objects, $\cdot:\mathbb{K}\times \mathbf{V}\to\mathbf{V}$.
  • An addition operator among our geometrical objects, $+:\mathbf{V}\to\mathbf{V}$.

So far the above define your vectors (or vector space). But as before, you can provide a group of transformations, $G$, which preserves the desirable properties of your vectors,${}^{\dagger\dagger}$ and $$\vec{V}\to \vec{V}^\prime = \Lambda\vec{V},\quad \text{for } \Lambda\in G.$$

Therefore, one says that our objects are in fact vectors under the group of transformations of $G$. So, the group of transformation should be included in the ingredients list.

  • A group of transformations $G$.

TENSORS

A tensor (by extension of the previous construction) is a geometrical object which under the group $G$ transforms homogeneously,$$\mathbf{T} \to \mathbf{T}^\prime = \Lambda\;\cdots\Lambda \mathbf{T}.$$

The number of transformation elements define the rank of the tensor.

NOTE 1: A vector is a tensor of rank one!

NOTE 2: Since tensors are defined under a group of transformations, a tensor under a group might not be a tensor (or at least not the same type of tensor) under other group.

Do we need all possible coordinate systems and transformations to define a vector space or just a few of them?

From all the above, you can conclude that you need ONE coordinate system, and the group of transformations. Immediately any other coordinate system related with the previous one is considered.

(note I didn't understand your last doubt, so I end it here!)


${}^*$ Coordinate transformation is another valid group of transformations.

${}^\dagger$ $\mathbf{R}$ represents the rotation transformation.

${}^{\dagger\dagger}$ The set of transformations is not necessarily a rotation.

$\endgroup$
1
$\begingroup$

Mathematically, the idea of a vector is prior. You could define objects that fulfill all properties of a vector space without referring to components or anything.

From the notion of a vector one can derive that there exist a maximum number of linearly independent vectors and any vector in your vector space can be represented uniquely by a linear combination of these basis vectors. You then represent the vector as the array of coefficients of this linear combination.

Since these components encode a more abstract object, they indeed have to change according to some rules if you choose a different set of basis vectors.

In schools and introductory physics classes one usually skips the more abstract steps to reach the above conclusions. One simply defines vectors as a set of coefficients in euclidean $R^3$ space, that behave in a specific way under rotations. This turns out to be equivalent to more rigorous definition in the easiest applications and it has the advantage that it is very relatable - you can actually imagine the vector and its behavior under a change of basis.

Also, this approach does not require a lot of abstract thinking and you can already solve most problems in classical mechanics with this. Thus, in schools or engineering classes, a refinement of the concept is barely needed. Physicists will learn the more abstract way to work with vectors anyways, as soon as you start to learn about quantum mechanics.

So in summary, the mathematical approach is to define vectors abstractly and then derive all the properties that we know - while the physical approach is to define a vector through its properties and learn about further relations only once you actually need to expand the framework.

$\endgroup$
-2
$\begingroup$
How to define the concept of basis and coordinate transformation without completing the very notion of vector?

We cannot. Bases are to vectors as $0$ and $1$ are to natural numbers (as in Peanos axioms). We can start with $0$, or $1$, or even $376$, but we need to have something to start with.
Without the basis vectors, we cannot define a vector space (and hence coordinate transformation), and hence cannot define any other vector or a tensor. Check this question and answer.

Do these (bases) represent reference frames?

Yes, bases represent reference frames.

Do we need all possible coordinate systems and transformations to define a vector space or just a few of them?

In order to define and measure a scalar or vector (or a tensor), we need at least one coordinate system, and that is sufficient.

While not a formal way to define vector/tensor, this is correct. A vector or a tensor is a collection (array) of bases and, while this collection is different in different coordinate systems, for the same vector or tensor, these collections are related to one another by coordinate transformations.


The following is an attempt to clarify the last point above, and the need behind coordinate transformations, starting with the example of a scalar.

If I measure a $6m$ long pole with my ruler (basis) that is $2m$ in length, my measure will be $3$ units.
Another choice of basis (in another system) can be a $1m$ long ruler, and then the length of the pole will be $6$ units.

The actual length of the pole - the scalar invariant - remains the same, while its measure in different frames (using different bases) differs. And we can have transformation rules between the different frames, so that knowing the measure in one frame, we can calculate it in another without actually doing the measurement. Also note that the length of the pole in any frame is a collection of its bases (defined by scalar addition).

Similarly, for vectors, we have not $1$ but $3$ bases. And the vector is a collection of these $3$ bases, however, this time, not defined by scalar but both scalar and vector addition. And while these bases can differ among different coordinate systems, the vector remains invariant. We have transformation rules between the different coordinate systems.


The three bases in a vector space give us $3$ components for any vector. In the case of vectors, this collection of 3 bases can be visualized as a single entity by applying vector addition. But the $3$ bases can also combine together in any number of ways giving more than $3$ components, with these components not adhering to the rules of vector addition. Additionally, we can have more than $3$ bases. That's how we get tensors.

As long as we are constrained to combine bases only according to scalar or vector addition, any number of bases will give us a vector, otherwise a tensor.

The entity denoted by the combination of the bases in different ways - the tensor - remains invariant, and this constraint gives us the transformation rules between different coordinate systems (having different bases).

Finally, any combination (or collection) of bases that doesn't follow scalar or vector addition cannot be visualized, and hence we cannot generally perceive tensors as a single coherent physical entity. But, both from physics and mathematics point of view, they are as much a single entity as scalars and vectors.

$\endgroup$
2
  • $\begingroup$ @VincentThacker Thanks for the explanatory feedback! I've modified the answer to tackle specific parts of the question one by one. If it still does not look like an answer, I'll be happy to delete my answer or vote for its deletion. $\endgroup$ – manisar Jul 8 at 0:39
  • $\begingroup$ It does look like an answer now, but it is full of technical errors and inconsistencies. I recommend reading up more before answering. $\endgroup$ – Vincent Thacker Jul 14 at 3:51

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.