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I Vector components, scalars & coordinate independence

  1. Jul 26, 2016 #1
    This question really pertains to motivating why vectors have components whereas scalar functions do not, and why the components of a given vector transform under a coordinate transformation/ change of basis, while scalar functions transform trivially (i.e. ##\phi'(x')=\phi(x)##).

    In my more naive days, coming from a physics background, my earliest introduction to vectors and scalars was in terms of so-called "Euclidean vectors", having the properties of magnitude and direction and having the pleasing intuitive picture of arrows pointing in particular directions. With this in mind, it always made perfect sense to me why vectors have components when represented with respect to a given basis, since one needs, in general, more than one coordinate to describe a particular direction with respect to a given coordinate system. The basis vectors (induced by the coordinate system) "point" along each of the coordinate axes, and so by describing a vector in terms of its components along each of these basis vectors (i.e. how much of the vector "points" along each coordinate axis) one can capture the direction in which the vector points (with respect to the particular coordinate system chosen). Furthermore, a scalar has a magnitude, but no direction associated with it, hence it can be fully described by a single number at a given point in space and does not require components with respect to a basis. In terms of a scalar function, one simple specifies a coordinate system and for each set of coordinate values within this system, the function has a single value.
    When it comes to basis transformations, one seems to usually consider coordinate bases, induced by a particular choice of coordinate system, so under coordinate transformations the components of a vector will change. Again, this makes sense, since a vector has an independent existence from any particular basis, with its direction and magnitude being intrinsic properties. Hence, under a coordinate transformation the vector itself must remain unchanged, which requires that its components change in an appropriate manner. Furthermore, since a scalar function (evaluated at a particular point) is just a number, with no directional dependence, its value at a particular point should remain invariant under any given coordinate transformation, since its value should not depend on an (essentially arbitrary) choice of coordinates.

    This was all well and good (if I've understood it correctly), until one considers vector spaces in an abstract sense. Here, a vector is simply an element of a given set that satisfies a prescribed set of axioms, and a scalar is simply an element of the underlying field, ##\mathbb{F}## associated with the given vector space, ##V##. One can use any basis one likes to describe vectors within a given vector space, and a given vector can be described in terms of its components with respect to the basis vectors of a particular basis. Since all (finite) ##n##-dimensional vector spaces are isomorphic to a (##n\times##) Cartesian product of their underlying field, ##\mathbb{F}^{n}## one can represent a vector as a ##n##-tuple of scalars with respect to a given basis. Of course, the vectors themselves exist independently of any given basis, and so again one ends up with an appropriate transformation law for vector components under a change of basis. So far, so good - this tallies up with the previous notions of vectors in a general sense.

    The problems arise for me, in that one can have vector spaces in which there is no notion of a vector having a direction, or a magnitude. Although, having said this, I get that, intuitively, by describing a vector in terms of its components with respect to a given basis, each component quantifies "how much" of the given vector "points" along the direction of each of the basis vectors.
    Furthermore, a set of functions can be considered as a vector space, and so any given function can be described in terms of its components. How does one argue that, in general, if one chooses a particular basis (or coordinate system) a vector is then described in terms of its components, whereas a scalar (or a scalar function) simply has a value and no components?

    Essentially, my questions are:

    1. What is the general argument for why one describes vectors in terms of their components (with respect to a given basis), and why does a scalar not have components? (Are they simply defined that way? Is the point that a scalar is fully specified by a single value, i.e. a magnitude, and therefore does not require additional information of coordinates, since this would imply that it has components along particular directions, and thus a direction associated with it. Whereas, to fully specify a vector quantity, one must describe its magnitude and direction, and to do this with respect to a given basis requires one to specify "how much" of the vector is "pointing" along each of the basis vectors, and this is quantified by the components of the vector along each of the basis vectors?!)

    2. In physics, scalars are described as having magnitude, but no direction, hence they are invariant under rotations of coordinate systems, but why should they be invariant under (essentially) arbitrary coordinate transformations?

    3. Following up on questions 1 and 2, if one considers a (scalar) function, why does this not have components, and why should it remain invariant under general coordinate transformations, i.e. why should it transform as ##\phi'(x')=\phi(x)##?

    (I have some ideas on the answers to these questions, but would like to here other people's opinions on the matter).

    Sorry for the long-windedness of this post, I'm hoping to convey the point that I'm at in my understanding to aid anyone who might answer. Any help would be much appreciated.
     
    Last edited: Jul 26, 2016
  2. jcsd
  3. Jul 26, 2016 #2

    fresh_42

    Staff: Mentor

    We've had a similar discussion yesterday.
    https://www.physicsforums.com/threads/coordinate-independent-version-of-gradient.880014/

    I think it at least sheds some light on the usage of coordinates versus components.

    Edit:
    This is not true in this generality. As for the first part: scalars in connection with transformations can, e.g. be seen as multiples of the identity which (usually) commutes with all transformations or simply by the definition of linearity, so it doesn't matter where you write it. The second part is where you're going wrong. The point ##4 \in \mathbb{R}## according to the coordinate ##x## is different from the point ##4 \in \mathbb{R}## according to the coordinate ##-x## or ##2x##.
     
    Last edited: Jul 26, 2016
  4. Jul 26, 2016 #3
    I'm not really sure what you're saying here, do you mean that given two different coordinate systems, different points will be mapped to the value ##4\in\mathbb{R}##, depending on which coordinate system you're in? Surely though, if I have a (scalar) function, ##\phi##, then if one considers two (one-dimensional) coordinate systems, then changing coordinate system will mean that the algebraic form of the function will change, i.e. in the first coordinate system, at coordinate ##x## it will have value ##\phi(x)##, whereas in the second coordinate system, at coordinate ##x'## it will have value ##\phi'(x')## (the change from ##\phi## to ##\phi'## is what I mean by the algebraic, coordinate representation, of the function changing), but these two values should be equal, i.e. ##\phi'(x')=\phi(x)##, since we have simple relabelled the points will different coordinates, the (scalar) function itself hasn't changed?!
     
  5. Jul 26, 2016 #4

    fresh_42

    Staff: Mentor

    Yes. And that's the point. The nature of a function, a vector or whatever doesn't change if coordinates change. Only the way we work with them.
    And if we draw them, we have to use coordinates and the picture will change. Scalar fields aren't affected by linear transformations by construction: the linearity. If you drop this requirement, things become different.

    And in general to answer your questions it might be helpful if you could define "component" properly first, as yesterday's discussion shows.
     
  6. Jul 26, 2016 #5
    How do scalar fields change under more general coordinate transformations then? With linear transformations, is the point that, like scalars, scalar functions have no direction associated with them and so are fully specified by a single number at each point in a given coordinate system. As they have no directional dependence, their value at a particular point should be unaffected by rotating or translating (by a constant shift) the coordinate system one evaluates it in. Then, with vectors, one requires more than a single value to specify them as they have a direction associated with them, thus, in a particular coordinate system, we must describe them in terms of the amounts they point along each coordinate axis, i.e. by specifying their components with respect to the coordinate basis. Thus, under a coordinate transformation, these components must transform to compensate for the change in basis, and ensure that the vector itself remains invariant (as after all its direction and magnitude is coordinate/basis independent).

    By component I mean, the vectors components with respect to a given basis, for example, given a basis ##\lbrace\mathbf{e}_{i}\rbrace##, a given vector ##\mathbf{v}## can be represented in this basis in terms of its components ##\lbrace v^{i}\rbrace## with respect to this basis, as $$\mathbf{v}=\sum_{i}v^{i}\mathbf{e}_{i}$$ Thus, each component, ##v^{i}## quantifies the amount the vector ##\mathbf{v}## "points" along the direction each basis vector, ##\mathbf{e}_{i}##.
     
  7. Jul 26, 2016 #6

    fresh_42

    Staff: Mentor

    To be exact: ##v^{i}## is the ##i##-th coordinate and ##v^{i} \mathbf{e}_{i}## the ##i##-th component (according to a composition / linear combination by basis vectors!).

    The field? Not at all. Who said this?

    You must stop to mix up coordinates and the transformation of two systems of them, and a transformation applied to a scalar or vector field as such. The latter hasn't least to do with the former.

    If you change coordinates, like we do when producing street maps, it doesn't change the nature of your city at all. It only enables you to store it in the compartment.
    If a heat wave strikes your city, then it changes the scalar temperature field for real.
     
  8. Jul 26, 2016 #7
    Fair enough. I was being a little lax in my terminology.

    I was referring to your comment in your previous post:




    What I meant here was that if one has two given coordinate systems, where one can be mapped into the other by a rotation, then the value of a scalar field (at a given point) should not change, since it is simply a number and has no direction associated with it. The two different coordinate systems amount to two different ways of observing the same object, and so viewing the object from a different angle should not affect the object itself. In the case of vectors, one represents it in terms of its components along each of the coordinate axes (of the coordinate system), thus a change of coordinate systems results in the coordinates of the vector transforming in such a way that the vector itself doesn't change, i.e. such that ##\mathbf{v}=\sum_{i}v^{i}\mathbf{e}_{i}=\sum_{j}v'^{j}\mathbf{e}'_{j}##.
     
  9. Jul 26, 2016 #8

    fresh_42

    Staff: Mentor

    I don't see an issue. Your general coordinate transformations include stretchings. These will affect the coordinate representation of scalar fields, too, like it does if you switch between Fahrenheit and Celsius, whereas a rotation (of an object!) wouldn't. It would, however, change the representation of the domain. So if we regard a scalar field instead of ##\mathcal{M} \rightarrow \mathbb{R}## as ##(\mathcal{M} , \mathbb{R}) \subset \mathcal{M} \times \mathbb{R}## then the entire representation by coordinates does change.
     
  10. Jul 26, 2016 #9
    Yes, good point. This is what I meant by ##\phi'(x')=\phi(x)##, the coordinate representation of the function changes under a change of coordinates, but the value at a particular point (labelled by ##x## in one coordinate system and ##x'## in the other) is the same (if one keeps the same system of units, e.g. keeping Celsius to measure temperature in both coordinate systems).

    My original confusion arose over how I should argue (at least heuristically) that in any given coordinate system, a vector will have components and a scalar (or a scalar field) won't, it will simply have a value? (is any of what I put in previous posts about this part correct at all?)
     
  11. Jul 26, 2016 #10

    fresh_42

    Staff: Mentor

    Well, in terms of components, we might say that a scalar field is a ##1##-dimensional vector field, i.e. there is exactly one component to talk about. Therefore all vectors point in the same or opposite direction and we are left by talking about their lengths, which are numbers. In case of real scalar fields, a positive or negative number. As soon as we have a complex scalar field, it may as well be seen as a two-dimensional real vector field. (I know that this is not really the same thing.)
    The whole debate then melts down to the point, where we lose the distinction between the scalars of our linear world (an embedding of ##\mathcal{M}## in a Euclidean space, the transformations we apply on objects or coordinates, or otherwise involved vector spaces) and the scalars of a scalar field, which - if we take it to be exact - is a mapping to a one-dimensional space ##\mathbb{R}^1## or ##\mathbb{C}^1##, i.e. ##\mathbb{K} \cdot \mathbf{e}_1##.

    Maybe scalar fields can be viewed as a picture where the scalars are the colors of the pixels. Changing the coordinate system might then be as if we look at it or a dog does. But this analogue is very edgy in my opinion.
     
  12. Jul 26, 2016 #11
    I guess what confuses me is that, if for example we have two coordinate bases ##\lbrace\frac{\partial}{\partial x^{i}}\rbrace## and ##\lbrace\frac{\partial}{\partial y^{j}}\rbrace##, induced by the coordinate systems ##\psi_{1}:U_{1}\subset M\rightarrow V_{1}\subset\mathbb{R}^{n}## and ##\psi_{2}:U_{2}\subset M\rightarrow V_{2}\subset\mathbb{R}^{n}## then in the intersection ##U_{1}\cap U_{2}##, the coordinates of a vector transform as $$V'^{i}(y)=\frac{\partial y ^{i}}{\partial x^{j}}V^{j}(x)\qquad\qquad (1)$$ (using Einstein notation for brevity)

    However, given the same transformation between the two coordinate systems, then a scalar field ##f:M\rightarrow\mathbb{R}## transforms as $$\phi'(y)=\phi(x)$$ where ##\phi=f\circ\psi_{1}##, ##\phi'=f\circ\psi_{2}##, and ##x=\psi_{1}(p)##, ##y=\psi_{2}(p)##.
    Now, in this case, the coordinate representation of the scalar field changes, but it's actual value at ##p## does not. If we view a scalar field as a 1D vector field, then wouldn't one have a transformation law of a similar form as (1)?!
     
  13. Jul 26, 2016 #12

    fresh_42

    Staff: Mentor

    Yes, absolutely. The equation ##\phi'(y)=\phi(x)## should only hold for those transformations for which ##\frac{\partial y^{1}}{\partial x^{1}} = \frac{dy}{dx} = 1## hold like isometries or rotations, i.e. no stretches allowed. It cannot be true for changing, e.g. between hPa and Pa.

    Edit: But you can always hide such a factor in the definition of ##\phi'##.
     
  14. Jul 26, 2016 #13
    So, in special relativity, a scalar field is invariant under Poincaré transformations, but not necessarily so under conformal transformations then? What about general relativity in which one has general coordinate transformations, how does a scalar field change then?

    By the way, for an active transformation, in which one actually rotates an object and keeps the coordinate system the same, is the point that, for scalar quantities, the scalar is invariant under such an active transformations, i.e. it remains the same scalar, unaffected the rotation (since it has no direction associated with it), whereas, if one actually rotates a given vector then it is no longer the same vector as it was before the rotation (since a vector has a direction associated with it and so even if its magnitude is kept the same, its direction is changed by the active rotation and so it is a different vector)?!
     
  15. Jul 26, 2016 #14

    fresh_42

    Staff: Mentor

    I'm not fit enough in SR and GR to answer this without likely making mistakes.

    You change the point representation where a scalar applies to. The point is part of the field, too. But yes, the temperature field of a box remains unchanged if you turn it upside down.

    I highlighted the crucial part above because I think it holds the answer. If you interpret a scalar field as a one-dimensional vector field, as ##\mathbb{R}## is one-dimensional, then you disregard this dimension by your spatial transformations. All one-dimensional (scalar) vectors still point in the same direction, yet another one as embedded somewhere. E.g. we only care about temperature, pressure or whatever. It doesn't matter whether its simply a value or a real height of many tiny pipes of mercury (which would be turned around, too!).
    Let's consider an ##n##-dimensional manifold with an ##m##-dimensional vector field. This is a mapping ##\phi: \mathcal{M}^n \rightarrow \mathbb{R}^m##. Now what do you actually transform? ##\mathcal{M}^n##, its embedding in some ##\mathbb{R}^N##, the coordinates of this ##\mathbb{R}^N##, or do you transform the entire structure ##\mathcal{M}^n \times \mathbb{R}^m \subset \mathbb{R}^N \times \mathbb{R}^m##?
    This makes the difference, because in the case of scalar fields we simply don't care about the dimension ##m=1##, resp. only consider those transformations which have "no direction associated with it".

    Edit: I would go even further. We tend to see a scalar field on ##\mathcal{M}## as being part of it as the colors in a picture are, or the temperature is where we are.
     
    Last edited: Jul 26, 2016
  16. Jul 26, 2016 #15
    How does the notion of a scalar, using the physics definition of a scalar as an object that simply has a magnitude, tally up with the notion of a scalar in the abstract sense, in which it is an element of the field associated with a given vector space? As in physics, a particular scalar quantity, such the temperature at a given point, is not necessarily associated with a vector space?!
     
  17. Jul 26, 2016 #16

    fresh_42

    Staff: Mentor

    Well, it's both a number. Scalar simply emphasizes that it could be many different kind of numbers, or even no numbers at all, but this isn't important here.

    The scalars as scaling (sic!) factors of vectors, transformations, functions and so on, as they belong to vector spaces, are a crucial part of the construction like linearity is.

    Scalar fields on the other hand are a property to the points of a manifold. (I would have written weight in the sense of weighted summands, but that would have only invited someone to insist that a weight is a vector and not a scalar.) Usually the scalars defining a scalar field are taken from the same underlying range which is involved in the definition of the related vector spaces. Thus they behave nice since in the end they are only numbers from the same range, e.g. ##\mathbb{R}##.
    But as you might have seen above, until one isn't really used to deal with them, it can't be wrong to distinguish the two. Especially when it comes to transformations and it's not clear by the context which vectors are meant to be transformed: points of the manifold, points of the space in which the manifold may be embedded, the coordinates of this space, or the combined space of embedding plus attached vectors of a vector field.
    Confusion only arises from the imagination, that an attached vector at a point of an embedded manifold is rotated with the manifold. Like the velocity vector changes by the change of coordinates or by rotation.
    However, if one would consider a vector field made of (temperature, pressure, moisture) at points on earth, nobody would even think about rotating it when the street map is turned. One should be more careful with a velocity vector when driving through a curve, though.
     
  18. Jul 27, 2016 #17
    Ok, thanks for your help.
    I don't know why I got such a "bee in my bonnet" about it - I guess it stemmed from me thinking how I would explain to a layman what vectors and scalars are and why one can have components and the other doesn't (and how this affects how they transform), and I wasn't convinced my answer to this was satisfactory enough.
     
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