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Requesting Clear Description of Contravariant vs Covariant vectors

  1. Sep 5, 2012 #1
    Ok, so here's my problem. I just graduated with a mathematics degree and am going full force into a physics graduate program. I'm taking a course called mathematical methods for physicists, in which the first subject is tensors. Everyone else seems to be comfortable with the material, but me, I'm totally lost. I had a couple linear algebra courses in my undergrad, but it was nothing like this.

    First, there's the idea of how to use the superscript and subscript in a transformation/vector. These "index gymnastics" are quite mindboggling to me.

    For example, in our notes, the transformed basic vectors [itex]\hat{e'}_{i}=\sum_{j=1}^{N}{e}_{j}A_{j}^{i}[/itex] with A_{j}^{i} being the transformation matrix. The placement of the j's and i's baffle me, and I feel as if I don't get this figured out, I'll be completely lost from here on out (Right now one of our homework problems involves creating these transformation matrices to an oblique system).

    See this link for the lecture material (you'll find slides 1-5 describe what I'm having trouble with).
  2. jcsd
  3. Sep 5, 2012 #2
    Do you need a description of what contravariant and covariant vectors are in general?

    In an abstract sense, you have something like this:

    [tex]a' = \underline A(a)[/tex]

    where [itex]\underline A[/itex] is some linear function (map, transformation, operator--all equivalent words to use here), and [itex]a, a'[/itex] are vectors.

    We can expand this out in terms of some basis.

    [tex]a' \cdot e_i = \underline A(a) \cdot e_i \\
    {a'}_i = \sum_j a \cdot e_j \underline A(e^j) \cdot e_i = \sum_{j} a_j {A^j}_i[/tex]

    In general, if an index appears twice in Einstein notation, it must appear as both a down index and an up index (the expression on slide 1 is sloppy for this reason). Also, you'll notice this is written in terms of components of the vector [itex]a[/itex], not in terms of the operator acting on the actual basis vectors. This is typical in physics.

    It's somewhat difficult to discover what exactly you're having trouble with. Perhaps there's a problem we can work together here?
  4. Sep 5, 2012 #3
    I'm having a difficult time describing what I'm having trouble with because I'm very lost in the notation. Here is one of our homework problems. I've figured out the actual transformations, but I am lacking the ability to express these in the correct notation. For example, take part b. I found that the new basis vectors are:
    [itex]\hat{e'}_2=-e_1cos(\beta )+e_2cos(\beta)[/itex]

    Now, what would be the proper way to express this in terms of the transformation, the old basis, and the new basis (and how does each component in the transformation matrix correspond to the correct i and j in [itex]A_i^j[/itex] in the formula that the professor wants us to get, [itex]e'_{i}=A_{i}^{j}e_{j}[/itex])?

    For part c, I got the same transformation (which makes sense, counting that it transforms like the basis vectors). Would this [itex]A_i^j[/itex] be the same?

    Part d is really where I realized that I don't understand this notation. To solve this, I first expressed r as a vector sum of [itex]x^{1/}[/itex] and [itex]x^{2/}[/itex] and projected these onto the [itex]\hat{e}_1[/itex] and [itex]\hat{e}_2[/itex] axis to get:
    Now what would the indices on the matrix M look like? When I get the inverse of M in order to find the matrix A in order to solve for the [itex]x^{i/}[/itex], what will the new indices be on that? And again, how do these indices tell you how to multiply the matrix times the coordinates? How do I know whether to express the coordinates as column vectors, row vectors, and on the left or right side of the matrix?
    Last edited: Sep 5, 2012
  5. Sep 5, 2012 #4


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    Well, you start out with a vector space. People usually envision vectors as little arrows, but their abstract properties are pretty much defined by the fact that they can be added together and multiplied by scalars.

    The map from a vector to a scalar is also a vector space, but it's known as a dual space. It transforms differently than a vector when you change your basis (or, for physical applications, your coordinates).

    Dual vectors can be represented by a set of parallel lines (in 2d) or surfaces (in 3d). Note that if you have a vector and a dual vector, there is a natural scalar, the "product", that occurs because a dual vector is a map from a vector to a scalar, so if you specify both, the dual vector operating on the vector gives a scalar.

    This natural product is represented by the number of surfaces that the vector (represented as a directed line, the "arrow" with a head and a tail) piercies.

    Textbooksk such as MTW's gravitation go into this graphical representation more - if you find it helps you to visualize them.

    The dual of a dual vector (i.e the map from a dual vector to a scalar) transforms in the same way as an ordinary vector, that's why there are only two types of vectors.

    In engineering use, you might see the two types being called "column" vectors and "row" vectors.

    I hope this helps some?
  6. Sep 5, 2012 #5
    Okay, right, so going back to what I wrote and plugging in a basis vector [itex]e_i[/itex], we have

    [tex]e_i' = \underline A(e_i) = \underline A(e_i) \cdot e^j e_j \equiv {A_i}^j e_j[/tex]

    The [itex]i[/itex] can be either 1 or 2. The [itex]j[/itex] must range over both 1 and 2.


    [tex]e_1' = e_1 \cos \gamma - e_2 \sin \gamma \implies {A_1}^1 = \cos \gamma, {A_1}^2 = - \sin \gamma[/tex]

    Can you do the same sort of thing for the [itex]e_2'[/itex] equation? Sounds like you can and you did.

    For part (d), you're going to need to solve the linear system for [itex]{x'}^1, {x'}^2[/itex] (just rearrange the equations to solve for these variables). The matrix components should then be apparent. Physicists most often write in terms of column vectors on the right for contravariant components and row vectors on the left for covariant components. Honestly, though, writing the linear operator explicitly as a matrix is not something I often worry about. Just by looking at an equation like

    [tex]{x'}^i = A^i_j x^j[/tex]

    I know this means

    [tex]{x'}^1 = A^1_1 x^1 + A^1_2 x^2[/tex]

    and it means exactly that regardless of what matrix representation I choose.
  7. Sep 6, 2012 #6


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    One source of confusion is that some texts refer to "contravarient and covarient" tensors while others refer to "contravarient and covarient" components of tensors. Since there is an natural isomorphism from covarient vectors to contravarient vectors, either one will work- we can think of "covarient components" of a vector given in "contravarient components" as the components of the covarient vector isomorphic to the given contravarient vector.

    Also, the whole concept of "contravarient" and "covarient" components only arise when we have coordinate system with coordinate lines that are NOT orthogonal. So for ordinary Cartesian, polar, spherical, or cylindrical coordinates, there is no distinction between "contravarient" and "covarient" components (If we stick to only such coordinate systems we are said to be working with "Euclidean tensors" (a phrase that I am sure would have baffled Euclid!)). Of course, in curved spaces, such coordinate systems do not necessarily exist.

    Here is a very simple way of looking at it. In a two dimensional Cartesian coordinate system, we can measure the "x coordinate" of a point in either of two ways: (i) drop a perpendicular to the x-axis and measure the distance from the foot of that perpendicular to the origin or (ii) drop a perpendicular to the y-axis and measure the distance from the foot of the perpendicular to the point. In a Cartesian coordinate system, we would be measuring the length of two opposite sides of a rectangle and so get the same thing. If, however, we have a coordinate system in which the axes are straight lines but not perpendicular, we woud get two different results. The first would be the "covariant" x-component and the other the "contravariant" x-component.
  8. Sep 6, 2012 #7
    Thanks guys, you've been a great help. I think I was able to reach a (somewhat) basic understanding of the notation. I have figured out that the indices almost "cancel out", from what Muphrid was saying about the requirement of both an index up if it shows up down, and vice versa. Also, in Einstein notation, [itex]x_{i}=A_{i}^{j}x_{j}[/itex] is equivalent to [itex]x_{i}=x_{j}A_{i}^{j}[/itex] because the position of the [itex]x_{j}[/itex] doesn't change anything. The subscript j on the x implies that it is covariant, thus it will be a row vector on the left of the transformation matrix. Is this correct?
  9. Sep 6, 2012 #8
    Yes, basically, these equations are written as relations between components (not as matrix equations that you have to be careful about what's on the left or on the right). There is an equivalent matrix equation where you have to be careful about whether something's on the left or on the right and such, but equations written in index notation don't require that level of attention.
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