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Representation of covariant and contravariant vectors on spacetime diagrams

  1. Sep 2, 2010 #1

    How can we represent covariant and contravariant vectors on curved spacetime diagrams?

    How can we draw these vectors on a spacetime diagram?
    Contravariant vectors are really vectors,
    therefore we can represent them on the diagram with directed line elements.
    Covariant vectors are linear functions acting on contravariant vectors,
    therefore the basis of a covariant system
    can be represented also with directed line elements, I think,
    because these functions can be see as scalar products.
    The main question: what is the basis unit vectors on the diagram?
    I think that contravariant unit vectors are parallel
    to the x=const coordinate curves on the diagram,
    and covariant basis unit vectors are orthogonal to it.
    Am I right?
    But this is just its direction, what about its length (on the diagram)?
  2. jcsd
  3. Sep 2, 2010 #2


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    Since they are defined in the tangent spaces, you have to imagine them to be infinitesimally small.

    [PLAIN]http://img199.imageshack.us/img199/8721/vec5.jpg [Broken]

    Covariant vectors are represented by a directed pair of parallel planes. Contravariant vectors are represented by directed line segments. In this representation, the action of a covariant vector on a contravariant vector would be given by "the number of times the directed line segment pierces the parallel planes", interpreted in a continuous sense.

    In Minkowski space, the canonical basis for the tangent and cotangent spaces is derived from the coordinate functions xยต of an inertial frame.

    In the diagram above, dt is the gradient of the function t on spacetime, and it is one of the basis vectors in the canonical basis. The representation of covectors in the spacetime diagram as parallel lines has a scaling behavior which is opposite to that of vectors. The lines of 2dt are twice as close to each other as those of dt.
    Last edited by a moderator: May 4, 2017
  4. Sep 2, 2010 #3
    Contravariant and covariant vectors "live" in different spaces. If you want to draw arrows in a same diagram, draw a contravariant vector the usual way, while you can draw a covariant vector simply taking the spatial components with the opposite sign (same angle with the t-axis, but "on the other side"). This way the minkowsky scalar product becomes the usual euclidean product. For example, a covariant vector is orthogonal to a contravariant vector if and only if the two arrows are perpendicular on the drawing (in the usual euclidean sense). You see also that any contravariant vector on the light cone is orthogonal to it's own associated covariant vector.

    I don't know if I interpret your question correctly though.
  5. Sep 2, 2010 #4
    Yes I know, but the relative length to each other is relevant!
  6. Sep 2, 2010 #5
    Yes, I want a representation of contravariant and covariant vectors separately.
    And I also want to define the act of 1-forms on vectors on the diagram,
    which is some kind of geometrical definition.
    It would be nice if somebody can make a sample simple plot with the unit basis vectors
    on the Penrose-Carter diagram of the Schwarschild spacetime or other curved space.

    For example, how can we draw (represent) two orthogonal contravariant vector on the space-time diagram?
    Last edited: Sep 2, 2010
  7. Sep 2, 2010 #6

    George Jones

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    If you have access to Introducing Einstein's Relativity by Ray d'Inverno, take a look at Fig. 17.10, which is a Penrose diagram for the Kruskal extension of Schwarzschild.

    In general, to see the directions of a vector field, (plot the images of) the integral curves of the vector field.
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