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Some Questions

  1. Jul 7, 2003 #1
    In how many dimensions does a quark manifest?

    What is the difference between a contravariant tensor and a covariant tensor?
     
  2. jcsd
  3. Jul 8, 2003 #2

    HallsofIvy

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    What do you mean by "dimensions" of a quark? I suspect you means "quantum properties" such as mass, charge, hyper-charge, charm, color, spin,... I don't know the whole list so I will leave the answer to someone who does!

    As far as "covariant" and "contravarient" are concerned, it is, strictly speaking, incorrect to talk about "convariant tensor" and "contravarient tensor" (although it is regularly don't). One should talk about "convariant" and "contravarient" COMPONENTS of a tensor.

    Let g be the metric tensor. Then g_ij represents its contravarient coefficients and g^ij it's covariant coefficients. For the metric tensor, g_ij g^ik (the contraction) is the "identity" transform reprsented in any basis by the identity matrix.

    That is, ds= g_ik dx^i dx^i is the "differential of arc length" in the space. Choosing a coordinate system, we can write g_ik as a matrix and g^ik is the inverse matrix of that.

    "Euclidean" tensors are a special subset of tensors: we require that all coordinate systems have straight line axes and axes are all orthogonal: these are used in engineering problems. In Euclidean tensors, there is no difference between covariant and contravariant components.

    Here is probably the simplest way to look at covariant and contravariant components: Set up a Euclidean coordinate system in two dimensions (in other words, an ordinary x,y coordinate system).

    Now, set up a new x', y' coordinate system so that the x' and x axes are the same but the y' axis is at theta degrees to the x axis.

    There are TWO ways to define "coordinates" (or "components" of vectors) in such a coordinate system.

    1. Drop a perpendicular from a given point to each of the coordinate axes. Define x' to be the distance from the point to the y' axis measured along the perpendicular, define y' to be the distance from the point to the x' axis measured along the perpendicular.

    2. Draw a lines parallel to the axes through the given point. Define x' to be the distance from the point to the y' axis measured along this parallel to the x'-axis and define y' to be the distance from the point to the x' axis along the parallel to the y'-axis.

    In a Euclidean coordinate system (theta= pi/2) "parallels" and "perpendiculars" are exactly the same and so these coordinates are exactly the same. If theta is not pi/2, they are not the same.

    If you calculate the metric tensors in each of these, you will find that they are inverse matrices to each other! In other words, one (the second one, where you measured parallel to the axes), is "covariant" and the other (the first, where you measured parallel) is "contravariant".
     
  4. Jul 8, 2003 #3

    jeff

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    Not "components", indices: the component T01 of some tensor Tab has both contravariant and covariant indices.

    Rather than "coefficient", we usually say component, or in the case of a 2nd rank tensor, matrix element.

    Generally speaking, it's spaces and not the coordinate systems defined on them that are viewed as being endowed with geometrical properties.

    Tensoriality is defined in terms of transformation properties under coordinate changes, not in terms of the geometry of the spaces on which they're defined, so the term "euclidean" in the way you've used it doesn't apply to tensors. However, in the special case of the metric tensor, the term "euclidean" is used to indicate it's positive-definiteness as an inner product. But even with tensors on spaces with no geometry, contravariant and covariant indices can be differentiated in the sense of being dual to each other, like dual vector spaces.
     
    Last edited: Jul 8, 2003
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