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Contravaint and covariant

  1. Sep 19, 2005 #1
    Can anyone explain to me what is contravariant and covariant? I just know that they are tensors with specific transformation properties (from website of MathWorld), i also know that the relation between two is the -ve sign.
    Then dose it mean that:
    given a 4-velocity of a particle is the vector
    u' = dx^i/ds then how about the covariant of u'?


    given u' = { 1/sqrt (1-V^2/C^2) , v/c sqrt (1-V^2/ C^2) , then I wnat to ask am I right that the covaritant of u is { -1/sqrt (1-V^2/C^2) , -v/c sqrt (1-V^2/ C^2)?

    if yes i can't get the relation u'u_' =1 (u_' indicate a subscripts)

  2. jcsd
  3. Sep 19, 2005 #2


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    One property that is easy to spot is that contravariant tensors are like vectors, little arrows in the tangent space you know, while covariant tensors are like differential operators. Indeed I think it is true that all the classical covariant tensors, except the metric itself, come from differntial operations, especially covariant differentiation. Because the Riemann-Christoffel or curvature tensor can be constructed as the difference between the second covariant derivatives of an arbitrary contravariant vector with opposite orders of differentiation, and the Ricci and Einstein tensors derive their contravariant components from R-C.
  4. Sep 20, 2005 #3


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    While the specific answer depends on your metric, this looks wrong.

    On a practical level, in relativity you lower the index of your tensor with the metric, g_ab.

    This means that given a vector u^a = (a,b), the contravariant vector is

    (g_00*a + g_01*b, g_10*a + g_11*b)

    The dot product of u^a u_a, the covariant and contravariant vectors, is supposed to give the "length" of the vector. In relativity, the vectors are generally 4-vectors, and the "length" of the vector is its invariant Lorentz interval. Your vector is a bit odd, having only 2 components - I'm assuming that one of them is probably time, and the other is probably some spatial dimension.

    Your result would only work with a metric of
    -1 0
    0 -1

    which is unlikely. A much more likely candidate would be either
    -1 0
    0 1


    1 0
    0 -1

    (both sorts of sign convention are used). This would be for a flat "Minkowski" space-time.

    Because this is posted in the relativity forum, and not the math forum, I'm assuming you are asking about covariant and contravariant vectors in relativity. In other applications, the vectors could be 3-vectors, and the "length" the usual length.
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