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Derivatives of contravariant and covariant vectors

  1. Apr 25, 2014 #1
    Can someone explain why the derivative with respect to a contravariant coordinate transforms as a
    covariant 4-vector and the derivative with respect to a covariant coordinate transforms as a
    contravariant 4-vector.
     
  2. jcsd
  3. Apr 25, 2014 #2

    vanhees71

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    Take a scalar field. Then its differential is
    [tex]\mathrm{d} \phi=\mathrm{d} x^{\mu} \frac{\partial \phi}{\partial x^{\mu}}[/tex]
    is also a scalar. Thus, since [itex]\mathrm{d} x^{\mu}[/itex] transforms contravariantly the four-gradient must transform covariantly, i.e., the correct notation is
    [tex]\mathrm{d} \phi=\mathrm{d} x^{\mu} \partial_{\mu} \phi.[/tex]
    In the same way you can show that deriving with respect to the covariant components leads to a contravariant object.
     
  4. Apr 25, 2014 #3

    bcrowell

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    The title says "Derivatives of contravariant and covariant vectors," which would be stuff like [itex]\nabla_a v_b[/itex] versus [itex]\nabla_a v^b[/itex]. But #1 seems to be talking about [itex]\nabla_a v_b[/itex] versus [itex]\nabla^a v_b[/itex] , and #2 seems to be talking about the gradient of a scalar, [itex]\nabla_a\phi[/itex] versus [itex]\nabla^a\phi[/itex]. Which are we really talking about here?

    Not to be too pedantic, but we also don't have contravariant coordinates and covariant coordinates. Coordinates are always upper-index, and an ntuple of coordinates is not a vector or covector (at least not in GR). An infinitesimal *change* in the coordinates is an upper-index vector.

    Assuming that the question is really the one posed in #1, then an easy way to see this is in terms of scaling. For example, suppose you change your units from meters to centimeters. All of your coordinates (which are upper-index quantities) get bigger by a factor of 100. Now suppose you have a scalar such as the electrical potential, and you take a gradient in order to find the electric field. The electric field is *smaller* in units of V/cm than it is in units of V/m. So the coordinates transform in one way under scaling, while a gradient transforms in the opposite way. This is what we expect for covariant quantities compared to contravariant ones.
     
  5. Apr 25, 2014 #4

    vanhees71

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    Sorry, I thought the question is about special relativity.
     
  6. Apr 26, 2014 #5
    Thanks for your responses. I think my question should really have asked about the 4-gradient in SR.

    μ = ∂/xμ = [∂/∂t, ∇]

    and

    μ = ∂/xμ = [∂/∂t, -∇]

    c = 1

    So in these cases the indeces are just telling you that there is a change of sign in the spatial
    coordinates. What I don't understand is how the process of taking the derivative of the contravariant components results in a covariant vector and vice versa.
     
    Last edited: Apr 26, 2014
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