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Transformations of a vector in the active viewpoint

  1. Nov 18, 2015 #1
    In Peskin and Schroeder page 37,

    a diagram illustrates how, under the active transformation, the orientation of a vector field must be rotated forward as the point of evaluation of the field is changed.

    I understand that the change of the orientation of the vector field is the same idea as that of the transformation matrix mixing up the vector field components to form the new vector field components.

    Now, the textbook goes on to mention that:

    under 3-dimensional rotations, ##V^{i}(x) \rightarrow R^{ij}V^{j}(R^{-1}x)##
    under Lorentz transformations, ##V^{\mu}(x) \rightarrow \Lambda^{\mu}_{\nu}V^{\nu}(R^{-1}x)##

    I understand that because we are working in the active transformation, the vector field itself is rotated (or boosted) in space(time) and the coordinate system (or reference system) is held fixed.

    Can someone explain to me in simple layman terms why we are applying the a transformation matrix on the vector field when we are using the inverse transformation matrix on the coordinates?
  2. jcsd
  3. Nov 18, 2015 #2
    A vector field is a map with vector domain and range. Think about the classic ##\mathbb{R}^2 \rightarrow \mathbb{R}^2## map. If you just do ##V^i(x) \mapsto R^{ij} V^j(x)## you are rotating the range vector while keeping it in the same position. That means you don't get a rotated vector field, but a completely different field. Think about the classic curling field:


    If you do ##V^i(x) \mapsto R^{ij} V^j(x)## with a 90° rotation matrix, you obtain a central vector field! While if you actively rotate the whole field, sure the resulting vector is rotated 90° but referred to the old vector function in the old position.

    Edit: maybe I see where your doubt lies. For instance, you don't do ##V^i(x) \mapsto R^{ij} V^j(R x)## because the right hand side V would have to be a different function from the left hand side V to be the actively rotated field. Also, don't see in the correct formula ##V^i(x) \mapsto R^{ij} V^j(R^{-1} x)## the resulting field as a function of ##R^{-1}x##, see it as a function of ##x##. Then it will make sense.
    Last edited: Nov 18, 2015
  4. Nov 19, 2015 #3
    Thanks! I get it! :smile:
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