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Dual basis in Minkowski space

  1. Oct 5, 2009 #1
    Normally, if you have an orthonormal basis for a space, you can just apply your metric tensor to get your dual basis, since for an orthonormal basis all the dot products between the base vectors will boil down to a Kronecker delta. However, in Minkowski space, the dot product between a unit vector and itself may also be -1, rather than just 1, so a base vector like this does not meet the requirements if you're constructing your dual basis. Does that mean that simply applying your metric tensor need not produce your dual basis? Or does the definition of the dual basis change to allow the -1?
     
  2. jcsd
  3. Oct 5, 2009 #2

    dx

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    The definition of the dual basis is the same in any space (doesn't matter what the metric is). If you take the coordinate basis vectors {∂t, ∂x, ∂y, ∂z} as the basis for the tangent space, then the dual basis is {dt, dx, dy, dz}, which satisfy

    µ⋅dxγ = δµγ

    Notice that these are metric independent relations. The action of a 1-form on a vector is independent of metric.
     
    Last edited: Oct 5, 2009
  4. Oct 5, 2009 #3
    Alright, that clears it up, thanks.
     
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