18. More on Tensors (General Relativity)

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  • เผยแพร่เมื่อ 26 พ.ย. 2024

ความคิดเห็น • 2

  • @chrismaudsley127
    @chrismaudsley127 5 ปีที่แล้ว +1

    Many thanks. Very clear and helpful. BTW can I just check that g^ab is defined as the inverse of gab and it is unique to g because of the way it is defined in terms of dot products of basis vectors). ie the tensor T^ab (with both indices raised) is not the inverse of Tab (both lower). Not really sure how to justify this difference between g and T. Please could you clarify.

    • @PhysicsUnsimplified
      @PhysicsUnsimplified  5 ปีที่แล้ว

      g^ab is defined as the inverse of g_{ab}. That is, g^{ac} g_{cb} = delta^a_b. In terms of matrices, g^{ab} is the matrix inverse of g_{ab} and it is unique by the assumption that the metric itself is non-singular (non-zero determinant). The metric must be nonsingular because we assume that at each point the spacetime geometry is flat---it is the spacetime of special relativity. So at any point, there is a coordinate system in which the determinant of the metric is -1.
      Given a tensor T_{ab}, we define T^{ab} = g^{ac} T_{cd} g^{db}. There is no particular relationship between T^{ab} and the inverse of T_{ab}. In fact, in GR we never define the inverse of a tensor T_{ab} (the one exception is the metric tensor g_{ab}) .