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Metric and Theorema Egregium

  1. Jun 29, 2015 #1
    I have ##ds^2=\cos^2(v)du^2 + dv^2## , i take a coordinate transformation x=u and cos(v)=##\frac{1}{(cosh(y))}##, I have to find a new metric with this coordinate transformation and proof it is in agreement with Theorema Egregium. ##ds^2=\frac{dx^2}{(cosh^2(y)}) +\frac{dy^2 }{(y^2(1-y^2))}## is my new metric . I used the jacobian to proof the second point, ##J=\begin{pmatrix} 1 & 0 \\ 0 & \frac{-1}{y(1-y^2)^(1/2)} \end{pmatrix}##, so ##j^-1 * \begin{pmatrix} \frac{1}{cosh^2{y}} & 0 \\ 0 & \frac{1}{y^2(1-y^2)} \end{pmatrix} *j ## = ## \begin{pmatrix} \frac{1}{cosh^2{y}} & 0 \\ 0 & (\frac{1}{y^2(1-y^2)})^2 \end{pmatrix}##
    I know cos(v)=##\frac{1}{(cosh(y))}## so the first part is ok, the second ##(\frac{1}{(y^2(1-y^2)})^2##=1 i think is not true...
    What is my mistake? Does someone could help me?
     
    Last edited: Jun 29, 2015
  2. jcsd
  3. Jun 30, 2015 #2
    I did a mistake, ##ds^2=\frac{dx^2}{(cosh^2(y)}) +\frac{dy^2 }{cosh{y}}##, ##J=\begin{pmatrix} 1 & 0 \\ 0 & \frac{1}{cosh{y}} \end{pmatrix}##, so ##j^-1 * \begin{pmatrix} \frac{1}{cosh^2{y}} & 0 \\ 0 & \frac{1}{cosh{y}}\end{pmatrix} *j ## = ## \begin{pmatrix} \frac{1}{cosh^2{y}} & 0 \\ 0 & (\frac{1}{cosh{y}})^4 \end{pmatrix}##
    Does someone have any ideas?
     
  4. Jun 30, 2015 #3

    HallsofIvy

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    That matrix multiplication is wrong. Since these are diagonal matrices multiplication is commutative so [itex]J^{-1}MJ= (J^{-1}J)M= M[/itex] for any matrix, M.
     
  5. Jun 30, 2015 #4
    ##j^T * \begin{pmatrix} \frac{1}{cosh^2{y}} & 0 \\ 0 & \frac{1}{cosh{y}}\end{pmatrix} *j ## = ## \begin{pmatrix} \frac{1}{cosh^2{y}} & 0 \\ 0 & (\frac{1}{cosh{y}})^4 \end{pmatrix}## there was a mistake, it was the transpose not the inverse.
     
  6. Jun 30, 2015 #5
    I solved it.
    I solved the exercise.
     
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