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Proving that the cartesian metric is rotation invariant

  1. Oct 9, 2009 #1
    I'm trying to prove that the cartesian metric [tex]g_{mn}=\delta_{mn}[/tex] doesn't change under a transformation of coordinates to another cartesian coordinate set with different orientation.

    As a starting point I am using [tex]ds^2=\delta_{mn}(x)dx^m dx^n=\frac{\partial x^m}{\partial y^r}\frac{\partial x^n}{\partial y^s}\delta_{mn}(x)dy^r dy^s[/tex].

    Then I have to prove that [tex]\frac{\partial x^m}{\partial y^r}\frac{\partial x^n}{\partial y^s}\delta_{mn}(x)=\delta_{rs}(y)[/tex].

    However, I am unsure as to how I should tackle those coordinate derivatives. How can I show this equality?

    Any help is appreciated.
     
    Last edited: Oct 9, 2009
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  3. Oct 9, 2009 #2

    dx

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    Write the new coordinates as functions of the original ones, and then perform the differentiations. Do you know how to use rotation matrices? You can also draw a picture and work it out geometrically.
     
  4. Oct 9, 2009 #3
    Well, this is in D dimentions. I know the rotation matricies for two and three dimentions, but not the ones beyond theat. Is there a general expression?
     
  5. Oct 9, 2009 #4
    For any linear transformation A, <Ax, y> = <x,(A^t)y>. So <Ax,Ay> = <x,(A^t)Ay>. And the transpose of a rotation matrix is its...?
     
  6. Oct 9, 2009 #5
    The transpose of a rotation matrix is its inverse.

    I'm not familiar with your notation.

    What we know is that [tex]dy^n=\frac{\partial y^n}{\partial x^m} dx^m[/tex] where [tex]\frac{\partial y^n}{\partial x^m}=A_m^n[/tex] is a rotation matrix.

    What do you mean by [tex]\left< Ax,y\right>=\left<x,A^T y\right>[/tex]?
     
  7. Oct 9, 2009 #6

    quasar987

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    the brackets means "dot product". and A^t is the transpose of the matrix A. x, y are vector in R^n
     
  8. Oct 9, 2009 #7

    Hurkyl

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    It's standard notation of linear algebra. Index notation is not always the most convenient way to manipulate vectors, covectors, and the other objects you study in differential geometry.
     
  9. Oct 10, 2009 #8
    Okay. So this is showing that if you rotate two vectors by the same angular displacement, their dot product in unaffected. Is this equivalent to the metric being unaffected by the rotation?

    Also, if x and y are both column vectors, we have [tex](Ax)y^T=x(A^Ty)^T[/tex] ? If you have a proof or derivation of this, I would like to see it.
     
    Last edited: Oct 10, 2009
  10. Oct 10, 2009 #9

    dx

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    The metric components are [tex] g_{ab} = \langle e_i, e_j \rangle [/tex], where the ei are the basis vectors of the coordinate system you are in. To get the components in a rotated system, you just need to find out what [tex]\langle Ae_i, Ae_j\rangle[/tex] is. Use zhentil's hint in post #4.
     
    Last edited: Oct 10, 2009
  11. Oct 10, 2009 #10
    Using zhentil's hint in #4 I get [tex]\left< Ax,Ay\right>=\left<x,A^TAy\right>=\left<x,A^{-1}Ay\right>=\left<x,y\right>[/tex].

    Does this mean that the two sets of basis vectors are equivalent? So I can conclude that the metric is rotation invariant?
     
  12. Oct 10, 2009 #11

    dx

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    No, it doesn't mean the two basis sets are equivalent. What you've shown is that δ'ij = <Aei, Aej> = <ei, ej> = δij, i.e. the components in the new system δ'ij are equal to the components in the original: δij.
     
  13. Oct 10, 2009 #12
    Okay. Thanks for the help. :)

    By the way, but do I explain that [tex]\left<Ax,y\right>=\left<x,A^Ty\right>[/tex]?

    Also, is this approach valid for cartesian coordinates of all dimensions?
     
    Last edited: Oct 10, 2009
  14. Oct 14, 2009 #13
    If you write it down, it's immediate (i.e. write down <Ax,y> as a sum and move some numbers).

    Yes, it's valid for any finite-dimensional Euclidean space. (I believe it's also valid for any bounded linear operator on a Hilbert space, but someone may have to vouch for this)
     
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