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Sign of Levi-Civita Symbol in spherical coordinates

  1. Feb 5, 2015 #1
    Hi, I am going through the derivation of an instanton solution (n=1) in Srednicki Chp. 93.

    Specifically, I went through eqn.s 93.29-93.38.
    However the sign of the Levi-Civita Symbol is bugging me:

    It says that in 4D Euclidean space,
    [tex]\epsilon^{1234}=+1[/tex] in Cartesian coordinates
    implies
    [tex]\epsilon^{\rho \chi \psi \phi}=-1[/tex]
    in spherical coordinates
    below eqn 93.17, by computing the Jacobian of coordinate change.
    But from my calculation, the Jacobian is positive, simply
    [tex]\rho^3 sin^2 \chi sin\psi[/tex]

    In the simplest case, e.g. 2D or 3D, it seems to me that the Levi-Civita symbol doesn't change sign either under this coordinate transformation.
    Can anyone explain why the sign of the Levi-Civita symbol is changed?

    Thanks!
     
  2. jcsd
  3. Feb 5, 2015 #2

    ChrisVer

    User Avatar
    Gold Member

    Your Jacobian should be....

    [itex] J = \begin{pmatrix} \sin \chi ~ \sin \psi ~ \cos \phi & \rho \cos \chi ~ \sin \psi ~ \cos \phi & \rho \sin \chi ~ \cos \psi ~ \cos \phi& - \rho \sin \chi ~ \sin \psi ~ \sin \phi \\ \sin \chi~ \sin \psi~ \sin \phi & \rho \cos \chi~ \sin \psi~ \sin \phi & \rho \sin \chi~ \cos \psi~ \sin \phi & \rho \sin \chi~ \sin \psi~ \cos \phi \\ \sin \chi ~ \cos \psi & \rho \cos \chi ~ \cos \psi & - \rho \sin \chi ~ \sin \psi & 0 \\ \cos \chi & -\rho \sin \chi & 0 & 0 \end{pmatrix} [/itex]

    Then, are you sure about your calculated determinant? Because I found:

    [itex]- \rho^3 \sin \psi ~ \sin^2 \chi [/itex]

    Jac.jpg
     
    Last edited: Feb 5, 2015
  4. Feb 5, 2015 #3
    Ahh right, now I got it. Thanks a lot
     
  5. Feb 6, 2015 #4
    Followup question:
    I am struggling to move from eq. 93.16 to 93.17.
    Now that

    [tex]n=\frac{1}{16 \pi^2} \int d^4x Tr(F_{\mu \nu}\tilde{F}^{\mu \nu})[/tex]
    [tex]=\frac{1}{16 \pi^2} \int d^4x \partial _\mu Tr(\epsilon^{\mu \nu \sigma \tau} (A_\nu F_{\sigma \tau}+\frac{2i}{3} A_\nu A_\sigma A_\tau))[/tex]
    [tex]\equiv \frac{1}{16 \pi^2} \int d^4x \partial _\mu Tr(\epsilon^{\mu \nu \sigma \tau} K_{\nu \sigma \tau})[/tex]
    [tex]n=\frac{1}{16 \pi^2} \int dS_\mu Tr(\epsilon^{\mu \nu \sigma \tau} K_{\nu \sigma \tau})[/tex]
    This is eq. 93.17.
    So,
    [tex]n=\frac{1}{16 \pi^2} \int (abs|J|) d\rho d\chi d\psi d\phi (\frac{1}{-(abs|J|)} \epsilon^{\delta \alpha \beta \gamma}) \partial_\delta Tr(K_{ \alpha \beta \gamma})[/tex]
    [tex]=\frac{-1}{16 \pi^2} \int d\rho d\chi d\psi d\phi ~ \epsilon^{\delta \alpha \beta \gamma} \partial_\delta Tr(K_{ \alpha \beta \gamma})[/tex]
    where [tex]\alpha, \beta, \gamma, \delta [/tex] are spherical coordinates
    [tex]abs(|J|)\equiv
    + \rho^3 \sin \psi ~ \sin^2 \chi
    [/tex]
    But how precisely should I do the final step of changing the volume integral to surface integral to get 93.16?
    I know by divergence theorem, but it seems to me that
    from
    [tex]\epsilon^{\delta \alpha \beta \gamma}[/tex]
    [tex]\epsilon^{\rho \chi \psi \phi}=-1[/tex]
    to
    [tex]\epsilon^{\alpha \beta \gamma}[/tex]
    [tex]\epsilon^{\chi \psi \phi}=+1[/tex]

    Eq. 93.16 is:
    [tex]n=\frac{-1}{16 \pi^2} \int d\chi d\psi d\phi ~ \epsilon^{\alpha \beta \gamma} Tr(K_{ \alpha \beta \gamma})[/tex]

    the minus sign from the previous step gets cancelled out, giving no minus sign in the end (93.16).
    Can you illustrate how to do it?
    Thanks
     
    Last edited: Feb 6, 2015
  6. Feb 10, 2015 #5

    ChrisVer

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    Gold Member

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