Sign of Levi-Civita Symbol in spherical coordinates

  • #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!
 

Answers and Replies

  • #2
ChrisVer
Gold Member
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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:
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  • #3
Ahh right, now I got it. Thanks a lot
 
  • #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:

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