Sign of Levi-Civita Symbol in spherical coordinates

[cedricyu803]

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,
$$\epsilon^{1234}=+1$$ in Cartesian coordinates
implies
$$\epsilon^{\rho \chi \psi \phi}=-1$$
in spherical coordinates
below eqn 93.17, by computing the Jacobian of coordinate change.
But from my calculation, the Jacobian is positive, simply
$$\rho^3 sin^2 \chi sin\psi$$

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!

 

[ChrisVer]

Your Jacobian should be...

$J = \begin{pmatrix} \sin \chi ~ \sin \psi ~ \cos \phi & \rho \cos \chi ~ \sin \psi ~ \cos \phi & \rho \sin \chi ~ \cos \psi ~ \cos \phi& - \rho \sin \chi ~ \sin \psi ~ \sin \phi \\ \sin \chi~ \sin \psi~ \sin \phi & \rho \cos \chi~ \sin \psi~ \sin \phi & \rho \sin \chi~ \cos \psi~ \sin \phi & \rho \sin \chi~ \sin \psi~ \cos \phi \\ \sin \chi ~ \cos \psi & \rho \cos \chi ~ \cos \psi & - \rho \sin \chi ~ \sin \psi & 0 \\ \cos \chi & -\rho \sin \chi & 0 & 0 \end{pmatrix}$

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

$- \rho^3 \sin \psi ~ \sin^2 \chi$

 

[cedricyu803]

Ahh right, now I got it. Thanks a lot

 

[cedricyu803]

Followup question:
I am struggling to move from eq. 93.16 to 93.17.
Now that

$$n=\frac{1}{16 \pi^2} \int d^4x Tr(F_{\mu \nu}\tilde{F}^{\mu \nu})$$
$$=\frac{1}{16 \pi^2} \int d^4x \partial _\mu Tr(\epsilon^{\mu \nu \sigma \tau} (A_\nu F_{\sigma \tau}+\frac{2i}{3} A_\nu A_\sigma A_\tau))$$
$$\equiv \frac{1}{16 \pi^2} \int d^4x \partial _\mu Tr(\epsilon^{\mu \nu \sigma \tau} K_{\nu \sigma \tau})$$
$$n=\frac{1}{16 \pi^2} \int dS_\mu Tr(\epsilon^{\mu \nu \sigma \tau} K_{\nu \sigma \tau})$$
This is eq. 93.17.
So,
$$n=\frac{1}{16 \pi^2} \int (abs|J|) d\rho d\chi d\psi d\phi (\frac{1}{-(abs|J|)} \epsilon^{\delta \alpha \beta \gamma}) \partial_\delta Tr(K_{ \alpha \beta \gamma})$$
$$=\frac{-1}{16 \pi^2} \int d\rho d\chi d\psi d\phi ~ \epsilon^{\delta \alpha \beta \gamma} \partial_\delta Tr(K_{ \alpha \beta \gamma})$$
where $$\alpha, \beta, \gamma, \delta$$ are spherical coordinates
$$abs(|J|)\equiv + \rho^3 \sin \psi ~ \sin^2 \chi$$
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
$$\epsilon^{\delta \alpha \beta \gamma}$$
$$\epsilon^{\rho \chi \psi \phi}=-1$$
to
$$\epsilon^{\alpha \beta \gamma}$$
$$\epsilon^{\chi \psi \phi}=+1$$

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

the minus sign from the previous step gets canceled out, giving no minus sign in the end (93.16).
Can you illustrate how to do it?
Thanks

 

[ChrisVer]

I am sorry, I don't understand your question. What you do is write it as a surface integral:
I think though you can find your answer here?
http://learn.tsinghua.edu.cn:8080/2010310800/Webpage/files/Seminar2011/Instantons12_Xianyu.pdf
for Eq6 and below