Levi Civita 4 tensor as pseudotensor

[Mindscrape]

Homework Statement

Show that the anti-symmetric 4-tensor is a pseudotensor.

Homework Equations

$$\begin{eqnarray}<br /> x'^0 &=& \gamma x^0 - \beta \gamma x^1 \\<br /> x'^1 &=& \gamma x^1 - \beta \gamma x^0 \\<br /> x'^2 &=& x^2 \\<br /> x'^3 &=& x^3<br /> \end{eqnarray}$$

The Attempt at a Solution

Under LT
$$e'^{ijkl}=\frac{\partial x'^i}{\partial x^q} \frac{\partial x'^j}{\partial x^r} \frac{\partial x'^k}{\partial x^s} \frac{\partial x'^l}{\partial x^t} e^{qrst}$$
I got that
$$\begin{eqnarray}<br /> e'^{0123}&=&1 \\<br /> e'^{1023}&=& -1 \\<br /> e'^{0132}&=& -1 \\<br /> e'^{1032}&=& 1<br /> \end{eqnarray}$$
After doing these first few terms, I'm seeing through induction that $e'^{ijkl}=e^{qrst}$. Which is what we want for a tensor, right? A pseudotensor should depend on the determinate of $e'^{ijkl}$. What am I missing??

 

[Mindscrape]

Nevermind, I solved it. Just wondering, could we make a metric where the Levi Civita 4 tensor is a proper tensor? I think this may be possible.

 

[gda]

Hi! You can redefine the Levi-Civita in order to convert it a tensor. In curved space (including Minkowski space) you may define:

$$\epsilon_{\mu\nu\rho\sigma}=\left\{\begin{array}{c}<br /> 0~~\mbox{any two indices repeated}\\<br /> +1~~ \mbox{even permutation of indices}\\<br /> -1 ~~\mbox{odd permutation of indices},\\<br /> \end{array}\right.$$

then define

$$\epsilon_{\mu\nu\rho\sigma}=g\epsilon^{\mu\nu\rho\sigma}$$

where $g$ is the determinant of the metric you are using.

Then, define the usual standard tensor densities:

$$\tilde{\epsilon}_{\mu\nu\rho\sigma}=|g|^{1/2}\epsilon_{\mu\nu\rho\sigma}$$

$$\tilde{\epsilon}^{\mu\nu\rho\sigma}=|g|^{-1/2}\epsilon^{\mu\nu\rho\sigma}$$

and so, $\tilde{\epsilon}$ transforms like a tensor.

 

[Mindscrape]

Yeah, I was thinking something just like what you posted, thanks for the verification.