# Levi Civita 4 tensor as pseudotensor

1. Feb 1, 2012

### Mindscrape

1. The problem statement, all variables and given/known data
Show that the anti-symmetric 4-tensor is a pseudotensor.

2. Relevant equations
$$\begin{eqnarray} x'^0 &=& \gamma x^0 - \beta \gamma x^1 \\ x'^1 &=& \gamma x^1 - \beta \gamma x^0 \\ x'^2 &=& x^2 \\ x'^3 &=& x^3 \end{eqnarray}$$

3. 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
$$\begin{eqnarray} e'^{0123}&=&1 \\ e'^{1023}&=& -1 \\ e'^{0132}&=& -1 \\ e'^{1032}&=& 1 \end{eqnarray}$$
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??

Last edited: Feb 1, 2012
2. Feb 2, 2012

### 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.

3. Feb 2, 2012

### gda

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

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

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.

Last edited: Feb 2, 2012
4. Feb 2, 2012

### Mindscrape

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