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&#039;^0 &amp;=&amp; \gamma x^0 - \beta \gamma x^1 \\<br /> x&#039;^1 &amp;=&amp; \gamma x^1 - \beta \gamma x^0 \\<br /> x&#039;^2 &amp;=&amp; x^2 \\<br /> x&#039;^3 &amp;=&amp; x^3<br /> \end{eqnarray}<br />

The Attempt at a Solution

Under LT
<br /> e&#039;^{ijkl}=\frac{\partial x&#039;^i}{\partial x^q} \frac{\partial x&#039;^j}{\partial x^r} \frac{\partial x&#039;^k}{\partial x^s} \frac{\partial x&#039;^l}{\partial x^t} e^{qrst}
I got that
\begin{eqnarray}<br /> e&#039;^{0123}&amp;=&amp;1 \\<br /> e&#039;^{1023}&amp;=&amp; -1 \\<br /> e&#039;^{0132}&amp;=&amp; -1 \\<br /> e&#039;^{1032}&amp;=&amp; 1<br /> \end{eqnarray}<br />
After doing these first few terms, I'm seeing through induction that e&#039;^{ijkl}=e^{qrst}. Which is what we want for a tensor, right? A pseudotensor should depend on the determinate of e&#039;^{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.