Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Levi Civita 4 tensor as pseudotensor

  1. Feb 1, 2012 #1
    1. The problem statement, all variables and given/known data
    Show that the anti-symmetric 4-tensor is a pseudotensor.


    2. Relevant equations
    [tex]\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}
    [/tex]


    3. The attempt at a solution
    Under LT
    [tex]
    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}[/tex]
    I got that
    [tex]\begin{eqnarray}
    e'^{0123}&=&1 \\
    e'^{1023}&=& -1 \\
    e'^{0132}&=& -1 \\
    e'^{1032}&=& 1
    \end{eqnarray}
    [/tex]
    After doing these first few terms, I'm seeing through induction that [itex]e'^{ijkl}=e^{qrst}[/itex]. Which is what we want for a tensor, right? A pseudotensor should depend on the determinate of [itex]e'^{ijkl}[/itex]. What am I missing??
     
    Last edited: Feb 1, 2012
  2. jcsd
  3. Feb 2, 2012 #2
    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.
     
  4. Feb 2, 2012 #3

    gda

    User Avatar

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

    [tex]\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.[/tex]

    then define

    [tex]\epsilon_{\mu\nu\rho\sigma}=g\epsilon^{\mu\nu\rho\sigma}[/tex]

    where [itex]g[/itex] is the determinant of the metric you are using.

    Then, define the usual standard tensor densities:

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

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

    and so, [itex]\tilde{\epsilon}[/itex] transforms like a tensor.
     
    Last edited: Feb 2, 2012
  5. Feb 2, 2012 #4
    Yeah, I was thinking something just like what you posted, thanks for the verification.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook