3
$\begingroup$

I have the following minus sign problem:

Consider an infinitesimal Lorentz transformation for which $\Lambda^{\mu}_{\nu}=\delta^{\mu}_{\nu}+\lambda^{\mu}_{\nu}$, where $\lambda^{\mu}_{\nu}$ is infinitesimal small. Define the vector fields $M_{\mu\nu}=x_{\mu}\partial_{\nu}-x_{\nu}\partial_{\mu}$. Show that acting on $x^{\mu}$, we have

$\frac{1}{2}\lambda^{\rho\sigma}M_{\rho\sigma}(x^{\mu})=\lambda^{\mu}_{\nu}x^{\nu}$

If i make the derivations:

$\frac{1}{2}\lambda^{\rho\sigma}M_{\rho\sigma}(x^{\mu})=\frac{1}{2}\lambda^{\rho\sigma}(x_{\rho}\partial_{\sigma}-x_{\sigma}\partial_{\rho})x^{\mu}= \lambda^{\rho\sigma}x_{\rho}\partial_{\sigma}x^{\mu}=\lambda^{\rho\sigma}x_{\rho}\delta^{\mu}_{\sigma}=\lambda^{\rho\mu}x_{\rho}=-\lambda^{\mu\rho}x_{\rho}=-{\lambda^\mu}_{\rho}x^{\rho}$

I can't see how lose the minus sign.. Probably trivial, but it keeps me busy.

Correction: orignal question had in the last step $\lambda^\mu_\rho$ which should be ${\lambda^\mu}_\rho$

$\endgroup$
0

1 Answer 1

3
$\begingroup$

You get a sign ambiguity because of your notation, as you simplify both ${\lambda^\mu}_\nu$ and ${\lambda_\nu}^\mu$ (which differ by a sign) to the same symbol $\lambda_\nu^\mu$.

$\endgroup$
3
  • $\begingroup$ Thank you for your reaction and sorry for the small mistake in typing but a mistake which is essential to my question: The correction, i made the derivation: $\frac{1}{2}\lambda^{\rho\sigma}M_{\rho\sigma}(x^{\mu})=\frac{1}{2}\lambda^{\rho\sigma}(x_{\rho}\partial_{\sigma}-x_{\sigma}\partial_{\rho})x^{\mu}= \lambda^{\rho\sigma}x_{\rho}\partial_{\sigma}x^{\mu}=\lambda^{\rho\sigma}x_{\rho}\delta^{\mu}_{\sigma}=\lambda^{\rho\mu}x_{\rho}=-\lambda^{\mu\rho}x_{\rho}=-{\lambda^\mu}_{\rho}x^{\rho}$ I assume the mistake is in the last step which i can't figure out. $\endgroup$
    – BB73
    Commented Mar 26, 2012 at 18:55
  • $\begingroup$ This looks correct to me; the last step can be verified by inserting explicitly the metric. Maybe your source has a typo, and the leftmost $\lambda^{\rho\sigma}$ should have its superscript interchanged? $\endgroup$ Commented Mar 26, 2012 at 19:20
  • 1
    $\begingroup$ @BB73, I'd suggest that you submit an edit to the question which fixes your mistake. $\endgroup$
    – David Z
    Commented Mar 26, 2012 at 20:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.