Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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 ${\lambda^\mu}_\rho$

share|improve this question

migrated from theoreticalphysics.stackexchange.com Mar 26 '12 at 6:49

This question came from our site for scientific theorists and academic scholars interested in theoretical, research-level physics.

add comment

1 Answer 1

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 $\lambda_\nu^\mu$.

share|improve this answer
    
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: $\frac{1}{2}\lambda^{\rho\sigma}M_{\rho\sigma}(x^{\mu})=\frac{1}{2}\lambda^{\rho‌​\sigma}(x_{\rho}\partial_{\sigma}-x_{\sigma}\partial_{\rho})x^{\mu}= \lambda^{\rho\sigma}x_{\rho}\partial_{\sigma}x^{\mu}=\lambda^{\rho\sigma}x_{\rho‌​}\delta^{\mu}_{\sigma}=\lambda^{\rho\mu}x_{\rho}=-\lambda^{\mu\rho}x_{\rho}=-{\la‌​mbda^\mu}_{\rho}x^{\rho}$ I assume the mistake is in the last step which i can't figure out. –  BB73 Mar 26 '12 at 18:55
    
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? –  Arnold Neumaier Mar 26 '12 at 19:20
1  
@BB73, I'd suggest that you submit an edit to the question which fixes your mistake. –  David Z Mar 26 '12 at 20:07
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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