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Infinitesimal Lorentz transformation is antisymmetric

+ 6 like - 0 dislike
121 views

The Minkowski metric transforms under Lorentz transformations as

\begin{align*}\eta_{\rho\sigma} = \eta_{\mu\nu}\Lambda^\mu_{\ \ \ \rho} \Lambda^\nu_{\ \ \ \sigma} \end{align*}

I want to show that under a infinitesimal transformation $\Lambda^\mu_{\ \ \ \nu}=\delta^\mu_{\ \ \ \nu} + \omega^\mu_{{\ \ \ \nu}}$, that $\omega_{\mu\nu} = -\omega_{\nu\mu}$.

I tried expanding myself: \begin{align*} \eta_{\rho\sigma} &= \eta_{\mu\nu}\left(\delta^\mu_{\ \ \ \rho} + \omega^\mu_{{\ \ \ \rho}}\right)\left(\delta^\nu_{\ \ \ \sigma} + \omega^\nu_{{\ \ \ \sigma}}\right) \\ &= (\delta_{\nu\rho}+\omega_{\nu\rho})\left(\delta^\nu_{\ \ \ \sigma} + \omega^\nu_{{\ \ \ \sigma}}\right) \\ &= \delta_{\rho\sigma}+\omega^\rho_{\ \ \ \sigma}+\omega_{\sigma\rho}+\omega_{\nu\rho} \omega^\nu_{{\ \ \ \sigma}} \end{align*}

Been a long time since I've dealt with tensors so I don't know how to proceed.


This post imported from StackExchange Physics at 2014-03-05 14:49 (UCT), posted by SE-user user82235

asked Nov 21, 2013 in Theoretical Physics by user82235 (35 points) [ revision history ]
edited Mar 6, 2014 by Dilaton
@Qmechanic: Why the homework tag? "[...] any question where it is preferable to guide the asker to the answer rather than giving it away outright." - If it's not actual homework, shouldn't the OP decide what kind of answer he'd prefer? If I asked the question and needed the answer for actual work, I'd be very unhappy if given a pedagogical answer.

This post imported from StackExchange Physics at 2014-03-05 14:49 (UCT), posted by SE-user jdm
@jdm: The homework tag does not relate to whether it is actual homework or not; it relates to the content of the question. See Phys.SE homework policy for details.

This post imported from StackExchange Physics at 2014-03-05 14:49 (UCT), posted by SE-user Qmechanic
Sorry, I need to remember to add homework tags >_<

This post imported from StackExchange Physics at 2014-03-05 14:49 (UCT), posted by SE-user user82235
Related: physics.stackexchange.com/q/28535/2451

This post imported from StackExchange Physics at 2014-03-05 14:49 (UCT), posted by SE-user Qmechanic

2 Answers

+ 6 like - 0 dislike

Note that if you lower an index of the Kronecker delta, it becomes the metric:

$\eta_{\mu\nu}\delta^{\mu}_{\rho}=\delta_{\nu\rho}=\eta_{\nu\rho}$

And in your last step you got a wrong index. It should be $\omega_{\rho\sigma}$, not $\omega^{\rho}_{\sigma}$.

Then, the metric terms cancel and you neglect cuadratic terms.

That should be enough to solve it.

This post imported from StackExchange Physics at 2014-03-05 14:49 (UCT), posted by SE-user jinawee
answered Nov 21, 2013 by jinawee (115 points) [ no revision ]
+ 4 like - 0 dislike

Since the Lorentz transformation is valid for any $x\in M_{4}$, it can be rewritten as $\Lambda_{\rho}^{\mu}\eta_{\mu\nu}\Lambda_{\sigma}^{\nu}=\eta_{\rho\sigma}$. Substituting the infinitesimal form of the Lorentz transformation into the previous formula we get

$$(\delta_{\rho}^{\mu}+\omega_{\rho}^{\mu})\eta_{\mu\nu}(\delta_{\sigma}^{\nu}+\omega_{\sigma}^{\nu})+o(\omega^{2})=\eta_{\rho\sigma}$$

after expanding

$$\eta_{\rho\sigma}+\omega_{\rho}^{\mu}\eta_{\mu\nu}\delta_{\sigma}^{\nu}+\omega_{\sigma}^{\nu}\eta_{\mu\nu}\delta_{\rho}^{\mu}+o(\omega^2)=\eta_{\rho\sigma}$$

and from this we can see that

$$\omega_{\rho\sigma}+\omega_{\sigma\rho}=0\Rightarrow\omega_{\rho\sigma}=-\omega_{\sigma\rho}$$

This post imported from StackExchange Physics at 2014-03-05 14:49 (UCT), posted by SE-user Leonida
answered Nov 21, 2013 by Leonida (130 points) [ no revision ]

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