If we view the components of \Lambda as functions of six parameters, representing a velocity difference and a rotation, chosen so that we get the identity transformation when all parameters are 0, we can Taylor expand Lambda like this:
\Lambda(\theta)=\Lambda(0)+\theta^a\frac{\partial}{\partial\theta^a}\bigg|_0 \Lambda(\theta)+\mathcal O(\theta^2)
where \mathcal O(\theta^2) represents all terms of second order and higher. Now let's define
\omega=\theta^a\frac{\partial}{\partial\theta^a}\bigg|_0 \Lambda(\theta)
and use this Taylor expansion in the equation that defines a Lorentz transformation:
\eta=\Lambda^T\eta\Lambda=(I+\omega+\mathcal O(\theta^2))^T\eta(I+\omega+\mathcal O(\theta^2))=\eta+\omega^T\eta+\eta\omega+\mathcal O(\theta^2)[/itex]<br />
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Since this holds for all \theta, the nth order terms on the right must be equal to the nth order terms on the left for all n. In particular, this means that<br />
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(\eta\omega)^T=-\eta\omega<br />
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Note that this is an exact equality, even though it came from a Taylor expansion. <br />
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Also, keep this in mind:<br />
<blockquote data-attributes="" data-quote="Fredrik" data-source="post: 0"
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Fredrik said:
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Recall that the components on row \mu, column \nu of the matrices<br />
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\Lambda, \Lambda^T, \eta, \eta^{-1}, \omega<br />
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are written as<br />
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\Lambda^\mu{}_\nu, \Lambda^\nu{}_\mu, \eta_{\mu\nu}, \eta^{\mu\nu}, \omega^\mu{}_\nu<br />
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and that \eta^{-1} and \eta and used to raise and lower indices.
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</blockquote>This means that \omega_{\mu\nu} are actually the components of \eta\omega, not \omega. That's why it's anti-symmetric.
