3
$\begingroup$

Members of the Lorentz group obey $$\eta=\Lambda^{T}\eta\Lambda$$ where $\eta=\textrm{diag}(1,-1,-1,-1)$ is the Minkowski metric.

First, in matrix form write $$\Lambda=I+T$$ where $T$ is an infinitesimal generator. The above condition implies to first order that $$T=-\eta T^{T}\eta$$ and so $T$ is anti-symmetric in the space-space and time-time components, and symmetric in the space-time components.

Next, in index form write $$\Lambda^{\mu}_{\,\,\nu}=\delta^{\mu}_{\,\,\nu}+\omega^{\mu}_{\,\,\nu}$$ The component version of the defining equation then implies that, to first order, $$\omega^{\mu}_{\,\,\nu}=-\omega^{\,\,\mu}_{\nu}$$ and so the generators are fully anti-symmetric.

As far as I can see, $\omega^{\mu}_{\,\,\nu}$ is just the index form of $T$. Therefore, I'm a little confused as to why they seem to obey different conditions. Could anybody please help?

$\endgroup$
  • 1
    $\begingroup$ Hint: Think how the notation $\omega_{\nu}{}^{\mu}$ (as opposed to $\omega^{\mu}{}_{\nu}$) is defined in the last equation. $\endgroup$ – Qmechanic Mar 29 '17 at 19:15
  • 1
    $\begingroup$ $T^\mu{}_\nu = - ( \eta T^T \eta)^\mu{}_\nu = - \eta^{\mu\alpha} (T^T)_{\alpha}{}^\beta \eta_{\beta\nu} = - \eta^{\mu\alpha} T^\beta{}_\alpha \eta_{\beta\nu} = - T_\nu{}^\mu$ which is the same as the condition on $\omega$. $\endgroup$ – Prahar Mar 29 '17 at 19:16
  • 1
    $\begingroup$ @Prahar In this derivation, why is it acceptable to use $(T^{T})_{\alpha}^{\,\,\beta}=T^{\beta}_{\,\,\,\alpha}$ in the third equality, but not acceptable to use $\omega^{\mu}_{\,\,\nu}=(\omega^{T})_{\nu}^{\,\,\mu}$ to argue that $\omega^{T}=-\omega$? Clearly I'm misunderstanding something quite fundamental, so it would be good to understand this. $\endgroup$ – klgklm Mar 29 '17 at 20:09
  • $\begingroup$ Can somebody tell me how do we derive $\eta=\Lambda^{T}\eta\Lambda$? $\endgroup$ – Naman Agarwal Dec 9 '17 at 4:36
1
$\begingroup$

I actually got confused by the third comment... The first equation in component form reads $$ \Lambda^{\mu}{}_{\alpha}\, \eta_{\mu\nu}\, \Lambda^{\nu}{}_{\beta} = \eta_{\alpha\beta} $$ so that with the expansion near the identity: $$ (\delta^{\mu}{}_{\alpha} + \omega^{\mu}{}_{\alpha})\, \eta_{\mu\nu}\, (\delta^{\nu}{}_{\beta} + \omega^{\nu}{}_{\beta}) + o(\omega) = \eta_{\alpha\beta} + \omega^{\mu}{}_{\alpha}\, \eta_{\mu\beta} + \eta_{\alpha\nu}\, \omega^{\nu}{}_{\beta} + o(\omega) = \eta_{\alpha\beta} $$

$$ \Longrightarrow\quad \omega^{\mu}{}_{\alpha}\, \eta_{\mu\beta} = - \eta_{\alpha\nu}\, \omega^{\nu}{}_{\beta} \quad \Longleftrightarrow\quad \omega_{\beta\alpha} = - \omega_{\alpha\beta}$$ or also $$\omega^{\gamma}{}_{\alpha} = - \eta_{\alpha\nu}\, \omega^{\nu}{}_{\beta}\, \eta^{\beta\gamma} =: -\omega_{\alpha}{}^{\gamma} \qquad (Eq)$$


Let $M$ be a matrix with component $M^i{}_j$ at line $i$ and column $j$. $$\textbf{Inconsistent notation}\qquad (M^T)^i{}_j := M^j{}_i$$ One has to understand the transposition as a "dual" map: $M: E\rightarrow F$ then $M^T: F^* \rightarrow E^*,\ \lambda \mapsto \lambda \circ M $.

Now let $(\mathbf{e}_1,\cdots, \mathbf{e}_n)$ be a basis of $E$ and $(\boldsymbol{\epsilon}^1,\cdots, \boldsymbol{\epsilon}^n)$ its dual in $E^*$. A vector $\mathbf{x}\in E$, resp. $\boldsymbol{\varphi}\in E^*$ decomposes as $$\mathbf{x}= x^{\mu}\, \mathbf{e}_{\mu}\quad \text{and}\quad \boldsymbol{\varphi} = \varphi_{\nu}\, \boldsymbol{\epsilon}^{\nu}$$

This motivates $$ (M^T)_i{}^j := M^j{}_i$$ this is the coefficient of the line $i$ column $j$ of $M^T$ matrix of $M^T: F^* \rightarrow E^*$ w.r.t. a basis of the domain and the target.

It thus seems that what you wrote in comment is true as an equality of maps from $F^* $ to $ E^*$ (I distinguish the domain and target on purpose because if one says $\mathbf{R}^4$ then... Map defined by (Eq))

Other thing to understand: $\eta_{\mu\nu}$ as the matrix component of a map $\eta: E \rightarrow E^*$ w.r.t. a base in the domain and its dual in the target space.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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