Given the Minkowski metric $\eta$ the Lorentz Transformation $\Lambda$ satisfies $$\eta=\Lambda^{T}\eta\Lambda$$ which in index form may be written $$\eta_{\mu\nu}=(\Lambda^{T})_{\mu}^{\,\,\alpha}\eta_{\alpha\beta}\Lambda^{\beta}_{\,\,\nu}$$$$\eta_{\mu\nu}=\eta_{\alpha\beta}\Lambda^{\alpha}_{\,\,\mu}\Lambda^{\beta}_{\,\,\nu}$$ How can I obtain an index expression for $\eta^{\mu\nu}$ starting from this expression? I have attempted to multiply through $\eta^{ab}$ using its index raising property, but this is not helping.

  • $\begingroup$ Sometimes, upper indices can be understood as inverse matrix, I think this might be one of those times. $\endgroup$ – Emil Feb 1 '17 at 17:00

Using the index-raising property

$ \eta^{\sigma \lambda} = \eta^{\sigma \mu} \eta_{\mu \nu} \eta^{\nu \lambda}$

Applying your formula

$= \eta^{\sigma \mu} \eta_{\alpha \beta} \Lambda^{\alpha}_{\,\mu} \Lambda^{\beta}_{\,\nu} \eta^{\nu \lambda}$

and using the index-raising property again

$ = \eta_{\alpha \beta} \Lambda^{\alpha \sigma} \Lambda^{\beta \lambda}.$

Is this what you had in mind? Alternatively, the $\eta_{\alpha \beta}$ of the final expression can also be exchanged for $\eta^{\alpha \beta}$ by appropriately lowering one of the indices of each of the transformation matrices:

$\eta_{\alpha \beta} \Lambda^{\alpha \sigma} \Lambda^{\beta \lambda} = \eta_{\alpha \gamma} \eta^{\gamma \delta} \eta_{\delta \beta} \Lambda^{\alpha \sigma} \Lambda^{\beta \lambda} = \eta^{\gamma \delta} \Lambda_{\gamma}^{\,\sigma} \Lambda_{\delta}^{\,\lambda}.$

  • $\begingroup$ This gives $\eta^{\mu\nu}=\eta^{\sigma\rho}\Lambda_{\sigma}^{\,\,\mu}\Lambda_{\rho}^{\,\,\nu}$. Is it possible to convert this into the form $\eta^{\mu\nu}=\eta^{\sigma\rho}\Lambda^{\mu}_{\,\,\rho}\Lambda^{\nu}_{\,\,\sigma}$ (I have this in my notes and am wondering it this is an error). $\endgroup$ – Watw Feb 1 '17 at 16:10
  • $\begingroup$ @Watw (sorry for the late reply, I clearly don't use SE frequently enough.) If you(r book) use(s) consistent notation, I'd read this condition as $\eta^{-1} = \Lambda \eta^{-1} \Lambda^T$. Of course, SO(1,3) matrices are invertible, so if we multiply this from left by $\Lambda^{-1}$ and from right by $(\Lambda^T)^{-1}$, we get $\eta^{-1} = \Lambda^{-1} \eta^{-1} (\Lambda^T)^{-1} = (\Lambda^T \eta \Lambda)^{-1}$, showing that the condition is equivalent (given invertibility) to the original one. $\endgroup$ – starless Feb 2 '17 at 10:27

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