1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Sometimes, upper indices can be understood as inverse matrix, I think this might be one of those times. $\endgroup$
    – Emil
    Feb 1, 2017 at 17:00

1 Answer 1

2
$\begingroup$

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:

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

$\endgroup$
2
  • $\begingroup$ 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, 2017 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, 2017 at 10:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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