vanhees71 said:
It's not that easy! You have ##(\eta^{\mu \nu})=\mathrm{diag}(1,-1,-1,-1)## (I'm in the west-coast camp, but there's no big difference when using the east-coast convention). An ##\mathbb{R}^{4 \times 4}##-matrix is called a Lorentz-transformation matrix if,
$${\Lambda^{\mu}}_{\rho} {\Lambda^{\nu}}_{\sigma} \eta_{\mu \nu}=\eta_{\mu \nu}.$$
In matrix notation (note that here the index positioning gets lost, so you have to keep in mind that the matrix ##\hat{\Lambda}## has a first upper and a second lower index while the matrix ##\hat{\eta}## as two lower indices) this reads
$$\hat{\Lambda}^{\text{T}} \hat{\eta} \hat{\Lambda}=\hat{\eta}.$$
Since ##\hat{\eta}^2=\hat{1}## we have
$$\hat{\eta} \hat{\Lambda}^{\text{T}} \hat{\eta}=\hat{\Lambda}^{-1}.$$
In index notation that reads restoring the correct index placement (note that also ##\hat{\eta}^{-1}=(\eta^{\mu \nu})=\hat{\eta}=(\eta_{\mu \nu})##)
$${(\hat{\Lambda}^{-1})^{\mu}}_{\nu} = \eta_{\nu \sigma} {\Lambda^{\sigma}}_{\rho} \eta^{\rho \mu}={\Lambda_{\nu}}^{\mu}.$$
Just wanted to clarify as this is still confusing to me. If you write in matrix vector notation, everything flows clearly, and if you write in index notation everything flows clearly too. The connection between the two is what's confusing. For example,
In matrix vector notation,
$$\eta = \Lambda^T \eta \Lambda \quad \rightarrow \quad 1 = \eta \Lambda^T \eta \Lambda \quad \rightarrow \quad \Lambda^{-1} = \eta \Lambda^T \eta \Lambda \Lambda^{-1} \quad \rightarrow \quad \Lambda^{-1} = \eta \Lambda^T \eta$$
In index notation,
$$\eta_{\rho\sigma} = (\Lambda^T)^{\mu\;}_{\;\rho} \Lambda^{\nu\;}_{\;\sigma} \eta_{\mu\nu} \quad \rightarrow \quad \delta^\alpha_\rho = (\Lambda^T)^{\mu\;}_{\;\rho} \Lambda^{\nu\;}_{\;\sigma} \eta_{\mu\nu} \eta^{\sigma\alpha} \quad \rightarrow \quad \delta^\alpha_\rho ((\Lambda^T)^{-1})^{\mu\;}_{\;\rho} = \Lambda^{\nu\;}_{\;\sigma} \eta_{\mu\nu} \eta^{\sigma\alpha} \quad \rightarrow \quad \delta^\alpha_\rho ((\Lambda^{-1})^T)^{\mu\;}_{\;\rho} = \Lambda^{\nu\;}_{\;\sigma} \eta_{\mu\nu} \eta^{\sigma\alpha} \quad \rightarrow \quad ((\Lambda^{-1})^T)^{\mu\;}_{\;\alpha} = \Lambda_{\mu\;}^{\;\alpha} \quad \rightarrow \quad (\Lambda^{-1})^{\alpha\;}_{\;\mu} = \Lambda_{\mu\;}^{\;\alpha}$$
The possible misconceptions here are,
1. Is it correct in index notation to write ##(\Lambda^T)^{\mu\;}_{\;\rho}## which is just the counterpart for the matrix vector notation? So that ##(\Lambda^T)^{\mu\;}_{\;\rho} = \Lambda^{\rho\;}_{\;\mu}##. This makes a lot of sense actually.
2. In the expression ##\;\delta^\alpha_\rho = (\Lambda^T)^{\mu\;}_{\;\rho} \Lambda^{\nu\;}_{\;\sigma} \eta_{\mu\nu} \eta^{\sigma\alpha}\;##, it is correct to just multiply both sides by the inverse of ##(\Lambda^T)^{\mu\;}_{\;\rho}## even there are indices present such that ##\; \delta^\alpha_\rho ((\Lambda^T)^{-1})^{\mu\;}_{\;\rho} = \Lambda^{\nu\;}_{\;\sigma} \eta_{\mu\nu} \eta^{\sigma\alpha} \;## right?
*For #1 I have not seen anybody write the transpose explicitly like I did above so I think we just follow the index placement obeying the transformation rules and make sure we have the corresponding matrix vector version in our head and just make sense of the resulting index version equation and say, "hey, we should transpose this matrix to have the correct row-column operation...". For #2 I believe even in index notations the symbol we use can also follow the matrix version as to not confuse the correspondence between the two versions right?
3. In
@vanhees71 SR notes (
Special Relativity by van Hees) in Appendix A.6, the steps are missing so maybe someone could fill in
all the steps so as to lift this long-time confusion already. For example, in eq. A.6.3 it has a term ##\Lambda^{\mu\;}_{\;\rho}## which is the term when written in matrix vector notation, the term with the transpose ##\Lambda^T##, but then when he multiplied by the inverse both sides, the indices in the resulting equation in eq. A.6.4 just suddenly flipped (with consideration to ##\delta^\alpha_\rho##), i.e. ##(\Lambda^{-1})^{\alpha\;}_{\;\mu}##. However there are no step which indicated why that is so.
I hope these may be clarified.
