For the Lorentz transfomation matrix elements ${\Lambda^\mu} _\nu$ and inverse Minkowski metric elements $\eta^{\mu\nu}$, how can the relation

$${\Lambda^\mu}_\rho\Lambda^{\nu\rho}=\eta^{\mu\nu}$$ be shown to be true?

I first tried to create a $\eta$ element:$${\Lambda^\mu}_\rho\Lambda^{\nu\rho}={\Lambda^\mu}_\rho(\eta^{\nu\beta}{\Lambda_\beta}^\rho)={\Lambda^\mu}_\rho{\Lambda_\beta}^\rho\eta^{\nu\beta}.$$ The next step is to show that ${\Lambda^\mu}_\rho{\Lambda_\beta}^\rho=\delta^\beta_\mu$, so that $${\Lambda^\mu}_\rho{\Lambda_\beta}^\rho\eta^{\nu\beta}=\delta^\beta_\mu\eta^{\nu\beta}=\eta^{\nu\mu}=\eta^{\mu\nu}.$$ One relationship of the form ${\Lambda^\mu}_\rho{\Lambda_\beta}^\rho=\delta^\beta_\mu$ I can think of is $${(\Lambda^T)^\beta}_\rho{\Lambda^\rho}_\mu={\Lambda_\rho}^\beta{\Lambda^\rho}_\mu=\delta^\beta_\mu,$$ which is the relationship that says $\Lambda$ is an orthogonal matrix (i.e. $\Lambda^T\Lambda=1$).

But how can I show ${\Lambda^\mu}_\rho{\Lambda_\beta}^\rho=\delta_\beta^\mu?$

  • 1
    $\begingroup$ As a comment: I don't recommend moving the indices of a Lorentz matrix. Since a transformation matrix is not a tensor, this is strictly speaking not something you can do. It's a notational shortcut you can take, but you should probably only do so once you're very comfortable with indices. $\endgroup$ – Javier Oct 20 at 15:52
  • $\begingroup$ @Javier Thanks. Didn't know that. Always thought I am doing the usual lower/raising of indices of a tensor... $\endgroup$ – TaeNyFan Oct 20 at 15:58

The relation $\Lambda^{\mu}_{\ \ \sigma}\,\Lambda^{\nu\sigma}=\eta^{\mu\nu}$, or more properly

$$ \Lambda^{\mu}_{\ \ \sigma}\,\eta^{\sigma\tau}\,\Lambda^{\nu}_{\ \ \tau}=\eta^{\mu\nu}\ , $$

actually is the definition of a Lorentz transformation $\Lambda^{\mu}_{\nu}$. As such it cannot be proven true, unless you rest on some alternative definition.

In case you were wondering about where the definition comes from, it comes from the requirement that the Minkowski norm $\eta_{\mu\nu}x^{\mu}x^{\nu}$ of a four-vector be invariant under such transformations, i.e.

$$ \eta_{\mu\nu}\,\Lambda^{\mu}_{\ \ \sigma}\,\Lambda^{\nu}_{\ \ \tau}\,x^{\sigma}x^{\tau}=\eta_{\mu\nu}x^{\mu}x^{\nu}\ . $$

In order to be verified for all $x$, $\Lambda$ must have the above-mentioned property (then and only then it is called a Lorentz transformation). This requirement justifies but does not prove the property.

| cite | improve this answer | |
  • $\begingroup$ Nice! Didn't think of it that way! $\endgroup$ – TaeNyFan Oct 20 at 15:44
  • 1
    $\begingroup$ $+1$: I once wrote an answer spelling out a little bit more explicitly as to how $\Lambda^{\text{T}}\eta\Lambda = \eta$ follows from the invariance of the Minkowskian spacetime interval: physics.stackexchange.com/a/339719/20427 $\endgroup$ – Dvij D.C. Oct 20 at 16:29

After some more thought:

${\Lambda_\beta}^\rho$ is the inverse Lorentz transformation matrix and also satisfy the orthogonal matrix requirement:

$${(\Lambda^T)_\beta}^\rho{\Lambda_\rho}^\mu={\Lambda^\rho}_\beta{\Lambda_\rho}^\mu=1$$ Taking the transpose of both sides, $${\Lambda^\mu}_\rho{\Lambda_\beta}^\rho=1=\delta^\mu_\beta$$ which is what I needed to show that ${\Lambda^\mu}_\rho\Lambda^{\nu\rho}=\eta^{\mu\nu}$ is true.

| cite | improve this answer | |
  • $\begingroup$ I am pretty sure this is circular. In order to use the property regarding the inverse of the Lorentz transformation, you would need to use $\Lambda^{\text{T}}\eta\Lambda=\eta$ which is what you are actually trying to prove in the first place. Do you have an independent derivation of the inverse property? $\endgroup$ – Dvij D.C. Oct 20 at 19:02

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.