# Lorentz transformation

1. Jan 11, 2015

### TimeRip496

As known, any Lorentz transformation matrix
$\Lambda$
must obey the relation
$\Lambda^μ{}_v$$\Lambda^ρ{}_σ$gμρ=gvσ
. The same holds also for the inverse metric tensor
g
which has the same components as the metric tensor itself (don't really understand why every tex formula starts from a new line), i.e.
$\Lambda^v{}_μ$$\Lambda^ρ{}_σ$g=gμρ
. Putting this all as a matrix relation, these two formulas are
ΛTgΛ=g , ΛgΛT=g
, where g is the metric tensor (and also the inverse metric tensor, as they are both the same).

I dont understand why is the lambda transpose and why the two different metric tensor suddenly become the same g. Is there something that I am missing out? And I a bit unsure of the inverse metric tensor stated above.

Last edited: Jan 11, 2015
2. Jan 11, 2015

### Staff: Mentor

A Lorentz transformation transforms between different inertial coordinate systems in flat spacetime. The metric tensor of flat spacetime is the same as the inverse metric tensor--i.e., the numerical values of the components are the same.

As for the transpose of the Lambda matrices, I'm not sure you have that right, because your Lorentz transformation matrices should have one unprimed and one primed index (unprimed for one frame, primed for the other), and the two expressions for the metric should also differ similarly. Using the standard symbol $\eta$ for the metric of flat spacetime, the relationships you wrote down would be:

$$\Lambda^{\mu}{}_{\nu '} \Lambda^{\rho}{}_{\sigma '} \eta_{\mu \rho} = \eta_{\nu ' \sigma '}$$

$$\Lambda^{\nu '}{}_{\mu} \Lambda^{\sigma '}{}_{\rho} \eta^{\mu \rho} = \eta^{\nu ' \sigma '}$$

Notice the primed and unprimed indices and how they are exchanged between the first and second equations. Also note that, although the $\eta$ matrices are all the same, in the sense that each individual component is the same, the equations refer to different components because of the different indices, so they are not saying the same thing.