Given the defining property of Lorentz transformation (which is linear) $\eta_{\mu\nu}\Lambda^{\mu}{}_{\rho}\Lambda^{\nu}{}_{\sigma} = \eta_{\rho \sigma}$, prove the following identities
(i) $$ (\Lambda k) \cdot (\Lambda x) = k \cdot x$$
(ii) $$ p \cdot (\Lambda x) = (\Lambda^{-1}p) \cdot x$$
(iii) $$ \left(\Lambda^{-1}k \right)^{2} = k^{2}$$
I understand that (i) can be proven as follows:
Given:
$$k' = \Lambda k, \ \ \text{where} \ \ k'^{\mu} = \Lambda^{\mu}{}_{\rho} k^{\rho}$$
And:
$$x' = \Lambda x, \ \ \text{where} \ \ x'^{\nu} = \Lambda^{\nu}{}_{\sigma} x^{\sigma}$$
Then:
$$k' \cdot x' = (\Lambda k) \cdot (\Lambda x) = \eta_{\mu \nu} k'^{\mu}x'^{\nu} = \eta_{\mu \nu}\Lambda^{\mu}{}_{\rho}\Lambda^{\nu}{}_{\sigma} k^{\rho} x^{\sigma} = \eta_{\rho \sigma} k^{\rho} x^{\sigma} = k \cdot x$$
My issue is proving (ii) and (iii) based on the same approach
- For (ii) I tried:
We now have
$$p' = \Lambda p, \ \ \text{where} \ \ p'^{\rho} = \Lambda^{\rho}{}_{\lambda} p^{\lambda}$$
$$x' = \Lambda x, \ \ \text{where} \ \ x'^{\nu} = \Lambda^{\nu}{}_{\sigma} x^{\sigma}$$
We need to solve for $p$
$$p^{\rho}= \Lambda_{\lambda}{}^\rho p^{'\lambda}$$
Thus
$$p \cdot x' = p \cdot (\Lambda x) = \eta_{\rho \nu} p^{\rho}x'^{\nu}=\eta_{\rho \nu} \Lambda_{\lambda}{}^\rho \Lambda^{\nu}{}_{\sigma} p^{'\lambda} x^{\sigma} =\Lambda_{\lambda}{}^\rho p^{'\lambda} x_{\rho} = (\Lambda^{-1} p') \cdot x$$
Note I simply assumed that $\Lambda_{\lambda}{}^\rho p^{'\lambda} x_{\rho} = (\Lambda^{-1} p') \cdot x$; how to prove such equation though?
- For (iii) I tried:
We now have
$$k' = \Lambda k, \ \ \text{where} \ \ k'^{\mu} = \Lambda^{\mu}{}_{\rho} k^{\rho}$$
We solve for $k$
$$k^{\mu}= \Lambda_{\rho}{}^\mu k^{'\rho}$$
Thus we get
$$k \cdot k = \eta_{\mu \nu} k^{\mu} k^{\nu} = \eta_{\mu \nu} \Lambda_{\rho}{}^\mu \Lambda_{\sigma}{}^\nu k^{\rho} k^{\sigma}$$
But here I do not see how to show that
$$\eta_{\mu \nu} \Lambda_{\rho}{}^\mu \Lambda_{\sigma}{}^\nu k^{\rho} k^{\sigma} = (\Lambda^{-1} k)^2$$
Any help is appreciated.