Weinberg Vol 1: Understanding Index Arrangement in (2.4.8) LT Transformations

  • Context: Graduate 
  • Thread starter Thread starter shehry1
  • Start date Start date
  • Tags Tags
    Weinberg
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
3 replies · 4K views
shehry1
Messages
39
Reaction score
0
Can anyone explain to me why in going from (2.4.7) to (2.4.8) the indices on the LT are arranged in the way they are. Why is mu the first index (lower) and rho the second (upper)?

Could they have been arranged in any other way? From the rules that I know, they can.
 
Physics news on Phys.org
Recall that the components on row [itex]\mu[/itex], column [itex]\nu[/itex] of the matrices

[tex]\Lambda, \Lambda^T, \eta, \eta^{-1}, \omega[/tex]

are written as

[tex]\Lambda^\mu{}_\nu, \Lambda^\nu{}_\mu, \eta_{\mu\nu}, \eta^{\mu\nu}, \omega^\mu{}_\nu[/tex]

and that [itex]\eta^{-1}[/itex] and [itex]\eta[/itex] and used to raise and lower indices. The components of [itex]\Lambda^{-1}[/itex] are

[tex](\Lambda^{-1})^\mu{}_\nu=(\eta^{-1}\Lambda^T\eta)^\mu{}_\nu=\eta^{\mu\rho}\Lambda^\sigma{}_\rho\eta_{\sigma\nu}=\Lambda_\nu{}^\mu[/tex].

Let's use all of the above to evaluate the first term on the right-hand side of (2.4.7).

[tex](\Lambda\omega\Lambda^{-1})_{\mu\nu}=\eta_{\mu\rho}(\Lambda\omega\Lambda^{-1})^\rho{}_\nu =\eta_{\mu\rho}\Lambda^\rho{}_\sigma\omega^\sigma{}_\lambda(\Lambda^{-1})^\lambda{}_\nu =\Lambda_{\mu\sigma}\omega^\sigma{}_\lambda\Lambda_\nu{}^\lambda=\Lambda_{\mu\rho}\delta^\rho_\kappa\omega^\kappa{}_\lambda\Lambda_\nu{}^\lambda[/tex]

[tex]=\Lambda_{\mu\rho}\eta^{\rho\tau}\eta_{\tau\kappa}\omega^\kappa{}_\lambda\Lambda_\nu{}^\lambda =\Lambda_\mu{}^\tau\omega_{\tau\lambda}\Lambda_\nu{}^\lambda[/tex]
 
Thanks a lot. Now I have just two short questions:

(1) In the last expression with the string of equalities, could you have expanded the bracket differently. Meaning that instead of [tex]\eta_{\mu\rho}\Lambda^\rho{}_\sigma\omega^\sigma{ }_\lambda(\Lambda^{-1})^\lambda{}_\nu[/tex] , would it had been correct to put that rho on the Lambda inverse and the nu on the Lambda? I tried it and it didn't turn out correct.

(2) After posting nearly 2300 messages, does one become more natural at the notation (like adding or subtracting integers) or do you still have to think about all the indices. :)
 
shehry1 said:
(1) In the last expression with the string of equalities, could you have expanded the bracket differently. Meaning that instead of [tex]\eta_{\mu\rho}\Lambda^\rho{}_\sigma\omega^\sigma{ }_\lambda(\Lambda^{-1})^\lambda{}_\nu[/tex] , would it had been correct to put that rho on the Lambda inverse and the nu on the Lambda? I tried it and it didn't turn out correct.
That doesn't work. Note that the only thing I'm using in this step is the definition of matrix multiplication:

[tex](\Lambda\omega\Lambda^{-1})^\rho{}_\nu =\Lambda^\rho{}_\sigma\omega^\sigma{ }_\lambda(\Lambda^{-1})^\lambda{}_\nu[/tex]

shehry1 said:
(2) After posting nearly 2300 messages, does one become more natural at the notation (like adding or subtracting integers) or do you still have to think about all the indices. :)
I still have to think about it. Probably took half an hour to remind myself about the things I needed to know before the actual calculation seemed trivial. Once I had written down the first equality in the last two lines (and knew why I was doing it), the rest was like adding integers. I try to avoid this notation when I can. I prefer an index free notation (e.g. [itex]\mbox{Tr}(AB)[/itex] instead of [itex]A^i{}_j B^j{}_i[/itex]), and my second choice is to write all the indices downstairs (Example).