- #1
binbagsss
- 1,291
- 12
- TL;DR Summary
- considering directly coordinate transformations and how to consider boost-translation
This is probably a stupid question but, I want to show that a Lagrangian written in field theory is Lorentz invariant WITHOUT using the Lorentz transformation representation / generators. I know we know that a Lorentz scalar is automatically Lorentz invariant, but, I want to show this by considering the coordinate expressions directly.
I.e to plug in :
##t’=\gamma (t-\frac{vx}{c^2}),##
##x’=\gamma(x-vt) ##(1)
and expanding out the fields.
So I expand out ##L[\phi(x,t), \partial_{mu}\phi(x,t)] ->L'[\phi(x',t'), \partial_{mu}\phi(x',t')]##
(where I just wrote ##\phi(x,t),## rather than ##x^{\mu}## just because of the transformation written as (1)).
1 )And then I think, for everything to be consistent, it should then come out that for ##L ##to be Lorentz-invariant ##\phi## would have to satisfy the known transformation laws for field theory (Since, in contrast to a Galilean invariant Lagranigan, where one has to uniquely decipher the way a wavefunction needs to transform in order to get Galilean invariance for the Lagrangian), the way fields transform are already predetermined).
(So a Lorentz transformation is defined by: ##g'^{\mu \nu}=\Lambda^{\mu}_{\alpha}\Lambda^{\nu}_{\beta}g^{\alpha \beta}=g^{\mu \nu}##, and where a vector must satisfy: ##x'^{\mu}=\Lambda^{\mu}_{\alpha}x^{\alpha}##). So, I think, I would expect to find that for ##L## to be Lorentz invariant, this should give an expansion for ##\phi## in terms of ##\Lambda^{\mu}_{\alpha}## expanded out for a boost- it would agree with ##\phi## in terms of ##\Lambda^{\mu}_{\alpha}## expressed in terms of boost generators and expanded out
2) How would I show that Lorentz boost and translation do not commute for a Lagrangian in field theory when we have Lorentz scalars so everything is invariant w.r.t boost translations and boosts . I want to consider performing a boost then a translation and vice versa..
I.e to plug in :
##t’=\gamma (t-\frac{vx}{c^2}),##
##x’=\gamma(x-vt) ##(1)
and expanding out the fields.
So I expand out ##L[\phi(x,t), \partial_{mu}\phi(x,t)] ->L'[\phi(x',t'), \partial_{mu}\phi(x',t')]##
(where I just wrote ##\phi(x,t),## rather than ##x^{\mu}## just because of the transformation written as (1)).
1 )And then I think, for everything to be consistent, it should then come out that for ##L ##to be Lorentz-invariant ##\phi## would have to satisfy the known transformation laws for field theory (Since, in contrast to a Galilean invariant Lagranigan, where one has to uniquely decipher the way a wavefunction needs to transform in order to get Galilean invariance for the Lagrangian), the way fields transform are already predetermined).
(So a Lorentz transformation is defined by: ##g'^{\mu \nu}=\Lambda^{\mu}_{\alpha}\Lambda^{\nu}_{\beta}g^{\alpha \beta}=g^{\mu \nu}##, and where a vector must satisfy: ##x'^{\mu}=\Lambda^{\mu}_{\alpha}x^{\alpha}##). So, I think, I would expect to find that for ##L## to be Lorentz invariant, this should give an expansion for ##\phi## in terms of ##\Lambda^{\mu}_{\alpha}## expanded out for a boost- it would agree with ##\phi## in terms of ##\Lambda^{\mu}_{\alpha}## expressed in terms of boost generators and expanded out
2) How would I show that Lorentz boost and translation do not commute for a Lagrangian in field theory when we have Lorentz scalars so everything is invariant w.r.t boost translations and boosts . I want to consider performing a boost then a translation and vice versa..
