LT & Translation Symmetry: Problem Analysis

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
4 replies · 1K views
HaoBoJiang
Messages
5
Reaction score
2
TL;DR
Does Lorentz transformation have the symmetry of time and space translation? Why?
As we all know, for the reference frame S' and S of relative motion, according to Lorentz transformation, we can get
As we all know, for the reference frame S' and S of relative motion, according to Lorentz transformation, we can get​
As we all know, for the reference frame S' and S of relative motion, according to Lorentz transformation, we can get
 

Attachments

  • ST.png
    ST.png
    4.1 KB · Views: 249
Physics news on Phys.org
@HaoBoJiang first of all, please use the PF LaTeX feature to post equations:

https://www.physicsforums.com/help/latexhelp/

We do not allow equations posted as images, since they cannot be quoted in replies.

Second, the transformations you are calling "Lorentz transformations" are different transformations from space and time translations. The transformations you are calling "Lorentz transformations" are more precisely called "boosts". Boosts are not translations, so of course you should not expect them to have the properties of translations.

The full group of Lorentz transformations contains boosts and spatial rotations, but not translations. If you add space and time translations, you get a larger group called the Poincare group.
 
  • Like
Likes   Reactions: Orodruin and berkeman
PeterDonis said:
Second, the transformations you are calling "Lorentz transformations" are different transformations from space and time translations. The transformations you are calling "Lorentz transformations" are more precisely called "boosts". Boosts are not translations, so of course you should not expect them to have the properties of translations.
I don’t know if I would say that he quoted boosts. It looks as if he has just quoted length contraction and time dilation.
 
Orodruin said:
I don’t know if I would say that he quoted boosts. It looks as if he has just quoted length contraction and time dilation.

Yes, you're right, the actual equations given in the OP are not the full Lorentz transformation (boost) equations.
 
The full symmetry group of Minkowski space as an affine pseudo-Euclidean space is the group ##\text{ISO}(1,3)^{\uparrow}##, i.e., the proper orthochronous Poincare group (I leave out the non-continuously connected parts, including time reversal and space reflections). They are the semidirect product of the proper orthochronous Lorentz group, represented by ##\mathrm{R}^{4 \times 4}## matrices, ##{\Lambda^{\mu}}_{\nu}## fulfilling
$$\eta_{\mu \nu} {\Lambda^{\mu}}_{\rho} {\Lambda^{\nu}}_{\sigma} =\eta_{\rho
\sigma}, \quad \mathrm{det} \hat{\Lambda}=+1, \quad {\Lambda^0}_{0} \geq 1.$$
with the pseudo-metric components ##(\eta_{\mu \nu})=\mathrm{diag}(1,-1,-1,-1)##, and translations ##x^{\mu} \rightarrow x^{\mu}+a^{\mu}## with ##a^{\mu}=\text{const}.##

The group element is ##(\hat{\Lambda},a)##, acting on the spacetime components as
$$(\hat{\Lambda}, a) \boldsymbol{x}=\hat{\Lambda} \boldsymbol{x} +a.$$
The group product thus is given by
$$(\hat{\Lambda}_2,a_2)(\hat{\Lambda}_1,a_1)=(\hat{\Lambda}_2 \hat{\Lambda}_1,a_2+\hat{\Lambda}_2 a_1).$$