An excerpt from my lecture notes on relativity (translated from Dutch):

"Special (special in the notes indicates that the determinant of the representation matrix equals +1) Lorentz transformations with arbitrary velocities don't form a group. Special Lorentz transformations form a group (an Abelian subgroup of the Lorentzgroup) only when the boosts are parallel."

I don't see this. Why don't arbitrary boosts in arbitrary directions form a group? What criterion (closed under the operation, existence of the inverse, containing neutral element etc) of forming a group is not satisfied here?

  • 2
    $\begingroup$ If by special you ean has determinant=1 then we have a group pf course, but I think the word "special" in regard to Lorentz tranformations means a "boost." Composing two boosts in non-parallel directions does not result in a a boost. $\endgroup$ – mike stone Jan 19 '20 at 21:59
  • 1
    $\begingroup$ Related: physics.stackexchange.com/q/395036/2451 $\endgroup$ – Qmechanic Jan 19 '20 at 22:02
  • 1
    $\begingroup$ Please edit this question so that it is clear whether you are talking about boosts or Lorentz transformations with determinant +1 when you say "special Lorentz transformation". Currently it's inconsistent. $\endgroup$ – ACuriousMind Jan 19 '20 at 22:10

Why don't arbitrary boosts in arbitrary directions form a group?

They don’t form a group because the composition of two non-parallel boosts is not a boost. Instead, the composition is a boost combined with a rotation. This rotation is known as a Wigner rotation.

See my answer to a related question for an explicit example.

The best way to understand the origin of the Wigner rotation is by considering infinitesimal Lorentz transformations and their generators. As mentioned here, the commutator of two boost generators is a rotation generator.

The “special Lorentz transformations”, which are those having a determinant equal to 1, include boosts, rotations, and compositions of these, and do form a group.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.