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I know that two consecutive Lorentz Boosts in different directions produce a rotation and therefore Lorentz Boosts don't form a group. But, my intuition tells me that, Lorentz Boosts in the same direction should form a group. Two boosts along the x axis should produce another boost along the x axis. Is that correct?

Abhimanyu Pallavi Sudhir
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MadScientist
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  • Boosts in different directions form a subgroup of the Lorentz group? If yes, how can I check it? –  Aug 25 '12 at 19:59
  • No they dont, because their commutator is a rotation. Boost in x, then in y, then backward in x, and backward in y, and you get a rotation. You should delete this answer, as it doesn't answer the question. – Ron Maimon Aug 25 '12 at 21:28

2 Answers2

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Yes, it is a one-dimensional subgroup generated by exponentiating an infinitesimal boost. Every one dimensional exponentiation of a generator forms an abelian group, because $e^{aG} e^{bG} = e^{(a+b)G}$, there is nothing to not commute. This result is the addition of velocities, you can explicity check that this is associative (it is always manifestly commutative).

Ron Maimon
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    Right, +1. And as you (and I) recently discussed in a different answer, http://physics.stackexchange.com/questions/23625/how-to-deduce-the-theorem-of-addition-of-velocities , the relationship of $a,b$ which are additive and form the group $({\mathbb R},+)$ to the velocities $v$ defining the boosts is $v/c=\tanh(a)$ or $\tanh(b)$, respectively. – Luboš Motl May 21 '12 at 19:40
  • Thank you for these answers. I can't +1 or check this now because my browser isn't letting me register an account. It is interesting that noncommutativty produces a rotation via the Baker Campbell formula and the fact that the commutator of two Boost Generators produces a Rotation Generator. But the first term is still e^{(a+b)G} in such a case, so I suppose it is actually a boost and a rotation simultaneously. – MadScientist May 21 '12 at 20:03
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Firstly, an excellent question. Never considered this before.

As has already been said, combining boosts along difference directions clearly doesn't form a group as the closure axiom is not met.

However, Lorentz boosts are nothing more than (hyperbolic) rotations in Minkowski space. So I think the set of boosts along the same axis should form a rotation group. I hope someone else can explain the reasoning without resorting to generators, as they're not my strong point.

See this YouTube video for some basic derivation.

  • Don't be scared about generators--- the exponential parameter is just the (hyperbolic) angle of the (hyperbolic) rotation, as Lubos Motl commented on my answer. If you follow the link, you can see that the angles add up in relativity, just as they do in geometry, so that the group is the additive real numbers (although not modulo $2\pi$ like in geometry).. – Ron Maimon May 22 '12 at 09:14