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I've recently learned about the Thomas precession. To shallowly summarize the Thomas precession takes place when you have a boost of first system to second system and then from the second system you have another non-collinear boost to the first one to a third system. If you now look at the relation between the first system and third system, the relation you get is just a different boost but you also have a rotation part of the transformation. Mathematically we could say $$ \Lambda_{1 \rightarrow 2} \Lambda_{1 \rightarrow 2} \neq \Lambda_{1 \rightarrow 3} $$ where we denoted $\Lambda$ as the boost transformation and the subscript of $\Lambda$ denotes from what to which system we do the transformation. It can be shown rather that $$ \Lambda_{1 \rightarrow 2} \Lambda_{1 \rightarrow 2} = R \Lambda \tag{1} $$ where $R$ denotes a rotation. This is all well and fine the problem that I have is that I've always believed Lorentz transformations $\Lambda$ are part of a (Lorentz) group. But if the relation (1) hold, this would break the definition of a group , specifically two objects in the group acting on each other would produce a object, which is not part of the group (Closure).

Could anyone please clear this up for me? Are perhaps rotations also part of the Lorentz group?

Links:

Thomas precession: https://en.wikipedia.org/wiki/Thomas_precession

Groups: https://mathworld.wolfram.com/Group.html

Nitaa a
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1 Answers1

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rotations are part of the Lorentz group, which is true to its name and is really a group (this is not always the case in physics, e.g. the renormalization group is not a group at all).

In fact, a boost can be thought of as a rotation in the plane spanned by one spatial direction and the time axis, with an imaginary angle.

Rotations form a subgroup of the Lorentz group, but boosts do not form such a subgroup, exactly due to Thomas precession (boosts along one specific axis do form a subgroup).