It would be nice to have a cute method that uses Lorentz transformations of basis...

Brody Collins

Brody Collins

Answered

2022-05-20

It would be nice to have a cute method that uses Lorentz transformations of basis vectors by exponential transformation using gamma matrices. To avoid confusion, let's assume + + + signature. Given γ μ as gamma matrices that satisfy t r ( γ μ γ ν ) = 4 η μ ν , then we have t r ( γ μ γ ν ) = 4 η μ ν if we put: γ μ = exp ( A ) γ μ exp ( + A )where A is any matrix.
Boosts and rotations use A as a bivector. For example, with α a real number, A = α γ 0 γ 3 boosts in the z direction while A = α γ 1 γ 2 gives a rotation around the z axis.
Solving for the value of α that gives a boost with velocity β = v / c appears to be straightforward. But how to do the rotations?
And by the way, what happens when you generalize A to be something other than bivectors?

Answer & Explanation

Elyse Huff

Elyse Huff

Expert

2022-05-21Added 15 answers

The most general Lorentz transformation that is connected to the identity is given by the conjugation by exp ( A ) where
A = 1 2 ω μ ν γ μ γ ν
and ω μ ν is an antisymmetric tensor containing D ( D 1 ) / 2 parameters. The group of all such transformations is isomorphic to S p i n ( D 1 , 1 ). If ω only contains one component 0 μ, then it is a boost, and the nonzero numerical value of ω is the rapidity - the "hyperbolic angle" η such that v / c = tanh η.
If only one doubly spatial component of ω is nonzero, then this component ω μ ν = ω ν μ is obviously the angle itself. Note that the spatial-spatial terms in A are anti-Hermitean, producing unitary transformations; the mixed temporal-spatial terms in A are Hermitean and they don't product unitary transformations on the 4-component space of spinors (but they become unitary if they're promoted to transformations of the full Hilbert space of quantum field theory).
In 4 dimensions, a general antisymmetric matrix 4 × 4 contains 6 independent parameters and has eigenvalues ± i a , ± i b, so in 3 + 1 dimensions, one can always represent a general Lorentz transformation as a rotation around an axis in the 4-dimensional space followed by a boost in the complementary transverse 2-plane. This is the counterpart of the statement that any S U ( 2 ) rotations in 3 dimensions is a rotation around a particular axis by an angle.
If you allowed A to contain something else than γ μ ν matrices which generate the Lorentz group, you could get other groups. Only for a properly chosen subset of allowed values of ω, you would get a closed group from the resulting exponentials (under multiplication). In particular, if you allowed A to be an arbitrary complex combination of any products of gamma matrices, well, then you would allow A to be any complex 4 × 4 matrix, and its exponentials would produce the full group G L ( 4 , C ). It's not a terribly useful groups in physics because actions are usually not invariant under this "full group", are they?
Also, there are not too many groups in between S p i n ( 3 , 1 ) and G L ( 4 , C ) - I guess that there's no proper group of G L ( 4 , C ) that has a proper S p i n ( 3 , 1 ) subgroup. Obviously, there are many subgroups of S p i n ( 3 , 1 ) - such as S p i n ( 3 ), S p i n ( 1 , 1 ) × S p i n ( 2 ), and others.

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