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?

Question

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&#039;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?

Elyse Huff · Accepted Answer

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 &quot;hyperbolic angle&quot;   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&#039;t product unitary transformations on the   4-component space of spinors (but they become unitary if they&#039;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&#039;s not a terribly useful groups in physics because actions are usually not invariant under this &quot;full group&quot;, 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&#039;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.