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 ${\gamma }_{\mu }$ as gamma matrices that satisfy $\mathrm{t}\mathrm{r}\left({\gamma }_{\mu }{\gamma }_{\nu }\right)=4{\eta }_{\mu \nu }$, then we have $\mathrm{t}\mathrm{r}\left({\gamma }_{\mu }^{\prime }{\gamma }_{\nu }^{\prime }\right)=4{\eta }_{\mu \nu }$ if we put:${\gamma }_{\mu }^{\prime }=\mathrm{exp}\left(-A\right){\gamma }_{\mu }\mathrm{exp}\left(+A\right)$where $A$ is any matrix.
Boosts and rotations use $A$ as a bivector. For example, with $\alpha$ a real number, $A=\alpha {\gamma }_{0}{\gamma }_{3}$ boosts in the $z$ direction while $A=\alpha {\gamma }_{1}{\gamma }_{2}$ gives a rotation around the $z$ axis.
Solving for the value of α that gives a boost with velocity $\beta =v/c$ 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

Expert

2022-05-21Added 15 answers

The most general Lorentz transformation that is connected to the identity is given by the conjugation by $\mathrm{exp}\left(-A\right)$ where
$A=\frac{1}{2}{\omega }_{\mu \nu }{\gamma }^{\mu }{\gamma }^{\nu }$
and ${\omega }_{\mu \nu }$ is an antisymmetric tensor containing $D\left(D-1\right)/2$ parameters. The group of all such transformations is isomorphic to $Spin\left(D-1,1\right)$. If $\omega$ only contains one component $0\mu$, then it is a boost, and the nonzero numerical value of $\omega$ is the rapidity - the "hyperbolic angle" $\eta$ such that $v/c=\mathrm{tanh}\eta$.
If only one doubly spatial component of $\omega$ is nonzero, then this component ${\omega }_{\mu \nu }=-{\omega }_{\nu \mu }$ 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 $±ia,±ib$, 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 $SU\left(2\right)$ rotations in $3$ dimensions is a rotation around a particular axis by an angle.
If you allowed $A$ to contain something else than ${\gamma }^{\mu \nu }$ matrices which generate the Lorentz group, you could get other groups. Only for a properly chosen subset of allowed values of $\omega$, 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 $GL\left(4,C\right)$. 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 $Spin\left(3,1\right)$ and $GL\left(4,C\right)$ - I guess that there's no proper group of $GL\left(4,C\right)$ that has a proper $Spin\left(3,1\right)$ subgroup. Obviously, there are many subgroups of $Spin\left(3,1\right)$ - such as $Spin\left(3\right)$, $Spin\left(1,1\right)×Spin\left(2\right)$, and others.

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