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 $\mathrm{t}\mathrm{r}({\gamma}_{\mu}{\gamma}_{\nu})=4{\eta}_{\mu \nu}$, then we have $\mathrm{t}\mathrm{r}({\gamma}_{\mu}^{\prime}{\gamma}_{\nu}^{\prime})=4{\eta}_{\mu \nu}$ if we put:${\gamma}_{\mu}^{\prime}=\mathrm{exp}(-A){\gamma}_{\mu}\mathrm{exp}(+A)$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?

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 $\mathrm{exp}(-A)$ where

$A=\frac{1}{2}{\omega}_{\mu \nu}{\gamma}^{\mu}{\gamma}^{\nu}$

and ${\omega}_{\mu \nu}$ is an antisymmetric tensor containing $D(D-1)/2$ parameters. The group of all such transformations is isomorphic to $Spin(D-1,1)$. 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 $4\times 4$ contains $6$ independent parameters and has eigenvalues $\pm ia,\pm ib$, 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 $SU(2)$ 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 $4\times 4$ matrix, and its exponentials would produce the full group $GL(4,C)$. 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 $Spin(3,1)$ and $GL(4,C)$ - I guess that there's no proper group of $GL(4,C)$ that has a proper $Spin(3,1)$ subgroup. Obviously, there are many subgroups of $Spin(3,1)$ - such as $Spin(3)$, $Spin(1,1)\times Spin(2)$, and others.

$A=\frac{1}{2}{\omega}_{\mu \nu}{\gamma}^{\mu}{\gamma}^{\nu}$

and ${\omega}_{\mu \nu}$ is an antisymmetric tensor containing $D(D-1)/2$ parameters. The group of all such transformations is isomorphic to $Spin(D-1,1)$. 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 $4\times 4$ contains $6$ independent parameters and has eigenvalues $\pm ia,\pm ib$, 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 $SU(2)$ 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 $4\times 4$ matrix, and its exponentials would produce the full group $GL(4,C)$. 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 $Spin(3,1)$ and $GL(4,C)$ - I guess that there's no proper group of $GL(4,C)$ that has a proper $Spin(3,1)$ subgroup. Obviously, there are many subgroups of $Spin(3,1)$ - such as $Spin(3)$, $Spin(1,1)\times Spin(2)$, and others.

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