A How is the invariant speed of light enocded in SL(2,C)?

jakob1111
Gold Member
Messages
24
Reaction score
16
In quantum field theory, we use the universal cover of the Lorentz group SL(2,C) instead of SO(3,1). (The reason for this is, of course, that representations of SO(3,1) aren't able to describe spin 1/2 particles.)

How is the invariant speed of light enocded in SL(2,C)?

This curious fact of nature, is encoded in SO(3,1), because this is exactly the group that leaves the Minkowski metric invariant. In contrast, SL(2,C) is just the group of complex 2x2 matrices with unit determinant.
 
Physics news on Phys.org
jakob1111 said:
representations of SO(3,1) aren't able to describe spin 1/2 particles

They aren't? SO(3,1) has spinor representations, doesn't it? SU(2) certainly does, and SO(3,1) is isomorphic to SU(2) x SU(2).
 
PeterDonis said:
They aren't? SO(3,1) has spinor representations, doesn't it? SU(2) certainly does, and SO(3,1) is isomorphic to SU(2) x SU(2).

No, SO(3,1) has no spinor representations. It is the complexified Lie algebra of the Lorentz group so(3,1)C which is isomorphic to the Lie algebra su(2) x su(2). This process off complexification is a Lie algebra deformation and changes something fundamental. Other names for this complexification are Weyl's unitary trick or Wick rotation. We map the boost generators Ki to iKi and this way we get the Lie algebra so(4), which is isomorphic to su(2) x su(2).

When we then do representation theory of the complexified Lie algebra so(3,1)C or equivalently of su(2) x su(2) and use the exponential map to get the corresponding group representations, we do not only get representations of SO(3,1), but instead the representations of SL(2,C). Some of these representations are also representations of SO(3,1), but we get more than that. For example,the scalar and vector representation are also representations of SO(3,1), but the spinor represntations aren't.
 
  • Like
Likes dextercioby
Indeed the answer is given in the stackexchange posting.

Just to clarify the thing with spin 1/2. If you consider only rotations, the SO(3), the covering group is SU(2), and the fundamental representation of SU(2) describes spin 1/2. To extend this to representations of the, proper orthochronous Lorentz group, ##\mathrm{SO}(1,3)^{\uparrow}## you are led to its covering group ##\mathrm{SL}(2,\mathbb{C})##. However, now there are two two-dimensional non-equivalent representations corresponding to two sorts of Weyl spinors.

To also be able to describe space reflections (parity) you have to add these two representations, which leads to the four-dimensional Dirac-spinor representations, and the two Weyl spinors these are made of are precisely the states of left and right-handed chirality. For details, see Appendix B of

http://th.physik.uni-frankfurt.de/~hees/publ/lect.pdf
 
Thread 'Can this experiment break Lorentz symmetry?'
1. The Big Idea: According to Einstein’s relativity, all motion is relative. You can’t tell if you’re moving at a constant velocity without looking outside. But what if there is a universal “rest frame” (like the old idea of the “ether”)? This experiment tries to find out by looking for tiny, directional differences in how objects move inside a sealed box. 2. How It Works: The Two-Stage Process Imagine a perfectly isolated spacecraft (our lab) moving through space at some unknown speed V...
Does the speed of light change in a gravitational field depending on whether the direction of travel is parallel to the field, or perpendicular to the field? And is it the same in both directions at each orientation? This question could be answered experimentally to some degree of accuracy. Experiment design: Place two identical clocks A and B on the circumference of a wheel at opposite ends of the diameter of length L. The wheel is positioned upright, i.e., perpendicular to the ground...
According to the General Theory of Relativity, time does not pass on a black hole, which means that processes they don't work either. As the object becomes heavier, the speed of matter falling on it for an observer on Earth will first increase, and then slow down, due to the effect of time dilation. And then it will stop altogether. As a result, we will not get a black hole, since the critical mass will not be reached. Although the object will continue to attract matter, it will not be a...
Back
Top