4 ms·

Also worth reading: http://geocalc.clas.asu.edu/pdf/SpacetimePhysics.pdf http://geocalc.clas.asu.edu/pdf/SpacetimePhysics.pdf https://slehar.wordpress.com/201

Also worth reading: http://geocalc.clas.asu.edu/pdf/SpacetimePhysics.pdf http://geocalc.clas.asu.edu/pdf/SpacetimePhysics.pdf https://slehar.wordpress.com/2014/03/18/clifford-algebra-a-visual-introduction/ https://slehar.wordpress.com/2014/03/18/clifford-algebra-a-v...

drostie 10y ago
Thanks for that first link, it looks very promising. My problem is that I already idiosyncratically try to understand spacetime in terms of spin-vectors,* so this is going to take a bunch of trying to map out a Rosetta stone to translate between them! * In brief, you can convert a world vector v^μ into a Hermitian 2D complex matrix V by adjoining σ_0 = I to the standard Pauli matrices σ_{1,2,3} and then V = σ_μ v^μ, when you do this you find that actually det V = v_μ v^μ under the (+ − − −) signature. This means that for nullvectors det V = 0 and therefore V is a projection V = φ φ^†, else V is merely Hermitian with eigenvalues v_0 ± √(v_1² + v_2² + v_3²) (because it has trace 2v_0) and there's not much more you can say. Lorentz transforms are linear and preserve Hermitianness and the determinant, with the parity-reversing stemming from V → -V and V → (det V) V^{-1} and the rest being V → L V L^† for some L such that det(L) = 1. The special case where L^† L = 1 (which is SU(2)) of course corresponds to 3D rotations because it preserves trace hence v_0, so if it's also preserving det V then it must be preserving the eigenvalues and hence the Euclidean metric √(v_1² + v_2² + v_3²). In turn L can be seen to act directly on the φ of the null vectors, φ → L φ, so call those "spin vectors" and confirm that they are spinors by finding out that L = -I for any 360-degree rotation. This is the most direct way I know of to see that the Lorentz transforms are the group SL(2, C) with this nice SU(2) group embedded in there. Then you can square-root the metric into an orientation tensor on the C^2 vector space as [0, -1; 1, 0] and build up the rest of the algebra in reverse from that starting point.
- jacobolus 10y ago
  You might find this clearer than other sources: http://arxiv.org/abs/1205.5935 http://arxiv.org/abs/1205.5935 Also check out http://geocalc.clas.asu.edu/pdf/LGasSG.pdf http://geocalc.clas.asu.edu/pdf/LGasSG.pdf

drostie 10y ago

Thanks for that first link, it looks very promising. My problem is that I already idiosyncratically try to understand spacetime in terms of spin-vectors,* so this is going to take a bunch of trying to map out a Rosetta stone to translate between them! * In brief, you can convert a world vector v^μ into a Hermitian 2D complex matrix V by adjoining σ_0 = I to the standard Pauli matrices σ_{1,2,3} and then V = σ_μ v^μ, when you do this you find that actually det V = v_μ v^μ under the (+ − − −) signature. This means that for nullvectors det V = 0 and therefore V is a projection V = φ φ^†, else V is merely Hermitian with eigenvalues v_0 ± √(v_1² + v_2² + v_3²) (because it has trace 2v_0) and there's not much more you can say. Lorentz transforms are linear and preserve Hermitianness and the determinant, with the parity-reversing stemming from V → -V and V → (det V) V^{-1} and the rest being V → L V L^† for some L such that det(L) = 1. The special case where L^† L = 1 (which is SU(2)) of course corresponds to 3D rotations because it preserves trace hence v_0, so if it's also preserving det V then it must be preserving the eigenvalues and hence the Euclidean metric √(v_1² + v_2² + v_3²). In turn L can be seen to act directly on the φ of the null vectors, φ → L φ, so call those "spin vectors" and confirm that they are spinors by finding out that L = -I for any 360-degree rotation. This is the most direct way I know of to see that the Lorentz transforms are the group SL(2, C) with this nice SU(2) group embedded in there. Then you can square-root the metric into an orientation tensor on the C^2 vector space as [0, -1; 1, 0] and build up the rest of the algebra in reverse from that starting point.

jacobolus 10y ago
You might find this clearer than other sources: http://arxiv.org/abs/1205.5935 http://arxiv.org/abs/1205.5935 Also check out http://geocalc.clas.asu.edu/pdf/LGasSG.pdf http://geocalc.clas.asu.edu/pdf/LGasSG.pdf