How to obtain the transformation of Null Tetrad ?

Karmerlo
Messages
14
Reaction score
0
Capture1.jpg


Currently, I meet with the so-called null rotation in my study. I cannot understand why it has a mathematical form like that? Is there anyone familiar with this? Can anyone give a lucid explanation of it or provide steps to derive it.

See the image above on the null transformation (in Newman-Penrose Formalism)

Thanks.
 
Physics news on Phys.org
These are all the transformations that preserve the orthonormality relations among k, ℓ, m. For example, ask yourself, what tetrad rotation preserves ℓ? The most general form is

ℓ → ℓ
m → m + α k + β ℓ
k → k + γ ℓ + δ m + ε m*

and try to determine the coefficients:

k = k* ⇒ δ = ε*
ℓ·m = 0 ⇒ α = 0
k·m = 0 ⇒ β - ε
k·k = 0 ⇒ γ - ε ε*
 
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...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. The Relativator was sold by (as printed) Atomic Laboratories, Inc. 3086 Claremont Ave, Berkeley 5, California , which seems to be a division of Cenco Instruments (Central Scientific Company)... Source: https://www.physicsforums.com/insights/relativator-circular-slide-rule-simulated-with-desmos/ by @robphy
In Philippe G. Ciarlet's book 'An introduction to differential geometry', He gives the integrability conditions of the differential equations like this: $$ \partial_{i} F_{lj}=L^p_{ij} F_{lp},\,\,\,F_{ij}(x_0)=F^0_{ij}. $$ The integrability conditions for the existence of a global solution ##F_{lj}## is: $$ R^i_{jkl}\equiv\partial_k L^i_{jl}-\partial_l L^i_{jk}+L^h_{jl} L^i_{hk}-L^h_{jk} L^i_{hl}=0 $$ Then from the equation: $$\nabla_b e_a= \Gamma^c_{ab} e_c$$ Using cartesian basis ## e_I...
Back
Top