I Lorentz Transf. of Complex Null Tetrads: Formula (3.14-17)

ergospherical
Science Advisor
Homework Helper
Education Advisor
Insights Author
Messages
1,097
Reaction score
1,384
For a complex null tetrad ##(\boldsymbol{m}, \overline{\boldsymbol{m}}, \boldsymbol{l}, \boldsymbol{k})##, how to arrive at formulae (3.14), (3.15) and (3.17)? The equation (3.16) is clear as is. (I checked already that they work i.e. that ##\boldsymbol{e}_a' \cdot \boldsymbol{e}_b' = 2m'_{(a} \overline{m}'_{b)} -2k'_{(a} l'_{b)}##.)

1640199241420.png
 
Last edited:
Physics news on Phys.org
ergospherical said:
formulae (3.14), (3.15) and (3.17)?
What reference are these from?
 
PeterDonis said:
What reference are these from?
These are from the second edition of "Exact Solutions of Einstein's Field Equations" by Stephani et al. I have have an elaboration on (3.17), which I have started to type in, but my wife is pulling me away to watch someone get murdered ... er, to stream a show, so it will be a couple of hours before I get back to it.
 
  • Like
  • Haha
Likes vanhees71, ergospherical and martinbn
I have looked at (3.17). I suppose that it is not enough to show that
$$2m'_{(a} \overline{m}'_{b)} -2k'_{(a} l'_{b)} = 2m_{(a} \overline{m}_{b)} -2k_{(a} l_{b)},$$
as this is obvious for transformation (3.17).

I will proceed in a pedestrian way, i.e., I will show that (3.17) defines a boost. Inverting (3.12) gives (using notation that I dislike)
$$\begin{align}
\boldsymbol{E}_4 &= \frac{1}{\sqrt{2}} \left( \boldsymbol{k} +\boldsymbol{l} \right) \\
\boldsymbol{E}_3 &= \frac{1}{\sqrt{2}} \left( \boldsymbol{k} - \boldsymbol{l} \right) .
\end{align}$$
Now define
$$\begin{align}
\boldsymbol{E}'_4 &= \frac{1}{\sqrt{2}} \left( \boldsymbol{k}' +\boldsymbol{l}' \right) \\
\boldsymbol{E}'_3 &= \frac{1}{\sqrt{2}} \left( \boldsymbol{k}' - \boldsymbol{l}' \right) ,
\end{align}$$
with ##\boldsymbol{k}'## and ##\boldsymbol{l}'## given by (3.17). Then, by (3.17),
$$\begin{align}
\boldsymbol{E}'_4 &= \frac{1}{\sqrt{2}} \left( A\boldsymbol{k} +A^{-1} \boldsymbol{l} \right) \\
&= \frac{1}{\sqrt{2}} \left[ \frac{A}{\sqrt{2}} \left( \boldsymbol{E}_4 + \boldsymbol{E_3} \right) + \frac{A^{-1}}{\sqrt{2}} \left( \boldsymbol{E}_4 - \boldsymbol{E_3} \right) \right] \\
&= \frac{1}{2} \left( A + A^{-1} \right) \boldsymbol{E}_4 +\frac{1}{2} \left( A - A^{-1} \right) \boldsymbol{E}_3
\end{align}$$
Since
$$\left[ \frac{1}{2} \left( A + A^{-1} \right) \right]^2 - \left[ \frac{1}{2} \left( A - A^{-1} \right) \right]^2 = 1, $$
we can set
$$\begin{align}
\cosh w &= \frac{1}{2} \left( A + A^{-1} \right) \\
\sinh w &= \frac{1}{2} \left( A - A^{-1} \right)
\end{align}$$
Something similar holds for ##\boldsymbol{E}_3##, so we have a boost with rapidity ##w## and speed ##v = \tanh w##.
 
  • Like
Likes ergospherical
In this video I can see a person walking around lines of curvature on a sphere with an arrow strapped to his waist. His task is to keep the arrow pointed in the same direction How does he do this ? Does he use a reference point like the stars? (that only move very slowly) If that is how he keeps the arrow pointing in the same direction, is that equivalent to saying that he orients the arrow wrt the 3d space that the sphere is embedded in? So ,although one refers to intrinsic curvature...
I started reading a National Geographic article related to the Big Bang. It starts these statements: Gazing up at the stars at night, it’s easy to imagine that space goes on forever. But cosmologists know that the universe actually has limits. First, their best models indicate that space and time had a beginning, a subatomic point called a singularity. This point of intense heat and density rapidly ballooned outward. My first reaction was that this is a layman's approximation to...
So, to calculate a proper time of a worldline in SR using an inertial frame is quite easy. But I struggled a bit using a "rotating frame metric" and now I'm not sure whether I'll do it right. Couls someone point me in the right direction? "What have you tried?" Well, trying to help truly absolute layppl with some variation of a "Circular Twin Paradox" not using an inertial frame of reference for whatevere reason. I thought it would be a bit of a challenge so I made a derivation or...
Back
Top