- #1
lalbatros
- 1,256
- 2
The Rindler geometry and its horizon can be obtained by a simple succession of Poincaré transformations to match the frame of an accelerated observer. By combining this SR result and the equivalence principle it follows that a uniform gravitational field is represented by the Rindler metric and inherits all its properties, specially the horizon.
Considering this, I have the feeling that:
- the Schwarzschild geometry does not need much more than the SR and the EP
- the additional need is merely related to the spherical symmetry
I would like to learn more about the additional physics to be understood in the Schwarzschild geometry as compared to the Rindler geometry.
I would also like to know how much it tells us about GR.
Thanks,
Michel
Considering this, I have the feeling that:
- the Schwarzschild geometry does not need much more than the SR and the EP
- the additional need is merely related to the spherical symmetry
I would like to learn more about the additional physics to be understood in the Schwarzschild geometry as compared to the Rindler geometry.
I would also like to know how much it tells us about GR.
Thanks,
Michel