- #36
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Let me try again to explain why I think that the conclusion that light should bend in a gravitational field can be made two different ways and that those two ways are equivalent (no pun intended).
The figure below shows four different reference frames:
For light-bending, we need to mention three coordinates: ##x, y, t##, where ##y## measures location in the vertical direction, and ##y## measures location in the horizontal direction.
The two different, but equivalent, ways to get to light bending in frame A are:
Path 1: D to C to A
The figure below shows four different reference frames:
- A: The reference frame of someone (inside an elevator, or not) at rest on the surface of the Earth. Whether in an elevator or not, his frame only describes a small region of spacetime.
- B: The reference frame of someone (inside an elevator) in free-fall near the surface of the Earth.
- C: The reference frame of someone inside an accelerating rocket in gravity-free space.
- D: The reference frame of someone (inside an elevator, or not) drifting unaccelerated in gravity-free space.
For light-bending, we need to mention three coordinates: ##x, y, t##, where ##y## measures location in the vertical direction, and ##y## measures location in the horizontal direction.
The two different, but equivalent, ways to get to light bending in frame A are:
Path 1: D to C to A
- Start with D. Then we know that light initially aimed horizontally will follow the path: ##x_D = c t_D##, ##y_D = y_0##.
- Transform to C. Assume that if everything is moving slowly relative to the speed of light, then we can relate the coordinates via: ##t_C \approx t_D##, ##x_C \approx x_D##, ##y_C \approx y_D - \frac{1}{2} g t_D^2##.
- This implies that the path of the light will be given by: ##x_C \approx c t_C##, ##y_C \approx y_0 - \frac{1}{2} g t_C^2##
- Go to A using the equivalence principle. So conclude: ##x_A \approx c t_A##, ##y_A \approx y_0 - \frac{1}{2} g t_A^2##
- Again, start with D. Again we know that light initially aimed horizontally will follow the path: ##x_D = c t_D##, ##y_D = y_0##.
- Go to B using the equivalence principle. So conclude: ##x_B = c t_B##, ##y_B = y_0##.
- Transform to A. Assume that: ##t_A \approx t_B##, ##x_A \approx x_B##, ##y_A \approx y_B - \frac{1}{2} g t_A^2##.
- This implies that the path of the light will be given by: ##x_A \approx c t_A##, ##y_A \approx y_0 - \frac{1}{2} g t_A^2##