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Light falls in a gravitational field with the same acceleration as cannonballs. Does this imply that, in a gravitation-free space, the speed of light (as measured by the observer) varies with the speed of the observer as predicted by Newton's emission theory of light, that is, in accordance with the equation c'=c+v?
ANSWER: The speed of cannonballs shot downwards with initial speed V (relative to the shooter) varies with the gravitational potential (gh) in accordance with the equation V'=V(1+gh/V^2) (it is assumed that V>>(V'-V) and air friction is ignored). If the cannonball is shot from top to bottom in an elevator of height h accelerating, in gravitation-free space, with constant acceleration g, then the bottom has acquired speed v=gh/V when it meets the cannonball. Accordingly, the speed of the cannonball as measured at the bottom is V'=V(1+gh/V^2)=V+v.
If, in a gravitational field, the speed of photons varies exactly as the speed of cannonballs does, then the speed of a light signal emitted downwards with initial speed c (relative to the emitter) varies with the gravitational potential (gh) in accordance with the equation c'=c(1+gh/c^2). If the signal is emitted from top to bottom in an elevator of height h accelerating, in gravitation-free space, with constant acceleration g, then the bottom has acquired speed v=gh/c when it meets the signal. Accordingly, the speed of the signal as measured at the bottom is c'=c(1+gh/c^2)=c+v.
The equation c'=c+v is fatal for Einstein's relativity. In the context of the above argument, its truth entirely depends on the PREMISE:
"Light falls in a gravitational field with the same acceleration as cannonballs"
If the PREMISE is true, the equation c'=c+v is true. If not, not.
Pentcho Valev pvalev@yahoo.com