# Rocket or elevator?

I realize the gravitational effect in a rocket can be equivalent to the gravity on the surface of Earth. I always hear there is no way to tell the difference between the two but if there were what are the implications?

Einstein's light elevator thought experiment

In this situation light entering the cabin on one side would appear to bend as it moves across the room to the other wall.

For a rocket to maintain the equivalent of one G it would need to accelerate continuously going faster and faster and faster.

Eventually the rocket would be going so fast that the lights spot on the other wall would appear to be moving further and further down the wall as a rocket went faster and faster. Standing in a cabin on the surface of the earth would not have this same moving light spot.

Another way to tell the difference is if you were standing in a cabin on the surface of the earth and you had two plumb bobs. They would both point toward the center of the earth and technically there would be a big difference between the two plumbs if you measured at the top compared to the bottom.

Could these observations make a difference in how we interpret gravity? Bending of space or deflection

• You don't have to construct an elaborate experiment to tell how far you are going relative to something else, Doppler shift will give it away quite easily. You just measure the frequency/wavelength of a monochromatic light source. How this relates to elevators/rockets is not clear to me. Dec 24, 2015 at 18:28
• It's a plumb bob not a plum bob - from the Latin plumbum meaning lead. A plum bob sounds like some form of baked dessert :-) Dec 24, 2015 at 18:58
• Thanks John I know that it's just the way my iPhone interprets it. Dec 24, 2015 at 19:01
• @JohnRennie: Google Books yields numerous references to the "Plumb-Pudding Model" of the atom... Dec 25, 2015 at 2:53
• physics.stackexchange.com/questions/277688/…
– Muze
Sep 2, 2016 at 15:42

• @BillAlsept: The spot does not move down the wall in the accelerating lift. If the light ray takes a time $t$ to cross the lift ($t = w/c$, where $w$ is the lift width) then the displacement of the spot on the far wall is $\tfrac{1}{2}at^2$, where $a$ is the acceleration of the lift. The spot displacement is not a function of the lift velocity. Dec 24, 2015 at 19:35