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As I understand it, the Bell's spaceship paradox thought experiment says that if you have two identical rockets aligned in the direction of travel, which start accelerating simultaneously (in frame S) with the same acceleration, a taut string tied between them will break because it experiences length contraction in the stationary frame S, and it will break because the distance between the rockets increases in the accelerated frame S′. To prevent the string from breaking, the rear rocket must accelerate harder than the front rocket.

Does this mean that for a single rocket under constant proper acceleration from its engine (in the absence of gravity), the acceleration measured at the front/top of the rocket would be less than the acceleration measured at the rear/bottom of the rocket? In the same way that for a rocket standing stationary on the surface of a planet (with gravity as the only force acting upon it), the acceleration due to gravity would be greater at the base of the rocket (closer to the center of the gravitational field) and lesser at the top of the rocket (farther from the center of the gravitational field)?

Basically, if you are inside a windowless rocket, experiencing a proper acceleration of about 1 g, could you determine, using only an accelerometer (of sufficient precision), if the acceleration was due to being stationary inside a gravitational field or due to a rocket engine by comparing the accelerometer measurements taken at the top and bottom of the rocket?

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The acceleration is actually lower in the front, not just apparently lower. This happens for a simple geometric reason. In a turn in a road, like this one:

you can see that the inner edge of the turn is shorter (fewer stones) but tighter (lower radius of curvature) than the outer edge. Due to time being much like another dimension of space, the same thing happens when you accelerate linearly in spacetime: the acceleration of one end of the spaceship is shorter (lower elapsed proper time) but tighter (higher acceleration) than the other, although the sign of the effect is reversed.

You can't tell the difference between a rocket ship and a gravitational field this way because the same thing happens in a gravitational field. To tell a real gravitational field from acceleration in vacuum, you have to measure spacetime curvature, which is a second-order effect. This acceleration difference is a first-order effect.

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  • $\begingroup$ Doesn't the equivalence principle mean that one cannot distinguish uniform gravitational field from uniform acceleration. I suppose one has to be careful with the word 'uniform'. Can I assume uniform means that a test mass released into freefall locally measures the same acceleration no matter where it is released in the spacecraft? For the gravitational field, an example would be the field given by GM/r^2 as M and r^2 approach infinity together. I'm not sure of an example for acceleration, other than saying the spacecraft is rigid and the dimensions of parts don't depend on position $\endgroup$
    – Roger Wood
    May 27, 2021 at 18:57
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Uniformity is not the problem, because you can just stipulate: A uniform gravitational field. It is a thought experiment, after all. With that, the answer is "no", you cannot distinguish between said gravitational field and uniform acceleration. A measurement of local $g$ anywhere in the rocket will be the same.

This doesn't mean the front and back are going the same speed (in an external frame), since time ticks faster at the front of the rocket, in the rocket's frame. Moreover, when comparing the speed of the front and back, it's important to define a plane of simultaneity. From the launch pad's frame, the rocket Lorentz contracts as it accelerates, so the back has accelerated more than the nose.

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  • $\begingroup$ I want to stipulate a non-uniform gravitational field, like we find in real life. We do seem to agree that the back of the rocket accelerates more than the top of the rocket in the launch pad's frame. Does that also hold true in the rocket's frame? $\endgroup$
    – Lawton
    May 31, 2021 at 16:27
  • $\begingroup$ @Lawton Try to answer this: What is the length of an accordion in accordion's frame when someone is playing the accordion. I mean, as we seem to agree that an accelerating rocket is like an accordion. : ) I mean, try to make your question such that it has some definite meaning. $\endgroup$
    – stuffu
    Jun 1, 2021 at 16:04
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The acceleration you will measure at the top of the rocket will be exactly the same as on the bottom of the rocket. On Earth, the acceleration measure on the top of the rocket will very very smaller than on the bottom. As a consequence, two aligned spaceships accelerated with the same force in the direction of alignment will not drift away from each other (which would make the rope snap).
As seen in the stationary frame, the length of the rope is contracted, but also the rockets are. All lengths are more and more contracted due to the accelerated motions. The lengths of the rockets shrink in time as well as the length of the rope. So all lengths contract in a way that no (extra) tension appears in the rope. Even if only the first rocket was accelerated. The tension in the rope would always be equal to the mass of the second rocket times the acceleration.
In the frame of the rockets time would go faster when moving towards the top of the first rocket. As in a gravitational field. But a mass that falls freely according to a passenger in the rocket will accelerate everywhere the same (namely with the same acceleration of both rockets). There will be no tidal effect driving the two free masses apart (which can be the test to see if the rockets are accelerating through outer space or if they are hanging still in a gravity field (one rocket hanging from the other by means of a rope).

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Does this mean that for a single rocket under constant proper acceleration from its engine (in the absence of gravity), the acceleration measured at the front/top of the rocket would be less than the acceleration measured at the rear/bottom of the rocket?

Yes.

Basically, if you are inside a windowless rocket, experiencing a proper acceleration of about 1 g, could you determine, using only an accelerometer (of sufficient precision), if the acceleration was due to being inside a gravitational field or due to a rocket engine by comparing the accelerometer measurements taken at the top and bottom of the rocket?

No.

Uniform gravity field is such that a lower hovering accelerometer shows a larger reading than an upper hovering accelerometer.

Free falling accelerometers read zero in such field, meaning that there is no tidal force. Absence of tidal forces is the definition of an uniform gravity field.

I mean free falling accelerometer with non-zero height is not stretched in an uniform gravity field.

Oh, I did not give any justifications. Well it's trivial to know what goes on in the accelerating rocket. Then the trivial equivalence principle tells us that the same thing is observed in a such gravity field where tidal forces can be ignored.

A locomotive whose acceleration is $a$, pushes a train whose length is $L$. It's evening and the air is getting cooler, so the train becomes 0.01% shorter each minute because of thermal contraction. What is the acceleration at the front of the train?

Next calculate the acceleration at the front of a Lorentz-contracting rocket the same way. By acceleration I mean coordinate acceleration. At low speeds proper acceleration is the same as coordinate acceleration, so we can get the proper acceleration inside the rocket by this simple calculation, if we choose to make the calculation at low speed.

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  • $\begingroup$ Your statement, "Uniform gravity field is such that a lower hovering accelerometer shows a larger reading than an upper hovering accelerometer" ,seems to contradict the idea that the gravity/acceleration field is uniform (constant) $\endgroup$
    – Roger Wood
    May 28, 2021 at 3:34
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    $\begingroup$ @RogerWood Uniform electric field: free falling formation of identical charged particles retains its shape according to an inertial observer, but not according to the particles themselves. Uniform gravity field: free falling formation of particles retains its shape according to the particles themselves, but not according to an inertial observer. $\endgroup$
    – stuffu
    May 28, 2021 at 20:14
  • $\begingroup$ That's an important statement, thanks. So for a rocket in a uniform gravitational field held stationary with respect to some distant observers, both accelerometers will locally measure the same acceleration, but to a distant observer, the lower accelerometer will register a higher value due to time dilation? That still doesn't sound right? $\endgroup$
    – Roger Wood
    May 29, 2021 at 3:33
  • $\begingroup$ @RogerWood You can't actually deduce anything about accelorometers from what I said. I just said it because it sounds cool, or something : ) Okay so those accelerometers, well they become more sensitive in a lower gravitational potential. And clocks become slower. The former kind of follows from the latter. Inside a speedometer there must be some kind of clock, although it may not be obvious where it is there. Because speed=distance / time. Same is true for force meters. F=p/t. Common kind of accelerometers are just force meters. $\endgroup$
    – stuffu
    May 31, 2021 at 2:46
  • $\begingroup$ @RogerWood An accelerometer consisting of a person holding a string with a weight at the other end of the string says that the force field is uniform, if the person stays in one position while moving the weight. I mean assuming that the gravity field is uniform, which can be defined as such a gravity field in which this kind of accelerometer says that the field is uniform. This accelerometer works the same way in an accelerating rocket too. I mean there too an uniform field is measured. $\endgroup$
    – stuffu
    May 31, 2021 at 3:03

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