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- Thread starter ryan albery
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Nugatory

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How are you slowing down the traveling spaceship at the end of its journey? I'm assuming that you mean that the space ship accelerates outbound for 2.5 years at 1g, then reverses its thrust so that it is experiencing 1g in the inbound direction and takes 2.5 years to slow to zero speed at the turnaround point then 2.5 years accelerating at 1g back towards earth before reversing thrust again to spend the last 2.5 years of the 10 year journey decelerating at 1g to end up at rest at the end of the journey.

And with that said: The traveling clock will be behind the stay-at-home clock. However, any experiment performed by an observer sitting in either elevator will produce the same result, so there is no way of telling them apart and the equivalency principle is not violated. The discrepancy in the clock readings will only be apparent when the two clocks are compared at the end of the journey.

(BTW, you might want to try calculating just how much energy it would take to accelerate an elevator at 1g for 2.5 years. It's pretty impressive).

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PAllen

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Say the elevator turns around basically by v^2/r=g, such that it doesn't 'feel' a change in direction. A cool historic video about 1 g:

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Vanadium 50

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Say the elevator turns around basically by v^2/r=g, such that it doesn't 'feel' a change in direction.

Impossible.

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V, please explain...

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PAllen

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V, please explain...

Drop an object. It will follow a curved trajectory. Or use a pendulum.

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Not in general, no.

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Nugatory

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Say the elevator turns around basically by v^2/r=g, such that it doesn't 'feel' a change in direction.

You can arrange to make the turnaround and keep the acceleration at a constant 1g in the same direction by doing a tight hairpin orbit around a sufficiently massive object at the turnaround point. However, if you do that, you're still accelerating as you turn back towards earth, and if you keep up the 1g all the way back to the earth you'll be moving seriously fast when you get there, you won't be stopping to shake hands with your stay-at-home twin and compare clock readings.

Of course you can report your clock reading by radio as you go zooming by past the earth on the return leg. The earth-bound observers can receive your message and compare the clock time you report with the time on their earth-bound clock. The time you report will be behind the time on the earth-bound clock.

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Vanadium 50

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V, please explain...

Forget General Relativity.

Forget Special Relativity.

Go back to freshman physics.

You are asking two objects to experience the same acceleration profile, and yet traverse two different paths. That's impossible.

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Forget General Relativity.

Forget Special Relativity.

Go back to freshman physics.

You are asking two objects to experience the same acceleration profile, and yet traverse two different paths. That's impossible.

Not with gravity. Every orbit has the same "acceleration profile", which is no acceleration.

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PAllen

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Not with gravity. Every orbit has the same "acceleration profile", which is no acceleration.

The OP had the object's start at rest colocated. That means Vanadium 50 is correct. When orbits intersect (colocation) they have different velocity vectors.

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The exact details aren't necesssary to say "no, they won't".

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The OP had the object's start at rest colocated. That means Vanadium 50 is correct. When orbits intersect (colocation) they have different velocity vectors.

But not different accelerations. The acceleration is zero for an orbit.

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Nugatory

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But not different accelerations. The acceleration is zero for an orbit.

We may be derailing this thread....

Pervect made the important point in #13: Despite the more than usually complicated trajectories through spacetime, this is just a routine Twin Paradox question.

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