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Could two spaceships which had initially accelerated to 200mph away from each other then coasted except to turn in a wide arc to be in a reference frame at rest with each other know they were moving. Maybe by turning away every so often but not adding any forward accelleration, to examine if their separating velocity was still about 200mph. I'm wondering can two spaceships know they are moving without any extra reference.

What is meant by extra reference frame...
As motion is subjected to acceleration cricular or straight the Laws of GR gives a different treatment.
The question is not yet clear to me.

Spaceships do not have any gravity "on board". When moving with constant velocities it is not possible to say one is moving and the other is not. This is the most basic principle of relativity which has already beem framed by Galileo, and mathematically formalised by Newton when he postulated his three laws of motion. This is still valid today. When a spaceship accelerates, one will experience a force on every item on the spaceship, and one will then know that the spaceship is accelerating. This force will also manifest when you follow a circular orbit with the spaceship; acceleration will then be directed to the centre of the orbit. If you do not know you are on a spceship, you will conclude that the force is a gravitational force (Einstein's principle of equivalence).

If a person lived inside an expanding balloon spaceship could they then say they were moving. If there was anything outside the spaceship it would eventually reach it. This probably sounds more like a joke and I'm sure of no practical value. Since you cannot conduct any experiment inside a spaceship to see if you're moving I persume this cheats by expanding.

You have an interesting point there. Our universe is like the expanding spaceship; however if the spaceship expands without any perceptible acceleration of its centre of mass, one could argue that the spaceship is stationary, or moving with a constant velocity. Expansion, however, still means that there are parts of the spaceship moving relative to the centre of mass, and special relativity should apply between these moving parts provided they are moving apart at constant speeds.

Thanks for the reply. I was musing on the twin paradox and wondered what was the difference for the twin who traveled. It seemed to me the twin who travelled having to return to the same point in space as the twin who didn't was part of the difference, but then no point in space is unique.
Three rockets: A B C. C rocket does nothing, A and B accelerate off and having reached four-fifths the speed of light, B turns back to come to rest at C rocket... while A continues away. A eventually turns as B did, and comes back after 10 years according to B's and C's clock but only 4.5 years is registered on A's clock. A and B had accelered and decelerated the same, A stayed at a high speed longer but also had to return to B's position. So it's speed relative to a fixed position that makes the twin younger. I think.

Something caught my eye in Dogrock's post. It seems that account of the acceleration time of the rockets is not properly accounted for.

In Special Relativity the coordinate frames are not accelerating with respect to one another. In other words, they look inertial to one another and there is a symmetry. An observer in each frame can treat the other as moving with self holding still. Each will see the other as distorted: mass, length, time, etc.

The resolution to the twin paradox is to realize that one twin, the traveller, has to accelerate with respect to the other and so is not in an inertial frame. After the travelling twin has gotten up to speed all goes according to Special Relativity. To turn around for the return voyage the traveller must de-accelerate and then accelerate toward home. On arriving home the traveller must, again de-accelerate. It is in these accelerations that the "time slipping" occurs.

General relativity can account for accelerating reference frames, but special relativity can deal with the sort of thing that is going on in the twin paradox. Briefly, the idea is to use a collection of moving reference frames each moving a little faster than the other. One then imagines the travelling twin as occupying these frames succesively during the acceleration.

Hope I have understood the topic the way you meant it. (I'm new here.)

What if the travelling twin's rocket merely completes a big loop, rather than slowing down and reversing? Nowhere in the setup does it say the rocket he/she is on has to stop somewhere and turn around.

Just a thought.

dr-rocket: you have the gist of the situation. In special relativity it only seems to the one twin as if the other's clock is running slower, just as when the two twins pass each other in two trains and each assumes his train is stationary; each twin will be under the impression that it is the other train that is moving. The time rates each perceives within the other's reference frames are not "real" differences, but only seem like it as long as they are moving relative to each other. If you could have generated movement and return the travelling twin without any acceleration, they will both be the same age; however, acceleration is required and therefore the twin that experienced the most acceleration will be the youngest.

Amaranth Rose: The rocket that completes a loop still accelerates towards the centre of the loop. Thus it still experiences an effective gravity, and its clock will thus really (not apparantly) tick slower. The twin completing the loop will be the youngest.

Dogrock,
You attempted to create a 'privileged observer`, by leaving 'C` stationary,
and then assumed that the clocks of 'A` & 'C` would agree.
This is not so. 'A` spent some time at relativistic velocity. Direction doesn't count.
Do all the calculations, not just some.
'C` is "oldest", then 'A`, then 'B`, when they are re-united.
Hint: when reunited their 'space/time` distances will be equal.
Their 'time` distances and their 'space` distances will not.
Pragmatist.

C is oldest but there's little difference because B might have turned back after an hour (or a day), so there can't be more than an hour or a day's difference in the C and B clock. The real difference is in A who is years younger and has accelerated, turned and decelerated the exact same as B. The C rocket isn't needed at all but I wanted to remove gravity and the earth from any consideration. I now realise it's the limited speed of light causes the paradox. If the A rocket stopped after five years and B then rushed towards A at four-fifths light speed, it's B would be younger. It's the return journey is the problem. Who ever is moving back towards the other will see the clock going faster for longer. The one who's stationary will see four more years of the outgoing journey and only one year of the return, and because light speed is limited, one year isn't enough to convey five years of travel at four-fifths light speed. So you might say the rocket is back before they know it, and younger. But I'm sure there are a few more twists and turns to be examined.

The whole basis of Einstein's special theory of relativity is that nobody is "stationary"; i.e. there is no experiment possible which can determine a "stationary" reference frame. So if you want to argue, do not use "the one who is stationary". If you want to do that, then join the "stationary flat-earth" society!