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Quote from: Halc on 21/10/2018 14:59:23Your ship is no different from a building sitting on a planet with a gravitational field identical to the acceleration of the ship. The upper floors accelerate less (you can tell because you weigh less up there),It is very plainly different.
Your ship is no different from a building sitting on a planet with a gravitational field identical to the acceleration of the ship. The upper floors accelerate less (you can tell because you weigh less up there),
I can measure the gravitational field as I go up + down building, and I can work out from those reading how big the planet is.
But on a ship, in space there's no planet nearby.So there's nothing to calculate the change of acceleration with distance.
Fundamentally, you are saying that my ship falls apart as I watch , but no matter how hard I look on my ship, I can find no source of the force that causes it to break up.
The force is the front of the ship pulling too hard, trying to get further ahead of the rear.
Please comment on the contradiction I pointed out in my prior reply to you.
No you cannot.
Einstein says they're identical, except that the building would need to be in a uniform gravitational field
For one, the different parts of the ship are not accelerating at the same rate, else the ship would break apart.
They're your claims.
Quote from: Halc on 21/10/2018 23:48:08They're your claims.Don't be silly. The title of the thread is your claim not mine.
Therefore the ship must be modelled as an array of infintesimal elements, each with its own engine and some means of ensuring that they work together in complete synchronism. Thus the entire ship must accelerate as a single entity. There being no change in length, there can be no relative velocity or acceleration between the front and the back of the ship and thus no change in perceived clock rates between observers on the ship.
So I know that the acceleration of the two ends are the same (and the clocks , which are stationary from my PoV, run at the same rate).
Quote from: Halc on 20/10/2018 14:27:54Clocks forward of a given observer will appear to run faster, and clock behind a given observer will appear to run slower. No, because you have stipulated that they are all accelerating at the same rate. You can't have your cake and eat it!
Clocks forward of a given observer will appear to run faster, and clock behind a given observer will appear to run slower.
If the front and back of the ship are not accelerating at (at least very nearly) the same rate, you are tearing your ship apart.
As the string all speed up all the rulers shorten. All the ships shorten and all the gaps between the ships shorten And they all shrink to exactly the same extent.So the rulers all still fit exactly into the gaps.
The gaps do shorten from someone else's perspective. But those people don't see anything fall apart, they just see the ship shrink slightly along its length
I carefully set the engines to produce the same acceleration.Each section of the ship has the same mass.So, they all are subject to the same forces.All that force (for each section) goes into moving that bit of the shipSo there's none left over to pull my ship apart.
Quote from: Halc on 21/10/2018 14:32:25For one, the different parts of the ship are not accelerating at the same rate, else the ship would break apart.That is obviously incorrect.
There is no relativistic effect within a body subject to uniform acceleration because there is no external comparator for it to be "relative to".
What is left over is the ships pulling apart from each other in their own frames. They remain equally spaced only in the original frame.
OK, that seems to be my list of quotes where one or the other of you seems to claim that the front of a ship accelerates identically (same g force) as the rear.
Your initial condition, that the ship is fragile, demands it. There is nothing to discuss, otherwise.
Relative to the clock in the middle, the clock at the back of the cluster racks up less elapsed proper time.By the same token, relative to the clock in the middle, the clock at the front of the cluster racks up more elapsed proper time.
What’s far more serious is that at the end of the maneuver, the cluster is not the same shape as when it started out! The length between sub-rockets has increased.
This was pointed out by Dewan and Beran (reference 6) and eventually became known as Bell’s Spaceship Paradox.
Both spaceships start accelerating simultaneously and equally as measured in the inertial frame S, thus having the same velocity at all times in S. Therefore, they are all subject to the same Lorentz contraction, so the entire assembly seems to be equally contracted in the S frame with respect to the length at the start. Therefore, at first sight, it might appear that the thread will not break during acceleration.This argument, however, is incorrect as shown by Dewan and Beran and Bell.[1][2] The distance between the spaceships does not undergo Lorentz contraction with respect to the distance at the start, because in S, it is effectively defined to remain the same, due to the equal and simultaneous acceleration of both spaceships in S. It also turns out that the rest length between the two has increased in the frames in which they are momentarily at rest (S′), because the accelerations of the spaceships are not simultaneous here due to relativity of simultaneity. The thread, on the other hand, being a physical object held together by electrostatic forces, maintains the same rest length. Thus, in frame S, it must be Lorentz contracted, which result can also be derived when the electromagnetic fields of bodies in motion are considered. So, calculations made in both frames show that the thread will break; in S′ due to the non-simultaneous acceleration and the increasing distance between the spaceships, and in S due to length contraction of the thread.
So discuss it. I took exactly that premise and drove it to inconsistency in the example in the end. I asked that you find the flaw in the example, else your assertion is worthless.
My method of moving the ship is quite simple. You have a length L that is the distance between the starting point of the tail of the ship and the destination point of the nose of the ship. Accelerate the tail as quickly as possible (instantly?) to whatever speed is required to contract L down to the length of the ship. That brings the nose to its destination (or actually brings the destination to the nose). Now we instantly stop the nose, which springs L back to its original length, bringing the tail to its final destination. We're done. The time it takes to do that is the same as the amount the clocks get out of sync between the nose and the tail.
Can someone help me out here please?I'm meant to be leading this "ship"- it's actually a flotilla of little ships.All the pilots know that they can accelerate their craft at well defined rates and they all move at the same speed WRT the launch pad (which they left long enough ago that its local gravity isn't a factor) I know that I can hold the fleet together simply by making sure all the little bits of my ship accelerate at the same rate.But someone is now saying that , in spite of being deliberately held together, it will fall apart.He refuses to give a mechanism, but my crew are still starting to get jumpy.What should I tell them?Do I tell them to rely on common sense, or do I tell them that magic gremlins are pulling the ship apart?(These are experienced spaceship pilots. they consider relativity to be common sense)
What stops you accelerating it faster than that?
If you can individually accelerate each atom to a tiny fraction below c and get the timing right, all of them can then move at that speed with a delay until the front of the ship is moving too,
With the right kind of launch and catch system, this could be done in such a way that nothing breaks despite the astronomical acceleration force because the arrangement of atoms isn't broken in any way - they are just momentarily the wrong distance apart, but that's put right again at a rate that propagates along the ship at a fraction below the speed of light.
If on the other hand the observed accelerations observed by the pilots are the same, then the readings of the accelerometers are not the same.
What are the pilots reading if not the accelerometers? How do they otherwise decide that they're the same?Second note is, same as what? Identical to the value measured on other ships, or just identical from moment to moment? Nobody seems to be proposing that a particular ship vary its acceleration during the process (except David just now), but it isn't off the table either.An accelerometer on a ship will measure proper acceleration. Not sure what meter the lauchpad guy is reading, but that one will read acceleration in his frame, not the ship's proper acceleration. The former falls off as speed grows, while proper acceleration should be constant for the duration.
Agree to all but the last one. Our pilot would need an accelerometer bolted to either end of his ship, and if he looked at them, they'd read a different value. If the ship is short as most are, they'd not read very different, but it gets quite apparent with longer ships. They're getting shorter in launchpad frame, so the front isn't getting up to the same velocity in that frame. In ship frame, the front clock is running faster, so it takes more time to do the same acceleration. In both frames that spells different reading on the accelerometers at either end of the ship.