Physics, Astronomy & Cosmology / Re: Bell's string paradox, but string doesn't break if constant thrust instead?« on: 16/09/2022 01:59:50 »
As I stated, I assume the string to have a negligible mass. In the Python program, I used real steel cable technical data, with 1 mm diameter and a break tension of 840 N. It's a very lightweight string.Not that lightweight. At 10 km, the tension needed to accelerate it at 1 g is already over half the rated load on it. Drag 20 km of it with nothing tied to the far end and it will break. But if it's long enough, no load rating is enough to prevent it from breaking. Hence my suggestion for the string to have independent thrust applied to it along its length, but that's a different scenario than the realistic one you're going for.
Is the 50kg force the string puts on S2 subtracted from the force being applied to S2, or is the force on S2 higher than the force on S1 to compensate for the fact that it's dragging 50 kg of string?
And yes, the string initially extends from leading ship S2 to trailing spaceship S1, and is being pulled by S2 with a tension force enough to accelerate the mass of the string, and it unwinds because there's a relativistic effect that makes the relative distance between the spaceships to increase with time, and that translates into an "unwinding force". The distance from spaceships will increase, but the string adapts to the required length.Right. I got that from the last post.
To consider the mass of such string, and the stretching due to that string mass will add nothing to the solution.If you use pre-stressed string, the stretching of it doesn't come into play during the reeling-out phase. Simplifies the simulation of it. Either way, the string exerts force on S2, and that very much does alter the solution.
The effects are going to be hard to see if the acceleration and separation are both so low. It’s almost a Newtonian exercise at those numbers. I tend to go for high g and big separations to make the effects obvious.
As I said previously, the relative speed at 24 hour is 1,66 mm/hour ... You can easily measure that just with a wrist watch, a normal ruler, an ink marker and some patience.[/quote]From a spaceship putting out a g for a whole day? No, I don't think you'd get the thrust precision accurate enough to being able to measure a couple mm per hour.
Put S2 out at Mu Pegasi (a star just over 100 LY away) and as measured by S1, after 24 hours, S2 will be reeling out that string at a rate higher than c. Of course your steel wire won't like that, being limited to about 160 km. But it's a thought experiment. Like I said, I like to make my scenario larger to see the effects more clearly.
In the Bell's original scenario, the Earth observer always sees the distance between the spaceships to be a constant, so being both spaceships identical it would deduce that the CM is always located halfway between.Sure, but only in that frame. The CM is elsewhere in a different frame, including the frame of a third ship in the middle that also has that identical constant proper acceleration. In the frame of that middle ship, the rear ship will recede slower than the lead ship. The center of mass of the system would be elsewhere. It seems wrong to measure the location of the CoM of a system in a frame other than the frame of the CoM of the system. The CoM of a rotating symmetrical wheel for instance is off-center in all other frames, but is centered on the axis in the frame of the axis.
In the constant proper acceleration scenario, or the constant thrust scenario, I would say that the CM is located such it observes the same outwards simultaneous relative speed for both spaceshipsAgree, but that's not where the Earth guy puts it, which is why I pushed back on using that somewhat arbitrary frame.
I'm pretty sure that if its not centered, the offset is not more than a tenth of a milliliter or so - compared with the length of 10,000 meters of the string, the offset is negligibleOf course it's negligible. Make the string longer or the g forces higher and it won't be negligible. Like I said, the effects become obvious if you scale it up, otherwise you might as well just use Newtonian physics, which was fine for putting guys on the moon. Nothing ever move fast or far enough to matter.
That's right, the whole exercise is to see if the string breaks or not.OK, I guess my 100 LY string is unrealistic then. Pure hypoThe exercise isn't about relativity, it's about the ability to drag a long string. It breaks at about 170 km at 1 g without anything attached to the other end. 500 N to drag it, leaving around 300 N for the 'gentle braking'. If the ships are massive enough, that 300N will not be enough to halt the increasing separation, or cancel this apparent repulsive force between them.
My coordinate system starts where I want - this is a democratic forum -, and I can arbitrarily set ...[/quote]No, the choice of origin can't be arbitrary, else the line passing the arbitrary origin and the corresponding event-point in the CM world-line at time tc will not correspond to the line of simultaneous events for the CM frame. It would be an arbitrary line, not one specific to the CM frame.To get the corresponding simultaneous proper time in the leading S2 spaceship, we know that the line of simultaneous events for the CM is given by a line that passes by the (0,0) origin of the space-time diagram and the corresponding event-point in the CM world-line at time tc.This seems wrong. The origin of your spacetime diagram seems arbitrarily assigned along the x axis.
Again, I don't understand... which is the "rigid scenario"? When the string can freely unwind from the spool, the tension in the string is due to: 1) the inertia of the mass of the string 2) the relative speed between ends, which appears as a relativistic and counter-intuitive effect.If the (ideal) string is being reeled out with only enough tension to accelerate the string to a speed matching S1 at the rear, then the entire string (except the part still on the reel) exhibits rigid motion, and one can apply rigid rules to anything stationary relative to the string, which neither S2 nor the CoM is.
The tension due to the inertial mass is negligible as stated.Over half the breaking tension is negligible?
The force diagram describes what happens when both ends of the string are fixed. The starting time=0 of the "simulation" is when this event happens, arbitrarily 24 hours after departure (or whatever value you want to start with). This lets us play with an initial condition for the relative velocity between the CM and the S2 spaceship.OK, so not gentle braking. You clamp on and let the string take up the strain until the two speeds are matched. Then it 'bounces' and for a bit the rear of the string goes slack.
Not if all these interactions are completely elastic, no. It will bounce forever without somewhere for the heat to dissipate.
And this is the result I get if the spool brakes at proper time 30 days after departure:Oh... No slack time. There's always negative force (pulling S2 back) exerted by the string except that wee hump you show in the closeup. What is that? Why would the string push the ship away for a brief moment when it clamps onto the string?
Does the positive hump also exist at each peak like the one at time 5¼?
What are the time units in the pictures? Surely not all the same.
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