Thought Expirement, Tethered Spacecraft travelling near light speed

[Mr.Bigg]

I have two spacecraft originating from the same point each at velocity .9c in opposite directions. One spacecraft has a chord attached to it rigidly. The other has a spindle attached that can freely unwind the chord with no friction. We can measure the speed of the chord unwinding on this spacecraft . I allow the spacecraft s to travel for some unit of time x (maybe 1 year). Then after they have each traveled for x time they very quickly (say .5 s), come to a stop and have no relative velocity.

Measured from their starting point, they will both have traveled many light years of distance and their separation to each other will be many light years. If the chord is still attached, and the ships are at rest, many light years of chord will be stretched across space, yet it will have been impossible for the people inside spacecraft 2 to have dispensed that amount of chord in the allotted time as they would not be able to achieve a mass flow rate of the chord inside the ship that dispenses faster than the speed of light.

Blow my mind physics ninjas.

 

[JesseM]

yet it will have been impossible for the people inside spacecraft 2 to have dispensed that amount of chord in the allotted time as they would not be able to achieve a mass flow rate of the chord inside the ship that dispenses faster than the speed of light.

There would be no need for the rate at which the cord was dispensed to be FTL, see the relativistic velocity addition formula. If after being dispensed the cord is at rest relative to the ship it's tied to, then in the frame of the ship with the spindle that's dispensing it, a point on the cord will be moving away at (0.9c + 0.9c)/(1 + 0.9*0.9) = 1.8c/1.81 = 0.994475c immediately after being dispensed.

 

[Mr.Bigg]

There would be no need for the rate at which the cord was dispensed to be FTL, see the relativistic velocity addition formula. If after being dispensed the cord is at rest relative to the ship it's tied to, then in the frame of the ship with the spindle that's dispensing it, a point on the cord will be moving away at (0.9c + 0.9c)/(1 + 0.9*0.9) = 1.8c/1.81 = 0.994475c immediately after being dispensed.

Right, the relative velocity is that on ship 2 (the one with the spindle) yes? But... if I have dispensed chord at a rate of .99447c, then my spindle has released .99447c*x length of chord yes? (x being the the time measured on the ship). Maybe I am measuring the weight of my spindle as it dispenses and am using that to verify velocity through mass flow rate. When I get out and measure, however, my amount of chord between the two is many times that.

Thinking about it, I suppose any each chord position would be distributed equally across the range. But the then the chord would be forced to stretch thinner even approaching infinite stiffness in the chord, meaning approaching infinite force...?

 

[JesseM]

Right, the relative velocity is that on ship 2 (the one with the spindle) yes? But... if I have dispensed chord at a rate of .99447c, then my spindle has released .99447c*x length of chord yes? (x being the the time measured on the ship)

I guess it kind of depends if the speed of bits of the cord is changing during the dispensing process (then you have to take length contraction into account), but if we imagine a dot drawn on the cord would first be moving in a circle at 0.99447c as the spindle was spinning, then is "released" from the spindle on a tangent to its former circular motion at the same speed, in that case you're right that the amount dispensed would be 0.99447c * t (btw I changed the variable because the normal convention is to use some form of t for time rather than x, since x is often used for distance along the x-axis). For example if as the cord is spinning on the spindle we see that dots are drawn on it at regular intervals, and the distance between successive dots is 1 light second as the cord rotates at 0.99447c (i.e. the length would be larger in the cord's rest frame), then after 1 year=31536000 seconds, we'll see 0.99447*3.1536*10^7=3.1362*10^7 dots have been dispensed, so 3.1362*10^7 light-seconds worth of cord if we're using the contracted length of the cord as it rotates.

Maybe I am measuring the weight of my spindle as it dispenses and am using that to verify velocity through mass flow rate. When I get out and measure, however, my amount of chord between the two is many times that.

Well, the number of dots that have been released at any given instant doesn't depend on your choice of frame, but the length of the cord does, for example if the distance between dots is 1 light-second in the frame where it's moving at 0.99447c, then in the frame where the cord is at rest the distance between dots is 9.5219 light-seconds, so the length of cord that's been released in this frame is 9.5219 times greater. But keep in mind that distance between atoms in the cord also changes in different frames, atoms are like the dots painted on the cord in that all frames agree how many atoms' worth of cord has been released at any given point on the journey of the ship with the spindle.

Thinking about it, I suppose any each chord position would be distributed equally across the range. But the then the chord would be forced to stretch thinner even approaching infinite stiffness in the chord, meaning approaching infinite force...?

I don't quite follow what you're saying here, but if it helps you can just think of the fact that the average distance between atoms in the cord when it's relaxed (and we can assume it's relaxed after being dispensed since it's not experiencing any acceleration) should be fixed in the cord's rest frame, while other frames moving relative to the cord see the atoms being more compressed, not more stretched. But "compression" due to length contraction is not like "compression" due to forces which causes stresses, fractures, etc., if there are no physical issues like fracturing and breaking in the rest frame then there won't be in other frames either.

 

[Dale]

I have two spacecraft originating from the same point each at velocity .9c in opposite directions. One spacecraft has a chord attached to it rigidly. The other has a spindle attached that can freely unwind the chord with no friction. We can measure the speed of the chord unwinding on this spacecraft . I allow the spacecraft s to travel for some unit of time x (maybe 1 year). Then after they have each traveled for x time they very quickly (say .5 s), come to a stop and have no relative velocity.

Measured from their starting point, they will both have traveled many light years of distance and their separation to each other will be many light years. If the chord is still attached, and the ships are at rest, many light years of chord will be stretched across space, yet it will have been impossible for the people inside spacecraft 2 to have dispensed that amount of chord in the allotted time as they would not be able to achieve a mass flow rate of the chord inside the ship that dispenses faster than the speed of light.

Blow my mind physics ninjas.

This is easiest to consider in the frame of the ship where the cord is attached rigidly. In this frame the cord is at rest, and it is simply being unspooled at .99c. In that frame the spindle is undergoing relativistic rolling, which is a rather complicated motion, but completely self consistent with different degrees of length contraction on the top and bottom. I believe that if you search for relativistic bike tire you will find some old threads on the topic.

 

[russ_watters]

Do two planes flying in opposite directions at 500mph create a sonic boom...? (speed of sound is about 756 mph)

 

[Mr.Bigg]

This is easiest to consider in the frame of the ship where the cord is attached rigidly. In this frame the cord is at rest, and it is simply being unspooled at .99c. In that frame the spindle is undergoing relativistic rolling, which is a rather complicated motion, but completely self consistent with different degrees of length contraction on the top and bottom. I believe that if you search for relativistic bike tire you will find some old threads on the topic.

The spindle wasn't really my concern. I was just trying to describe a way to freely release chord that would instantaneously be at the relative velocity of the two ships. I was trying to reconcile the fact that after the two ships come to a complete stop and one got walked the chord you would have 2 miles of chord that weighed 100 kg (absurdly hyperbolic simplification), whereas the ship that was dispensing the chord only saw 1 mile of chord leave the spindle and it only weighed 50 kg. Then I was wondering if the chord would be forced to be 2 miles in total length at rest, but weight only 50 kg. ie stretched.

 

[Mr.Bigg]

Well, the number of dots that have been released at any given instant doesn't depend on your choice of frame, but the length of the cord does, for example if the distance between dots is 1 light-second in the frame where it's moving at 0.99447c, then in the frame where the cord is at rest the distance between dots is 9.5219 light-seconds, so the length of cord that's been released in this frame is 9.5219 times greater. But keep in mind that distance between atoms in the cord also changes in different frames, atoms are like the dots painted on the cord in that all frames agree how many atoms' worth of cord has been released at any given point on the journey of the ship with the spindle.

Ok so if I follow what you are saying, in ship two I observe the chord dispensing at .99447c, but I also observe many times the number of dots pass by than I would expect to at rest or in low speed. IE at .1c I would observed 1 dot (not correct with your numbers but I don't feel like doing math), per second passing. But at .9c I would observe many more than 9 dots per second passing? By the same token I dispense 1 kg of mass per second at .1c but at .9c I dispense many more than 9 kg per second?

 

[JesseM]

Ok so if I follow what you are saying, in ship two I observe the chord dispensing at .99447c, but I also observe many times the number of dots pass by than I would expect to at rest or in low speed. IE at .1c I would observed 1 dot (not correct with your numbers but I don't feel like doing math), per second passing. But at .9c I would observe many more than 9 dots per second passing? By the same token I dispense 1 kg of mass per second at .1c but at .9c I dispense many more than 9 kg per second?

Well, only if what you expected was based on the distance between dots when the cord is at rest, and you hadn't thought about the issue of length contraction...if you had never seen the cord moving at any speed other than 0.99447c, then if the distance between dots in your frame is 1 light-second, it makes perfect sense that the time between dispensing successive dots would be (1 light-second)/(0.99447c) = 1.0056 seconds, this would be just as true in Newtonian physics as in relativity (BTW, just a little spelling nit, should be cord not chord).

 

[Meir Achuz]

Do two planes flying in opposite directions at 500mph create a sonic boom...? (speed of sound is about 756 mph)

Not unless they collide.

 

[Dale]

The spindle wasn't really my concern. I was just trying to describe a way to freely release chord that would instantaneously be at the relative velocity of the two ships. I was trying to reconcile the fact that after the two ships come to a complete stop and one got walked the chord you would have 2 miles of chord that weighed 100 kg (absurdly hyperbolic simplification), whereas the ship that was dispensing the chord only saw 1 mile of chord leave the spindle and it only weighed 50 kg. Then I was wondering if the chord would be forced to be 2 miles in total length at rest, but weight only 50 kg. ie stretched.

But the spindle is fairly important in this. Let's assume that the spindle is revolving fast enough to lay the cable out without tension. In that case the spindle and the cable is length contracted. So you may lay one mile of length-contracted cord, but then when you accelerate to be at rest wrt the cord you will find that it is two miles in length. This is standard length contraction applied to the spindle and the cord.

 

[pervect]

Any physical material will have a finite stiffness - otherwise the speed of sound through the material will be infinite. So physically, any such rope on a spool will be stretched because of the centrifugal force.

You can avoid this by having the rope lying on the floor, but of course that spoils the symmetry of the problem.

There's a rather detailed analysis of an elastic relativistic rotating hoop at http://www.gregegan.net/SCIENCE/Rings/Rings.html - the topic gets tricky quickly.