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Closing speed and length contraction paradoxes 
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#1
Mar1112, 01:36 AM

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A google search revealed an often asked question (including a locked thread here) about two ships approaching with closing speed > c, say each going .75c. And there's always multiple answers from the experts about relative frames and Lorentz this and that and you have to use the special formula etc.
I get that the spaceship measures the speed of the other spaceship at some speed less than c given by whatever that formula is, I won't argue that (well I don't TRULY get it, but I won't attempt to refute it). My question though, is what if the spaceships are rigged with superpowerful airbags and other safety devices such that the collision at the speed they measure the incoming spaceship at gives it a kinetic energy below the threshold of immediate death? What about the earth observer that sees them crashing with a KE = 1/2 * m * .75c^2 per ship? The earth guy knows they are going too fast for the safety equipment and correctly predicts they die... But on the ship they correctly predict the equipment will save them? Well they either die or don't as if they actually didn't they would fly back and talk to the guy on earth and call him crazy... All I can think of is that the KE is increased by the m factor... apparently the incoming ship is not merely going faster but it has somehow e.g. quadrupled in size or density? The second sort of paradox I have is a variation on the barn and pole paradox: http://math.ucr.edu/home/baez/physic...barn_pole.html What happens if the barn has the doors removed and is rigged with a laser "trip wire" in each open doorway, such that if the trip wires are both cut, a bomb goes off? Let's presume that the contraction makes the pole just barely fit from the stationary observer's perspective. The stationary observer thus watches the runner safely pass through the barn. The runner unfortunately watches the pole pass through the rear laser, breaking line of sight with the detector at least until the front of the pole breaks LOS on the second laser and detector, so both lasers are cut and the bomb goes off. The paradox in the link talks about how the doors didn't close and open at the same time from the runner's point of view, which I can accept, and therefore the runner never smashed into one of them. But what happens in this case where the "doors" (trip wires) are required to remain "closed" (not cut)? In the original version, the front door closes to the runner's view, the pole gets to it, it opens, the pole passes far enough through that the rear door can then barely close, then the rear door opens. But the pole did actually pass through a very small barn, even if the when of the doors closing is disagreed upon. So just change the *temporary* door closing idea with a *permanent* requirement to not have two broken lasers at ANY POINT... The runner breaks both lasers at some point from their point of view, even though individually they broke at different times and at different times from when the guy on the roof saw them break (and he didn't even know they were both broken... until the bomb goes off). 


#2
Mar1112, 07:32 AM

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Hi Seanner! Welcome to PF!
The observers in both the earth and the spaceship will do their calculations on the basis of hitting a wall at the position and velocity of the centre of mass. 


#3
Mar1112, 10:12 AM

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#4
Mar1112, 10:16 AM

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Closing speed and length contraction paradoxes



#5
Mar1112, 10:33 AM

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#6
Mar1112, 11:08 AM

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to be honest, i was being even more ambitious than adding the mass of the Earth … i was assuming my wall had infinite mass! 


#7
Mar1112, 11:14 AM

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I’m going to take a few liberties, and modify your proposal for ease of analysis.
Here’s a bunch of things I’ll suppose: 1) The barn is 10m long in its rest frame, and the pole is 20m long in its rest frame. 2) The relative velocity is √3/2 c. To observers in the barn frame, the pole is Lorentz contracted to 10m – just enough to fit in the barn. 3) There is a series of clocks along the pole and along the inside of the barn. All the clocks on the pole are synchronised with each other (in the pole’s rest frame), and all the barn clocks are synchronised in the barn frame. 4) The bomb is replaced by a series of firecrackers, strung along the length of the barn. These firecrackers are triggered by readings on clocks immediately next to them. We don’t want anyone getting hurt. 5) The experiment has been trailled, so we know in advance various timings, readings on clocks etc. In the trial, the front end of the pole enters the front near end of the barn when clocks on each read zero. In the barn frame, the front of the pole reaches the rear of the barn at t = 10/(√3/2 c) = 20/√3c. So, knowing this information and in preparation for the rerun, the barn owner resets all his clocks, and arranges for the firecrackers to go off simultaneously in his frame, at a preset time t = 20/√3c. He should then see, in the rerun, all the firecrackers go off simultaneously when the pole is entirely contained in the barn. And he will. Now from the frame of the pole carrier: it’s the barn that is approaching her at a velocity √3/2 c, and the barn is Lorentz contracted down to 5m. In the rerun, she sees, again, the front of the pole enters the barn when clocks at the front of the pole and near end of the barn both read zero. But she also notices that the clocks along the barn wall aren’t synchronised – the far ones are running ahead of the one at the near end. Nevertheless, as the barn continues to move along her pole (as she sees it), the front of the pole reaches the far end of the barn at a time 5/(√3/2 c) = 10/√3c. This is the reading on her clock at the front of her pole. But the far end barn clock reading for this event, you’ll recall, 20/√3c. Nevertheless, the firecracker at that position goes off. Their clock readings might disagree, but there’s no doubting the event happens. But all the firecrackers go off only when the barn clock next to them reads 20/√3 c – and since, to her, they aren’t synchronised with each other, the other ones in the barn haven’t reached this reading yet. However, they inexorably will – one after another. As the process continues, one firecracker after another, in succession (working from the one at the far end of the barn to the one at the near end) will go off, as each barn clock next to it eventually reaches t = 20/√3c. Finally, as the front end of the barn reaches the rear end of her pole, the firecracker there goes off. Both sets of observers can agree that every part of the pole witnessed a firecracker going off alongside it. If we wished the firecrackers to be powerful enough to destroy anything next to them, both sets of observers would agree that the pole had been reduced to ashes (assuming the pole carrier is wearing a bombproof suit!). It’s just that what appeared to be a simultaneous destruction of the pole to the barn owner appeared, instead, to the pole carrier to be a gradual destruction, starting at the front of the pole and working backwards until the back of it entered the barn. 


#8
Mar1112, 04:26 PM

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Thanks for the answers. For the first part, I don't get the bit about two things with opposing velocities colliding with the same energy as one of them colliding with a stationary wall. That would violate conservation of energy. That is assigning a KE of 0 to the other car. If your point is merely that the other car's apparent closing velocity relative to the first car is what determines the energy and ultimately force experienced, then okay. This would imply that the observer on earth gets the wrong answer, unless he uses a calculation to adjust for what one ship sees the other ship's speed as, which he can do, but it's unfortunate that this is not what he is literally observing with his eyes. He intuitively knows the ships are going too fast and won't survive, but relativity "kicks in" and the ships aren't aware that they are supposed to be going faster. Good news for them.
For the second part, I have another question about EXACTLY what happens in the runner's frame with my bomb setup. Let us first suppose that the laser trips use fiber optics to transmit their arming status to the centralized bomb. Here is what I see happening from the runner's view: 1. front of pole breaks first laser beam...signal sent at light speed to bomb...bomb is half armed 2. front of pole breaks second beam...signal sent to bomb...bomb is armed and detonated The answer as you all would put it, based on your replies, is that the breaking of the second beam happens so far into the future that by the time the rest frame bomb actually gets the signal, the rear laser is reestablished. That would be mixing frames though. The laser is NOT reestablished in the runner's frame. The runner is supposed to be staring at both lasers being broken. If the runner had exceptional vision, they could even watch the breaking of the beams, followed by the light signal in the fiber optic traveling to the bomb, watch it process the detonate command, watch the actual detonator fire, etc... There's a lot of cause and effect here and the runner should not get any weird surprises... The above implies, since the official answer given is that they were not armed simultaneously, somehow the signal from the front detector takes longer to reach the bomb than the time that the back detector is armed for, i.e. with the runner's supervision they watch a very slow photon in the fiber optic cable head towards the bomb, and by the time the photon says to fully arm the bomb, the rear arming device already switched back to safe. So things in front of the runner happen in slow motion. 


#9
Mar1112, 04:45 PM

P: 450

have an increased relativistic mass right? At least when they observe each other. If you observe them from earth, and you see that at the speed they are going to collide they should die without taking into account the relative velocity they have between each other, then your still not going to be accounting for their relativistic mass? Otherwise, the two observers heading for collision see not only that they are going slower than the earth observer thinks, but also that they are more massive. So even though they collide at a velocity which is safe not accounting for relativistic mass, the addition of relativistic mass means that the force is still the same either way? 


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Mar1112, 05:15 PM

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#11
Mar1112, 05:22 PM

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#12
Mar1112, 05:27 PM

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#13
Mar1112, 06:06 PM

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My answer was based on: if you want to do it in the frame of one of the ships, you must first compute the relative velocity using the velocity additions formula: (u+v)/(1+ uv/c^2). Then, with this relative velocity, you compute the KE using the correct formula I gave earlier  the only one that is true at all speeds. You also must account for the fact that, in this frame, the total momentum before collision is that of the the other rocket, and this total momentum must be conserved after collision. For this, you must use the correct momentum formula, which is: mv/sqrt(1v^2/c^2). If you do all of this correctly, you find that everything works out the same as in the COM frame  only it is a lot more computation. 


#14
Mar1112, 09:27 PM

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First, I assume it is clear that the barn observer never sees or detects, by any means, both beams interrupted at the same time as he perceives it. The runner, equally clearly, sees both beams interrupted at the same time (as runner perceives it) for a short time. Let's organize this into events: E1  front of pole crosses beam 1; arm1 signal sent to detonator at center of barn (slightly displaced from pole path. E2  back of pole reopens beam 1; disarm1 signal sent to detonator. E3  front of pole interrupts beam 2; arm2 signal sent. E4  back of pole passes, reopens beam 2; disarm2 signal sent. (detonator doesn't fire, because the first arm signal is canceled before the second is received). Let us also specify some events at the detonator: R1  arm1 signal received R2  disarm1 signal received R3  arm2 signal received R4  disarm2 signal received We need not specify the relation of the R events and the E events; that is, it does not matter how fast we assume the signals are traveling, as long as all are going the same speed per barn observer, and the detonator is centered (signal paths same length). The above describes the order of E events for the barn observer. The R events are observed to be in the same order by both runner and barn observer  they have to be, because that are all events on 'history' of one object  the detonator receiver. For the Runner, the E events occur in the order E1, E3, E2, E4. So does this imply the runner sees asymmetry in transmission speed of signal? Yes definitely. This follows directly from the velocity addition formula you indicate you have trouble accepting, but it is fundamental. In Galilean relativity, if v is the speed of the barn (per the runner), and u is the signal speed per the barn observer, then for the runner, the signals would propagate at simply v+u and vu, converging symmetrically on the detonator moving at v. However, the correct velocity addition formula (for all speeds  just that it only differs noticeably from Galilean at high speed) gives: (v+u)/(1 + uv/c^2) and (vu)/(1  uv/c^2) Note that these speeds are, indeed, asymmetric relative to detonator speed of v per the runner. This speed difference allows for the runner to agree on the order of the R events while disagreeing on the order of the E events. 


#15
Mar1212, 02:17 AM

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Suppose, there is pulse generator in middle of the barn. Barn observer sees the pulse reaches to both ends simultaneously. But, pole observer doesn't see simultaneously. What would be the order in pole's frame? Pole observer sees barn is coming towards him. So, light signal fired from middle of the barn reaches to front (right of barn observer) ends first, and then it reaches to rear (left of barn observer) end. Now, suppose the light signals can close the doors. So in barn frame door is closed simultaneously. And in pole frame front door would be closed first, then rear door would be closed. Which matches with situation that when front of the pole reaches to the front end of the barn. The door would be closed. Still rear end of the pole is out of rear end of barn. Now, front door opened and pole come out from front end, and pole's rear end comes inside the rear end of the barn. Now rear door would be closed. Now, suppose light signal fired and reaches at front end at [itex]t_f[/itex] in pole frame. And light signal reaches at rear end at [itex]t_r[/itex]. This time gap created in pole's frame. Now, barn is contracted in frame of pole. Pole's front end have to come out from front end of barn in [itex]\Delta t = t_r  t_f[/itex]. Pole's access length [itex]\Delta x[/itex] relative to barn which have to come out from barn in [itex]\Delta t[/itex] time. So, pole's observer conclude that front door closed/opened at [itex]t_f[/itex], [itex]\Delta x[/itex] of pole come out of front end at [itex]t_r[/itex] and rear door closed/opened. Now, we can come to your situation. Here light signal takes more time to go from front end to middle, and less time to go from rear end to middle. Now, pole's front end cuts line of sight of rear end. Then pole going to front end. Now pole's front end cuts LOS of front end of barn. But, the signal takes long to reach to middle. Suppose, it takes [itex]\Delta t_f[/itex]. Rear end's signal takes [itex]\Delta t_r[/itex]. And, we can conclude that the time gap should be [itex]\Delta t = \Delta t_f  \Delta t_r[/itex]. In this time gap pole's [itex]\Delta x[/itex] part should be come out of barn. Here [itex]\Delta t < \Delta t_f[/itex]. So, when signal reaches from front end of barn to middle the enough pole already passed thorough barn and it again establish LOS at the rear end. The rear end signal of establishment takes less time to reach to middle. And both signal simultaneously reaches to middle same as barn frame. And also same as barn frame bomb would not go off. 


#16
Mar1212, 05:37 AM

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#17
Mar1212, 08:18 AM

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vu, v, v+u ; closing speed of each signal to detonator is u, as in barn frame. Using the correct formulas for all speeds, you have (vu)/(1  uv/c^2) , v , (v+u)/(1 + uv/c^2) There are two interesting limits here. For v (detonator = barn speed) > 0, you approach the Galilean relations. For u>c (for any v < c  which is required), you get: c, v, c ; closing speed v+c for s1, and cv for s2. For v close to c and u=c case, then even though disarm signal is sent from E2 which occurs after E3, it's closing speed approaching 2c allows it to reach R before the arm signal from E3 (with vanishing closing speed). [Edit: as an example of slower signal speed with relativistically moving barn e.g. u=.1c, v=.9c, you get s1 closing speed of .021c and s2 closing speed of .0174c. Thus a significant asymmetry even for fairly slow signals]. 


#18
Mar1212, 09:10 AM

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One final note is that while length contractions is important for the full calculation of either barn or runner frame, it has no part in explaining the inversion of transmission and reception events for the runner, because it introduces no asymmetry in signal distances. It is only the relativistic velocity addition that introduces asymmetry  in closing speed, and thus can explain inversion between transmission and reception events.



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