# Closing speed and length contraction paradoxes

by Seanner
Tags: closing, contraction, length, paradoxes, speed
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Hi Seanner! Welcome to PF!
 Quote by Seanner The earth guy knows they are going too fast for the safety equipment and correctly predicts they die...
Nope, if two equal cars with equal and opposite momentum collide simultaneously with a wall, the forces and damage on one car are the same if you remove the wall, or if you fix the wall and remove the other car.

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.
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 Quote by tiny-tim Hi Seanner! Welcome to PF! Nope, if two equal cars with equal and opposite momentum collide simultaneously with a wall, the forces and damage on one car are the same if you remove the wall, or if you fix the wall and remove the other car. 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.
Unless I misunderstand, I have a tiny, in principle, correction. If you fix the wall and remove the other car, the relevant COM is earth + car, in which frame the car has a tiny bit less KE than in COM including opposite moving car. In practice, this difference is obviously unobservable. I guess the the wink implies you know this, and were focusing on the reality.

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## Closing speed and length contraction paradoxes

 Quote by Seanner 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?
At high speed, KE = 1/2 m v^2 is wrong. Instead KE approaches infinity as speed approaches c. This removes your paradox, which is purely a paradox of mixing relativistic and non-relativistic formulas in a relativistic problem. FYI, the correct KE formula for all speeds is: m c^2 (1 - 1/sqrt(1 - v^2/c^2)) . If you calculate this for low speeds, or Taylor expand it, you will see it is indistinguishable from 1/2 m v^2 for low speeds, but differs a lot for e.g. .9c.
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 Quote by Seanner 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).
You have just disguised the difference in simultaneity. The barn circuit for detecting beam breakage at the same time in the barn frame becomes a circuit timed to detect beam breakage with a time offset in the runner's frame. It will fail to trip in the runner's frame because the pole will get through second beam before the time offset is reached. Simultaneity is relative for all purposes - including timing circuits. If in one frame, you detect simultaneity, in another frame you have a time offset detector.
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 Quote by PAllen Unless I misunderstand, I have a tiny, in principle, correction. If you fix the wall and remove the other car, the relevant COM is earth + car, in which frame the car has a tiny bit less KE than in COM including opposite moving car.
he he

to be honest, i was being even more ambitious than adding the mass of the Earth …

i was assuming my wall had infinite mass!
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 Quote by Seanner 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?
This is not an answer; Just a question. The people heading for collision with each other, will
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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 Quote by PAllen You have just disguised the difference in simultaneity. The barn circuit for detecting beam breakage at the same time in the barn frame becomes a circuit timed to detect beam breakage with a time offset in the runner's frame. It will fail to trip in the runner's frame because the pole will get through second beam before the time offset is reached. Simultaneity is relative for all purposes - including timing circuits. If in one frame, you detect simultaneity, in another frame you have a time offset detector.
But the runner will see both lasers tripped at the same time.
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 Quote by jreelawg But the runner will see both lasers tripped at the same time.
So what? Either the circuit fires the bomb based on runner simultaneity or barn simultaneity. One circuit can't do both. If it is based on barn simultaneity, it won't fire; if runner simultaneity it will. But this is no paradox at all: barn observer sees rod fit in barn, runner sees it fail to fit in barn. So the laser tripping is perfectly consistent with other methods of observation. Whoever sets up the circuit will find it consistent with all other observations they make.
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 Quote by jreelawg This is not an answer; Just a question. The people heading for collision with each other, will 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?
Relativistic mass is an outdated concept. Instead see my post relativistic KE. In fact, KE shows exactly why relativistic mass is a dubious concept: plug relativistic mass into 1/2 m v^2 and you with get the wrong answer for relativistic KE. Instead, you must use the relativistic KE formula I posted earlier, which is expressed in terms of rest mass.
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 Quote by Seanner 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.
Tiny Tim's answer was based on actual practice - such calculations are done in COM of colliding objects, irrespective of how either object is moving relative to scientist - simply because this is much easier and guaranteed to be consistent with any harder method of calculation. He also proposed that an infinite mass wall would produce the same damage as hitting a ship moving in the opposite direction. In either case, the total KE of the ship would be available for damage (the infinite mass wall would carry away no momentum or KE).

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(1-v^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.
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 Quote by Seanner 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.
In the following, I am assuming everything is set up by the barn observer, and no bomb goes off. The issue is to relate what the runner sees that makes sense to you.

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 re-opens 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

Let us also specify some events at the detonator:

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 v-u, 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 (v-u)/(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.
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 Quote by Seanner 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).
Let us think reverse...
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 $t_f$ in pole frame. And light signal reaches at rear end at $t_r$. 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 $\Delta t = t_r - t_f$. Pole's access length $\Delta x$ relative to barn which have to come out from barn in $\Delta t$ time. So, pole's observer conclude that front door closed/opened at $t_f$, $\Delta x$ of pole come out of front end at $t_r$ 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 $\Delta t_f$. Rear end's signal takes $\Delta t_r$. And, we can conclude that the time gap should be $\Delta t = \Delta t_f - \Delta t_r$. In this time gap pole's $\Delta x$ part should be come out of barn. Here $\Delta t < \Delta t_f$. 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.
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 Quote by Seanner 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.
You don't seem to understand what it means for energy to be conserved. "Conserved" doesn't mean "the same in all reference frames." (That is what "invariant," as in "the invariant speed of light" means.) "Conserved" means that in a single reference frame, the energy remains constant with time. But the actual amount of energy that an observer calculates for a body is and has always been frame-dependent, even in Newtonian physics. A sidewalk observer measures the energy of a car on the freeway to be 1/2 mv^2. An observer in the car measures it to be zero. Both observers correctly calculate the impulse when it hits a brick wall.

 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
No, it only implies that the Earth observer gets the wrong answer if he naively uses the nonrelativistic kinetic energy formula instead of the correct formula K=(gamma-1)mc^2.

 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.
No, the runner observes the pulses in the fiber optics to travel at c. All observers observe all light rays to travel at c. But while the official observes the bomb in the middle to be at rest, the runner does not; the bomb for him is traveling backwards at nearly c. So one fiber-optic impulse has a closing speed to the bomb of nearly zero, while the other has a closing speed of nearly 2c. So of course one gets there long before the other, safing the bomb.
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 Quote by PAllen In the following, I am assuming everything is set up by the barn observer, and no bomb goes off. The issue is to relate what the runner sees that makes sense to you. 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 re-opens 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 v-u, 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 (v-u)/(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.
Just to be sure there is no confusion, the asymmetry in the runner's frame does not involve the signal's themselves traveling very slow for either signal (in general), but the closing speed between signal and detonator is what becomes very asymmetric. Also, I do not assume light travels at c in a fiber optic cable (it does not; not even close), and further, using any type of signal would not change the result. That is why I introduce signal speed u, as a separate variable. As noted, for Galilean relativity, you would have signal 1 speed (-u in barn frame), detonator (0 in barn frame), signal 2 speed (u in barn frame) as follows, in runner frame:

v-u, 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

(v-u)/(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 c-v 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].
 Sci Advisor PF Gold P: 4,863 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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