# Length contraction experiment results

1. Dec 21, 2015

### name123

I was looking at the barn door paradox, http://math.ucr.edu/home/baez/physics/Relativity/SR/barn_pole.html

But supposing that instead of a barn, there was a piece of measuring apparatus which was in two halves, with a gap in between through which the pole would pass. One half of the measuring apparatus has a laser at each end (these correspond to the barn doors), and that the beam of each would be broken by the pole as it passed. The other half (the receiver) has a light detector at each end, each of which detects the beam from one of the lasers. These detectors are linked by fibre optic cable to a NOR gate that is situated in between them, equidistant from each in the measuring equipments rest frame, and the NOR gate is connected to a light bulb, such that if in the rest frame of the measuring apparatus you were to break both beams simultaneously there would be a later point in time where the NOR gate simultaneously did not receive a signal from either light detector, and would turn the light on. At first glance I would expect the light to not be expected to come on by an observer from the measuring equipments rest frame but be expected to come on from an observer from the poles rest frame, but I realise that would mean there would be an experimental difference, so was wondering what the expected results would be? Would it be something like the NOR gate not being equidistant from each detector in the poles rest frame?

2. Dec 21, 2015

### Ibix

The NOR gate is equidistant from the barn doors. However, since it is moving in any frame except its rest frame, it is not equidistant from the places the doors occupied when the rod went through them. Thus the signals from the detectors are emitted at different times, but take different amounts of time to reach the NOR gate. They end up arriving simultaneously.

3. Dec 21, 2015

### name123

So are you saying that if there was an observer at the NOR gate (in the measuring equipment's rest frame), and an observer in the middle of the pole (in the pole's rest frame) that if they passed each other at t = t' = 0 that the observer on the pole would measure one detector to be closer to it at t' = 0 than the other detector (both detectors are at rest with respect to the NOR gate)?

4. Dec 21, 2015

### Samy_A

I wonder if one can look at this this way:
In the measuring equipment rest frame, the signals all arrive at the same location, the same x.
The Lorentz transform for the time variable is $t'=\gamma (t-vx)$.
So in the frame of the pole, the order in which the signals will arrive will be the same as in the other frame, as for two events at x with times $t_1$ and $t_2$, $t_1'-t_2'=\gamma (t_1-t_2)$. So the lamp will not come on in the frame of the pole either.

5. Dec 21, 2015

### Ibix

No - I explcitly said that the NOR gate is equidistant from the barn doors. This is true in all inertial frames. However, the only frame in which the rod triggers both sensors simultaneously is the rest frame of the barn. In all other frames, the sensors are triggered at different times, and the barn is moving.

According to the rod observer, then, at the time t=t'=0, the sensor at the back of the barn has already tripped. But the barn is moving. Both the NOR gate and the sensor that tripped have moved since it tripped - so the NOR gate is more than half the (contracted) length of the barn from the place where the sensor tripped. A little while later, the sensor at the front of the barn trips. But the barn and the NOR gate are moving, so at any time after it trips, the place where it tripped is between the sensor and the NOR gate.

To summarise, the NOR gate is always equidistant from the sensors. It is not equidistant from the place where one of the sensors was at one time and where the other sensor was at another time. The difference in place and time always compensate so that the NOR gate receives signals from the sensors simultaneously, whatever the frame.

6. Dec 21, 2015

### JVNY

The original post suggests that the one would not expect the light to come on in the measuring equipment's frame. If the relative speed is great enough that the pole is length contracted in the measuring equipment's frame to less than the distance between the two sensors in that frame then the suggestion is correct. The light will not come on, because the sensors will not be tripped at the same time in the measuring apparatus' frame. Therefore the light will not come on in the pole's frame either, so there is no experimental difference.

It is true that the sensors will be tripped at the same time in the pole's frame. But this is not relevant. You have programmed the light system to turn on the light based on events that are simultaneous in the measuring apparatus' frame, not those that are simultaneous in the pole's frame.

If the length-contracted pole is nonetheless long enough (despite its contraction) that it does trip both sensors simultaneously in the apparatus' frame then it will cause the light to come on. Again, this is because the system is programmed to turn on the light based on simultaneity in that frame.

7. Dec 21, 2015

### name123

Ok, thanks I think I get it. From http://math.ucr.edu/home/baez/physics/Relativity/SR/barn_pole.html as I understand it, adapting it to the light sensor example where there was an observer at the NOR gate (in the measuring equipment's rest frame), and an observer in the middle of the pole (in the pole's rest frame) that passed each other at t = t' = 0. At t = 0 from the observer at the NOR gate's perspective the light sensors are 40m apart (20m away in either direction) , but the pole is less than 40m (less than 20m in either direction) so that neither end of the pole is breaking a light beam, but at t' = 0 where the observer in the middle of the pole is opposite the observer at the NOR gate, the light sensors are 10m in either direction, but the pole is 40m in either direction, and so both light beams would be broken simultaneously from that perspective.I assume by the phrase "because the system isn't programmed" you mean that (from the pole's pespective) the NOR gate doesn't fire because the light down the fibre optics wouldn't progress along the fibre optic cable at the same rate because the fibre optic cable is moving in one direction (from the poles perspective) at near the speed of light. So the light signal progresses quicker down the 10m of fibre optic cable from the head sensor than it progresses down the 10m of fibre optic cable from the tail sensor (from the pole's perspective). And it would be the same even if it was some lever system, as the "push" or whatever couldn't be observed to propagate faster than the speed of light (which it would be from the pole's) perspective if it took the same time to travel from the head sensor to the NOR gate as from the tail sensor to the NOR gate. I assume even if the signal were to propagate very slowly, so that it's propagation speed plus the velocity would still be less than the speed of light that it would still be a similar situation.

8. Dec 22, 2015

### JVNY

By "programmed" I meant that you set up the NOR gate system in a particular way. The gate is placed equidistant from the sensors. The gate turns the light on if two events occur. First, the gate receives a signal from each side at the same place at the same time. Second, at that same place the gate subsequently receives a signal from only one side.

I think that your description of what happens is correct with one exception. You specify more than 50% length contraction (80m own length pole contracted to less than 40m in the apparatus frame), so the sensors are less than 10m away in each direction from the pole observer you describe (not 10m). Also, I think like you that other signals would work the same way, but perhaps the other posters can comment on this.

Last edited: Dec 22, 2015