# A question about special relativity theory

We have two sets: set No.1 and set No.2 as in this picture:

The observer is fixed to set No.1 . He sees set No.1 motionless and observes set No. 2 approaching with velocity 100,000 m/s.

Each set has one lamp and two, so called, touchers. Each set is designed so that if both touchers are touched simultaneously the lamp is turned on otherwise it remains off.

Set No. 2 is approaching set No. 1 so that each toucher in each set will be touched twice by the touchers of the other set.

The observer on set No. 1 observes the distance between touchers in each set 10 meters. He thinks the lamps on set No. 1 will be turned on because touchers in it will be touched simultaneously.

My question: Will the lamp on set No. 2 be turned on? How if an electrical current flowing from one toucher in a set to the other proves simultaneous touching?

## 3 Answers

If all the lengths in this diagram are as measured by the observer on set 1 then only lamp 1 will light up, but not lamp 2. When the left "touchers" on each set align in frame 1 then the right touchers will too, because they are both 10 meters separated from the left ones. However, the touchers on set 2 are further than 10 meters apart according to observer 2 while the touchers on set 1 are less than 10 meters apart, so they can never simultaneously align in that moving frame.

Can you clarify the set-up of the lamp system? You say that the lamp turns on if the two touchers are touched simultaneously. But that implies that in frame 1, the two space-time events (event L1L2, pair of touchers on the left touching, and event R1R2, pair of touchers on the right touching) L1L2 and R1R2 are simultaneous. Which means that they are space-like separated, and they are space-like separated from wherever the bulb is at that time. Which means that the bulb cannot instantaneously know if the two pairs of touchers are touching. Instead, a signal will have to propagate from the touchers, and reach some comparison circuit that decides if the signals were simultaneous or not. I think we will need a few more details to decide which bulbs light up. If the circuit just does a comparison of the time-stamps, according to the frames' local clocks, then the bulb will light in set 1 but not in set 2. This question features a related paradox.

• what I wrote is that the lamps are designed to be turned on whenever the two touchers are touched simultaneously. Clearly I mean the observer on the set will determine whether or not it has been simultaneous. – Minimus Heximus Jul 29 '14 at 5:58
• When you say frame 1 and frame 2 in your second to last sentence, do you mean set 1 and set 2? If so, I agree with everything you say. – BMS Jul 29 '14 at 6:20
• I should read your answer a few times more to understand your point! btw, I think everything is clear. I did not say how the sets work and how the system in a set understands simultaneous touchings. But it is assumed that the system works perfectly and can identify simultaneity correctly. In my comments in the other answer I asked how if an electrical current triggers the lamps! that is more clear! my problem was the flow and how it is seen in the two frames, – Minimus Heximus Jul 29 '14 at 8:06
• @BMS You're absolutely right, that should have been 'set'. I've edited it to make the change. – G. Paily Jul 29 '14 at 17:36
• @MinimusHeximus, if you want to think in terms of electric flow, you will first need to think about what happens when you connect a piece of wire to (just) the positive terminal of a battery. We know that initially the wire is at zero potential, and then some time after attaching it to the battery, the wire is at the higher potential of the battery's positive terminal. What happened?The electric field (of the battery) spread from one end of the wire to the other, so there was a pulse of charge flow as the charges rearranged themselves. This pulse(and the electric field) move at light speed. – G. Paily Jul 29 '14 at 18:33

No, they will not. The transformation between frames is $$x'=\gamma(x-vt)$$ So, the endpoints 0 and L in frame F transform into :

$$x'_1=\gamma(0-vt)$$ $$x'_2=\gamma(L-vt)$$

in frame F'. But the location of the "touches" in frame F' is

$$x'_1=-L$$and $$x'_2=0$$, respectively

So:

$$-L=\gamma(0-vt_1)$$ $$0=\gamma(L-vt_2)$$

i.e.

$$t_1=\frac{L}{\gamma v}$$ $$t_2=\frac{L}{v}$$

So, $$t_1 \ne t_2$$, i.e. the circuit NEVER closes. Not in frame F, neither in frame F'. No "paradox".

On the other hand, if you arrange such that :

$$x'_1=-\gamma L$$ and $$x'_2=0$$, respectively, then

$$t_1=t_2=\frac{L}{v}$$

so, the circuit closes in BOTH frame F and frame F'. No "paradox".

• Did you notice that 10 m is not the proper length for set 2? 10 m is what set 1 observes. – Minimus Heximus Aug 10 '14 at 17:23
• Then the circuit closes in BOTH frames: Still no "paradox" – xyzt Aug 10 '14 at 17:43
• No. Because proper length of set 2 is longer than the proper length of set 1. for set 2 set 1 is evern shorter. – Minimus Heximus Aug 10 '14 at 17:48
• Do the math, you'll find out that you are wrong. You have been wrong all along. You just can't put your words into math, this is why you don't understand that your "paradox" is just your lack of basic understanding. – xyzt Aug 11 '14 at 2:36
• well I think wikipedia has done all the math needed it it has suggested what I say: en.wikipedia.org/wiki/Length_contraction – Minimus Heximus Aug 11 '14 at 11:02