# Analysis of train in tunnel problem using an AND gate

1. Dec 27, 2015

### Alfred Cann

This analysis uses an AND gate, the very embodiment of simultaneity, for clarity.
A 1km long train travels through a 1km tunnel at a speed of 0.6c=180m/microsecond. At each end of the tunnel is a photo-electric beam sensor, which can be blocked by the train. They are connected to an AND gate at the center of the tunnel, which drives a light. The light will be on whenever both beams are unblocked. (We can even dispense with electrical transmission by using 45° mirrors and placing the photocells at the AND gate.)

In the tunnel reference frame, the train is shrunk to a length of 800mhttps://www.physicsforums.com/file:///C:/Users/fred/Desktop/Fredmisc/Train%20in%20tunnel.docx#_ftn1 [Broken]. Let us say the rear beam is unblocked at t=0. Then the front beam is blocked at t=200m/180m/us=1.11..us. Since the transmission times from the two beam sensors are equal, the light is turned on for a duration of 1.11..us.

Assume the train has an observation car so that an observer can see the output light. In the train reference frame, the tunnel is shrunk to a length of 800m. Say the front beam is blocked at t=0. As seen from the train, that signal travels toward the AND gate at a speed of 300m/us. But the gate is traveling away from it at a speed of 180m/us, so the net speed is 300-180=120m/us (even though an observer at the gate would measure a speed of 300m/us). The distance is half the shrunken tunnel length, 400m. Thus, the travel time is 400/120=3.33..us.

The rear beam is unblocked at t=200m/180m/us=1.11..us. That signal travels toward the gate at a speed of 300m/us. But the gate is traveling toward it at a speed of 180m/us, yielding a total speed of 480m/us (even though an observer at the gate would measure a speed of 300m/us). The distance is400m; thus, the travel time is 400m/480m/us=0.833..us, and the arrival time is 1.11..+0.833..=1.944..us. The light is on (both beams unblocked) for a duration of 3.33..-1.944..=1.388..us.

As viewed from the train, all tunnel clocks run slow by a factor of 0.8. The light is such a ‘clock’, and its ‘on’ time is stretched to 1.11../0.8=1.388..us, in agreement with the previous paragraph.

https://www.physicsforums.com/file:///C:/Users/fred/Desktop/Fredmisc/Train%20in%20tunnel.docx#_ftnref1 [Broken] Using L'=L√(1-(v/c)Λ2)

Last edited by a moderator: May 7, 2017
2. Dec 27, 2015

### Ibix

Without having checked your maths in detail, it looks reasonable. Do you have a question?

3. Dec 27, 2015

### Alfred Cann

Do you and Peter Donis think this is a useful pedagogic approach?

4. Dec 27, 2015

### Staff: Mentor

Pedagogically, I see a few potential issues with your presentation. I'll phrase them as comments on your OP.

No, it isn't, because signals still take time to travel to the AND gate. See below.

I assume that you mean the train is 1 km long in its rest frame, and the tunnel is 1 km long in its rest frame. But you should not leave such things for the reader to assume. You should specify them. One of the most common ways of going wrong in relativity problems is to fail to state which frame a frame-dependent quantity (like length) is relative to; pedagogically, you want to avoid that.

No, it will not. It will be on whenever a signal is reaching both inputs of the AND gate. But the signal takes time to travel from where the beams are or are not blocked, to the AND gate. You have ignored that signal travel time in your statement above, which invites the reader to ignore it, and pedagogically, that's not good. (You didn't ignore it in your actual analysis, which is good; but pedagogically, you should be sure to phrase everything in a way that makes clear that signal travel time needs to be taken into account.)

5. Dec 28, 2015

### Ibix

Further to Peter's comments, I'd be a bit wary of the way you calculated your signal delays in the train frame.

"Speed" can mean two different things - the closure rate as seen from some frame and the closure rate as seen from the rest frame of one of the objects. These are the same thing in Galilean relativity but not in Einsteinian relativity. For example, a pedestrian sees two cars doing 30mph towards each other. According to the pedestrian they are coming together at 60mph. In Galilean relativity the drivers also see themselves coming together at 60mph, but in Einsteinian relativity they see 59.99999999mph (ish).

These speeds have different properties. One cannot exceed c, and the other cannot exceed 2c. You are using the latter. You may wish to call it something other than "speed" to avoid questions about how speed can be greater than c. Alternatively, structure your calculation as "there are 400m to cover; the gate is doing 0.6c and the signal is doing c. They meet after $\Delta t$, where $400=0.6c\Delta t+c\Delta t$.

Also, you are using the same symbol (t) for times in both frames. I'd tend to think this would be confusing. Typically people use t for one frame and t' for the other. I'd recommend following the convention.

6. Dec 28, 2015

### Alfred Cann

You people are very sharp and I thank you for sharpening my argument. How does the following grab you?

"This analysis uses an AND gate to avoid the difficulty of needing to know the train’s speed to decide where to site the sensor for slamming the doors, and may be clearer for some people.

A train which is 1km long at rest travels through a tunnel which is also 1km long at rest, at a speed of 0.6c=180m/microsecond. At each end of the tunnel is a photo-electric beam sensor, which can be blocked by the train. They are connected to an AND gate at the center of the tunnel, which drives a light. The light will be on whenever both inputs to the AND gate are high. Further, let us dispense with electrical transmission by using 45° mirrors and placing the photocells at the AND gate. Then all signal transmissions occur at light speed.

In the tunnel reference frame, the train is shrunk to a length of 800mhttps://www.physicsforums.com/file:///C:/Users/fred/Desktop/Fredmisc/Train%20in%20tunnel.docx#_ftn1 [Broken]. Let us say the rear beam is unblocked at t=0. Then the front beam is blocked at t=200m/180m/us=1.11..us. Since the transmission times from the two beam sensors are equal, the light is turned on for a duration of 1.11..us.

Assume the train has an observation car so that an observer can see the output light. Now, in the train reference frame, the tunnel is shrunk to a length of 800m. Say the front beam is blocked at t’=0. (We now use t’ because we are in a different reference frame.) As seen from the train, that signal travels toward the AND gate at a speed of 300m/us. But the gate is traveling away from it at a speed of 180m/us, so the net closing rate is 300-180=120m/us (even though an observer at the gate would measure a light speed of 300m/us). The distance is half the shrunken tunnel length, 400m. Thus, the travel time is 400/120=3.33..us.

The rear beam is unblocked at t’=200m/180m/us=1.11..us. That signal travels toward the gate at a speed of 300m/us. But the gate is traveling toward it at a speed of 180m/us, yielding a total closing rate of 480m/us (even though an observer at the gate would measure a light speed of 300m/us). The distance is 400m; thus, the travel time is 400m/480m/us=0.833..us, and the arrival time is 1.11..+0.833..=1.944..us. The light is on (both gate inputs high) for a duration of 3.33..-1.944..=1.388..us.

As viewed from the train, all tunnel clocks run slow by a factor of 0.8, the same factor by which the tunnel is shrunk. The light is such a ‘clock’, and its ‘on’ time is stretched to 1.11../0.8=1.388..us, in agreement with the previous paragraph."

https://www.physicsforums.com/file:///C:/Users/fred/Desktop/Fredmisc/Train%20in%20tunnel.docx#_ftnref1 [Broken] Using L'=L√(1-(v/c)Λ2)

Last edited by a moderator: May 7, 2017
7. Dec 31, 2015