# Einstein thought experiment, different outcome of a same event

1. Jan 30, 2012

### rajeshmarndi

One person on the ground and the other traveling in a train.

Two light detector X and Y are placed in the train at equal distance from the observer in the train.

If only one is hit first by the light, only that detector will display its name else if both detector are hit simultaneously both detector will display their name.

When both are just exactly facing each other, a light source from the observer in the train is switch on.

Since in relativity two person will not agree on simultaneity, person on the ground will found that the detector to the rear end of the train only display its name while the person on the train will found both the detector display their name because in his frame both the detector are hit by the light simultaneously.

Wouldn't the person on the ground find his observation wrong when the train come to rest.

Isn't the outcome of a same event would be different if both are correct.

2. Jan 30, 2012

### ghwellsjr

How does each detector know if it gets hit first by the light or if it gets hit at the same time as the detector on the other side of the train observer gets hit?

3. Jan 30, 2012

### HallsofIvy

The person on the ground would find that he and the person on the train disagreed as to what would happen. It would not follow that either was "wrong".

4. Jan 30, 2012

### tiny-tim

if that is decided in the train frame, then the result will be the same in either frame

5. Jan 30, 2012

### elfmotat

This is very misleading. Both detectors will display names in both frames. They events can't differ, only their order.

Last edited: Jan 30, 2012
6. Jan 30, 2012

### rajeshmarndi

isn't that could be made possible, something like if the detector are hit by the light they would record the current time of that frame, and this way whichever detector was hit first would show up the display.

7. Jan 30, 2012

### Matterwave

The two detectors have to communicate to each other via signals. The fastest signal being light.

I suppose you can set up the detectors to emit a photon towards the other detector when it receives an initial photon, and then if it receives a second photon a time t=l/c behind the first photon (l being the length of the ship) to detonate.

For problems like these, though, whether the thing detonates or not requires you to go into the frame of the spaceship because that's the frame where the clocks on the detectors are synchronized.

8. Jan 31, 2012

### ghwellsjr

I get the impression, based on your statement in your first post, "both are just exactly facing each other", that your idea is that each detector is just looking at the other detector to see which one gets the light first or if they see them at the same time.

The only problem with this is that just as the light takes time to travel from the source in the middle of the train to each of the detectors at the ends, it takes an additional time for the light reflected off each detector to travel all the way across the entire train to get to the other detector. So each detector will always see that it got the light first and display its name which means that both detectors will always display their names.

First, let's suppose the train is not moving so that both observers agree on the speed of the light. Can you see that both detectors will put up their names?

Now let's continue to suppose the train is not moving but this time let's put the light source near the detector at the front of the train. Obviously, the front detector will see the source before it sees the reflected light from the rear detector but what about the rear detector, what will it see? Well, the light coming from the source will be just ahead of the light reflected off the front detector so the rear detector will see the source light before it sees the reflected light and display its name. They both display their names even though we know that the front detector actually saw the source light before the rear detector.

I think you can see that if we put the source near the rear detector, the same thing will happen the other way around and both detectors will display their names.

Now let's go back to the condition where the light source is in the middle of the train but let's have the train travel at almost the speed of light in the forward direction. We will stipulate that this creates a tremendous "head wind" for the light such that it takes a very long time to go in the forward direction but almost no time to go in the reverse direction. Now when the light gets turned on it almost instantly arrives at the rear detector so it displays its name. Meanwhile the light slowly progresses toward the front of the train, but no matter how slow it is, the reflected light from the rear detector, although traveling at the same slow speed, will never catch up and so the front detector sees the light from the source first and displays its name.

Even it we try to even things out by putting the light source near the rear of the train, the reflected light will always be behind the source light and both detectors will display their names in all cases.

So now we get to the bottom line: if both detectors always display their names under all conditions, how do you propose to improve things so that under some conditions, only one will display its name?

9. Jan 31, 2012

### rajeshmarndi

I don't understand why the reflected light has to do anything here. Bcoz the moment it is hit first it will record the time(digital atomic clock) of first impact only and not of the reflected light, say the circuit cut-off on first impact and this time information is passed to another common circuit, where both time information are recieved and compared and only then show up appropriate display on those detector.

10. Jan 31, 2012

### elfmotat

There needs to be some way for the detectors to compare data and decide whether or not both were hit simultaneously. Let's say, for example, that there's a computer in the center of the train that records the time when the detectors are hit. We'll also say that the detectors communicate with the central computer wirelessly (using EM radiation).

Let's consider the rest frame of the train first: The light in the center is switched on and it reaches both detectors simultaneously. The detectors, upon being triggered, send a signal to the central computer. The computer receives signals from both detectors simultaneously, both displays are lit, and everything is dandy.

Now, from the ground frame: The light in the center is switched on. The light reaches the detector in the back of the train first (due to the motion of the train), and it reaches the front a short while later (we'll call the difference in time Δt). The detector in the back sends a signal to the computer first. The detector in the front emits a signal Δt seconds later. The signal from the back, due to the motion of the train, takes Δt seconds longer to reach the computer than the signal from the front. The time differences exactly cancel, and the computer is "tricked" into thinking both detectors were triggered at the same time. Both displays are again lit.

Last edited: Jan 31, 2012
11. Jan 31, 2012

### elfmotat

Whoops. What I meant was "The events can't differ, only their coordinates."

12. Jan 31, 2012

### ghwellsjr

Well, there's nothing faster than light to communicate at the fastest possible speed and even when using light, both detectors always put up their two names, no matter how you set up the scenario. I'm just wondering how you're going to set up your scenario in a meaningful way so that each detector can really tell if it received the light from the source before the other one or if they received it at the same time.
So you've got some complicated scheme in mind here which I have no idea what you're describing and you think this will be better than light signals. Let me ask you some questions:

Where is this digital atomic clock located?

Where is this circuit that does the cut-off?

Where is this other common circuit?

Where do the wires go?

Could you draw a diagram?

13. Feb 2, 2012

### rajeshmarndi

Two synchronous atomic clock(digital) are connected to the detector, whenever the detector is hit by the light, that atomic clock will stop its time.

What i mean is would the ground observer see both atomic clock time reading same at the time when light reach them.

OR

Would the ground observer see both atomic clock attached to the detector running synchronously?
If yes, would he find the front end detector showing more time when it is hit.

14. Feb 2, 2012

### ghwellsjr

If you mean that each detector has its own atomic clock and if you mean that the traveler on the train synchronizes them so that he sees the same time on both clocks, then the ground observer will not see the same time on both clocks.

Therefore, when the light pulse is sent by the traveler and the two clocks stop, he will see them having the same time on them and so will the ground observer. How could it be otherwise?

15. Feb 2, 2012

### _PJ_

The individuals perceive the events at differing times depending on their relative motions. This will be the case regardlerss of how many clocks and lights you have.
Since each observer has their own relative perspective, their observations will always differ.

16. Feb 2, 2012

### rajeshmarndi

You mean to say for the ground observer, both clock will not look synchronously, the front end clock will appear slow to him than the rear end clock.

So that when the light hit the front end clock, it will read the same time but after a short delay.

My confusion arises thinking the ground observer would also see both the clocks running synchronize and therefore the front end clock will read more time than the rear end clock.

Thanks for the assist.

17. Feb 5, 2012

### mangaroosh

Apologies for posting this in here; I PM'd rajesh, about this, but perhaps he hasn't been online since. I thought I saw a thread started by rajesh outlining a thought experiment involving the double slit experiment, but I can't seem to find it now - just wondering if anyone knows the thread and could PM me a link or anything.

18. Feb 5, 2012