Thought experiment on Simultaneity

1. Jul 9, 2012

kannank

I am having a variation of the classic simultaneity experiment (Explained in
http://en.wikipedia.org/wiki/Relativity_of_simultaneity#The_train-and-platform_thought_experiment)

From the midpoint of a train 2 laser beams are sent to both sides of the train. The beams reach the 2 end-points of the train. Since the train is moving, Observer inside the train (A) sees the beams reaching the 2 end-points as simultaneous & the Observer in the platform (B) sees it as non-simultaneous. Agree.

We have a train which is moving at a velocity where Observer B observes a non-simultaneity by 1 second.

Instead of a straight beams, let's send 2 very small pulses. Lets have 2 mirrors on both sides of the train which reflects in incoming laser beams. The beams go back to the center where there is a sensor. The sensor is wired to a bomb which explodes if the sensor receives 2 pulses within 1 second.

For observer A, both pulses reach the sensor at the same time and the bomb should take off. For observer B, pulses reach back with a delay of 2 seconds (travel back and forth) between them and bomb should not take off.

Will the bomb set off or not?

2. Jul 9, 2012

Drakkith

Staff Emeritus
Hmmm. Would observer B actually see a non-simultaneous result? Remember that the beams have to go out from the source and then back, so both beams are travelling with and against the motion of train at some point.

3. Jul 9, 2012

Staff: Mentor

Sure. And being a space-time coincidence, all observers will agree.
Don't assume this, actually figure it out. While observer B sees the pulse reflect off the rear mirror first, that pulse has a longer path to travel back to the center which is moving away from it.

4. Jul 9, 2012

ghwellsjr

Your question has nothing to do with the issue of Relativity of Simultaneity. Even before Einstein came up with his convention for establishing the definition of simultaneity in 1905 when virtually all scientists believed in an absolute ether and in absolute time, they realized that the bomb would go off.

You could even repeat your experiment using sound where a pop is emitted in the middle of a very long flat bed train car with two sound reflecting walls at the front and rear end. Obviously, if the train were not moving and there was no wind, the echoes of the pop coming from the front and the rear would arrive together at the center where the bomb is and it will go off, correct? (Let's narrow down the timing and require that the echoes are detected within a millisecond.)

But what if the train were moving and/or there was a head wind. Then you would expect the time for the sound of the pop to reach the rear wall would be less than the time for the sound of the pop to reach the front wall. This is because the sound traveling toward the rear experiences a tail wind which speeds it up and the sound traveling toward the front wall experiences a head wind which slows it down.

But now the echoes have to travel in the opposite directions to get back to the middle of the train car. The echo coming from the rear wall experiences a head wind and is slowed down while the echo coming from the front wall experiences a tail wind and is sped up. The sounds of the pop have the same experience in their round trips, it's just that the order of the two parts is reversed from one another and thus takes the same total time.

Another way to look at this is if there were always a constant wind blowing from the east to the west and you started on the west coast and flew at a constant air speed to the east coast taking five hours and on the way back it took three hours, then it would take the same round trip time if you started on the east coast and flew to the west coast and back. Both round trips would take eight hours even though each part took a different amount of time.

This is why when Michelson and Morley performed their famous experiment to detect the ether wind, they didn't reflect light in both directions along a straight line, they created two paths at right angles to each other. They realized that doing it the first way would yield a null result even if there were an ether wind. What surprised the scientists of the day was that even though they believed there still was an ether wind, they couldn't detect it and their explanation was that the path along the direction of the ether wind must have been shortened in such a way that the round trip times came out equal.

5. Jul 10, 2012

Vandam

Here's a quick sketch of a Loedel diagram showing what's going on.

Attached Files:

• train mirrors-.jpg
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6. Jul 11, 2012

Vandam

I got it right to show you my skectch directly in the post.
(many thanks, bobc2)

7. Jul 11, 2012

mananvpanchal

Front pulse has to travel more journey than rear pulse. Because, front pulse is travelling in the same direction, and rear pulse travelling against.
In return journey of pulse, rear pulse has to travel more distance than front.
Pulse reaches at center simultaneously for both observer.

8. Jul 11, 2012

bobc2

Very nice sketch, Vandam. It makes it so clear as to what is going on. You see clearly the distances and times involved in this scenario. And the explosion observed by both A & B is easily resolved. The Loedel diagram is good the way it renders the same space-time distance calibrations for both A & B inertial frames of reference.

9. Jul 11, 2012

ghwellsjr

These diagrams are only clear to people who have previously been trained in how to intrepret them which doesn't include me and probably doesn't include most people asking these types of question. Why don't you people make animations that show what happens in two dimensions as a function of time instead of these overcrowded diagrams packing in less information (only 1D)? I find these static diagrams to be very confusing and hard to interpret.

Here, I have made an animation that shows what happens in a stationary frame where an observer sets off a flash of light or a sound pop which reflects off of four omnidirectional reflectors and you will note that the observer detects all four reflections arriving at the same time:

Now here's an animation that shows what happens in a non-relativistic frame where the observer is moving and how the reflections arriving from front and back arrive together while the reflections arriving from top and bottom also arrive together:

The point of these animations is to show that even if there is a wind in any direction and you set your reflectors inline with yourself, the reflections will always arrive at the same time, whether or not the world is relativistic. This issue has nothing to do with when the wavefront hits the different reflectors. It has only to do with the fact that the wavefront makes a round trip and the two directions (front and back or top and bottom) total the same time.

Last edited: Jul 11, 2012
10. Jul 11, 2012

Austin0

Hi ,,,,very nice animations. Thanks.
i am curious as to why the second does not accurately show what would take place in the relativistic situation with light emissions?

Last edited: Jul 11, 2012
11. Jul 11, 2012

ghwellsjr

Because I'm focusing on the OP's issue and pointing out that it has nothing to do with Relativity.

12. Jul 11, 2012

Austin0

Let me rephrase. The animation looks perfectly correct.
But even at non-relativistic velocities all four signal paths you depicted should arrive back at the emission point simultaneously , no?

13. Jul 12, 2012

ghwellsjr

I'm sorry, I see why this was confusing. When I said a non-relativistic frame, I was referring to the sound wavefront created by a pop. The first animation applies for a flash of light or a sound pop while the second one applies only for a sound pop (or for what scientists prior to MMX expected of a flash of light). It might help to re-read post #4.

14. Jul 12, 2012

Austin0

Well in actuality i had not read post 4 at all as the answer to the OP question was self evident.

Again nice animation.

15. Jul 12, 2012

ghwellsjr

Thanks.

16. Jul 12, 2012

Vandam

I agree. But if somebody would show me a diagram I would react: "WAW, I must absolutely learn how to read this, because it might be a fantastic tool." ;)
Good luck.

17. Jul 12, 2012

bobc2

Originally Posted by ghwellsjr:
These diagrams are only clear to people who have previously been trained in how to intrepret them which doesn't include me and probably doesn't include most people asking these types of question.

Vandam, I think you are right that the Loedel diagram is a really good tool. It complements the use of the Lorentz transformation math with unique perspective and insight. However, I think ghwellsjr is being quite modest. He has demonstrated utility and command of the Lorentz transformation math as well as any on the forum and probably has a pretty good grasp of analytic geometry. I think he is speaking for a lot of forum visitors who haven't had that kind of background but are really interested in special relativity and are hoping for presentations that can bring it to them at their level.

Of course folks can Google "Loedel Diagram" -- and many would find it not so difficult to follow as they might think. Or, perhaps someone on the forum (such as yourself) could start a thread aimed at presenting a short tutorial on the Loedel Diagram (or perhaps a FAQ).

I think people would find it quite useful. In my first year in grad school the special relativity prof had us do most of the homework problems first with the Lorentz transformations, then do them again with space-time diagrams (quite often using Loedel Diagrams). I remember having trouble with the old pole-in-the-barn problem; I wound up understanding it first by doing it with a Loedel Diagram. It became so obvious to me then and led me to quickly understand how to represent the process with the Lorentz transformations.

Last edited: Jul 12, 2012
18. Jul 13, 2012

ghwellsjr

I'm not being modest. I'm sure I could master Loedel Diagrams if I was motivated but I don't see them as solving any problem in Special Relativity. In fact, I think they create a problem--the notion that each observer must have his own rest frame in order to observe the world around him. That's not what SR is all about. It's about picking any single frame so that we can analyze what all the observers observe of the world around them. It doesn't matter which frame we choose and since inertial frames are so much easier to work with, I always choose a single inertial frame, just because it's easy. A Loedel Diagram, by definition, combines two frames, so I'm not motivated. I have taken another look at Vandam's Loedel Diagram, and I just find it confusing, I don't know what it is trying to convey. Put it this way, if there was a mistake in it, I doubt that I would notice it.
OK, I did that. Vandam's Loedel Diagram doesn't appear to be a legitimate Loedel Diagram. It doesn't look anything like the Loedel Diagrams I see when I Google it. Is that a mistake?
If you would express the pole-in-the-barn problem in a single frame and if you believed that any single frame is all you need, then you wouldn't have seen it as a problem. Virtually all the so-called paradoxes in SR are a result of specifying different parts of a scenario in two different frames. Once you point that out and specify the entire scenario in a single inertial frame, which of course won't allow any paradox, then you can transform the entire scenario into a different inertial frame using the Lorentz Transformation process and again, there won't be any paradox. But then, that takes all the fun out of Special Relativity.

19. Jul 14, 2012

Vandam

Interesting question that I am not allowed to answer myself. :(

Maybe you recognize loedel in following diagram. I've only rotated the same diagram.

Last edited: Jul 14, 2012
20. Jul 14, 2012

DrGreg

As I understand it, a Loedel diagram is a Minkowski spacetime diagram, depicting two objects with equal-but-opposite velocities, in other words where the observer is chosen to be halfway between the two objects.

21. Jul 14, 2012

bobc2

That's correct, DrGreg. Here are a couple of sketches below to illustrate this. First (upper sketch) I show the basic Loedel diagram for red and blue observers moving at the same speed in opposite directions with respect to the black rest frame. The advantage of using this symmetric configuration is that the lines on the screen have the same distance calibrations for both red and blue observers. For example, you can use this feature to derive the Minkowski metric as shown here (not a true mathematical metric). A key feature of this diagram is that the green photon world lines always bisect the angle between the X1 axis and the X4 axis for all coordinate systems (note that this is true for the black, blue and red frames). Sometimes in preparing a sketch one has a little trouble getting all of the geometry to come out right on the display. I assumed Vandam had a little problem with his graphics construction but wasn't going to knit-pick that since he was telling a pretty good story with his graphics.

The sketches below depict red and blue rockets moving at the same speed in opposite directions with respect to the black rest frame. Again, the lengths of the rockets can be compared quantitatively since the distance calibrations in space-time are the same. The lower left sketch is the original standard configuration presented by Loedel. Some people like to rotate the sketch when they are emphasizing the orientation of one particular observer from the point of view of his rest frame. So, in the lower right sketch I've simpy rotated the diagram as suggested by Vandam in his earlier post.

Last edited: Jul 14, 2012
22. Jul 14, 2012

Staff: Mentor

You should have rotated it the other way in order to get it into the standard presentation. However, other than the non-standard rotation, it is a legitimate Loedel diagram. The time axis of one frame is perpendicular to the space axis of the other frame and pulses of light bisect the angle formed between the time and space axes of each frame.

23. Jul 14, 2012

cepheid

Staff Emeritus
I'm not sure what a Loedel diagram is. I was able to make sense of it using a standard spacetime diagram:

The unprimed (t,x) coordinate system is the coordinate system of Observer B, who is at rest w.r.t. to the track. The primed (t',x') coordinate system is the coordinate system of Observer A, who is at rest w.r.t. to the train.

The worldlines (paths through spacetime) of the sensor (and Observer A), and the two mirrors are shown. I arbitrarily decided that the train should be moving at speed +0.5 in the unprimed coordinate system (using c = 1 units of course). That's what sets the slope of those world lines. You can also see the worldlines of the forward-going (red) and rearward-going (blue) light pulses. These, of course, must always have slope +1 or -1. They are only colour-coded to distinguish them from each other. The colours are not meant to imply anything about the light itself.

So, as you can see, relative to Observer B, the forward-going (red worldline) photon travels a much farther distance (measured along the x-axis) on the way to the mirror than the rearward-going photon. However, this is exactly compensated for on the way back, when the photon indicated by the blue worldline now travels a larger distance than the one indicated with the red worldline. As a result, they both arrive back at the sensor simultaneously according to Observer B. (Their worldlines both intersect the worldline of the sensor at the same t-coordinate).

They also arrive back at the sensor simultaneously according to Observer A (their worldlines intersect the world line of the sensor at the same t' -coordinate). You can see that the distance travelled by each pulse on the way out is the same in the primed coordinate system. (Just measure the distance along the x'-axis).

The reason why the t' and x' axes are rotated w.r.t. to the t and x axes is because that's what the Lorentz transformation between the two coordinate systems does (in the case where one is moving at a constant speed relative to the other one). You can think of it as rotating the coordinate system in spacetime, so that what is experienced as space by one observer will be experienced partially as space and partially as time by another observer.

EDIT: Observer A gets blown up. Seems a bit unfair . Then again, maybe Observer A is just a camera connected to a computer, or something.

Last edited: Jul 14, 2012
24. Jul 14, 2012

bobc2

Very nice presentation of the situation, cepheid. And I certainly agree that you do not need the Loedel version of the space-time diagram to make the essential points, as you have done so clearly. The Loedel is good when you would like to have the same line lengths in reference frames for two different observers have the same actual calibration in terms of actual times and distances. I've tried to explain this in the sketches below. The hyperbolic equation was developed in post #21 above.

25. Jul 14, 2012

cepheid

Staff Emeritus
One thing I just realized while drawing these spacetime diagrams is this. Sure, it's true that two events that are simultaneous in one inertial reference frame are not necessarily simultaneous in another inertial reference frame. However, if two events have exactly the same spacetime coordinates (t,x,y,z) in one reference frame, then they'll have exactly the same spacetime coordinates in any other inertial reference frame, since the result of Lorentz transformation will be to assign the same new set of coordinates (t',x',y',z') to both events. So the conclusion is that if two events occur at the same place at the same time in one reference frame, then they'll occur at the same place at the same time in any reference frame. In hindsight, this is obvious, and it's the answer to the OP's question right there.