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Please be gentle to a layman

  1. Feb 19, 2008 #1
    Hi there

    I have a question that has possibly been covered before, however, I cannot find the answer that I seek in any of the threads much less understand the odd mathamatical symbols and explanations that go with most of them.

    I have no physics backgroud so most of the posts are filled with and explanations that I dont find easy to follow.

    However I have a question with regards to the famous train experiment that shows how observers can observe the same events out of sequence to one anouther.

    To clarify the experiment in question: a train speeding along has a pile of gunpowder in the middle of it. An observer/referee is standing next to the gunpowder and is going to light it at some point. When he does so a person at the front of the train will set his clock to 12:00 noon as will someone at the back of the train.

    The referee lights the gunpowder, sees the flash and watches as both people at front and back of the train set their clocks simultaniously.

    All is well until an observer watching from a platform points out that he saw the light reach the back observer first and saw him set his clock before the person at the front of the train.
    The platform observer correctly concludes that (for arguments sake) the time on the clock will be 12:01 at the back of the train while it still only reads 12:00 at the front.

    However the referee also cloncludesd correctly that this is not the case - he saw both of them set their clocks at the exact same time.

    I think sofar that I understand all of this (to a point) however what I dont get is who is ultimately right? if the train is stopped and the 2 observers walk over to each other with their timepieces what will each clock say? will they be in sync or out of sync?

    Am I missing something? Is this even a valid question to be asking?

    If you know the answer would it be possible to explain it in an easyly understood way?

    Kind Regards
    Pete
     
  2. jcsd
  3. Feb 19, 2008 #2

    George Jones

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    Hi Pete; welcome to Physicsforums!

    In relativity, everything has to be spelled out in detail. Your observers at the front and back of the train are moving along with the train, and so are their clocks F(ront) and B(ack). Imagine another series of clock stationary with respect to the ground, and laid out all down the tracks. Suppose further that all clocks involved have photocells sensitive to light from flashes of gunpowder, and that each clock stops running when a flash of gunpowder light is received.

    When the light reaches clock F, it will also reach clock F' that is instantaneously coincident with F, and that is stationary with respect to the ground. Similarly, Wwen the light reaches clock B, it will also reach clock B' that is instantaneously coincident with B, and that is stationary with respect to the ground.

    After the experiment, all clocks are collected together. Their reading show t_F = t_B and t_F' > t_B'. Simultaneity is relative.
     
  4. Feb 19, 2008 #3
    So to confirm: The clocks that the observers have on the train are in sync even when the train stops, but anyone who was not on the train, their clocks will yeild different results? So this means that as far as the referee is concerned there is nothing to worry about as any observers that were not on the train will have had their results skewed?

    Pete
     
  5. Feb 19, 2008 #4

    Janus

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    If the clocks on the train stop running before the train start to decelerate, then yes they will read the same time after the train has stopped. But then they will also read the same time according to the observer's on the bank. According to them, the clocks didn't read the same time while running, but stopped running at at different moments, with the difference between the time the clocks stooped running being equal to the difference in time between the clocks while running.

    If the clocks continue to run while the train slows, then the clocks will not be sychronized at the end of the experiment according to any observer, including those on the train. This is because in order to come to a stop, the train must undergo an acceleration, and acceleration adds a new wrinkle to the problem. While undergoing acceleration, the observers in the train will see the two clocks run at different rates. Theresult will be that at the end of the experiment, all observers will agree that the closk are out of sync by the same amount.
     
  6. Feb 22, 2008 #5
    answered

    Thank you for the reply. This answers my question perfectly
     
  7. Feb 22, 2008 #6
    So to continue this experiment. I am wondering if I understand the nature of simultaneous events with regard to time dilation.

    We could use two observers F(first) and S(second) 1000 miles apart beside the train tracks. The ref in the middle could sychronize each observer's clocks. We'll say via a laser.

    So our train is travelling at 200 mps and observer T(third) is on the train using the photocells sensitive to light, previously mentioned, and records the time elapsed between observers F and S.

    For a final measurement observer Fth(fourth) also on the train sends out light during the 1000 mile journey at 1 second intervals that is received by a series of photocells along the side of the track. Of course this would need to be calculated by a digital computer. The computer's time is set by each of the synchronized observers F and S and it calculates the time it takes to receive each of the photocell signals simultaneously and adjusts itself accordingly. Just to be pedantic.

    So we would be able to record the start and end time for both observers F and T and also compare the time difference between the one second intervals on the train and when they are received by the stationary photocells.

    Thus recording the speed of a moving clock comparatively to a stationary clock. We could if we so wished (hypothetically) make a recreation of this senario in a 3d environment.

    Have I said anything false so far?
     
  8. Feb 22, 2008 #7

    Doc Al

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    So you have two synchronized clocks in the track frame. F and S are now on the tracks, not on the train.

    What do you mean? At some point the gunpowder (at the middle between F & S) explodes and when the light reaches clocks F and S, they stop? The train observers also measure the times at which the light reached clocks F and S? The train has its own (synchronized) clocks all along its length and notes at what point F is along the train when the light reaches it and marks down the time showing on the coincident train clock?

    You have to be painstakingly clear when describing relativity thought experiments. It's not clear what experiment you are describing.

    Why don't you start over.
     
  9. Feb 22, 2008 #8
    Well my experiment works by synchronizing F and S clocks and I'm assuming that if they both record the exact time the train passes then you can work out how much time elapsed. T starts their clock at F and shuts it off at S. So we have two recorded times.

    Is that not feasible?
     
  10. Feb 22, 2008 #9

    Doc Al

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    OK, now it's clear. An observer T on the train starts her clock when she passes F and shuts it off when she passes S. Observers at F and S record the clock readings (on clocks F & S) as she passes each clock. We are now able to measure the time that the train takes to get from F to S according to two different frames. Excellent!

    What can you conclude?
     
  11. Feb 22, 2008 #10
    Well nothing that was just the beginning. But we now know how long it took for the train to get between two points from two perspectives. (I'm not exactly sure whether the two clocks should match.)

    Continuing. Firstly I we have observer Fth who is next to T on the train. When T starts their clock Fth starts sending a single flash of light every second thereby counting the time of T observers clock. I would also like a series of sensors along the side of the 1000 mile track to record the time of Fth's flashes. Problem is these aren't synchronized. So since we are no longer in the age of gunpowder I'm planning on using a computer. With fibre optic cables connected to the sensors and by calculating the amount of time a signal takes to travel along the cable (and measuring the length of the cable) we can adjust the clocks so everything is kosher. The computers clock is synchronized by observers S and F.

    So we now have a system of calculating the seconds on the train by a stationary clock. Can I go on?

    No animals were killed or injured during this mental experiment and all observers were required to have completed a course in occupation health and safety.
     
  12. Feb 22, 2008 #11

    Doc Al

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    I think the flashing lights and the sensors along the track will make things more complicated to analyze and will only obscure what's going on. I recommend that you first analyze the simple case without such add-ons.
     
  13. Feb 22, 2008 #12
    The "add-on" is exactly what I'm trying to record. Hmm? I understand that this does not equate to the same time a single stationary observer (for eg. F) would record if he was timing the seconds counted by observer Fth. I was going to get to this next.

    If we had a sensor located directly above observer F's head and Fth flashed his light towards this sensor every second, then we could record the clock on the train from the reference point of F.

    Conversely if there was a sensor above the train directly over Fth's head and observer F flashed a light beam every second (after the train passed) towards the Fth's sensor, then we could record F's clock as viewed by observer Fth.

    We could also use the string of parallel sensors to record the time on the train without the issue of distance created by movement. Each sensor is a exactly 2 ft from the train. So we have a third perspective of how fast T's clock on the train is compared directly to F. This would be like a side by side comparison of time only one is moving at a constant speed the other is stationary.

    So far that is five recordings of time from five different perspectives.
     
  14. Feb 22, 2008 #13

    Dale

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    Hi Joanna,

    I don't want to get into the specifics of your questions, I think that you need some overall framework in which to understand relativity rather than specific details which will naturally fall in place once the framework is in your mind.

    Please look at the attached image. It is fairly simple, but represents in one geometric figure all of the predictions of relativity. You can generate this figure yourself (in fact, I would encourage you to do so as it is educational), it is simply a Lorentz transform with one spatial dimension (horizontal) and one time dimension (vertical). What we have here is two reference frames, the black unprimed frame and the white primed frame. Here the units are not really specified but it is in any unit system where c=1. The primed frame is moving at 0.6c to the right wrt the unprimed frame.

    As you can see the geometry is fairly straight-forward and simple. More important it is self consistent. Any one pair of (t,x) coordinates maps linearly to one pair of (t',x') coordinates and vice versa.

    I have to go and hang out with my family, but let me leave you with this challenge: can you look into this diagram and see what is meant geometrically when we say that a moving clock ticks slower? Can you further see that from the perspective of both the primed and unprimed frames?

    I will follow-up later.
     

    Attached Files:

  15. Feb 22, 2008 #14

    Doc Al

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    What you'd be recording would be the apparent frequency of light signals from Fth's flasher. To use that to deduce the rate of Fth's clock (according to the track frame), you'd need to account for the changing distance between F and Fth. (This is the Doppler effect.)

    Sure you could. Would you expect that Fth would see the signals from F arrive at a different rate than F would see the signals arriving from Fth?

    In order to interpret these recordings, you'll need to account for light travel time.

    Keep it simple!
     
  16. Feb 22, 2008 #15
    "Would you expect that Fth would see the signals from F arrive at a different rate than F would see the signals arriving from Fth?"

    I don't know, but I am guessing that you might state they would not arrive at a different rate.

    The sensors along the track would eliminate the light travel time. The light ticks travel an equal 2 feet each time before hitting the sensors. I already adjusted the time on the central computer to account for the signal to travel along fibre optic cables. I did this because if F was recording the ticks how would you know what position the train was at when Fth flashed his light, in order to account for light travel time? If I could that would make it simpler sure.
     
  17. Feb 22, 2008 #16

    Doc Al

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    Take advantage of the symmetry of the situation. F is observing Fth (moving towards him at speed v) flashing at one flash per second. On the other hand, Fth is observing F (moving towards him at speed v) flashing at one flash per second. How can they see different things?

    Light flashes go everywhere and will eventually hit all sensors. The sensors in front of Fth will see the flashes arrive more rapidly (since the distance between successive flashes is getting smaller); similarly, sensors behind Fth will see them arrive less rapidly. Once light travel time is taken into consideration, all data taken in the track frame will confirm that the flashes were emitted at a rate slower than 1 per second (according to track frame clocks). F recording the ticks is no different than any other sensor recording the ticks. (By "accounting for light travel time" I mean the time it takes for the light to reach the sensors, not the time it takes for the signals to get to the computer--although that must be done also.)
     
  18. Feb 22, 2008 #17
    I can understand how this geometric figure shows moving clock ticks slower and I think the x/x1 coordinates shows length contraction. I don't have a problem understanding how it works, just why it works. That seems to be the most difficult explanation to formulate or even pose the question asking for an explanation.
     
  19. Feb 22, 2008 #18

    Doc Al

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    What do you mean you don't have a problem understanding how it works? That's what this entire discussion is about! Are you just teasing me? :wink: Do you really understand how it's possible that both observers can see the other's clock run slow?

    I don't know what you mean by why it works or what kind of explanation you are looking for. All I can do is help you understand how it works, and how it's a consequence of the principle of relativity and the constancy of the speed of light.

    Regarding Dale's spacetime diagram and understanding relativity: In my view there are three paths to understanding the basics of special relativity. (a) In terms of the Lorentz transformations, (b) in terms of the basic properties of moving clocks and metersticks, and (c) in terms of spacetime diagrams. You need all three. (In my opinion, only after you are comfortable with spacetime diagrams will you really understand how things work.)

    In this thread, I suspect we are trying to follow path (b)--an excellent way to go.
     
  20. Feb 22, 2008 #19
    So I have set up my experiment to record an observation of the clock on the train from a single stationary position not accounting for light travel time and also from a stationary position minus the light travel time.

    Please let us clarify something here. When a stationary person says they are observing the clock on a moving train which of the above methods would they be refering to?

    Just to cover all bases, I am also adjusting my computer clock to account for the time that Fth's light flashes takes to travel the 2 feet from the train to the row of sensors running parallel down the 1000 mile section of track. Ok Doc?

    Now everything is set up.

    F and S observer (who are 1000 miles apart and each have a clock and both are synchronized) both record the exact time the train passes them. Taking these two results they compare how much time elapsed.

    T starts their clock at F and shuts it off at S and records how much time elapsed.

    What should I expect the result to be when result T is compared to FS? Would these times match or not?
     
  21. Feb 22, 2008 #20
    Well I understood from Einstein's paper that if I compared a rocket travelling to a rocket that is stationary to myself the clock on the moving rocket will be slower and it's length will be shorter. If I switched positions then I would achieve the same results. That is how it works.

    I can easily understand english, but I don't know why this works.

    This experiment i'm setting up is designed to give me the answer. So if we could just continue.
     
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