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Thought experiment to test Relativity of Simultaneity

  1. Jan 21, 2013 #1
    The thought experiment starts with the basic experiment shown here:


    "A flash of light is given off at the center of the traincar just as the two observers pass each other. The observer on board the train sees the front and back of the traincar at fixed distances from the source of light and as such, according to this observer, the light will reach the front and back of the traincar at the same time.
    The observer standing on the platform, on the other hand, sees the rear of the traincar moving (catching up) toward the point at which the flash was given off and the front of the traincar moving away from it. As the speed of light is finite and the same in all directions for all observers, the light headed for the back of the train will have less distance to cover than the light headed for the front. Thus, the flashes of light will strike the ends of the traincar at different times."

    Now my modification:

    Problem is we have different simultaneities. So let's put photon detectors on each end A and B.

    Same photon that hits the back of the wagon in the ground frame it does before the first photon hits the front. In the wagon frame at the same time. I said to move the light source slightly to the front so that in the wagon frame the sequence from the detectors will get "01" and from the ground "10" ...

    I thought that if the information from both ends (from the detectors) is transferred to a sequence analyzer let's say in the middle of the wagon, we can argue that it may compensate the sequence as if this information is transmitted at "c". But then I thought of eliminating that by doing the following. When each detector triggers, a very slow mechanical device (let's say a plastic pin with Mickey face from the front and another one with Minnie face from the back) goes from each end at the same speed. So if the detector in the front triggers first, Mickey will reach the center of the wagon before Minnie. If Mickey wins "boom", if Minnie wins no boom.

    Some of my own math:

    In the frame of the wagon, light travels at "c". The time to reach each end from the middle is the b/2/c. Where b is the length of the wagon. Time is distance travelled/velocity.

    In the ground frame and noticing that b' will have some shortening due to the length compression if the speed of the wagon is high you have:

    Let's call A the left edge of the wagon, B the right edge of the wagon and C the center.

    Time from C to A (left) is = b'/2/ (c+v) (speed of light on this frame is also c). v is the velocity of the wagon.


    Time from C to B is tCB=b'/2/(c-v)

    You notice that tCB can be a very large number if v approaches c and tCA approaches to b'/4/c

    Now I said both photon detectors launch a slow device (not anything travelling at c), so after the triggering there is no relativity of simultaneity due to c being constant at both frames. In other words you won't have the (c+v) or (c-v) terms to compensate the time to reach the edges. The devices that run from the edges to the center go at a speed vD that is much slower than c and we can say than vW as well. The time for each device from the edge to the center will always be the same since there is no light effect. It is always b/2/vD.

    I still see this stands to force SR to be wrong and to have a winner frame of reference.

    I'd like opinions, see if I made mistakes, if it is a valid good experiment to force a preferred simultaneity.

    Thank you.
    Last edited: Jan 21, 2013
  2. jcsd
  3. Jan 21, 2013 #2


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    There can be no test of Relativity of Simultaneity. That would require the ability to measure the one-way speed of light. You don't gain anything by sending slow signals; that just increaes your errors.
    Last edited: Jan 21, 2013
  4. Jan 21, 2013 #3
    Can you expand on the one-way speed of light? Are you saying also that Wiki is wrong regarding the relativity of simultaneity?
  5. Jan 21, 2013 #4


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    Read the wiki article on the one-way speed of light. According to SR, we assume the one-way speed of light is c in order to establish simultaneity of remote events.
  6. Jan 21, 2013 #5
    Thank you ghwellsjr . I found it already and it's interesting. But the relativity of simultaneity differs mainly by the factors (c+v) and (c-v) that account for the difference in sequence of events (that from the ground it hits the back side before the front side), for the hypothesis that c is constant for both frames. That's independent on the difference between one-way speed and two-way speed, I think. I mean the difference is considerable if v approaches c, as I calculated. I take into account what you are saying, but don't just bow to it and forget about the rest. I would like more opinions. Thanks.
  7. Jan 21, 2013 #6

    Doc Al

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    What's your issue with the factors (c+v) and (c-v)?
  8. Jan 21, 2013 #7
    I just wanted to see why it gets to the back of the wagon first from the ground reference.

    I calculated the difference in time (as Wiki article says it arrives first).

    I got tCA=b'/2/(c+v) and tCB=b'/2/(c-v) so I can understand the Wiki example...

    v=velocity of the wagon.
  9. Jan 21, 2013 #8


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    The quickest and easiest and most successful way to learn and understand SR is to learn how to do Lorentz Transformations. This will allow you to set up and describe a scenario according to one Inertial Reference Frame and then convert its coordinates to another IRF, especially if you draw a spacetime diagram. It's really very simple, although tedious, but it always works correctly and will never fail you.
  10. Jan 21, 2013 #9
    in the train frame. But in the earth frame the device going backwards will take slightly less time. As ghwellsjr suggested, assign values (distances and speeds) in one frame and calculate the scene in the other frame through Lorentz transforms - and draw both spacetime diagrams. It will makes things clear for you in much less time than banging your head against it!
    Last edited: Jan 21, 2013
  11. Jan 22, 2013 #10


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    No, in fact the Wiki article that you linked to provides the Lorentz Transfomation but the link in it has some simplification which I would like to show you here. First, we use units where the speed of light, c, is 1 so that we represent speed as a fraction of the speed of light and represent it with the Greek letter beta, β:


    Next we calculate the value gamma, γ:


    Then we can use the simplified equations:

    t' = γ(t-βx)
    x' = γ(x-βt)

    I have taken the train spacetime diagram from the wiki article and eyeballed in some coordinates for the five events. I show the coordinates as (x,t):


    Notice that I was careful to make sure the paths of the light flash are on a 45 degree angle, meaning that the value of the x-distance traveled is equal to the time interval, which is 9 units of time and 9 units of distance in both directions. And notice that the light hits both ends of the train at the same coordinate time of 10 units, marked by the dashed grey line.

    Looking at the second diagram, I eyeballed in the speed of the train as 0.3c. Since we are transforming from the train's rest frame to the platform's rest frame and the train is moving to the right, the rest frame of the platform is moving to the left, so the speed we want to use is β = -0.3 which means γ = 1.048.

    Using these two factors in the equations above for t' and x', we get a new set of coordinates which I have drawn into the second Wiki spacetime diagram for the rest frame of the platform:


    Notice that this time, I didn't have to do anything special to make the red paths of the flash of light go along the 45 degree diagonal. The Lorentz Transformation takes care of that automatically. You can verify that the light takes 6.6 units of time over that same distance to hit the rear of the train and that it takes 12.3 units for the path to the front of the train. The dashed grey line shows that the light reaches the rear of the train much sooner than the front of the train.

    Now if we had drawn these diagrams on graph paper, we would be able to draw a horizontal line and show that the length of the train was contracted by the factor of 1/γ or 17.2 units.

    Attached Files:

  12. Jan 22, 2013 #11
    Thank you guys. I will review the calculations and plot them in the spacetime diagram.
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