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Train thought experiment confusion

  1. Aug 26, 2014 #1
    I'm seeing examples where for the person in the train the offset of both lights at the end of the train are not seen as simultaneous, while for the observer on the platform it is seen as simultaneous - and vice versa from other sources. What's going on?

    Furthermore, if my school textbook is correct in stating that it is not seen as simultaneous for the person in the train, theoretically, could they not set up an experiment of setting up two lights to flash simultaneously, stand in the middle, and if they are seen as non-simultaneous, conclude that their train must be traveling at relativistic speeds, thus breaking the law of relativity? What am I missing?
     
  2. jcsd
  3. Aug 26, 2014 #2

    pmr

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    I think we need to put mathematical labels on things in order to make sense of this. I'm going to propose the following scenario:

    Jack on the platform thinks that he is globally at rest. We will call his frame ##Q##.

    Ashley is on the train, and she doesn't care to look outside of it. From her perspective the seats, luggage, and all of the objects on the train appear to be at rest. We will call her frame ##Q'##.

    The catch is that ##Q'## is moving relative to ##Q## at speed ##u##. So Jack sits on his platform ##Q##, and he sees Ashley in her train ##Q'## speed past at speed ##u##.

    Now, Ashley sets up two lights on her train ##Q'## and stands perfectly in the center between them. One is at location ##x'_1## and the other is at location ##x_2'##. At some arbitrary time ##t_1## the light at ##x'_1## fires a photon in her direction, and at another arbitrary time ##t'_2## the light at ##x'_2## fires its own photon in her direction. Times ##t'_1## and ##t'_2## might be different, or they might be the same. Either way, we have two "events" here. The release of the one photon is event ##(x'_1, t'_1)##, while the release of the other photon is event ##(x'_2, t'_2)##

    Now, let us establish the notation
    [tex]
    \eqalign{
    \Delta x' &= x'_2 - x'_1 \cr
    \Delta t' &= t'_2 - t'_1 \cr
    }
    [/tex]

    Next let us say that Jack measures these events to be at ##(x_1, t_1)## and ##(x_2, t_2)## in his own frame ##Q##. Again we will say that
    [tex]
    \eqalign{
    \Delta x &= x_2 - x_1 \cr
    \Delta t &= t_2 - t_1 \cr
    }
    [/tex]

    The Lorentz transform tells us that
    [tex]
    \eqalign{
    \Delta x &= \gamma(\Delta x' - u\Delta t') \cr
    \Delta t &= \gamma\left(\Delta t' - {u\Delta x'\over c^2}\right) \cr
    }
    [/tex]
    where ##\gamma = 1 / \sqrt{1 + u^2/c^2}##.

    Now, let us say that it just so happens that in ##Q'## we have ##t_1' = t_2'##, or in other words ##\Delta t' = 0##, meaning that both photons leave their lights at the same time as measured by Ashley in ##Q'## on the train. What will be the time delta between the two events from Jack's perspective on the platform? Will be ##0## like for Ashley? The Lorentz transform tells us that the time delta for Jack will be
    [tex]
    \Delta t = -\gamma\left({u\Delta x'\over c^2}\right)
    [/tex]
    The important thing here is that this is clearly not ##0##, so Jack sees the photon emissions as non-simultaneous, even though from within the train Ashley saw them as simultaneous.

    This result does not contradict any postulates of relativity. If you think you have a hypothetical result that does then try to expand on the experiment that I've outlined here, using the same language that I've used here, and we can perhaps try to make sense of things.
     
  4. Aug 26, 2014 #3

    Nugatory

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    Staff: Mentor

    It can be done either way. If you set it up so that the flashes are simultaneous according to the train viewer they won't be simultaneous for the platform viewer, and vice versa. The point of the thought experiment is that if they are moving relative to one another (note that I did not say one is moving and one is at rest - see below) they won't agree about simultaneity.

    Flash simultaneously according to which observer? If we arrange for the two flashes to happen simultaneously according to the platform observer and everyone else at rest relative to the ground nearby, then the train observer will see them as non-simultaneous and you will be able to correctly conclude that the train observer is moving relative to the platform observer and the ground.

    However, you will not be able to conclude that the train observer is moving and the platform observer is not. We could just as easily say that train observer is at rest while the ground and the platform are moving rapidly backwards. We could further reinforce this point by flashing two lights simultaneously on the train, noting that they weren't simultaneous to the platform observer and therefore - by exactly the same logic you used above - that the platform "really" is moving.

    At first it may seem to you a bit silly to suggest that the ground is moving and the train isn't... but consider that the surface of the earth is moving at 1600 km/hr because of the earth's rotation, the earth is moving about the sun at some kilometers per second, the entire solar system is moving through interstellar space... Or you just just consider the entire experiment as if you were watching it through a telescope from Mars (you aren't moving, you have your feet propped up on the table in the observatory enjoying a cold drink as you watch)...
     
  5. Aug 26, 2014 #4

    TheDemx27

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    Gold Member

    I'm a bit confuzzled about this. I thought ##\Delta x## and ##\Delta t## were more simply

    [itex]\eqalign{\Delta x = \gamma\Delta x'}[/itex]

    [itex]\eqalign{\Delta t = \Delta t' / \gamma}[/itex]

    Why are we subtracting lengths from lengths (for instance) when we already have our ##\Delta x'##'s and our ##\Delta t'##'s?
     
  6. Aug 26, 2014 #5

    Nugatory

    User Avatar

    Staff: Mentor

    There are two similar-looking sets of formulas. One is the Lorentz transformations, which relate the coordinates in one frame of a single event to the coordinates in a different frame of the same event. They answer questions such as "This event happened at a point ten light-seconds behind me, twenty seconds ago. Where and when did it happen according to you who is moving relative to me?". These are:
    [tex]
    \eqalign{
    x'= \gamma(x-vt) \cr
    t'=\gamma(t-vx)
    }
    [/tex]
    where ##\gamma = 1 / \sqrt{1 - v^2}##
    (By measuring distances in light-seconds and times in seconds I've managed to get the speed of light to be equal to one, so I don't need to mess up the formulas with factors of ##c## and ##c^2##).

    The second set of formulas, which is what you're thinking of, are for calculating time dilation and length contraction: "If I say that there are ##\Delta{t}## seconds between two events, namely two consecutive ticks of my clock, how many seconds do you see between them?" and "If I say that the left-hand end of the rod is here and the right-hand end is ##\Delta{x}## meters away at the same time, how far apart do you say the two ends are?" Both of these formulas are derived from the Lorentz transformations (and it's a good exercise to try this yourself).

    Unfortunately, many pop-sci treatments of relativity focus on the time dilation and length contraction although the Lorentz transforms are more fundamental and more important. Einstein developed special relativity (google for "On the electrodynamics of moving bodies" to find the original paper) by starting with two postulates and deriving the Lorentz transformations from them. The length contraction and time dilation formulas then follow from there, but so do many other equally important results.
     
  7. Aug 26, 2014 #6

    TheDemx27

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    Gold Member

    *Slaps forehead*

    Thanks Nugatory.
     
  8. Aug 27, 2014 #7
    Thank you for the answers, it cleared up a lot of things. I am in high school so the maths is beyond me, however I can see that there is logical reasoning behind the solution. At least I'll have something to look forward to studying in the future haha.
     
  9. Dec 23, 2016 #8
    Someone should put together some rules for thought experiments telling what you are allowed to do. Are you allowed to pretend you are an observer inside the thought experiment? If so, are you allowed to change which observer you are in the thought experiment? The physicists never got together to establish a standard set of rules for us to follow.
     
  10. Dec 23, 2016 #9

    Ibix

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    Science Advisor

    What on earth do you mean? Getting together and figuring out the rules is pretty much all physicists do. It's a somewhat casual, but perfectly reasonable, definition of fundamental science. We even write the rules down in text books and papers for you to read if you like.

    For thought experiments, all you have to do is say to yourself: could I do this in reality? You asked if you can change from being the observer on the platform to being an observer on the train, I think. Well, can you get on a train? If so, then you can switch. There are implications to that, of course, because you are no longer an inertial observer, but that's the way the world is.

    Edit: Zombie thread...
     
    Last edited: Dec 23, 2016
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