Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Continuously confused by Special Relativity

  1. Aug 23, 2013 #1
    Dear Forum members,

    I have a question regarding Special Relativity. I think its answer may be a simple and obvious one but as a complete physics amateur I cannot put my finger on it.

    Imagine a long train travelling east at half the speed of light relative to a 'stationary' observer. For the first part of the experiment a beam of light is fired from the back of the train to the front. For the stationary observer the beam travels a greater distance than it does for a passenger on the train. Because of the constancy of the speed of light, this is accounted for by stating that, relative to the stationary observer, the train travels through time with less speed.
    Now, for the second part of the thought experiment, the beam of light is fired from the front to back, from east to west. For the stationary observer, the light beam travels a smaller distance. Do we then infer that (because of the constancy of the speed of light) the train is in this case travelling faster through time relative to the stationary observer? If so how can this be, how can the carriages speed through time depend upon the direction of a single light beam?

    Please let me know where I am going wrong in thinking this way,
  2. jcsd
  3. Aug 23, 2013 #2


    Staff: Mentor

    Hi phellen welcome to PF!

    There are three effects that you have to account for in special relativity. You already mentioned time dilation, but there is also length contraction and the relativity of simultaneity. Usually (including here) the relativity of simultaneity is the key.
    Last edited: Aug 23, 2013
  4. Aug 23, 2013 #3
  5. Aug 23, 2013 #4


    User Avatar
    Science Advisor
    Gold Member

    I think the easiest way to understand Special Relativity issues like this is to make a spacetime diagram including whatever you know from your description of the original Inertial Reference Frame (IRF), then use the Lorentz Transformation (LT) process to make another spacetime diagram to see what it looks like in another IRF and add more information, then transform back to the original IRF and see what a complete spacetime diagram looks like. The LT process will automatically take care of all three of the effects that DaleSpam mentioned.

    Here is a diagram depicting the scenario that you described:


    I am defining the speed of light to be 1000 feet per microsecond (usec). The stationary observer (let's call him the ground observer) is depicted as the blue line and since he is stationary in this IRF, the dots represent one microsecond increments of his time. The train is depicted as the thick black and red lines. The black line represents the black locomotive and the red line represents the red caboose. In the ground observer's IRF, the length of the train is 4000 feet and it is traveling at 0.5c or 500 feet per microsecond. Does this all make sense to you?

    When the caboose gets to the ground observer, the passenger in the caboose emits a flash of light which travels at c to the locomotive. As you can see, it takes 8 usec to get there. When it does, a second flash of light is immediately emitted back toward the caboose (or there could just be a mirror there to accomplish the same thing). This reflected flash only takes 2.6 usec to get back to the caboose.

    Now we will transform the scenario to the IRF in which the train and the passenger are at rest. We will also add in the dots that indicate one microsecond intervals of time for him:


    As you can see, the train is about 4600 feet long in its own rest IRF and so it takes 4.6 usecs for the light to get from the caboose to the locomotive and another 4.6 usecs to get from the locomotive back to the caboose for a total round-trip time of 9.2 usecs. Note also that time for the ground observer is dilated--it takes longer for each microsecond to pass than for the coordinate time.

    Now we can transform this new scenario back to the original IRF and add some more information:


    What we have added is a green ground marker at 8000 feet that marks where the light "turned around" and we have allowed the reflected light to make it all the way back to the ground observer. In the ground IRF, it takes 8 usecs for the light to traverse from its source to the reflection point and another 8 usecs to get back to the ground observer.

    Note also that time for the passenger is dilated so that he measures the same amount of time for the light to make the roundtrip as it did in the train's rest IRF. This emphasizes the point that any IRF is just as good as any other IRF for depicting all the information of a scenario.

    I couldn't figure out what you meant by "the train travels through time with less speed" but hopefully these spacetime diagrams will make it all clear to you.

    Attached Files:

  6. Aug 24, 2013 #5

    Thanks guys I really appreciate the help. Jhwellsjr thanks for the explanation it really helped, I hadn't been thinking about lights journey back to the grounded observer. I can see how this makes sense when light travels from one end of the train and returns to its starting point, but I'm still a bit confused. What happens if we asses a single journey, a light beam which only travels from the front to the caboose.

    I have added an attachment of the graphs for this. I moved the line for the observer to the middle of the train I hope that's OK. It seems, in this case that for the passenger observer, the interval for the journey (front to caboose) takes longer than for the grounded observer (which it shouldn't because the train and passenger are moving). Is this a feature of SR that the light must travel a return journey? I hope this makes sense and it isn't too silly a question.

    Thanks again

    Attached Files:

    • rel.jpg
      File size:
      28.2 KB
  7. Aug 25, 2013 #6


    User Avatar
    Science Advisor
    Gold Member

    Your scenario covered both directions but you can look at just one direction and see what happens for the journey from the locomotive to the caboose. In the ground rest frame, the journey takes 2.6 usecs, as I mentioned at the end of the second paragraph under the first diagram in post #4 and in the train rest frame, the journey takes 4.6 usecs as I said below the second diagram.

    It's OK but it really doesn't matter where the observers are for this kind of problem because it's a frame issue, not an observer issue.

    That is correct and you got the same answer for the ground observer that I got, 2.6 usecs, and you would have gotten the same answer that I got for the passenger if you had drawn the distance between the caboose and locomotive correctly (remember, the train gets length contracted in the ground frame in which it is moving).

    I don't understand why you think this.

    No, the light doesn't have to travel a return journey.

    You're welcome.
  8. Aug 25, 2013 #7
    I think that this pinpoints my major lack of knowledge. I thought that if an object travels more guickly through space, then it travels more slowly (time dilation) through time (because an object will always have the same speed through space-time). Yet in the example (whereby light travels toward the passenger and grounded observer) the object (train) which has greater speed in fact travels through time more quickly (time dilation-we would see his clock tick 4us in our 2.6 us).

    PS sorry I completely forgot about changing the length of the carriage

    Thanks again :smile:
  9. Aug 25, 2013 #8


    Staff: Mentor

    You always need to keep in mind all three relativistic effects: time dilation, length contraction, and relativity of simultaneity. Here is a brief example:

    Suppose we have a train car of proper length 20 (in units where c=1). In the center of this train car is a flash bulb and there is a detector in the front and in the rear of the train car, and there is a clock next to the bulb and each detector with all three clocks synchronized using the standard synchronization convention. The train is travelling at v=0.6 relative to the embankment.

    Now, in the embankment frame, when the clocks on the embankment read 0, the center clock of the train also reads 0, but the front clock reads -6 and the rear clock reads 6, this is due to the relativity of simultaneity. The front clock is located at 8 and the rear clock is located at -8, this is due to length contraction.

    So, if the bulb flashes at when the clock on the train reads 0 (which is time 0 in the embankment frame) then, because the front of the train is moving away from the flash, it will not reach the front detector until time 20 in the embankment frame. Because of time dilation, the front train clock will only have ticked 16 times, and since it started at -6 it will read 10 when it receives the light from the flash. This matches with the fact that the proper distance is 10.

    Similarly, because the rear of the train is moving towards the flash it will reach the rear detector at time 5 in the embankment frame. Because of time dilation, the rear train clock will only have ticked 4 times, and since it started at 6 it will also read 10 when it receives the light from the flash.
  10. Aug 25, 2013 #9


    User Avatar
    Science Advisor
    Gold Member

    Let me guess: you got the idea that "an object will always have the same speed through space-time" from Brian Greene, correct?

    If you look at the last two diagrams that I made for you in post #4, you can see in one that the passenger is moving and his time is dilated (his red dots are farther apart than the blue dots) while in the other one the ground observer is moving and his time is dilated (his blue dots are farther apart than the red dots).

    But time dilation doesn't apply to light (that's why there are no dots on the thin lines). The different frames establish different propagation times for the light signals. But you can't just say that light takes longer in one frame compared to the other frame because it depends on the specifics of where the light is going from and to. You compared the propagation times for the light going from the locomotive to the caboose but the comparison is opposite for the light going from the caboose to the locomotive. But in both cases, as I said, this has nothing to do with time dilation.

    Furthermore, neither observer can see the time dilation or the light propagation times that are a result of the different reference frames. Whatever they see or measure as depicted in one frame is identical to what they see or measure as depicted in any other frame. That's an important attribute of Special Relativity; all frames are equally valid, none is preferred over the others.
  11. Aug 25, 2013 #10


    User Avatar
    Science Advisor
    Gold Member

    I have made a couple spacetime diagrams to illustrate your scenario. First the train frame:


    And one for the embankment frame:


    Attached Files:

  12. Aug 25, 2013 #11
    Some authors describe relativity like this, although I think quite a few people on this forum would disapprove of this way of thinking. However, since the seed has been sown, we should clarify things a bit. In the ground frame the train is moving so the clocks on the train appear to tick more slowly relative to clocks in the ground frame so the train appears to travel through the time dimension more slowly. To observers at rest on the train, the train is stationary, so it is the ground that moving ('through space') and so clocks at rest in the ground frame tick more slowly ('travel more slowly through the time dimension') than clocks on the train, as far as the observers on the train are concerned. In your first post you asked:
    The trains 'speed through time' does not depend on the direction of any light been. It depends on the trains speed as measured in a given reference frame. I assume the train in the first and second parts of your original post is always travelling at the same speed relative to the ground from West to East, so its 'speed through time' is always the same (slower) irrespective of whether the light beam is going from back to front or front to back. If the train was travelling at 0.6c relative to the ground, then clocks on the train would always tick slower than clocks on the ground by the time dilation factor (0.8) for that speed, as far as the ground observers are concerned.
    The time dilation is not determined by the time it takes a light signal to travel one way. As you have already found out, you end up with different answers depending on which way the light is going. It is more useful to consider the the times for the two way trip of the light signal, from back to front and back again. In ghwellsjr's first diagram it takes the light approximately 10.6 usec for the round trip according to the ground observers and only 9.2 usecs for the round trip according to the observers on the train (second diagram).
    I am glad you noticed that. I have attached some diagrams for your original scenario with a train velocity of 0.6c relative to the ground that are accurate and take length contraction into account. [Broken]
    This first diagram is the point of view of the observers at rest with the ground frame and the train is travelling to the right. The green lines are the worldlines of observers and their clocks onboard the train with (a) at the back and (b) at the front. Two more observers, A and B (The red worldlines) are at rest with ground and have synchronised their clocks with each other. To synchronise clocks, A sends a signal to B which is reflected back to A. A notes hat it takes 4 seconds for the round trip and assuming the speed of light is the same in both directions works out that B is 2 light seconds away. He sets his clock to zero and sends a signal to B and tells B to set his clock to 2 seconds when the signal arrives. (B could do exactly the same with A and it would not change anything.) Because the clocks of A and B are synchronised in this reference frame, any horizontal line on the chart passes through the same time for A and B. At T=2 the light from the back of the train arrives at the front and the time of this event according to the observer at the front of the train (b) is t = 1 second. At T = 2.5 (according to the observers on the ground) the light signal arrives back at the back of the train but the time of this event according to the observer on the back of the train is only 2 seconds. The time dilation factor for 0.6c is 0.8 and we see that the ratio of the times measured is indeed 2/2.5 = 0.8. Note that the one way time from front to back is 2 secs according to the ground frame observers and only 0.5 seconds for the return trip according to the same observers, but those one way times are not directly proportional to time dilation. This second diagram is the point f view of the observers on the train:
    Note that a line drawn through the two T=2.5 events on the red lines passes through the t=2 event on the green worldline of observer a at the back of the train. This line connecting the two T=2.5 events is not horizontal in this reference frame and this is because the clocks in the ground rest frame are not synchronised as far as the observers on the train are concerned. If you look back to the first diagram you will notice that if you join equal time events on the green wordlines that they too are not horizontal and according to the ground obsevers the train clocks are synchronised in the ground rest frame. Also note that the horizontal length of the train is 4 squares in the first diagram and 5 squares in the second diagram. This is length contraction and the ratio 4/4.5 = 0.8 is the same as the time dilation factor mention earlier. The observers on the train synchronise their clocks in the same way as the observers on the ground. They time a two way signal using a single clock and then add half that time to the second clock when the synchronising signal arrives.

    I have not drawn a third ground based observer at the event t=2 when the light signal returns to the back of the train to avoid clutter. It is easy enough to imagine a third ground observer with a clock synchronised with the other ground observers who is right there at that event so he does not have to calculate radar travel times to work out what is happening. For these types of problems, it always easier to imagine there are as many observers as required in a given reference frame, (all with synchronised clocks) who are conveniently local to any event being analysed.

    One difficulty is that measuring a one way light trip requires two clocks. In order to synchronise those clocks an assumption of equal speed of light both ways has to made. Having synchronised clocks assuming the one way speed are equal it is pointless trying to measure the one way speed of light using those clocks. Essentially it is impossible to measure the one way speed of light independently without making circular arguments. However, once you accept the postulate that the speed of light is constant in all reference frames (and in all directions) then you can synchronise clocks using Einstein's method and then happily make measurements of distances etc using one way light signals and spatially separated clocks.
    Last edited by a moderator: May 6, 2017
  13. Aug 25, 2013 #12


    Staff: Mentor

    Thanks, those are very helpful.
  14. Aug 25, 2013 #13
    Wow, all this help is excellent. I'm gonna read through and process it all. Btw ghwellsjr, it was Brian Greene :rofl:

    Thanks very much for all the help
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Continuously confused by Special Relativity