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

A Time Dilation Paradox?

  1. Dec 25, 2013 #1
    Hello!

    I am currently learning the theory of relativity, but have a hard time understanding the time dilation concept. In my example down here, the subscript “A” will be used to denote “in the frame of reference of observer A”; the subscript “B” will be used to denote “in the frame of reference of observer B”

    Imagine that there are 2 trains, A and B, which move relative to each other. Let the time elapsed by an event that happens on train A be 1 second in A’s frame of reference (i.e. ∆tA=1). By Lorentz transformation, from B’s perspective, the time elapsed is
    ΔtB = [itex]\frac{Δt_{A}+vx_{A}}{\sqrt{1-v^{2}}}[/itex]=[itex]\frac{1+v*0}{\sqrt{1-v^{2}}}[/itex]=[itex]\frac{1}{\sqrt{1-v^{2}}}[/itex]. Thus ∆tB > ∆tA

    The derivation above depends on the fact that we assume that the train B is the stationary observer. But also by Einstein theory of relativity, it is impossible to tell which object is really moving. Thus, we can turn the problem around. Now let the event happen on train B instead; and the time measured by observer B is 1 second (i.e. ∆tB=1). With the same arguments, we can say that ΔtA=[itex]\frac{Δt_{B}+vx_{B}}{\sqrt{1-v^{2}}}[/itex]=[itex]\frac{1+v*0}{\sqrt{1-v^{2}}}[/itex]=[itex]\frac{1}{\sqrt{1-v^{2}}}[/itex]. Thus ∆tA > ∆tB

    But how can these both be true? Isn’t that a paradox that ∆tB > ∆tA and ∆tA > ∆tB? In reality, how do we know which clock would run slower than the other?

    I also have done some research and realized that this is kinda similar to the “twin paradox”. However, the problem presented above doesn’t include any accelerated frame of reference.

    So what have I done wrong here? please help me out! Thank you in advance!
     
  2. jcsd
  3. Dec 25, 2013 #2

    Nugatory

    User Avatar

    Staff: Mentor

    Despite the superficial similarity, this is not a Twin Paradox problem; it's much simpler. You've forgotten to consider the relativity of simultaneity.

    There are a number of good explanations already posted to this forum... I'll dig up a pointer to one in a few minutes if someone else doesn't beat me to it.
     
  4. Dec 25, 2013 #3

    Nugatory

    User Avatar

    Staff: Mentor

  5. Dec 25, 2013 #4

    ghwellsjr

    User Avatar
    Science Advisor
    Gold Member

    Hello to you and welcome to the forums.

    You're on the right track (pun intended) to use the Lorentz Transformation (LT) to learn about Special Relativity. However, you also need to learn the correct terminology. You are using the word "event" in a colloquial sense, that is, something that takes an elapsed time. However, in SR, the word "event" has a much narrower meaning. It refers to a specific instant in time at a specific location in space. We can apply coordinates (3 spatial and 1 temporal) to the event according to an arbitrarily chosen Inertial Reference Frame (IRF). Then we can use the LT to see what the coordinates of that same event are in a new IRF moving at a constant speed with respect to the first IRF. SR says that all IRF's are equally valid, none is preferred, not even one in which a particular observer is at rest.

    When we describe any particular scenario according to any particular IRF, we need to specify all the significant events for all the observers and objects. In your case, what you are calling an "event" is really two events, one marking the beginning of the interval and one marking the end of the interval. Since in the IRF under consideration, the events happen at the same location, you can consider the difference between the temporal coordinates to be the amount of the elapsed time.

    With that little bit of information, you have everything you need to understand what Time Dilation is all about and you don't need two trains. All you need is a single clock (or an observer with a clock) stationary in an IRF. Then pick out two times (it's easier if the first event is at the origin with all coordinates equal to zero) and transform to a new IRF moving along the x-axis. Make sure you transform both the x and the t coordinates. I would suggest that your second time be 10 units and that you transform to a speed of 0.6c (just to make the arithmetic rational). What is the time coordinate now? Just for fun, try it at 0.8c.

    I think it helps to plot the events on a diagram which means you will have to also calculate the x-coordinate.

    Just remember, Time Dilation is a coordinate effect. For inertial clocks that pass through the origin (all 4 coordinates equal zero) it is the ratio of the Time Coordinate to the Time on the clock. It should be no more confusing to you than the fact that the x-coordinate is different in different IRF's. Both the x- and t-coordinates for any particular event (except the one at the origin) have different values in each IRF. Don't try to make it any more complicated than that.
     
  6. Dec 25, 2013 #5

    Dale

    Staff: Mentor

    Hi haisydinh, welcome to PF!
    I am going to make a very small correction to your formula: [itex]\frac{Δt_{A}+v\mathbf{Δx_{A}}}{\sqrt{1-v^{2}}}[/itex]
    It isn't a very big correction, but it is important that if you are using coordinates you consistently use coordinates and if you are using displacements then you consistently use displacements and you don't mix up the two.

    Correct. If ##Δx_A=0## then ##Δt_B>Δt_A##.

    No. There is no need to label one observer "stationary". The term "stationary" has no meaning other than an arbitrary label. There is no dependence whatsoever on which observer is stationary and which is not.

    This is also correct. If ##Δx_B=0## then ##Δt_A>Δt_B##.

    They are both true because you have used the same labels to label two different things. In the first you worked a problem where you assigned ##Δx_A=0## and proceded to prove something about ##Δt_A##. In the second you assigned ##Δx_B=0## and proceded to prove the exact same thing about ##Δt_B##. You cannot have both ##Δx_A=0## and ##Δx_B=0## in the same scenario, so it should be clear that these are two different scenarios which are simply re-using the labels.

    The situation where x is stationary wrt A (##Δx_A=0##) is a distinct scenario from the situation where x is stationary wrt B (##Δx_B=0##). Because of the principle of relativity, we know that the same laws of physics will apply to both scenarios, but that doesn't mean that they are the same scenario. I hope that clarifies the symmetry, it is helpful if you don't re-use variables before and after a change of scenario like that.
     
  7. Dec 25, 2013 #6

    dvf

    User Avatar

    (Quibble)
    Almost correct. Mathematically that should be:
    If ##Δx_A=0## then ##Δt_B>=Δt_A##
    and
    If ##Δx_B=0## then ##Δt_A>=Δt_B##.

    And then, if they insist on having them both in one scenario, then clearly:
    If ##Δx_A=Δx_B=0## then ##Δt_B>=Δt_A## and ##Δt_A>=Δt_B##,
    and thus ##Δt_A=Δt_B##,
    which can only be if ##v=0## or ##Δt_A=Δt_B=0##.
    If we exclude the former (##v=0##), then the only remaining possibility is the latter, which can be trivially derived from the transformation equations. Of course this is a useless scenario, since ##Δt_A=Δt_B=Δt_A=Δt_B=0## means that we would only be referring to only one single event.
     
  8. Dec 25, 2013 #7

    Dale

    Staff: Mentor

    Excellent quibble :smile:
     
  9. Dec 25, 2013 #8
    I don't really understand what you mean by the relativity of simultaneity. Also, I can't get my head around the concept that both clocks can both run slower compared to one another, simply because i don't really get the problem in the link that you posted. Im just a beginner in SR :cry:

    Thank you for the correction. It seems that i have really misused the term "event" in the case of SR

    This is really what i suspected when i first thought about the problem. However, i then ask myself what would really happen in reality. How does the nature know which clock to slow down? I mean, if what you are saying are right, then the mathematics can only predict that there's time dilation between the 2 trains, but cannot predict in which train time would actually slow down.
     
  10. Dec 25, 2013 #9

    jtbell

    User Avatar

    Staff: Mentor

    How would you decide by experiment in which train time would "actually slow down?"
     
  11. Dec 25, 2013 #10

    ghwellsjr

    User Avatar
    Science Advisor
    Gold Member

    Do you have a problem with nature knowing which clock is traveling and which is at rest? Or do you readily accept the idea that it's optional? In fact, can you think of both of them moving in opposite directions at some "average" speed? Do you feel like nature has to have an absolute rest state?

    You shouldn't think of Time Dilation as being between 2 trains, that will always lead to confusion. Time Dilation is between each single train and whatever frame you decide to reference it to. No train can see its own Time Dilation nor the Time Dilation of any other train.
     
  12. Dec 25, 2013 #11

    Nugatory

    User Avatar

    Staff: Mentor

    Relativity of simultaneity is one of the basic concepts of relativity, something that you have to understand before you will be able to make sense of length contraction and time dilation. If you google for "einstein train thought experiment" you will find a number of good explanations - the first hit is a link to the wikipedia article on "relativity of simultaneity".

    You will be lost until you understand that concept.
     
  13. Dec 25, 2013 #12
    Yes, i've read about that concept before. It was actually the first thing in my textbook. However, my question was that how would i use this concept to solve the problem i asked in the beginning?

    Well, I couldnt think of such an experiment, but are you suggesting that there's no way we can know in which train time would actually be slower than in the other train?

    Is what you are talking about somewhat related to quantum mechanics, more specifically the double-slit experiment (that the photon can actually pass both holes at the same time, unless we observe it)?, i.e. the time on both trains can be slower to each other. Cuz i find these two problems really similar.

    Anyway, I now have another question. Consider the twin paradox, but this time, the twin on the spaceship will not return to the Earth, but continues to travel in a straight line. Does the twin on the spaceship still age younger than the one on Earth? or is it impossible to know if we use the theory of special relativity?
     
  14. Dec 25, 2013 #13

    Nugatory

    User Avatar

    Staff: Mentor

    It is essential to understanding the apparent paradox of both clocks being slow compared to the other one. (And I've bolded the phrase "at the same time" to make sure that you notice it - most relativity "paradoxes" are constructed by trying to sneak that idea in without you noticing).

    First, let's think about about what it really means to say that a clock B is slower than clock A. First, we set the two clocks to the same time (say, noon) and start them running. Then, we start one of the clocks moving relative to us, wait a while and compare the readings of the two clocks at the same time. If clock A says that it is 13:00 when clock B says that it is 12:40, we say that clock B is running slow.

    In other words, we say that clock B is running slow because the two events "clock A reads 13:00" and "clock B reads 12:40" are simultaneous.

    But thanks to the relativity of simultaneity, these two events are not simultaneous for an observer who is moving relative to us. Indeed, that observer might find that "Clock B reads 12:40" and "clock A reads 12:20" are simultaneous events, and then conclude that clock A is the slow one.

    The easiest case to analyze is the one in which each observer is at rest with respect to one of the clocks; that's the one that you'll usually see in introductions to relativity. Unfortunately, that's also the case that is most likely to confuse the beginner by making them believe that the time dilation is caused by the motion of the clock rather than the observer's motion relative to the clock.
     
  15. Dec 25, 2013 #14

    Dale

    Staff: Mentor

    Exactly. The words "really" and "actually" don't really actually mean anything.

    There is no "real actual" velocity. Velocity is only relative to a given reference frame.

    Similarly, there is no "real actual" slow-down of a clock. The time-dilation is only relative to a given reference frame.

    I hope that I didn't overstate it and make it more confusing than helpful, but the point is that frame-dependent quantities don't have any "real actual" value, just their value wrt a given reference frame.
     
  16. Dec 26, 2013 #15

    ghwellsjr

    User Avatar
    Science Advisor
    Gold Member

    No, it's not related to QM.

    OK, we'll take this problem. We'll start with the twin on the spaceship and I'm going to do what I suggested that you do in post #4. First, I'll show the spaceship twin at rest in his own Inertial Reference Frame (IRF). The dots show 1-second increments of his time:

    attachment.php?attachmentid=65103&stc=1&d=1388049615.png

    Next I'm going to use the Lorentz Transformation to see what he would look like when he is traveling at 0.6c to the right (by transforming to -0.6c):

    attachment.php?attachmentid=65104&stc=1&d=1388049615.png

    Now you can see his time is dilated according to this new frame. Gamma at 0.6c is 1.25 which means that his 1-second increments of time have been stretched out to 1.25 seconds of Coordinate Time.

    Next I'm going to add in the earth which is stationary in this frame:

    attachment.php?attachmentid=65105&stc=1&d=1388049615.png

    As you can see, earth's time is not dilated because its speed is zero and gamma is 1.

    But now I want to add in one more feature and that is how each twin views the other twin. To do this, I add in light rays that go from the dot of each twin to the other twin along 45-degree diagonals:

    attachment.php?attachmentid=65106&stc=1&d=1388049615.png

    As you can see, each twin sees the other ones clock ticking at one-half the rate of his own. This is called the Relativistic Doppler Effect (RDE) and shows us what they can really actually see. They cannot see the Time Dilation because it is frame dependent but the RDE is not dependent on the frame.

    To illustrate, let's go back to the rest frame of the spaceship twin:

    attachment.php?attachmentid=65107&stc=1&d=1388049615.png

    Note that all the light rays are traveling at c and they arrive at the same Doppler-shifted rate as before but the Time Dilation has shifted from the spaceship twin to the blue earth twin.

    And to make the point even clearer, I'm going to transform the original diagram to a speed of 0.33333c:

    attachment.php?attachmentid=65108&stc=1&d=1388049615.png

    In this diagram, the Time Dilations of both twins are the same because they are traveling at the same rate (0.333c) in the frame (in opposite directions) and they also continue to see each others time going at one-half the rate of their own. At 0.333c, gamma is 1.06 so that is the Time Dilation factor.

    The point of all this is to show how Special Relativity successfully incorporates the observations of each observer into any arbitrary IRF and how it assigns frame-dependent parameters (such as Time Dilation) that are a result of the speed of light being c in any IRF.

    So the answer to your question is that just like we assign the speeds arbitrarily according to our IRF, the Time Dilation follows from the formula for gamma based on speed. In other words, it depends on the IRF. But what always happens, independent of the IRF, is that each twin sees the other one aging at half his own rate.
     

    Attached Files:

    Last edited: Dec 26, 2013
  17. Dec 26, 2013 #16
    Thank you very much for such a detailed explanation, Ghwellsjr! That is exactly what that puzzled me when i first studied time dilation. Now i understand that there's no such a paradox with time dilation because it is frame-dependent. RDE is certainly a new concept that i really need to do some further reading. So thank you again! :smile:
     
  18. Feb 11, 2014 #17
    Thanks for the nice simplified explanation.

    Does that mean, when frame A observer clock reads 13:00 , simultaneously frame B clock reads 12:40. Similarly in frame B when its clock read 13:00, simultaneously frame A clock will read 12:40.

    In this way, both see the other clock run slow and both are correct.

    Is that right?
     
  19. Feb 11, 2014 #18

    Nugatory

    User Avatar

    Staff: Mentor

    yes.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: A Time Dilation Paradox?
Loading...