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Need help understanding Einstien's Light Clock experiment

  1. Jul 24, 2014 #1
    A general summary of the experiment is that when the light clock is stationary on the platform it runs as normal, up and down, but when someone is looking at a clock on a moving train, it appears to move slower because the light has further to travel, and thus, time runs slower.
    What confuses me is that there seems to be a general agreement that it doesn't matter what type clock you use, it will always appear to run slower on a moving train. Take the example of a pendulum clock. Just as the reason for the light clock running slower is because the light has further to travel, what would be the reason for a pendulum clock running (or appearing to run) slower under the same conditions? Or an atomic clock, or a watch, or any type of clock? In my head, it makes sense that the experiment only works for light clocks, I don't see any reason why a normal watch would be slower when moving, I mean, it doesnt have a greater distance to travel just as the light in a light clock does. If I'm going to be honest (and I know this is going to sound stupid) I don't get how the light clock experiment proves that time runs slower for moving objects at all. I mean, it proves that light in a moving light clock appears to travel a greater distance, but i still don't think it proves anything about time. If a blue car travels 100m and a red car travels 50m, if they both have the same speed (just as the speed of light is constant) it is clear that the red car is going to have a shorter journey. That still proves nothing about time. It doesn't show that time somehow runs faster for the red car. It just shows that one had a greater distance to travel, just as the light in the moving light clock had a greater distance to travel.
    So my questions are:
    1. How does the experiment prove that time runs slower for moving objects?
    2. Why would other clocks behave the same way (other than simply "because time runs slower")? What is the actual physical reason?
     
    Last edited: Jul 24, 2014
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  3. Jul 24, 2014 #2

    Nugatory

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    We use a light clock in thought experiments because it is drop-dead simple to understand; as you say, "it makes sense" that the experiment just has to work with a light clock. Now what happens if both train and platform observers also have pendulum clocks? They can observe the speed of light by taking the distance between point of emission and point of reflection (different when they're moving or at rest with respect to the light clock) and divide it by the time interval measured by their pendulum clocks (this is just speed equals distance traveled divided by time to travel). The distances are different, so the only way they can get the same speed of light (and it is both a postulate of relativity and an experimentally confirmed fact that they will) is if the time measured by the pendulum clocks between the two events is also different.

    Good analogy, but you need to add three additional constraints to make it match the light clock thought experiment:
    1) The two cars leave together from the same point on their separate journeys.
    2) The two cars arrive together at the same point at the end of their separate journeys.
    3) The two cars travel at the same speed throughout their journey.
     
    Last edited: Jul 24, 2014
  4. Jul 24, 2014 #3
    I guess it depends on what you start off believing to be true, if you accept Einstein’s postulates (which seem to model reality very well) then you’d just need to consider another non-light clock that doesn’t move relative to the light clock and is synchronized with it. If a moving observer says the light clock slows down, but the non-light clock doesn’t then that would basically create a preferred reference frame and violate Einstein’s principle of relativity postulate.

    If you want to show that some particular type of clock slows down from first principles, then I guess you’d need to look at how electromagnetic fields (and anything else that is relevant to the construction of your clock) transform under boosts. Then show that an observer moving relative to your clock observes it to run slow. Of course that begs the question of how you know how things transform under boosts without already knowing about special relativity (not saying there aren’t other paths to get there).

    I’m not too familiar with the history of this, but at a high level this seems similar to the approach Lorentz took.
     
  5. Jul 24, 2014 #4

    Nugatory

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    It is extraordinarily difficult to pull this approach off. You end up having to postulate one mechanism that causes nuclear processes such as radioactive delay to slow down, others that cause biological processes as different as the souring of milk, the greying of my hair, and the growth of a fetus to slow down, yet another that would affect the design of mechanical clocks, and so on.... And somehow all of these different mechanisms would have to produce the exact same time dilation. And even more amazingly, every time that we encounter some new process, we find that it also comes with its own built-in mechanism that miraculously causes it to slow down with motion.

    That's getting to be a lot of baggage to carry around for a theory that isn't even experimentally distinguishable from relativity.
     
  6. Jul 25, 2014 #5

    stevendaryl

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    A pendulum clock involves gravity, which makes things more complicated. Instead, let's consider a windup clock (or an electric clock).

    We can reason as follows:
    1. Pick a rest frame, F.
    2. Choose a windup clock and a light clock so that when they are both at rest in frame F, they keep the same time.
    3. Assume (the principle of relativity) that all inertial frames are equivalent.
    4. Then it must be true in any other frame F', that if that light clock and that windup clock are at rest in frame F', then they keep the same time.
    5. We can prove that, as measured in frame F, a light clock that is at rest in another frame, F', must run slower than one at rest in F.
    6. Therefore, the same must be true of a windup clock (since they keep the same time in frame F')

    This doesn't say WHY the windup clock slows down, as measured in frame F, but it shows that it must happen; if it doesn't that would violate either the principle of relativity or the principle that light has a constant speed.
     
  7. Jul 25, 2014 #6
    Hi everyone,

    I’m not an expert on relativity but there is one important element that helped me understand the clock experiment and I think it could be useful for you too. The main point of the light clock is to compare the time it takes for the light to complete a loop. And as you said very well, it is normal to understand that light will take longer to travel a bigger distance, which seems to be the case of the clock in the train. But let’s not forget about the word RELATIVE. The person that is in the train could not be aware that he has a different speed than the person in the platform. He doesn’t even need to know that the clock he is using is a light clock. Let’s suppose for this experiment that he needs to take a pill every 15 seconds. The observer going on the train at a specific speed will see the watch acting normally, and for him the 15 seconds will be normal. On the other hand, an observer in the platform looking at his own watch will see that the “traveler” is waiting longer that 15 seconds to take the pills. In other words, only who is outside the train can see the difference of time due to the RELATIVE movement. However, this gets even more complicated when you introduce the fact that the observer in the platform is actually moving with the speed of the earth’s traveling + rotation. So why should this observer think that he owns the correct time when looking at his clock (which happens to be a light clock that is also moving).

    In general, everything in the universe is moving, there is no stationary point. So relativity of time is always depending on how it is being watched and the frame of reference you are considering. This helped me understand the reason of fixing the speed of light as constant, as doing so allows us to measure things even when we know that all the frames of references are actually moving.
     
  8. Jul 26, 2014 #7

    ghwellsjr

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    In your head, does it make sense that the experiment would also work identically for a light clock where the mirrors were placed such that the light moves side to side or front to back?

    Let's say the passenger uses his ruler to measure the distance between the mirrors from floor to ceiling as ten feet. So he places two more mirrors upright on the floor side to side that are also ten feet apart and gets another light clock going between them. Finally, he places two more mirrors upright on the floor front to back, one closer to the locomotive and the other closer to the caboose such that they are ten feet apart and sets up a third light clock.

    Now the question is, will they all tick at the same rate?
     
  9. Jul 29, 2014 #8
    I don't know if I understood well, but I think the idea is to set a light clock in such an orientation that the direction of the light is the same as the direction of the train. I haven't done the math, but it quickly comes to my mind that if the orientation of the light clock is such as the beam of light travels in a line parallel to the movement of the train, then you will have a time clock working in two main regimes:

    a)The light leaves the "source mirror" and travels in the same direction than the train. As the train has a speed, to the beam it will take longer to reach the reflection mirror as it is moving away from the beam. (In the limit, if the train reaches the speed of light, the light beam will never reach the reflection mirror, making it impossible to measure time with it or "stopping the time".)

    b) The light reaches the reflection mirror and travels back, this time in the opposite direction of the train. As the "source mirror" is moving with the speed of train, it will catch the reflected beam sooner.

    As I said, without doing the math, my estimation indicates that the additional time it takes to the light in the regime (a) is the same time that is recovered on the regime (b). So they eliminate each other, having a total period time of the clock as if it was not moving. With this orientation, the person in the platform and the person in the moving train see both clocks "ticking" together.

    Is this idea correct? If it is... then I agree with our friend Deep_Thinker97 in the statement that this experiment depend on rather using a light clock on not. Moreover, I also believe now that this also depends on the orientation of the clock.

    Why is this wrong?

    Thanks and best regards.
     
  10. Jul 29, 2014 #9

    ghwellsjr

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    Yes, you understood quite well regarding setting up light clocks in different orientations.

    The train cannot reach the speed of light so "stopping the time" at the speed of light is not a correct conclusion. Instead, you should say that time does not apply for light. Do you see the difference?

    The OP stated that he believed the moving clock ticks at a slower rate than a stationary clock. It sounds like you don't agree.

    The question that I asked was only if all the moving light clocks in all orientations tick at the same rate.

    Deep_Thinker97 did not consider light clocks in different orientations. That is what I'm asking him to think deeply about.

    In your case, you first say that you think the orientation of the light clock doesn't matter and then you say it does depend on the orientation of the light clock so I don't know what you believe.

    Let's forget about other kinds of clocks until we determine whether of not the orientation of a moving light clock affects its tick rate.
     
  11. Jul 29, 2014 #10
    I don't actually see the difference. I understand that the train cannot reach the speed of light, but that is just an imaginary exercise to bring the idea of the light beam never catching the reflection mirror. In this case, the observer in the train will never see the clock ticking, which is why he could assume that time has stopped. But, is this just and observation of the clock? or is there more to it? "Time does not apply for light" What does that mean? Not even for the stationary clock at the platform?


    Before your reply I was trying to give my point of view on the relativity associated with time and speed...
    I actually thought that the light clock experiment was an easy way to demonstrate how moving frames of reference have slower time if you consider light speed to be constant.

    However, I had never thought about the orientation of the clock, and after your reply I changed my mind. Now I think that the orientation of the clock does affect its tick rate.

    More specifically, the ceiling-ground clock assembly and the left-right clock assembly have the same tick rate as the distance light has to travel is always longer when the clock moves. Moreover, the faster the train goes, the longer the distance for light to travel, the slower the “rate of time” or the ticking of the clocks.

    On the other hand, the front-back light clock assembly seems to be different, since the length the light has to travel increases when the beam is in the same direction of the train and it reduces in when the beam is in the opposite direction. According to my mental math, as a result of the canceling effect of the “forward + backward” movement, this clock will always tick with the same rate no matter the speed of the train.

    So basically you have brought an approach that confused me, and I sincerely thank you for that hehe…
     
  12. Jul 30, 2014 #11

    ghwellsjr

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    If you understand that the train cannot reach the speed of light, why would you imagine that a light beam can never catch the reflecting mirror?

    There is no case in which the observer in the train will never see the clock ticking. Why are you bringing up situations that you understand cannot happen?

    When we are talking about how different observers observe the ticking of clocks (which is the measure of time) as they progress from one event to another, they all see the same number of ticks whether the clocks are moving or stationary to any particular observer. There's no ticking going on with light beams because there can be no clock traveling with the light and therefore the concept of time does not apply.

    I think so too.

    If the orientation does affect the tick rate, then a light clock is not a very good clock. In fact, we could not call it a clock at all.

    That is correct.

    I don't understand how you come to this conclusion. Consider the train moving at just under the speed of light. In this case, the light takes just about as long to get to the reflecting mirror for all orientations of the light clocks, doesn't it? You might conclude that it is different in this orientation than in the others but I don't see how it wouldn't change at all with speed.

    But there's an even more insidious problem. If you are going to maintain that the orientation of a light clock changes its tick rate except when it is at rest, then you are tacitly promoting a preferred state of rest and that doesn't comport with experimental evidence, such as the Michelson Morley Experiment.

    Well, hopefully, you can think deeply about the situation some more and figure out what it would take for the moving light clock to tick at the same rate in all orientations.
     
  13. Jul 30, 2014 #12

    At this point, what confuses me the most is the geometry involved in the experiment. Let’s say that the train is traveling in the “X” axis. And let’s call “Z” the perpendicular to “X”, defining with both the ground plane. This leaves the “Y” axis to denote the height above the ground.

    Now here is my confusion: if you consider the classic light clock experiment, where you have a ground-ceiling clock assembly inside the train that is, for the moment, at rest, then the light beam travels only in the “Y” axis, up and down. When the train begins it's traveling in the “X” axis, the beam of light now has to decompose in two movements, in the “Y” axis and in the “X” axis, having a total magnitude of “the speed of light” in the resulting direction of the sum of the vectors of “speed of light” in “Y” and train speed in “X”. As a result, the “Y” component on the light, which is the one you are using to measure the ticking, is now smaller that “speed of light”.

    (By the way: why is the light leaving the source mirror forced to move also on the “X” when the train moves? Why doesn’t it just go in a straight line missing the other mirror?)

    When the orientation of the beam of light matches the orientation of the movement of the train, there is no decomposing of speed vectors. So the speed of light is always completely projected in the axis of measuring time, so no effect of slowing the clock is observed (?).

    So for this particular orientation, even for a train just under the speed of light, and for a tiny observer in the mirror (let’s say an ant), the light beam will be observed as leaving the source mirror at the speed of light. However, in the other orientations, if the ant is standing in the source mirror, the light will not be seen as leaving at the speed of light, but just a little slower instead, because the vector of velocity of the beam is decomposed by the movement of the train in other axis.

    Is this all completely wrong?

    Best regards,
     
  14. Jul 30, 2014 #13

    Nugatory

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    This isn't just a "by the way" question - it's essential to understanding what's going on here.

    First, imagine a man bouncing a ball up and down between the floor and his hands. No problem, the ball is going straight up and down on the y-axis, right? Now suppose that the floor is actually the deck of a cruise ship... Still no problem, the ball is moving straight up and down at right angles to the floor. But then some guy watching the ship sail by notices that the ball seems to be following a slanted path relative to him and says "now wait a moment! Why is the ball bouncing off the floor forced to move also on the x axis? Why doesn't it just go in a straight line up missing his hands?". What do you tell him? Furthermore, the outside observer says that the trajectory of the ball does not make a right angle with the deck, while the person bouncing the ball just as correctly says that it does.

    Second, try analyzing the train problem assuming that the train is at rest and the platform is moving backwards, in the negative x direction. It's the exact same problem, the velocity vector of the light projects on the two observers' x and y axes in the exact same way, the light follows a straight up and down path for the train observer and a slant path for the platform observer.... But now the question of what makes the light move on the x-axis never even arises.
     
  15. Jul 30, 2014 #14
    In the case of the bouncing ball in the deck of a cruise, I assume a classical mechanics scenario where the person holding the ball in his hand is already giving an horizontal velocity to the ball, this also probably happens when the ball touches the deck, due to friction forces. Actually, I also believe that if the cruise is not moving when the person throws the ball, and while it’s in middle air the cruise starts to move, he will feel that the ball get’s left behind. If the cruise is at rest, and the platform is moving backwards, the observer on the moving platform will see a similar pattern of movement, but this is due to relative movement and perceived trajectories.

    However, the bouncing ball has mass, and that is why I think the rules of classic mechanics can apply. It gets accelerated in the X axis by the person and by the moving ground forces.

    The light experiment is different (I think), because light cannot be accelerated (?). As soon as it leaves the source which is pointing fully in the “Y” axis, and in exits instantaneously with the speed of light, what is the reason for it to move forward with the train? Does the train transfer initial conditions to the light beam? I know this is just not a “by the way” question. And the more I think about this examples the more I get focused on classic mechanics rules, which I don’t think apply to the light clock experiment.
     
  16. Jul 30, 2014 #15

    ghwellsjr

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    Special Relativity has no ability to answer your question about how or even if the light will continue to bounce off the two mirrors when you go from a stationary train to a moving train. All it can do is if you describe a scenario in one Inertial Reference Frame (IRF), it can show you what happens in another IRF moving with respect to the first one. That is why I proposed three different light clocks in three different orientations, none of which accelerates.

    Yes, it is completely wrong.

    First off, we are discussing how an observer on the platform understands what is happening to the light clocks on the train, not how an observer on the train understands what is happening except to the extent that the observer on the train must see all three light clocks ticking at the same rate. That's what the Michelson Morley Experiment (MMX) indicated would happen and there is no point trying to explain the facts away.

    Secondly, no observer can observe the speed of light as you indicated. An ant on the source mirror cannot determine (by observation or measurement) that the light is leaving at c or some other speed.

    So assume that all three light clocks tick at the same rate. Now given that we assume that the light is traveling at c in the platform frame, you need to explain how the platform observer, not the train observer, concludes that all three light clocks tick at the same rate (we're not concerned what that rate is). This, by the way, was the problem that Lorentz addressed to explain MMX prior to Einstein explaining it.
     
  17. Jul 30, 2014 #16

    jtbell

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    Suppose the light source is a laser that can produce very short pulses. We put it at one end of a narrow tube, and align it so the pulses travel down the center of the tube without hitting the tube walls, and emerge from the other end. We put this apparatus on the deck of our cruise ship and point it vertically upwards.

    If we stand on the dock and watch the ship sail past us, the tube is still vertical, but it's moving horizontally. In order for the laser pulses to leave the tube, they must travel at an angle from our point of view. After they leave the tube, they continue traveling along the same straight line, at an angle.

    But maybe the pulses really don't leave the tube? Maybe the motion of the ship and tube causes them to run into the wall of the tube?

    Consider this: we can just as well produce the motion of the ship relative to us by leaving the ship stationary and having us ride along the dock in a car traveling at constant velocity. Nothing is "happening" to the ship or the laser apparatus themselves. How can whether the laser pulses leave the tube or not, depend on whether we are moving or not?
     
  18. Jul 30, 2014 #17
    I investigated a little about the Michelson Morley Experiment and, to be honest, I found that most of the content of the article that I read was to deep for me to try to make a real digestion of the causes and consequences of the result. However I think I understood that after the experiment one of the conclusion is that the orientation of the beam does not change the speed of light.

    In the same article I was also introduced to the Lorenz Contractions, which I have read about several times and I have a very vague idea of what their implications are. However, I believe that the platform observer concluding that all three light clocks tick at the same rate has something to do with this space and time contractions of relative moving frame of references.
     
  19. Jul 30, 2014 #18

    stevendaryl

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    One thing about the light clock that makes using it to reason about time dilation a little tricky is that to understand the light clock, you have to take into account length contraction, as well.

    If you have two mirrors on a train oriented perpendicular to the direction the train, and the mirrors are a distance [itex]L[/itex] apart (as measured by a "stationary" observer), then the time required for a light pulse to make a round-trip between the two mirrors can be calculated to be:

    [itex]T_{perp} = \dfrac{L}{c} \dfrac{1}{\sqrt{1-\frac{v^2}{c^2}}}[/itex]

    So that's the time-dilation prediction--the time for one round trip grows as the speed of the train increases.

    But now, turn the light clock so that the mirrors are facing the same direction the train is going. The time for a pulse to travel from the rear mirror to the front mirror is given by:

    [itex]T_1 = \dfrac{L}{c} \dfrac{1}{1 - \frac{v}{c}}[/itex]

    That's because the front mirror is moving away from the rear mirror, and so light takes longer to get to it from the rear mirror. For the reverse trip, the time is shorter, because the rear mirror is moving forward toward the light pulse.

    [itex]T_2 = \dfrac{L}{c} \dfrac{1}{1 + \frac{v}{c}}[/itex]

    The round-trip time is just the sum of these, which works out to be:

    [itex]T_{par} = \dfrac{L}{c} \dfrac{1}{1 - \frac{v^2}{c^2}}[/itex]

    That is not the relativistic time dilation formula. What went wrong?

    Well, since the time for a light pulse to make a round-trip between two mirrors doesn't depend on orientation for a light clock at rest, the principle of relativity says that it can't depend on orientation for a light clock in motion, either. So [itex]T_{perp}[/itex] must be equal to [itex]T_{par}[/itex]. The only way that is possible is if the distance between mirrors depends on the orientation. If we let [itex]L_{perp}[/itex] be the distance between the mirrors when they are oriented perpendicular to the train's motion, and [itex]L_{par}[/itex] be the distance when they are oriented parallel to the train's motion, then comparing the round-trip times tells us:

    [itex] \dfrac{L_{perp}}{c} \dfrac{1}{\sqrt{1-\frac{v^2}{c^2}}}
    = \dfrac{L_{par}}{c} \dfrac{1}{1-\frac{v^2}{c^2}}[/itex]

    So [itex]L_{par} = L_{perp} \sqrt{1-\frac{v^2}{c^2}}[/itex]

    So that tells us that there has to be length contraction in the direction of motion, in addition to the effects of light having to travel different distances.

    There is still one thing left to calculate, which is how [itex]L_{perp}[/itex] relates to [itex]L_0[/itex], the distance between the mirrors of the light clock, as measured in the frame in which the light clock is at rest. The answer turns out to be:

    [itex]L_{perp} = L_0[/itex]

    That is, there is no length contraction in the direction perpendicular to the motion. I don't know of a simple argument for why that would have to be true, though.
     
  20. Jul 30, 2014 #19

    PeterDonis

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    It has to be true because otherwise you won't get the right answer for ##T_{perp}##.
     
  21. Jul 30, 2014 #20
    I think symmetry and the principle of relativity works. Consider a meter stick moving relative to you that is perpendicular to the direction of motion. Does it expand or contract? Either would allow one to construct a preferred reference frame. For example, align the bottom end of your meter stick with the one that moves relative to you. Put a knife at the top of both meter sticks. As the one moving relative to you passes the one stationary with respect to you one of these must be true: it cuts your meter stick (it contracted), yours cuts it (it expanded), or neither (no contraction).

    In the first two cases you've found a preferred reference frame since an observer co-moving with the meter stick that moves relative to you will agree with your statement of which meter stick cut which. For example in the first case you'd say his contracted (moving meter sticks contract) and he'd say yours expanded (moving meter sticks expand).
     
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