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Time dilation/ light speed-thought experiments

  1. Jul 9, 2011 #1


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    There are a few things bothering me about time dilation, its derivation, and special relativity. Einstein started with the assumptions that light travels at a constant speed no matter the observer or their frame of reference. He then details the light box experiment in which two observers are moving relative to one another (thus with different reference frames) and one observer has a box that emits a beam of light from the bottom of the box to the top where it reflects back down. The observer not in the frame with the box would record that the beam of light traveled a greater distance and since light travels at a constant speed he would say the event took longer. This is the effect of time dilation. What I don’t understand is that if the box only emits a beam of light within itself then neither observer would be able to view the event. Light has to come to you for it to be detected by you. Further more, since the observer with reference frame not containing the box is at distance it would indeed take longer for the light to reach him.
    Also, I am having trouble with the assumption that light travels at the same speed relative to any observer/ reference frame. Consider the following; say we have two observers, starting in the same place, but one remains there and the other is moving away at .6 the speed of light. After 10 seconds the second observer is 6 light seconds away from the other. At this time the first observer emits a beam of light in the direction of the second. Now if the light wave is traveling at the same speed relative to both of them, it should reach the second observer in just 6 seconds (he’s only 6 light seconds away). But after another 10 seconds, the second observer is now 12 light seconds from the first and the wave front is only 10 light seconds away. So either the wave front would have to travel faster than the speed of light (and have some strange properties of extending itself to reference frames) or it must not be traveling at the speed of light relative to the second observer. Has it ever been noted that during the Michelson-Morley experiment the air, in which they conducted the experiment moved with the earth in orbit and rotation. Even if we were dragging through an ether, who’s to say that it permeates the earth and the atmosphere or doesn’t at least change within it? Obviously it has been experimentally shown that light travels at a constant speed, but I fear we may be misinterpreting the results.
    Thanks for reading.
    Think inversely.
  2. jcsd
  3. Jul 9, 2011 #2
    I find this easier to see if one observer moves with the box. Then suppose that every time the light hits the bottom of the box, a tiny fraction of the light triggers a counter on the box clicks up one. The observer moving with the box might see the counter (for a huge box) change once per second. An observer not moving with the box sees the light traveling a greater distance, making the counter change at a slower rate.
  4. Jul 9, 2011 #3
    The problem comes in with "at this time" because the two observers will not agree on the moment they are 0.6 light seconds apart.

    Many of the counterintuitive situations set up by relativity make sense when you take into account that events which are simultaneous in one reference frame are not simultaneous in another.
  5. Jul 10, 2011 #4
    Actually, what you are saying is incorrect, modern experiments are executed in vacuum, so the variation of medium properties with temperature, pressure, humidity, etc. is eliminated. See HERE, for example.
  6. Jul 10, 2011 #5


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    As Fewmet points out, you have not taken into account relativistic time dilation. Since they are moving relative to each other, their passage of time will not agree; they will not agree on when the light beam is 6 light seconds away. What will happen is that, for each observe, when they each individually measure 6 seconds having passed, they will measure the light beam as having travelled 6 light seconds.
  7. Jul 10, 2011 #6
    I too had the same notion as Mr. DeG about light speed being observed the same, no matter to what reference the observer belongs!

    This might sound stupid, apology before hand...

    This is my idea: When referring Einstein's postulate to this theory about light speed, i actually found no sites referring the observer. Which I wonder should be meaning both living and non-living, sensible and insensible, or else referring every matter. So if I'm right about the 'observer' thing, I could presume that a light photon could be an observer too. In this aspect, if the light photon should observe other light photon, the observing light photon should see the other light photon a light year distance away in a period of 1 second. Which is in accord with Einstein's postulate. But I wonder in a single beam of light photon following one another, are they really one light year away from each other? If not, is it that the observing photon observed something different that what I presumed? If it did, how good does it go with Einstein's postulate?
  8. Jul 10, 2011 #7


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    DeG, Welcome to the forum. I like the way you think.

    Let's start with the bottom line--your bottom line--where you said "it has been experimentally shown that light travels at a constant speed". I'm glad you affirm that. But think about how you do an experiment to measure the speed of light. You said something very important, "Light has to come to you for it to be detected by you." So let's say you are going to measure how long it takes for light to travel from you, where you have some sophisticated electronic timing gadgetry, to a target some accurately measured distance away from you. You start the timer when the flash of light is emitted and away it goes. Can you tell when it hits the target to stop the timer? I think you will agree the answer is "no". Once the light has left your vicinity, you have no knowledge of how it is progressing. What you can do is put a reflector at the target and have the light come back to you so that you can then detect it. Isn't that what you said earlier? Now you can calculate the average speed as being twice the distance divided by the time interval, correct? Does that make sense? But you will have to agree that you are not really measuring the speed of light because it could take longer to go from you to the target than it takes to get from the target back to you. How would you know?

    Well that was the quandry that Michelson and Morley had when they were thinking about doing their experiment. Now they, along with everybody else at the time, believed that light only traveled at a constant speed relative to an ether which they thought permeated all of space. They knew that they could not measure the one-way speed of light but they assumed that it would be the same for both halves of their round-trip measurement of the speed of light as long as they were stationary in the ether.

    But they also believed that the round-trip speed of light would change from the true speed as they moved through the ether and since they believed that the earth moved through this ether it would cause the measured speed of light to change depending on how fast they were moving relative to the stationary ether.

    The problem for them was that the technology of the time was not sufficiently precise enough to let them measure this very slight difference in the measured speed of light as the motion of the earth changed during the day and during the seasons. But they figured out an ingenious work-around. They figured out that by comparing the round-trip speed of light along one axis with the round-trip speed of light at right angles to that axis, they should be able to detect very minute differences as a result of the earth moving through the ether.

    But they had a problem since they didn't know the stationary state of the ether. If both axes of their experiment happened to be at a 45 degree angle to the motion of the earth through the ether, there would be no difference in the measured times no matter how fast they were moving through the ether. But if one axis were aligned along the motion of the earth through the ether, they would see a positive difference and if the other axis were aligned they would see a negative difference. So to take advantage of the way these differences would show up, they made their entire apparatus capable of rotating slowly to maximize the positive and negative differences.

    They were sure that during the course of a day and during the course of a year, they would be able to accurately identify the stationary state of the ether. But it always appeared that the earth was always stationary in the ether. They actually concluded that the earth was dragging the ether, just as you proposed, and suggested repeating the experiment at the top of a high mountain where presumably the drag would be smaller. But there were other reasons to reject their explanation and the one that held was that every non-accelerating person who measures the round-trip speed of light will get the same answer.

    But how can this be if it is only when you are stationary with respect to the ether that you will get the same answer to the measured round-trip speed of light? Well, several scientists figured out that if your clocks start running slower as they travel with you through the ether and if your rulers (and everything else traveling with you) get shorter along the direction of motion through the ether, this would explain how it is possible to always measure the round-trip speed of light to be the same.

    But they still had a problem: how to figure out the stationary state of the ether. Remember, it is only when stationary in the ether that the two halves of the round-trip of light will take the same time and the rulers are the correct lengths and the clocks run at the correct rate. This made for a very confusing and ambiguous science because there was no way of determining that stationary state of the ether.

    What Einstein did was say, "forget about trying to find the stationary state of the ether, as long as you are not accelerating, you can assume that you are at rest in it. Your clocks will run at the correct speed, your rulers will have the correct length and most of all, the time it takes for light to travel from you to a target a fixed, measured distance away will equal the time it takes for the light to get back to you from the target."

    You might recognize this last statement as Einstein's famous second postulate. It's very important in your understanding of Special Relativity to realize that Einstein was proposing something that could not and can not be measured, the time it takes for light to get from point A to point B. He is stating without any proof that it travels at c in any one direction. Now you should not take this as a true statement in and of itself, it is only true within the context of the Theory of Special Relativity. It's what makes the Theory useful.

    But you must also realize that it is true for only one assumed rest state of the ether at a time. But we don't call it that or refer in any way to ether, rather we call it a Frame of Reference.

    So now let's apply what we have learned to your scenario. We will use the Frame of Reference in which both observers start at rest and the first one, A, remains at rest. The second observer, B, starts off at 0.6c in some direction. After 10 seconds, A emits a flash of light in B's direction when B is 6 light-seconds away from A. After another 10 seconds, B will be 12 light-seconds away from A and the light will be 10 light-seconds away from A so the light will not have yet reached B. So far so good. But in between these two statements you say:
    Now if the light wave is traveling at the same speed relative to both of them, it should reach the second observer in just 6 seconds (he’s only 6 light seconds away).​
    I think what you are saying here is that relative to A, the light must be traveling at 1.6c in order for it to be traveling at c relative to B. And after the second 10-second interval, the light will have traveled 16 seconds and therefore will have passed B because he has only traveled 12 light-seconds.

    So now we have a contradiction in that after 20 seconds, the light has passed B from B's point of view but it has not passed B from A's point of view, is this what you are saying?

    The problem with your thinking is that you are jumping between Frames of Reference without properly applying the Lorentz Transform (look it up in wikipedia). You need to describe the entire scenario all within one FoR, which is what you did but if you want to see what things look like in a different FoR, you need to correctly apply the Lorentz Transform to obtain a new set of coordinates to your scenario.

    So in your original FoR, you had three events for A. We will describe these with a time coordinate followed by an x-dimension coordinate leaving off y- and z- since they are always zero. So the three events are:

    [0,0] (start of scenario with A at location 0)
    [10,0] (start of light flash at time 10 and location 0)
    [20,0] (end of scenario, 10 seconds later with A still at 0)

    You also had three events for B:

    [0,0] (start of scenario with B at location 0)
    [10,6] (when light flashes at time 10, B is at 6)
    [20,12] (end of scenario, 10 more seconds later with B at 12)

    And two events for the light:

    [10,0] (start of light flash at time 10 and location 0)
    [20,10] (end of scenario, 10 seconds later with the light at 10)

    Now when we transform all these events into the FoR in which B is at rest, we get a new set of coordinates:

    First for A:

    [0,0] (start of scenario)
    [12.5,-7.5] (start of light flash)
    [25,-15] (end of scenario)

    Three events for B:

    [0,0] (start of scenario)
    [8,0] (start of light flash)
    [16,0] (end of scenario)

    Two events for the light:

    [12.5,-7.5] (start of light flash)
    [17.5,-2.5] (end of scenario)

    The first thing we want to notice is that since B is at rest in this new FoR, his x coordinates are alway 0. Next we can note that what took 20 seconds for the whole scenario to last now takes 25 seconds for A and 16 seconds for B. Before, in A's FoR, B was traveling at 12/20 = 0.6c, now in B's FoR, A is traveling at -15/25 = -0.6c. And we can see that the light is traveling from -7.5 light-seconds to -2.5 light-seconds so it never reaches B just like in A's FoR.

    And finally, it is easy to see how the speed of light in B's FoR is c. The distance the light has traveled is -2.5 - (-7.5) = 5 light-seconds and the time in which this happened is 17.5 - 12.5 = 5 seconds. The speed of light is the distance divided by the time which is 5/5 = 1 light-second per second.

    Now this may be very unsatisfying to you because as you pointed out at the beginning, "Light has to come to you for it to be detected by you" and since the light never got to B, how's he supposed to observe it traveling at c? Well you could let the scenario go on a little bit longer and the light will reach B but no matter how long you let it go on, unless you put a reflector in your scenario at some positive distance away from B and let the light reflect back to him, your statement will not be affirmed.

    I think a more interesting scenario is to consider your same two observers but let the light flash at time zero when they are all together, then have a pair of reflectors, say 10 light-seconds away from each of them, one stationary with respect to A and one stationary with respect to B and watch how the round trip speed of the one light flash can be measured by both of them to be c. Note that you will have to pretend that A's reflector is semi-transparent in order for the same light flash to be both reflected and continue on to B's reflector. You can do this in the FoR in which A is stationary and then again in the FoR in which B is stationary and you will see how they both can measure the round-trip speed of light to be c, how they can both assume the one-way speed of light is c, how they each observe the other one to be length contracted and how they each can see the other one experiencing time dilation.
    Last edited: Jul 10, 2011
  9. Jul 10, 2011 #8


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    A photon does not have a valid frame of reference. Photons dont experience the passage of time at all. To try to attribute a FoR to it from which to observe events will result in non-sensical results.
  10. Jul 10, 2011 #9
    Thankyou, Sir. I was really looking forward to clarify this...
  11. Jul 10, 2011 #10


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    Thanks ghwellsjr for your very detailed reply. I understand what you are saying and it makes a lot more sense now, that you'd have to transform the events into the separate FoR. Now I wish to take a look at your bottom line, "how they each observe the other one to be length contracted and how they each can see the other one experiencing time dilation." So you are saying that, according to each observer, the other is experiencing time dilation? Shouldn't one experience a "faster" time and the other a "slower" time? How can they both observe a contracted length? Shouldn't one be contracted and the other expanded? Sorry if this seems trivial; I start my class on modern physics in the fall.
  12. Jul 10, 2011 #11


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    Never mind, I get that last bit now. The observers will both see time dilation and length contraction with respect to the event happening in the other observers FoR. Is this correct?
    P.S. Sorry for the double post.
  13. Jul 10, 2011 #12


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    Hi DeG. I'm still learning this stuff too and this was one of the things that really confused me at the start. How can both observers see length and the passing of time being less?

    I can't give you the mathematical explanation but I can offer you my way of thinking about it. I need to say up front that this is not based on fact, just how I have interpreted all that I have read so far on the subject.

    If you imagine an object in the distance. You can take a measurement of that object by holding up a ruler and derive a length. Now unknown to you, someone turns that object at a slight angle away from you. So when you measure it again, it measures shorter.

    Well I see travelling though 4 dimensional space-time as having a similar effect. The object is 'turned' slightly at an angle in 4D space. Now I am sure that is not what really happens, it is just my way of imagining it.

    The only difference is that the effect if not just visible for you watching an object move through space-time but is also the same for an observer on the object looking at you.

    I imagine it to be a bit like looking through blinds that are tilted slightly. I can't see all of the outside world and the outside world can't see the all of me.

    Obviously the other major difference is that it does this with time too and the faster something travels, the more the 'angle' shortens length and slows down time.

    So who is actually experiencing the effects then?

    There is a theory I have read that says that no one can actually be at rest in space-time due to the effects of gravity. So in a way, we are all moving and experiencing the effects of time dilation and length contractions to different degrees.

    But in order to see these physical effects first hand, two observers must meet up again. (I am assuming that they started off their journey from the same FoR.) In order to do this one or both must turn or speed up (accelerate) in order to meet up.

    So without doing the math, it is a combination of who has travelled the fastest/farthest and/or gone through the most acceleration.

    Again, this is just my way of thinking about it so I hope it helps.
  14. Jul 10, 2011 #13


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    Staff: Mentor

  15. Jul 10, 2011 #14
    The equation, -c²(t-t₀)² + (x-x₀)² + (y-y₀)² + (z-z₀)² =0, represents the characteristic surfaces of the 3D wave operator with apex at (t₀,x₀,y₀,z₀).

    When the charge density and electric current are both equal to zero in Maxwell`s eqns. we get the homogeneous 3D wave equation, which describes the time-evolution of light in vacuum.
  16. Jul 10, 2011 #15


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    There are many approaches to the the "Twin paradox". My current approach to the topic (and I'm looking for feedback on how helpful it is and what ways actually "work") is this. The approach uses space time diagrams and occurs in two parts. The first part is to suggest that if "A" compared his clock to B's clock using the green lines, and "B" compared her clock to A's clock using the red lines, there would be no paradox (see attached diagram).

    The second part (not shown here, but it also uses space-time diagrams) is to demonstrate that Einstein clock synchronization actually implies that A uses the green lines and B uses the red lines.

    I'm of the opinion that getting the people interested in the question in drawing the space-time diagrams themselves would be the best approach, but I'm not sure how to motivate them to do that.

    Attached Files:

  17. Jul 10, 2011 #16


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    I don't see any correlation between matphysik's post and the post by DaveC426913 that matphysik was replying to.
  18. Jul 10, 2011 #17
    "Photons dont experience the passage of time at all."
  19. Jul 10, 2011 #18


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    Humour me. Are you corroborating? Or refuting?
  20. Jul 10, 2011 #19
    Yes, this is nice. Now, when you show the second half (the return leg) you can point out a large time gap that is experienced only by inertial twin. The "returning" twin, simply "skips" over a time interval . The larger its speed at the direction reversal point, the larger the interval.
    Last edited: Jul 10, 2011
  21. Jul 10, 2011 #20
    Stating a fact.
  22. Jul 10, 2011 #21


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    Being evasive.

    The statement was made: how is it relevant to DaveC426913's post about the invalidity of a photon's FoR?

    So, is your post relevant because it is serving to support that claim, or because it is serving to refute that claim? If it isn't one or the other, then how is it relevant to what you quoted?
  23. Jul 10, 2011 #22
    Coordinates (t,x,y,z) satisfying the relation -c²(t-t₀)² + (x-x₀)² + (y-y₀)² + (z-z₀)² =0 are valid inertial coordinates for light.

    For example, in 1+1 and taking (t₀,x₀)=(0,0), -c²t² + x²=0 for which |x|=ct. Taking x>0, x=ct which Lorentz transforms to x`=ct`.
  24. Jul 10, 2011 #23


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    So. Refuting then. Thank you.

    Clarification was requested in post 16. In the future, here on PF, you'll need to provide it in a way that it requires somehat less than 5 back and forth posts to get there.
    Last edited: Jul 10, 2011
  25. Jul 11, 2011 #24
    So if photons don't have a frame of reference and do not experience a passage of time, what is this 'time-evolution of light in vacuum' that matphysik was referring to?
  26. Jul 11, 2011 #25


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    That is the unanswered question. I have a feeling it'll be a few pages yet before we have an answer...:wink:
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