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Couple basic time dilation / length contraction questions

  1. Nov 22, 2014 #1
    It pains me to even type these out. I realize how many threads there are with very similar questions and to someone well versed in these topics, these questions probably all seem the same. But after reading what seems like all the questions, I feel I'm still confused.

    1) In regards to the light clock experiment (and this may be more a classical physics question as opposed to a special relativity question). Due to the extended path (the 'c' in pythagorean's theorem) a light particle takes when traveling between a couple of mirrors at relativistic speeds, time must slow since light has a longer path to follow. The question then, is why is the light particle even following the path of the mirror? Why isn't it reflected perpendicular to the mirror and the other mirror moves out of the way by the time the particle reaches the opposite one (or at least the particle is off-center)? Thus rendering this experiment irrelevant?

    2) Hitting on the twin paradox here. Before I begin, any speeds/velocities listed here are using Bill as the inertial frame. So... Jebediah and Bill are at Cape Canaveral. Jebediah on a rocket and Bill at Mission Control (We're pretending there is one at the Cape still). Jebediah takes off in his super-fast in his rocket and hits .5c. Bill see's Jebediah's Rolex (Bill is straight up eagle eye status here,people) ticking slower (this is conceptual folks, I don't care at this point about the Lorentz factor for exact numbers) than his own. Jebediah reaches uh... wherever. Betelgeuse and turns around. Now he's headed back at .5c. This is where I am confused. Since he is still going .5c (wrt Bill ) his Rolex is still ticking in slow motion, correct? And Jebediah still see's Bill's Rolex (yes he's been watching this whole time, the wonders of auto-pilot... or mechjeb for those of you who already caught on to my reference here) ticking slow since Bill is moving at .5c relative to Jebediah. Now most explanations I've read involve invoking acceleration (i.e. General Relativity, right?) or like, Doppler Shift or something. How does one reconcile this using JUST special relativity?

    3) Length Contraction. The book I'm reading "How to teach your dog relativity" talks about a dog running along a steel barred fence at relativistic speeds. Bob, watching the dog, see's the dog contract some length to be able to fit between the bars of the fence. The dog, however, see's the bars of the fence shrink together some distance (again, not interested in numbers here, just the concept). The dog see's that he would be unable to fit through the fence if he were to change direction a bit. However the outside observer (in the same reference frame as the fence) see's the dog easily able to fit in the fence. So what if this outside observer pushed the dog through the fence? According to the outside observer the dog would fit, however from the dog's reference frame he would hit the fence. I'm clearly missing something here.

    I've got plenty more questions, but I'll leave it at this for now because A) These might answer a number of the rest of my questions, and B) My blood pressure is through the roof after becoming so frustrated by this. And lastly, I hope these all made sense, I just got off work and its 0345.
  2. jcsd
  3. Nov 22, 2014 #2
    The stays on it's path, because there's nothing to change it's path. As long as the clock doesn't change direction the light will keep reflecting. When the clock does change direction, the light will keep going straight, and therefore might miss the mirror.
    Imaging two mirrors which will be | and a photon which will be * the velocity v is <--

    <-- | * | (for some reason this didn't work out the way I wanted it to, it removes the extra spaces.)

    as long as the clock doesn't suddenly change it's velocity, the photon will keep reflecting. when the clock changes direction to for instance downwards, the mirrors will go down, but the photon won't and will eventually exit the clock.

    Not sure about this one, but the basic idea is that time dilation does not apply when accelerating or decelerating. So when Jebediah stops and turns around, he changes reference frame, and at that moment time dilation no longer applies. Maybe someone else can explain this more clearly

    Well here's a question I can answer, length contraction only applies same direction as the velocity. So the space between the bars doesn't contract, when the dog does pass through the fence, the fence would seem thinner, because that distance is in the same direction as the velocity. Imagine the dog is travelling at a velocity v in the x direction, only lengths in the x direction are contracted, the space between the bars is not in the x direction and therefore is not contracted.

    I hope this helps you
    Last edited: Nov 22, 2014
  4. Nov 23, 2014 #3


    Staff: Mentor

    It is reflected perpendicular to the mirror--in the rest frame of the mirror. In that frame it's obvious that the light hits both mirrors.

    But if the light hits both mirrors in one frame, it must hit both mirrors in all frames. You can't change whether the light hits the mirrors or not by changing frames. So in a frame in which the mirror is moving, the light path must be diagonal, so that it still hits both mirrors.
  5. Nov 23, 2014 #4


    Staff: Mentor

    Read the Usenet Physics FAQ article on the twin paradox. Then re-think your scenario and come back if you still have questions. I think that will work better than trying to answer your questions right now.
  6. Nov 23, 2014 #5


    Staff: Mentor

    This is just the barn and pole "paradox" in different guise. The missing piece (as it often is in relativity puzzles) is relativity of simultaneity. Read the linked article and then, if you still have questions, ask them.
  7. Dec 6, 2014 #6
    Thank you for the replies Peter.

    I get the light clock/mirror issue now. The easy way for me to figure that one out, is instead of having the mirror moving relative to me and the earth, is having the mirror stationary on the earth and me moving by it. That seemed to be an easier way to think about it for me.

    The barn and pole "paradox" makes more sense to me than the dog one that was used in this book. The one thing I am trying to figure out with regards to simultaneity, is why is the more distant event seen first wrt the runner/pole? Seems to me the closer door would be seen closing first as the light from it has less distance to travel. I feel I'm missing something here. The book also uses an example of a someone sitting in the middle of a 100ft yard next to a light emitter, with 1 human measurer on either side of the yard to make a mark to measure the yard. Someone comes running by this as the middle person hits a button to emit a flash of light. The someone running by will see the more distant measurer mark the yard before the closer one. Again, I'm caught up on this because with the closer measurer, the light has less distance to travel than from the further measurer. Is it to do with the fact that the light has to travel further to reach the closer measurer? Would the actual distance of the measurers from the light source not affect this? So whether they are 1ft or 1 light year the result is the same? I guess not since the ratio is the same. I hope that all made sense and isn't completely off base.

    As for the twin "paradox"... gimme a little bit more time on that. I wanted to nail down these two things first.
  8. Dec 6, 2014 #7


    Staff: Mentor

    The word "see" is misleading here. Note that the article I linked to explicitly asks the question of how the pole "looks" to the person stationary on the barn roof, separately from how the pole's length is measured by that person. (You might want to read the article linked to at that point, about whether you can actually "see" length contraction.) To the runner moving with the pole, the more distant door is measured to close first--i.e., when the runner takes the time he actually sees each door close, and then subtracts the time it took the light to get to him from each door, he finds that the more distant door actually closed first.
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