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Some thought experiments

  1. Oct 1, 2013 #1
    I've just started learning about special and general relativity and came up with a couple of questions that look like paradoxes to me. I just need a break from reading, thought I'd discuss this with someone to see the application of what I've read so far. Here are my examples:

    1. Say you leave earth on a spaceship capable of instantly accelerating to very near c and you're travelling to a garage that is 1 light-hour away. The second you leave earth, your butler activates a signal to close the garage door. It takes 1 hour for the signal to get there and he times it so it gets there 1 second after you. From his point of reference, looking into his telescope he would see light getting to the garage door 2 hours after he sent it out, a second after you flew through the open door. But from your reference point what do you see? If you look out of the ship, the signal would still be c in relation to you, not stationary, so it should pass you at c speed in relation to you, get to the door when you're half way (intuitively) and you will observe the door closing when you're 3/4ths of the way to the garage door, thereby causing you to crash. So which thing actually happens? Do you make it through the door or do you crash into it?

    2. You decided to build a death star, you build it on earth and then accelerate it to travel at 0.8c towards a planet on the other side of the galaxy that is stationary in relation to earth. Once you're on the death star, you are at rest in relation to it. And while on it, you decide to built a space station, which you then launch from the death star towards the target planet at 0.8c in relation to your death star. Then later, while on the space station, you launch a space ship from it towards the target planet, again at 0.8c. What speed are you traveling at in relation to the target planet? Also light from earth is still technically supposed to arrive at the target planet before you, since it's travelling at c in relation to you, right? Am I missing something?

    3. Say you were going to test out the twin paradox, but instead of travelling out an back, you got into a massive torus, where you got magnetically accelerated to near c. Does it still count that you are in motion at the speed of c in relation to an observer outside the tube, or does the fact that it's angular momentum completely change everything? If it doesn't, and you decide to synch clocks, and say, you were moving at 86% c, then would 2x time have passed outside the tube than inside, or because there is no "return journey" the effect would cancel out? At what interval would you receive the clock synch signals and how would they change while inside the tube if they were sent every second based on the outside frame?
  2. jcsd
  3. Oct 1, 2013 #2


    Staff: Mentor

    Hi Serioza, welcome to PF!

    You see the door open as you pass through it, and then the signal arrives; i.e., you do not crash. The key thing you have left out of your analysis is relativity of simultaneity; relative to you while you are in your ship (and moving close to c relative to your butler and the garage), the time at which the butler sends out the door closing signal is *later* (compared to the time at which you leave Earth), late enough that, even though the signal moves at c relative to you, it can't catch up to you before you pass through the door.

    Intuition is not a good guide here; you need to actually work the numbers, being sure to take into account relativity of simultaneity, as above.

    Use the relativistic velocity addition formula:

    w = \frac{u + v}{1 + uv}

    Here we have to do this twice; the first time, ##u = v = 0.8##, so we get ##w = 1.6 / 1.64##, which works out to about ##0.976##. The second time, we have ##u## equal to the first result we just got and ##v = 0.8##, so we get ##w = (0.8 + 0.976) / (1 + 0.8 * 0.976)##, which works out to about ##0.997##. (These are all in units where ##c = 1##.)


    Yes. Subatomic particles in cyclotrons do this routinely, and they are time dilated just as the formulas of special relativity predict.


    There is a "return journey": on every trip around the torus, you return to your starting point, so each revolution corresponds to one "twin paradox" journey out and back.
  4. Oct 1, 2013 #3
    Thank you Peter!
    Ah cool, well the first 2 make some sense now. I'll have to read up about simultaneity more to get the first one.

    As for the third one, what if the syncing clock was being measured from the inside of the torus, so the distance between the observer and the traveller remains constant? Does the clock simply continuously receive 1 tick for every 2 ticks it sends out? But then from the frame of reference of the traveller, he sees the same, 1 tick for every 2 ticks he sends out. In the "there and back" example, even though the traveller sees earth as moving slower, earth fast forwards a few years when he turns around, thereby catching up all that slowed time. But when does the catchup happen in a torus, while the observer is in the middle? During deceleration?
  5. Oct 1, 2013 #4


    Staff: Mentor

    I'm not sure what you mean. Do you mean the observer is at the center of the torus?

    In any event, what I said in my last post applies generally; it doesn't depend on the observer being in a particular location. The key point is that, relative to the observer, the traveller returns to the same point in space on each revolution; that's what makes the scenario analogous to the ordinary twin paradox.

    The precise location of the observer does affect the details of what he observes during each revolution; see below. But it doesn't affect the general point I made in my last post.

    First let's restate the fundamental fact that anchors everything else. Pick one location on the torus and call that point O. The traveller passes through point O once per revolution; call two successive passages of the traveller through point O events A and B. The fundamental fact is that the traveller's elapsed time between events A and B is half the observer's elapsed time between events A and B.

    Now: suppose the traveller sends out a light signal each time his clock ticks, and the observer sends out a light signal each time *his* clock ticks. Suppose, for definiteness, that the traveller's clock ticks exactly once per revolution, at the instant he passes point O. Then the traveller will send out a light signal from events A and B. Suppose also, for simplicity, that the traveller *receives* light signals from the observer at events A and B, at the same instants that he sends out his own signals. Call the events at which the observer sends out these two light signals events A' and B'.

    Since the traveller's clock ticks once per revolution, the observer's clock will tick twice per revolution. We know the observer's clock ticks at events A' and B' (because he sends out light signals from those events); but it will *also* tick at exactly one event in between them (because it ticks twice per revolution, and there is one revolution between A' and B'). Call that event C', and call the event at which the traveller receives this light signal event C. Then the exact time of event C, by the traveller's clock (i.e., exactly when during his revolution around the torus he receives that light signal), will depend on where the observer is located relative to the torus (because that will affect the light travel time from the observer to various points on the torus).

    Some simple examples:

    (1) If the observer is located exactly at the center of the torus, then event C will be exactly halfway between events A and B, by the traveller's clock (because the observer is at the same distance from all points on the torus).

    (2) If the observer is located exactly at point O, then event C will be *later* than halfway between events A and B (because the observer is co-located with the traveller when he passes point O, so events A and B are co-located with events A' and B'; but event C is *not* co-located with event C', and event C' happens when the traveller is halfway around the torus).

    (3) If the observer is located next to the torus at a point exactly opposite point O, then event C will be *earlier* than halfway between events A and B (because event C is co-located with event C', which happens exactly halfway between events A' and B'; but events A and B are *not* co-located with events A' and B', because the traveller is on the opposite side of the torus when A' and B' happen).

    You should be able to work out other examples from the above. In all these cases, it's still true that the observer's clock ticks twice per revolution while the traveller's clock ticks once per revolution, i.e., the observer's clock ticks twice as fast as the traveller's. But how that appears in direct observations depends on light travel time as well, which is what is affected by the location of the observer relative to the torus.
  6. Oct 1, 2013 #5
    Yes, sorry, that's what I meant.

    What you explained does actually make good sense, but according to STR the traveller must perceive the observer to be slower as well, since motion is relative and the observer is in motion in relation to the traveller just as much as the other way around, except when there is acceleration.
    Quote from wikipedia:
    So it would follow, taking your example, both of the observers would receive either A' or B' for every AB pair they send out, right?

    The reason I said the observer is in the center is because if he was at point O, then one revolution would be similar to a forward-and-back journey, since when the traveller travels the first half of the revolution, up to the farthest point from O on the torus, that would be like the first leg in a straight line from O's point of view. Then, the farthest point would correspond with the turn-around point as shown here:

    and there would be a jump in time at that point (from the traveller's point of view of the observer).

    However, if the observer is in the center, the traveller's speed is constant and the traveller's distance to the observer is constant as well, what happens?
  7. Oct 1, 2013 #6
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