• flannel jesus
    1.4k
    I want to see if you guys get the same answers as me. This is a question about how relativity deals with a race.

    We're sat in the stands watching a race. The two racers start at the center of a track (since they're only hypothetical racers for a mathematical puzzle, you may imagine them as occupying the exact same point in space at the start of the race, 0,0,0). You may also assume these racers run at relativistic speeds - some significant fraction of the speed of light. 10%? 25%? 50%? Who cares, you choose. The finish line for both of them is some relativistically relevant distance away - perhaps half a light second? A full light second? It doesn't matter, you choose, if you feel like dealing with numbers - you may answer this question without numbers entirely. One racer is running to the left, one racer is running to the right. At each finish line, one on the left and one on the right, is a person with a stop watch (a mathematically perfect person who clicks the stop watch as soon as he sees the light from some relevant event).

    So, the starting pistol fires (you may assume this is occupying the 0,0,0 point with the racers) and the racers immediately start running towards their respective finish lines. The stop-watch-dudes at each finish line start their watches as soon as they get the light from the starting pistol. After some time (a few seconds? less than a second? who cares?) both racers get to their finish lines, the stop watches are stopped, and it's a perfect tie. Both stop watches are stopped at the exact same time.

    The question in Relativity is, if there's an observer in some relativistic frame of reference, travelling at some significant fraction of the speed of light to the left or the right (you choose, it doesn't matter), how do they perceive the race? Do they also think it's a tie? Do they think the runners ran the same speed? What do they have to say about the stop watches?

    If it makes you more comfortable to deal with real numbers instead of just abstract questions, you may imagine:

    both runners are running at 25% of the speed of light relative to the people in the audience, they reach their finish lines after 4 seconds of running (so they ran 1 full light-second), and the observer watching the race in a relativistic frame of reference is travelling to the right at 50% the speed of light.

    You need not answer with specific numbers about anything, but feel free to.

  • noAxiomsAccepted Answer
    1.3k
    Things not specified:
    The starting point is presumed relatively stationary to both finish lines, in a frame we'll call S.

    OK, the guys with the stop watches at the finish lines measure the time from the light signal to the runners getting there.

    Both stop watches are stopped at the exact same time.flannel jesus
    Both are stopped at the same time in S, but not simultaneous relative to some other frame.

    If there's an observer in some relativistic frame of reference, travelling at some significant fraction of the speed of light to the left or the right (you choose, it doesn't matter), how do they perceive the race?flannel jesus
    They'd observe the closer runner finishing first. They'd also see that the stoped watches read identical values. All this is true even under Newtonian physics, and the observer in question doesn't have to be moving in S to observe any of this.
    The moving observer would probably not consider the race to be a tie in that moving frame S' because the stop watches were not started simultaneously in that frame. They'd consider the slower (in S') runner to have reached his finish line first (a bit unintuitive, I know)

    Do they think the runners ran the same speed?flannel jesus
    Not relative to S', of course not.

    both runners are running at 25% of the speed of light relative to the people in the audience, they reach their finish lines after 4 seconds of running (so they ran 1 full light-second), and the observer watching the race in a relativistic frame of reference is travelling to the right at 50% the speed of light.flannel jesus
    The guy running to the right (slow, at about -.286c relative to S') wins the race. The guy running directly left moves at 2/3 c relative to S' and gets to his destination some time after (in S') the first guy does.
  • flannel jesus
    1.4k
    All same answers as what I got. The slower runner runs a shorter distance and gets there first. The stop watch of the faster runner gets started later, and therefore is still able to show the same reading as the slower runners stop watch.

    I use this to visualise the Lorentz transformations

    https://www.geogebra.org/m/s7xsubde
  • flannel jesus
    1.4k
    All this is true even under Newtonian physics, and the observer in question doesn't have to be moving in S to observe any of this.noAxioms

    In Newtonian physics... I don't know how to do that version of transformation between S and S', but it would seem to me that once you as an observer account for the time it took the light to reach you for each of your observations, wouldn't your results look exactly like they would look from the perspective of a stationary observer?

    And if someone wasn't moving in S, wouldn't they just see the same thing as the original description? Both runners run the same speed, both runners arrive to their destination at the same time. The description of the scenario was from the perspective of someone in S to begin with.
  • noAxioms
    1.3k
    In Newtonian physics... I don't know how to do that version of transformation between S and S', but it would seem to me that once you as an observer account for the time it took the light to reach you for each of your observations, wouldn't your results look exactly like they would look from the perspective of a stationary observer?flannel jesus
    The part I said was true under Newtonian physics was the bit about which runner was first observed to finish, which is a function of where the observer is at the time of that observation and not at all a function of how fast or what direction the observer is moving. There's a set of events that the light from both runners finishing reaches simultaneously. That set of events forms a 3D hyperplane in spacetime. If the observation of the observer is made on one side of that plane, the one runner is first observed finishing, else the other runner is first observed finishing.

    The part asked at the end (which runner actually finishes first in S') is not the same in Newtonian physics, the latter which says that if the race is a tie, it's a tie in all frames since time is absolute. Frame rotations don't involve changing any time coordinate in Newtonian physics.

    And if someone wasn't moving in S, wouldn't they just see the same thing as the original description?
    The original description doesn't say where the observer is ("in the stands" but where in S are 'the stands'?). Anyone in S would compute that the race was a tie, but the nearer runner would first be observed finishing. For instance, each guy with the stopwatch would see the other runner finish 2 seconds later.
  • flannel jesus
    1.4k
    ah okay, I think we're on the same page for pretty much everything.
  • flannel jesus
    1.4k
    may I ask you another relativity question? This might be very basic, but I don't know the answer.

    Imagine you're on a train at a train station, and the train is 1 light second long.

    Now imagine the train accelerates, and it continues to accelerate until it's going half light speed, 0.5c, then it stops accelerating and stays at half light speed (from the frame of reference of the train station, of course).

    Through this process of accelerating, is there any difference in the length of the train *in the reference frame of the train itself*? Or, instead, is the train still 1 light second long in its own reference frame, from the perspective of someone who is on the train?
  • noAxioms
    1.3k
    Through this process of accelerating, is there any difference in the length of the train *in the reference frame of the train itself*flannel jesus
    Train has no wheels since relativistic wheels are a whole new problem. So it's sort of a mag-lev situation where the track is there but doesn't touch.
    There are three trains actually, called F, R, and U.
    Train F is like a normal train with the engine in front, pulling all the stuff behind it.
    Train R is like a rocket, with the thrust coming from the rear
    Train U has every car in the train self-powered, thus achieving uniform acceleration.

    F will stretch under tension. The front begins to accelerate at say 1g but the rear stays put for say a minute until the acceleration wave travels at the speed of sound to the rear when it finally begins to accelerate, and at more than 1g at first. The sound waves bounce back and forth and an accelerometer at the rear wavers up and down until dampers put a stop to that nonsense. While the bouncing is going on, the different parts of the train are not stationary in the frame of the engine, and the train has no proper length. Once the waves stop (a few hours maybe?), the train will have a longer proper length under tension than it did parked at the station unstressed. The accelerometer at the rear will read a slightly higher value that 1g. There's another set of bouncing waves when the acceleration ceases and the rest of the train takes time to notice. The people in the rear get kind of seasick at the start and end of the acceleration.

    R is similar. We accelerate the rear at the slightly more than 1g level. The train compresses, but eventually stops bouncing with a shortened (under strain) proper length. The train is accelerating at 1g as measured at the front.

    U is powered everywhere, and independently. Each thruster knows when to start and when to cut off. Hence there is no stress, and thus no strain ever on this train. It exhibits rigid motion unlike the other two trains. We don't even need couplers between the cars. The proper length of U never wavers and is constant always. The front accelerates at 1g and the rear a bit more.

    At 1 light second and 1g, the difference in acceleration between front and rear isn't much. If the train was a light year long, the difference grows to the point of impossibility.
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