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Close to light speed visualization question

  1. May 24, 2009 #1
    So I'm having a problem actually visualizing what close to light speed travel would look like, both from the viewpoint of an observer on earth and from the viewpoint of an on-board traveller.

    It doesn't seem that any of the visualizations in the movies fit my understanding of it, so maybe I've just done something wrong.

    Anyway, here goes.

    By my understanding, the classic thought experiment is of two twins, one of whom leaves earth in a spaceship flying close to the speed of light, the other staying on earth. When the first returns, he appears only a few hours older, while his twin has aged many years.

    But as the ship always flies off into space, we never see how each perceives the event itself, only the final outcome.

    So I tried changing it a bit. There's a hyper-space Concorde, and it's going from NY to London. It'll be traveling at 2.998 x 108 m/s and will take 15 min. to complete it's journey (ok, so it's a thought experiment. I had to artificially slow it down just to be able to see it. Actually, it would of course take less than a sec.). At the end, about 1000 years will have passed on earth. There are observers at both ends of the journey, one in the middle of the Atlantic, as well as on-board.

    Since the on-board clocks will only show 15 min. have passed, but by standard earth time, it will have been 1000 years, to me, it looks like the ship should continue to appear in the sky for all those 1000 years, going at a rate of, say, 2 feet per minute. In fact, to my mind's eye, the trajectory should be something like:

    The Concorde takes off from JFK. From the point of view of the observers on earth, the plane would at first appear to accelerate, then hit some "magic point" where relativity effects would kick in. At that point, the plane would seem to be moving slower and slower, the closer it gets to light speed. At some point, it would hit cruise speed, and continue on, clearly visible in the sky to the mid-Atlantic observer, say, for most of the next 1000 years, going about 2' per minute.

    At Heathrow, as the plane decelerates, the observers on earth would see the plane accelerating faster and faster. Again, a "magic point" where relativity effects would no longer be apparent would come into play, and the plane would then seem to decelerate normally until it came to a stop.

    On-board, starting from JFK, the observers would at first see normal acceleration. As the effects of relativity would kick in, the world outside would seem to suddenly slow down, getting slower and slower, until it seemed the world was, again, only moving about 2' per minute. Most of the 15 min. would pass, with the observers seeing only about 30' pass by their windows.

    Then, as they approached Heathrow, and began their deceleration, the world would suddenly start to speed up around them. As they decelerated further, the world would begin rushing at them, faster and faster, catching up with the "real" time around them, a thousand years of history in the blink of an eye. Finally, as relativity no longer had an apparent effect, they would see themselves actually decelerate to normal (earth) speed, and stop, now a 1000 years in their own futures.

    But let's go even further. Let's say this was a really awesome Concorde, and could actually decelerate beyond the speed of the rotation of the earth and and even further, slowing down further and further relative to the speed of light (sorry, I couldn't think of how this would work -- just a thought experiment again :)).

    From the perspective of the observers at Heathrow, the plane would suddenly seem to accelerate as it hit the apparent relativity effects. It would go faster and faster, until finally, it disappeared altogether. A mere few seconds later, a fast moving Concorde would suddenly appear, going slower and slower, hitting that magic point, and then seeming to accelerate again. Finally decelerating to actually slow, and coming to a stop.

    From the point of view of the on-board observers, the world would seem to accelerate as they slowed, faster and faster, until it was just a frenetic rush of history. Their on-board clock would show many years as passing. Finally, as they started to accelerate to reach normal space/time, the world would appear to slow down, until they finally reached the point at which there was no apparent relativity effect, and they would appear to accelerate normally, then decelerate and come to a stop, now just 15 min. into their world's future, but their ship an ancient 1000 year-old ruin full of corpses.

    This is like nothing I've ever heard of before. Have I just got this all wrong?

    Thanks.
     
    Last edited: May 24, 2009
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  3. May 24, 2009 #2

    HallsofIvy

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    Why do you say that? If the ship is at speed v, each sees the other as slowed down by [itex]\sqrt{1- v^2/c^2}[/itex].

    I cannot imagine why you think the "speed of the rotation of the earth" has anything to do with this!

    No, to people on the plane, the on-board clock would just show normal time.

    Yes, basioally, you have it all wrong. Why would you come to such a conclusion?

     
  4. May 24, 2009 #3

    diazona

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    Yeah, sorry to break it to you but you did get it wrong. (Perhaps your first mistake was believing what you see in movies)

    Let's look at the situation you've developed. Say that your "hyperspace Concorde" travels at a maximum speed of 2.997 x 108 m/s (99.969% light speed), relative to Earth (that is, relative to the ground, the air, the New York and London airports, etc.) Now, relative to Earth, a.k.a. in the Earth's reference frame you say the plane takes 1000 years to complete its journey. That means that New York and London are, in the Earth's reference frame, 5.879 x 1015 miles, or just over 1000 light-years, apart. No way around that. The distance, time, and velocity are related by the equation d = vt, so you can't arbitrarily say that the flight takes 1000 years without adjusting the plane's velocity and/or the distance between the airports to match.

    Anyway, we have two airports that, from the point of view of the observers at either one, are just over 1000 light-years apart. The observers at JFK would see the plane take off and accelerate to 2.997 x 108 m/s, nearly the speed of light. They would then see it continue traveling at that speed for the next 1000 years, after which they would see it decelerate and land at Heathrow. The observers at Heathrow would see the same thing - the plane takes off from JFK, accelerates to 2.998 x 108 m/s, remains at that speed for 1000 years, then decelerates and lands. And the mid-Atlantic observer would see the same thing. (I'm not sure where you got that 2 feet per minute from)

    The observer on the plane, however, would see a different story. In special relativity, all observers in inertial systems would be equally qualified to think that they are at rest, and for the majority of the trip, while the Concorde is flying at its constant cruising speed, it is in an inertial frame. So if you were on the airplane, you would feel it accelerating as it took off and for some time thereafter, as it worked its way up to 2.997 x 108 m/s. You could look out your window, back at JFK, and see the events there happening more and more slowly as the plane accelerated. Once the plane reached its top cruising speed, events at JFK would be about 40 times slower than you would be used to seeing them. (Actually, they would probably appear even slower than that due to the Doppler shift, but I can't calculate that offhand) If you looked forward toward Heathrow, events also appear to happen at a different rate, but more importantly you would see the distance between the plane and Heathrow shrinking as the plane accelerates, far faster than you would expect based just on the acceleration. In fact, once the plane reached its top cruising speed, you would see the distance shrunk by a factor of about 40, so it would look like you had just a 25-light-year journey ahead of you. Also, while at cruising speed, you would see Heathrow airport approaching you at 2.997 x 108 m/s. It's the same speed at which the people on the ground see you moving. After about 25 years of travel, you would feel the plane start to decelerate as it approached its destination. You would see the activities of people at Heathrow (or at JFK, if you happened to be looking back) gradually approach the speed at which you are used to seeing them, and by the time you taxi up to the gate and stop, everything would appear perfectly normal.

    Hopefully one thing you can take from this explanation is that there is no "magical point" at which relativistic effects kick in. They are always there, it's just that they're too small to notice unless you're moving really fast. (How fast is really fast? It depends, how small is too small to notice? If you have an atomic clock or some other super-precise measuring instrument, you can even observe relativistic effects at the speed of a real-life Concorde.)
     
  5. May 24, 2009 #4
    Ok, maybe this is where I'm confused. How would they see something going that fast for 1000 years? Wouldn't it appear that it was simply hanging in space -- or more to the point, traveling very slowly? Also, it wouldn't be traveling for 1000 years of on-board time. That would be just a fraction of a sec., in fact. It's only 1000 years on the ground, no?

    EDITED: I think I see, you've corrected my example. Let's not do that. Just leave NY to London at the distance they are, the flight on-board then takes a fraction of a sec. At 2.998 x 108 m/s that would still be 1000 years of ground time, I think. So what would they see at JFK and Heathrow under those conditions?
     
    Last edited: May 24, 2009
  6. May 24, 2009 #5

    DaveC426913

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    No, at 99.97% of the speed of light, the trip, according to Earth observers, would take about 1/60 of a second - slightly longer than it would for a beam of light to travel the same distance. So the time discrepancy between Earth-based and onboard observers would be the difference between a very small fraction of a second and a VERY small fraction of a second.

    If I'm not mistaken, you seem to think that a trip of ~3000 miles could result in a giant time difference; it couldn't, at any speed.

    This is why you need to use long distances. A thousand year discrepancy requires a journey of at least a thousand light years. Either that, or accept time dicrepancies in the fractions of seconds.
     
    Last edited: May 24, 2009
  7. May 25, 2009 #6
    Ah, ok.

    As you can see, I'm having a hard time getting my head around these concepts.

    So here's what I was thinking. If there's a discrepancy of 1000 years, allowing for a 1000 light year journey minimum as you've said, wouldn't the journey necessarily appear to be slower to the ground observer than on-board? I'm just really confused as to why that wouldn't happen.
     
  8. May 25, 2009 #7
    Hi,

    You seem to be using the figures that I gave you, but I think you have overlooked something important.

    Note that I started with an equation with a vx/c2 and then I eliminated it by breaking the journey into two legs. But the journey still includes a significant spatial component (a total of vt). So if you travel at close to lightspeed for 1000 years (earth time) you will have covered close to 1000 light years.

    It's not enough to travel at close to lightspeed, the equations I gave had you travelling at close to lightspeed for 1000 years (earth time).

    What you are confused about (I think) is that you have a velocity, a smidgen under light speed, you have a distance (a smidgen under 1000 light years) and you have two times (50 years shipboard and 1000 years earth time). How does that work?

    It works because the distance between the departure point and the arrival point (or turnaround point) contracts for the shipboard observer. Remember I broke the journey up into two legs, so for an earth observer, there would be a distance of about 500 light years between the departure point and the turnaround point. For the shipboard observer at a smidgen under lightspeed (enough to make the outward journey 25 years), the distance would be contracted to about 25 light years.

    So really you have two times (50 and 1000 years), two distances (25 and 500 light years, for totals of 50 and 1000 light years travelled) and one velocity, a smidgen under light speed.

    hopefully this clears things up a little,

    cheers,

    neopolitan
     
  9. May 25, 2009 #8

    DaveC426913

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    The onboard observer would experience a journey of about 25 years, making him 975 years younger than Earth upon his return.
     
  10. May 25, 2009 #9
    Ok, I think I see where I went wrong.

    Thanks again, guys.
     
  11. May 25, 2009 #10

    DaveC426913

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    We are very happy to illuminate the curious. Please continue asking.
     
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