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Time Dilation

  1. May 3, 2015 #1
    Hello PF,
    A question on special relativity I've not found an answer to,I tried to google it but no luck, so here is it :
    If you were to observe a moving body whose speed is very near to the speed of light, will you see it slowly due to time dilation, or you'll observe it as it really is, very fast.
    In theory: why not, clock will record different intervals of time and you'll be in a messy situation where time depends on speed and vice-versa
    However, In real life, no! If so, light itself, will have struggles in reaching, and all of the things will cease to exist.
    I hope you answer me as soon as possible, and if I have a wrong understanding Of SR, I hope you help me out .
     
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  3. May 3, 2015 #2

    Orodruin

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    Anything moving close to light speed is moving really really fast.

    Time dilation relates to how you would consider an internal clock in the moving body to be ticking. It does not relate to the speed of the object.
     
  4. May 3, 2015 #3
    What I actually mean about this, If i boost my self with jet pack that can accelerate me to very hight speed (say 0.999*c) would it take few seconds to go to my distination or several years, not from my perspective, but how will my father wait for me to join him in mars ,?
     
  5. May 3, 2015 #4

    Ibix

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    In the situation you are describing, you would experience less time on your journey than someone who stays at home would experience. At 0.999c, you'd experience around one day for every three weeks on earth calendars.

    Note that this is a slight simplification. It was actually an early test of relativity, though. Unstable particles called muons are created by cosmic rays striking the upper atmosphere. Even at the speed they go they don't have time to reach the Earth's surface before decaying - yet they do get here. That turns out to be because muons experience less time on the journey than our clocks measure, so they don't decay as fast as they would if they travel slowly.
     
  6. May 3, 2015 #5
    So we, observers, can never see something going at high speed. If I understood, travelling 3*10^8 at a speed near c will take approximately 1 second for the person in the rock but minutes for observes on earth, or observers will wait 1 sec and the traveller will experience few microsec .
     
  7. May 3, 2015 #6

    Ibix

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    I'm not sure what you mean. You will certainly see the rocket going very quickly through space. People on board it (once you correct for light speed delay) will look like a slow-motion film.
     
  8. May 3, 2015 #7

    phinds

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    "See" is a bit confusing in this regard because light doesn't travel instantaneously so what we "see" is something that has happened in the past but I can see that you have the right idea. Let's think about "deduce" or "compute" and think about this scenario:

    A spaceship traveling at .999c passes earth at T1 and at that time, the clock on the spaceship is synchronized with one on Earth, so they both show the time as T1. On Earth, we don't know how fast the ship is going, just that the clock synchronization occurred in the instant that they passed by. By prearrangement, the spaceship is going to keep going at the same speed and when it is one light year away it is going to send back to Earth a signal encoding the time shown on the spaceship clock. Since the signal is sent from 1 light year away, we on Earth could not possibly get it sooner than T1 + 1 year, but what you need to work out is, when DO we get it and what does it say and what does that tell us about the speed of the spaceship?

    Well, since the Lorentz transform works out to 20:1 (rounded for simplicity, but basically it's the 3 weeks to one day quoted by Ibix) we know that the signal is going to tell us that the spaceship clock at one light year out from Earth reads T1 + .05 years.

    Now, we think to ourselves, WHOA DUDE! If they had gotten there instantaneously, the signal would have said T1 and it only said T1 + .05 years so they must be moving at near the speed of light!

    So we know how fast they are moving. What I've left for you is to figure out when we will GET the signal. Based on your post, I think you already know the answer.

    After you've got that, think about this: at the instant that the first spaceship passes the 1 light year mark, a second spaceship traveling at exactly the same speed but heading straight FOR Earth instead of away from Earth synchronizes clocks with the first space-ship so that the both their clocks read T1 + .05 years.

    When the second ship gets to Earth, its clock will read T1 + .1 years

    The question for you then is what does the Earth clock read at that time? We know from the arrival of the signal from the first ship and from the reading on the clock of the second ship that they were both traveling at .999c so we don't exactly "see" the fast travel but we certainly deduce it. The question I've left for you in both cases is, when do we find that out?

    Also, we know from the readings on the two spaceships' clocks that when the second ship arrives at Earth, the captain of that ship will correctly deduce that in the time-line of the spaceship clocks, the first ship left Earth 1/10th years ago. Knowing when that ship arrives at Earth will tell you how long ago we on Earth think it left.

    There will be a radical difference in the two so then you ask yourself, but how can that BE? Well it can be, and is, because the spaceships traveled through one spacetime path and we on Earth traveled through another and although both of our clocks were ticking at one second per second the number of ticks differs between spacetime paths (also called "world lines").

    The scenario I've described is, of course, a form of the "twin paradox".
     
  9. May 3, 2015 #8
    Thank you very much for your answers, so we can now stop wishing to travel into other galaxies unless we were on that rocket, because, if I understood, instead of waiting one year, the traveller only felt 0.05 year, he feels that he's taking less time to go the same distance therefore experiences more speed,
    As a reply for the "problem" you gave me, I'd say that when the Cap of the ship send the message that he's coming back, he'll arrive just few microsec after we, people on earth, recieve it thus is 1 year and everythings become messy, taking one year to send every message to the earth and synchronization becomes difficult,
    Plz let me know if i did any mistake here and again thanks :)
     
  10. May 3, 2015 #9

    phinds

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    Well, you have to ask "less time than WHAT?" Certainly less time than what is perceived by people back on Earth, and yes you have the right idea, but when you say "more speed" you have to think about "more speed relative to what?" The spaceship does not experience any speed in its frame of reference. NOTHING experiences speed in its own frame of reference. He's stationary and the distant point is rushing towards him at .999c and will arrive at his location at T1 + .05 years.

    No, you're wrong on this. Think about the trip outbound. How long after T1 did the signal ARRIVE back at Earth. According to your calculation in this paragraph it would have arrived at T1 plus microseconds. Does that sound right? Did it take him microseconds from Earth's perspective for him to get there? Figure out when THAT signal arrived at Earth and it will be clear when the second signal arrived.
     
  11. May 3, 2015 #10
    What I mean is, the signal is moving at c, the rocket a .999c so, when we have the message, the rocket is likely to arrive in few secondes, right ?
    Thnks very much :) :)
     
  12. May 3, 2015 #11

    phinds

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    You have the right idea, but do the calculation. What is .001 of 1 year? Is it a few seconds?
     
  13. May 3, 2015 #12
    Yes, you're right, Thank you again for making it much simpler for me, :)
     
  14. May 5, 2015 #13
    Now that signal cannot be sent (by pre-arrangement) after 1 year's travel as the clock on that ship only reads 0.05 years...
    So it mast have been sent after the ship has travelled 1 light-year. OK so far?
    But we know that, on board the ship, they know they have travelled 1 light-year (as they have sent the pre arranged signal), yet their clock is telling them that it has only taken 0.05 years.
    So those on board the ship will have to calculate that they have travelled 1 light-year in 0.05 years...

    Can you explain please?
     
  15. May 5, 2015 #14

    PeterDonis

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    No, they won't. They will calculate that the Earth has moved away from them by a little less than 0.05 light-year in 0.05 years. In the ship frame, the distance to Earth is length contracted.
     
  16. May 5, 2015 #15

    Ibix

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    There is no general sense in which the ship has travelled one light year. Say that (viewed in the Earth's rest frame) it travelled from the Earth to a space station one light year away. In the ship's own rest frame, it hasn't travelled anywhere. Instead, the Earth moved away at 0.999c and the space station came towards it at 0.999c. Since these are the moving objects, the distance between earth and the station is length contracted to 0.05ly viewed in this frame. In both frames, the distance travelled (by whatever is moving) is consistent with the time taken in that frame.

    I get the impression that you think the effects of relativity are rather like a magic trick, and if you look at just the right angle you'll be able to see what's really going on. But relativity is pretty much the mathematical expression of the notion that nothing is "really going on". There are no hidden wires or trap doors. Distances and times really are different in different frames, so there is no "travelled one light year" without specifying a frame.
     
    Last edited: May 5, 2015
  17. May 6, 2015 #16
    In the original proposal of this scenario phinds wrote:
    and:
    I have no problem with the relativity, I am questioning the validity of the example and how it would work.
    For those on the space ship must be aware they have reached the 1 light year mark in order to send the signal with their clock reading at that time. Therefore, they know how far they have travelled and from their clock how long it has taken.
    It is OK to say that in their frame they have only travelled 0.05 light years, but that doesn't work either does it? For unless the length contraction is only measured by the Earth-bound observer, the spaceship is travelling to a marker only 0.05ly away in its frame...
     
  18. May 6, 2015 #17

    Orodruin

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    This right here is your problem. You are assuming that there is a well defined thing as "how far they have travelled". There is not, you can only define this with respect to a given inertial frame and different inertial frames will not agree. Sure, aboard the ship you can calculate how far it will correspond to in the Earth's rest frame, but this does not change the fact that in the ship's rest frame, the distance is length contracted.
     
  19. May 6, 2015 #18
    No. I am not assuming anything, I am simply analysing the scenario advanced above.

    But there is another intriguing aspect to this.
    The time slows. Only 0.05 years passes on the moving ship as measured from the Earth.
    Yet Einstein wrote in Chapter 12:

    "As judged from K, the clock is moving with the velocity v; as judged from this reference-body, the time which elapses between two strokes of the clock is not one second, but
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    seconds, i.e. a somewhat larger time. As a consequence of its motion the clock goes more slowly than when at rest. Here also the velocity c plays the part of an unattainable limiting velocity."

    ∴ for our thought experiment, as K is the Earth system, the time measured by the Earth-bound observer should be γt = 20ly rather than t/λ = 0.05ly?
    For if more time passes between two ticks of the clock, then time must be passing much quicker for the travelling clock, for the clock in its own frame is still ticking at the same rate one second at a time and as we all know a second is a second is a second...
    And by the time the second hand has swept round the clock, 1 minute = 60seconds will have passed on the ship yet γ seconds will have passed for each second as measured by the Earth observer.

    So do we call that faster or slower other tan by convention?
     
  20. May 6, 2015 #19

    Ibix

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    Although he does not say so, it is clear from phinds' maths that he's talking about a destination one light year away in the Earth's rest frame. Viewed from a frame moving at 0.999c, this destination is indeed only 0.05ly away, and moving towards Earth at 0.999c.

    The full picture in the frame of the Earth is that there is a 5m long spaceship with clocks ticking once every 20s travelling at 0.999c to a destination 1ly away.

    The full picture in the frame of the ship is that they see a 100m long spaceship with clocks ticking every 1s, with the destination starting at 0.05ly and approaching at 0.999c.

    You seem to be attaching special importance to the fact that the destination point is 1ly away in the rest frame of the Earth. The fact that the rocket passes its destination means that it (and everybody else) knows that it has travelled one light year in the rest frame of the Earth. It does not mean that there is any special significance to the 1ly.
     
  21. May 6, 2015 #20

    Ibix

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    I think you are mixing frames, rather than assuming something. You seem to take a distance measured in one frame (usually the first one mentioned) and use that distance measurement as if it were "the" distance that everyone agreed on.

    Time is not measured in light years. You are also misinterpreting this. When the traveller's clock reads 1 year, the earth's clocks will, indeed, read 20 years (a 20:1 ratio). That's not really relevant to this experiment, however, since at that point the traveller will be 19.98ly away and will have long passed the destination.

    At the point the rocket passes the destination point, it's clock will read 0.05y while the Earth's clocks read 1y - that same ratio of 20:1.
     
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