TIME DILATION. WHY do clocks that are

  1. moving closer to light speed relative to another clock tick slower? I understand that the waves take longer to reach the stationary observer on the turn around, but that's just appearance. It still seems the clocks would be in sync upon the return. What is making time actually slow down (comparatively in that frame) by moving faster??
     
  2. jcsd
  3. HallsofIvy

    HallsofIvy 40,310
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    I'm not clear on what kind of answer you want. My reaction would be to say "that's the way the universe is". But you seem to want some kind of "mechanistic" answer.
     
  4. Janus

    Janus 2,367
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    It's a consequence of the speed of light being the same in all reference frames.

    For example: Sam and Tom are have a relative speed with respect to each other. As they pass each other, they both set off a flash of light. According to Tom, both flashes of light travel away from him at c and he always remains at the center of the expanding sphere of light, while Sam moves away from the center.

    However, according Sam, he is one that stays at the center of the sphere of light and its Tom that moves away from the center.

    Now imagine that Tom and Sam have placed mirrors an equal distance apart, spaced perpendicular to their relative travel. According to each, the light pulse goes straight up, hits their mirror and comes straight down. This is one tick. But also, according to each, the light path follows a different path while traveling back and forth between the other person's mirrors. It must follow a longer diagonal path. since this light takes a longer path but travels at the same speed, it must take longer for the light to make the round trip. Thus according to Sam, his clock tick faster than Tom's, and according to Tom, his clock tick faster than Sam's

    Something like this:

    [​IMG]

    So according to each, nothing happens to his clock, but the other person's clock runs slow.

    Okay, now why is it that, if one of them were to travel off, turn around and come back, that they would find that his clock has accumulated less time?

    There are two other effects besides Time dilation to consider. One is length contraction. According to Tom a meter stick laying parallel to the relative motion and traveling with Sam will be shorter than his and vice versa.

    So let's say that Sam travels off at 99% of c to a distance of 1 light year from Tom as measured by Tom and returns. According to Tom, this trip takes a little over 2 years. During which Sam's clock runs 1/7 as fast as his and reads ~2/7 of a year on returning.

    According to Sam, however, the distance measured as 1 light year by Tom is only 1/7 of a light year, and thus it takes him only ~2/7 of a light year to make the trip and he Returns to Tom with only 2/7 of a years ticked away on his clock, which agrees with the time that Tom expects to have ticked away on Sam's clock.

    The third effect is the Relativity of Simultaneity, and it explains why even though Sam expects Tom's clock to tick more slowly than his while he is on his outbound and return trips, he finds that Tom's clock has ticked away more time upon his return.

    Basically, events that are simultaneous to Tom, won't be for Sam, and vice versa.

    So let's say that there is a clock sitting at that 1 light year mark that reads the same as Tom's clock according to Tom. According to Sam, as he travels outward towards this clock it will actually read ahead of Tom's clock by 48/49 of a year. The outbound trip takes 1/7 of year according to Sam, during which the clock at the turn around point ticks off 1/7 as fast and accumulates 1/49 of year reading 1 year upon Sam's arrival.

    Now here comes the tricky part. In order for Sam to return to Tom, he has to change velocity. And when he does so, he changes reference frames from one going away form Tom to one heading towards Tom. So now, according to the Relativity of Simultaneity, it is Tom's clock that must read ahead of The turn around point clock, This means, that according to Sam, when he turns around, Tom's clock must "jump ahead" to read 1 & 48/49 year, and ticks away 1/49 year on the return trip to read 2 years upon Sam's arrival.
     
  5. ghwellsjr

    ghwellsjr 5,058
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    Janus, I really like your great animation. I wish I knew how to make animations like that which appear right on the webpage. I have had to use youtube for mine which makes them inconvenient to see.

    I also like your explanations up until the last few paragraphs:
    Can you explain where you got these numbers from? I cannot replicate your results.
     
    Last edited: Sep 4, 2011
  6. DaleSpam

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    Actually, all of the relativistic effects like time dilation are what remains after you take "appearance" into account.
     
  7. Janus

    Janus 2,367
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    The frame are drawn with POV-Ray, then assembled by a GIF animator (I use Animation Shop 3).
    To be quite frank, I cheated a bit on those numbers. I knew what times each clock needed to read upon Sam's arrival, how long the trip took according to Sam and the fact that time dilation is reciprocal. Since it took ~1/7 of a year to make the trip according to Sam, reciprocal time dilation says that Tom's clock and the turn around clock run ~1/7 year on his clock equal ~1/49 year on the other clocks. Since the turn around clock must read ~1 year when Sam arrives, then it has to read ~48/49 of a year when Sam leaves Tom according to Sam.

    That ~48/49 of a year is due to the Relativity of Simultaneity. In the rest frame of Tom and the turn around clock, both clocks read the same value. In Sam's frame, they differ by an time equal to

    [tex]\frac{\frac{xv}{c^2}}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]

    Where x is the distance between Tom and the turn around clock as measured by Sam and v is their Relative velocity.

    At 0.99c, x = 0.1411 ly (1 ly length contracted)

    Since the bottom half of the equation also equals 0.1411, this leaves us with

    [tex]\frac{(1 ly)(0.99c}{c^2}[/tex]

    If I use 1 ly/yr for c, I end up with an answer of 0.99 yr as the difference between the two clocks.

    So when Sam leaves Tom at 0.99c and Toms clock reads 0, according to Sam, the turn around clock already reads 0.99 yr.

    Now according to Tom. it takes 1ly/0.99c = 1.01 years for Sam to reach the turn around point, so this is what the turn around clock reads when Sam arrives.

    According to Sam, it take 1.01*0.1411 = 0.1425 years to complete the first leg of the the trip. During which time the turn around clock runs at a time dilated rate of 0.1411, and accumulates 0.020 years. 0.99 +.02 = 1.01 yr, meaning that the turn around clock reads the same upon arrival according to both Sam and Tom.

    Now Sam turns around and heads back towards Tom. We will assume for eases sake that this happens instantly. Now he still finds himself at the turn around clock but with his velocity in the opposite direction. But since he is now heading towards Tom and away from the turn around clock, their roles, according to the Relativity of Simultaneity, are reversed. The difference in their times remain the same, but now it is Tom's clock that must read 0.99 yr later than the turn around clock.

    One thing that everyone must agree to is that the turn around clock reads 1.01 yr upon Sam's arrival, turn around and leaving the turn around clock. So if the Turn around clock reads 1.01 years when Sam leaves it, and Tom's clock reads 0.99y later, then according to Sam, Tom's clock now reads 2 yr.

    The return trip is a mirror of the outbound trip, with Sam experiencing 0.1425 yrs, and expecting Tom's clock to accumulate 0.02 yrs and reading 2.02 yr upon arrival.
     
  8. ghwellsjr

    ghwellsjr 5,058
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    Janus, I appreciate your expanded explanation but I'm still very confused. Are you considering just two clocks, both stationary with respect to Tom, one you call Tom's clock, colocated with Tom and the other one that is 1 ly away that you call the turnaround clock, correct? Sam doesn't have any clocks, correct? You are only explaining how Sam will analyze both of these two clocks that are stationary with respect to Tom during this scenario, correct?
     
  9. Halls of Ivy has the key idea as far as I am concerned.

    yes it does SEEM that way upon a first look...but that is classical not relativistic reasoning. Physics requires some new "logic" and ways of thinking. When traveling at high speed, who would have guessed space is shortened...and that space and time are interchangeable entities (in Lorentz transforms for example)??

    It would also "seem" that particles can be measured (observed) with infinite precision subject to the cleverness our measuring devices, but that too is not how our universe works. And who would have guessed in the subatomic world that objects/particles have discrete not continuous values?? It's a crazy world!!
     
  10. Janus

    Janus 2,367
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    Sam does have his own clock by which he can make comparisons

    He makes two comparisons:

    One is comparing the times on the two clocks at rest with respect to Tom. This comparison reflects the Relativity of Simultaneity. It depends on not only the separation of the two clocks, but also Sam's relative velocity(speed and direction) with respect to them

    The other is the Rate at which these clocks run with respect to his own clock during the two legs of the trip. This comparison reflects time dilation and depends on only the relative speed between Sam and the other two clocks.

    Here's the situation drawn up as space-time diagrams (I reduced the relative speed to 0.866c in order to keep the second pair of diagrams from being too stretched out, but the ideas are the same).

    Here's The S-T diagram from the rest frame of Tom. T1 is Tom's clock, T2 is the clock at the turn around point and S is Sam's clock.

    [​IMG]

    Here's the same scenario from Sam's rest frame during the outbound leg. The inset shows the details during the outbound leg and upon reaching turn around more clearly:


    [​IMG]

    Notice the offset in times of T1 and T2

    And here's what it looks like from the rest frame of Sam during the return leg. The inset shows events right after Sam has reversed direction to when he returns to Tom.

    [​IMG]

    Notice that the time on T2 remains unchanged at this instant, but T1 now reads a later time.
     
  11. Noob here. Sorry for the stupid question. I tried many times to understand relativity but it seems my brain is like a black hole; it absorbs but it doesn't shine off.

    Does this have anything to do with time passing slower when one does nothing and vice versa?

    For example, Tom and Sam are in a room. Tom is playing a game on PlayStation 3, while Sam is staring at a wall. Tom is enjoying the game and doesn't notice as time passes. Sam however, is very bored staring at the wall, and every minute seems like an hour to him. After an hour, when Tom and Sam stop their activities, they face each other again. To Tom, it seems like "time flied", and only minutes passed. To Sam, it seems like many hours have passed.

    In our uneducated simple reality though, according to an "outside observer at rest", only an hour has passed. But according to theoretical physics, Tom and Sam experienced time differently and therefore there is no "real" time, and Sam has aged by several hours while Tom by only minutes.
     
  12. HallsofIvy

    HallsofIvy 40,310
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    Where did you get that idea? No matter how bored Sam may be, no matter how long he may feel like he stood there, "theoretical physics" says that, since he and Tom were stationary relative to one another, he actually ages no more than Tom. How much time has passed is measured on a clock, not by "feelings".
     
    Last edited: Sep 5, 2011
  13. It was my dumb attempt at demonstrating in a humorous way how everyday people unlearned in physics "understand" relativity. Pardon me.

    On a serious note tho, relativity says time would "stop" if we were to approach the speed of light (hypothetically), and for light itself, time doesn't exist. But the speed of light is not infinite. It's only 300,000km/sec. So then it needs two seconds to reach a point at a distance of 600,000km. So how can there be no time at the speed of light? What am I missing?
     
  14. DaleSpam

    Staff: Mentor

    https://www.physicsforums.com/showthread.php?t=511170
     
  15. HallsofIvy

    HallsofIvy 40,310
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    You gotta be careful, people are liable to take you seriously!

    You are not thinking "relatively". Light would take two second to reach a distance of 600,000 km relative to a person stationary with respect to the light. Of course, since the speed of light is the same relative to any observer, that would be true for any observer. In the frame or reference of light, if there were such a thing, it would take no time at all to go any distance. That is why we say that there cannot be a 'frame of reference' at the speed of light relative two any observer.
     
  16. The simple answer, if it is simple, is that clocks moving relatively follow different paths in spacetime. A clock measures this distance. It is similar to getting different odometer readings in your car if you take a different route to the supermarket.
     
  17. ghwellsjr

    ghwellsjr 5,058
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    Janus, in your first scenario, you has a speed of .99c which resulted in a gamma of about 7 or a time dilation of about 1/7. You then did some kind of calculation which resulted in squaring 1/7 to get 1/49.

    In your second scenario, the speed was .866c for a gamma of 2 and a time dilation of 1/2. Where is the similar squaring of 1/2 to get 1/4?
     
  18. Janus

    Janus 2,367
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    Yes, but I just noticed that there was some errors when I wrote down the numbers.( for instance, where I have T1 = 0.284 it should be 0.288) So some of the other numbers might be a tad skewed.

    The 1/4, as did the 1/49, come from the fact that time dilation is reciprocal.

    Just as Tom measures Sam's clock as running 1/7 the rate of his own when they are moving at 0.99c with respect to each other, Sam measures Tom's clock as running at 1/7 the rate his own. Thus for every second that ticks away on his clock, he would expect 1/7 sec to pass on Tom's clock. Since we established that Sam measured measures 0.1425 yrs as having passed by his clock on the trip out, he would expect ~1/7 of that amount of time to have passed on Tom's clock.

    This means that when Tom reads 1.01 yrs by his clock, then according to him, at the same time, Sam's clock reads 0.1425 yrs. Sam however, say's that when his clock reads 0.1425 yrs, at the same time, Tom's clock reads 0.020 yrs.

    This is just another way in the Relativity of Simultaneity comes in to play: "At the same time" is different for Tom than it is for Sam.
     
  19. I have a few questions about the clock which is on the moving ship. What would happen if the clock is parallel to the ship instead of perpendicular? In this example i would remind you that special relativity says that no matter your speed the photon in that clock would still move at its constant speed so the same slowing effect that is seen in its perpendicular position could not be observed in a parallel position.

    Also why do we consider that time itself is slowing down, instead of the mechanism of the clock changing? I mean if we change the wheel inside a mechanical clock to a bigger one then the arrow would start moving slower than with a smaller wheel, does that mean that we are effectively slowing down time just by changing the wheel? I fail to see the logic of a clock slowing down = time slowing down.

    And finally you say that if the ship is moving at half the speed of light, then the beam from the stationary car would pass it by at "full" speed (its constant speed). Well i want to imagine that the car is moving at the speed of light as well, what would happen then? Would the beam still pass it by with the same speed?

    It's just a few questions which i've been pondering for a while now. I'm not a physicist so the math of this escapes me, but i like to think that a logical explanation would be sufficient. Thanks in advance.
     
  20. I have a few questions about the clock which is on the moving ship. What would happen if the clock is parallel to the ship instead of perpendicular? In this example i would remind you that special relativity says that no matter your speed the photon in that clock would still move at its constant speed so the same slowing effect that is seen in its perpendicular position could not be observed in a parallel position.

    Also why do we consider that time itself is slowing down, instead of the mechanism of the clock changing? I mean if we change the wheel inside a mechanical clock to a bigger one then the arrow would start moving slower than with a smaller wheel, does that mean that we are effectively slowing down time just by changing the wheel? I fail to see the logic of a clock slowing down = time slowing down.

    And finally you say that if the ship is moving at half the speed of light, then the beam from the stationary car would pass it by at "full" speed (its constant speed). Well i want to imagine that the car is moving at the speed of light as well, what would happen then? Would the beam still pass it by with the same speed?

    It's just a few questions which i've been pondering for a while now. I'm not a physicist so the math of this escapes me, but i like to think that a logical explanation would be sufficient. Thanks in advance.
     
  21. HallsofIvy

    HallsofIvy 40,310
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    What do you mean by this? What part of the clock is parallel to the ship?
     
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