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FTL and time dilation

  1. Jan 27, 2015 #1
    EDIT: split from "twin paradox with a twist" thread
    Re: https://www.physicsforums.com/threads/twin-paradox-with-a-twist.791673/page-5#post-4985729

    one question: on the charts I understand the reflections as being at the speed of light correct? what would change in the case of FTL communications? would it negate the time dilation for the twin observers? or just make it more obvious if we include their clocks.
     
    Last edited by a moderator: Jan 27, 2015
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  3. Jan 27, 2015 #2

    ghwellsjr

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    If things were different, things would be different. In Special Relativity, IRF's are defined to have light propagate at c, not less than c, not more than c.

    In my non-inertial frames, Time Dilation doesn't apply, even though light still propagates at c. You can define a different non-inertial frame in which light doesn't always propagate at c.
     
  4. Jan 27, 2015 #3
    I take it that FTL communications would alter the perception of A & B yet both would still age differently than C who never traveled from the start point now I'm wondering how a super gravity well which has a measurable affect on light would affect actual aging if one of the twins flew through its area of affect during their trip? would it stretch the flight?
     
  5. Jan 27, 2015 #4

    PeterDonis

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    What theory do you want to use to determine the effect of FTL communications? That's not a part of standard relativity.

    Yes. This is called the Shapiro time delay. The usual derivation is for light, but the general concept applies to other objects too.

    Yes. (At least, if we adopt the split of spacetime into space and time that is usually adopted for a static source. A different split of spacetime into space and time could give a different answer.)

    Note, however, that in both of these cases (the Shapiro delay and the "stretching" of the flight), the implicit comparison that is being done, between the actual measurements and what they would have been without the gravitating body present, can't actually be done. There is no way to remove the source, make measurements, then put it back, make further measurements, and compare the two sets of measurements.

    The comparison that is really being made is between the actual measurements, and calculations of what would be expected if spacetime were flat. But that comparison requires some way of determing the correspondence between the two cases. For example, in the case of the Shapiro delay, we might compare (I'm describing an idealized measurement here) the measured time for a round-trip light signal to travel between two observers, who are both in the same circular orbit about the central mass, just on opposite sides of the orbit, with the time we would calculate for a round-trip light signal to travel between two observers on opposite sides of a circle with the same circumference in flat spacetime. The circumference of the circle is what is "held fixed" between the two cases, to give a basis for comparison. But it's not always easy to determine what should be held fixed for a meaningful comparison. (In this idealized case, the circumference of the circle is an obvious choice because the spacetime is spherically symmetric.)
     
  6. Jan 27, 2015 #5
    in the case of FTL communications it'd be easiest to assume instant communications without delays just to keep it simple. what they see of their opposite is whats happening as they see it. I was wondering if the physical time dilation was observable without the influence of the communications delay.

    and thanks for clearing the rest up too!
     
  7. Jan 27, 2015 #6

    PeterDonis

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    "Instant" with respect to which frame? If they are moving relative to each other, "instant" for one is not "instant" for the other.

    What is usually called "time dilation" is what is left over after the communications delay has been factored out. However, this time dilation is not "physical" in the sense that it is coordinate-dependent; you can change "time dilation" in this sense merely by changing coordinates, without changing anything physical at all.

    What is directly observable is better termed "differential aging": if you take two clocks, start them out together with both set to the same time, let them separate and follow different trajectories for a while, then bring them back together, they might not read the same time; one might have aged more than the other. The term "time dilation" can be used to describe this, but that can easily cause confusion, which is why I suggested "differential aging" as a better term for it. Notice that in the scenario I just described, there is no communications delay; the two clocks can be compared directly at the start and finish of the scenario.
     
  8. Jan 27, 2015 #7
    i may be confused now. isn't time dilation based on speed differential? A is traveling at a opposite/other speed or not at all compared to B who is moving at speed.
    an example of what i meant by no delay communications would be the third observer who never moved from the starting point in the OP's scenario would be able to see a seeded plant on the twins vessels grow as if it was in fast forward or the twins facial hair sprout unusually fast as they watch each other while the twins would see normal growth on each other and slowed growth for the stationary person.
     
  9. Jan 27, 2015 #8

    PeterDonis

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    On relative speed, yes. Note that "speed" is always relative; there is no such thing as absolute motion or absolute rest.

    Which still requires you to specify with respect to what frame (which observer's coordinates) there is no delay.
     
  10. Jan 27, 2015 #9
    each observer is seeing the others in reference to their own frames (real time video feeds) they can see each other constantly.
     
  11. Jan 27, 2015 #10

    PeterDonis

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    By means of light signals, yes. But you are trying to add "instant" viewing, which is not allowed in standard SR. If you're trying to speculate about how it might work, then you have to specify with respect to whose frame the viewing is "instant". You can't just say it's like normal "seeing", because it isn't; that's the whole point.
     
  12. Jan 27, 2015 #11
    i wouldn't know where to start trying to work out how FTL would be made possible. I wanted to understand if the individual observers would be able to see a distinct change in the passing of time during the trips on the other members if there was no delay between when they looked and the object being observed as it moved at speed.
     
  13. Jan 27, 2015 #12

    Nugatory

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    The problem with this question is that "no delay" is equivalent "time of reception is the same as time of emission" which is equivalent to "sees at the same time that it happens". But because of the relativity of simultaneity, there can be no universally applicable notion of "at the same time"; a question that assumes the existence of something that cannot exist will lead to the same sorts of problems that you'll see if you ask "Suppose that 2+2=5; then what would happen if ....?"
     
  14. Jan 27, 2015 #13
    but math does this quite frequently by assigning a value to X. X=0 so what would be....
    my question is similar in assigning a 0 to part of the equation.
    this is not to be contrary. the point of the question was to understand if the observers would be able to see an appreciable difference in the time passing for the observed target without needing to wait for the return of the object of observation. IE twin A goes three light years away and back at .8c twin B stays at the starting point. twin B is older when twin A has returned.
    but in my question would twin B being able to see twin A during the trip see twin A as being in fast forward and twin A would be seeing twin B as moving slowly.
     
  15. Jan 27, 2015 #14

    Nugatory

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    If A and B were equipped with super-powerful telescopes that would allow them to watch each other's clocks, they would both see the other clock running slow if they were moving away from one another and fast if they were moving towards one another.

    Time dilation only becomes apparent when they also allow for the light travel time: If at 11:00 AM I see in my telescope a distant clock reading 9:00 AM and that clock is one light-hour distant, I know that I am seeing light that left that clock one hour ago so conclude that that clock read 9:00 when it was 10:00 according to my clock. After making those corrections, both A and B will conclude that the other clock is running slow.

    Google for "relativistic Doppler" for more information about what they actually SEE in their telescopes.

    When you have A turn around and come back, you've changed the problem: for a while they are moving away from one another and for a while they are moving towards one another, and they both see through their powerful telescopes the other clock running slow on the outbound leg and fastbon the return leg. They also both conclude that for them to see this, the other's clock must be running slow at all times. And despite the apparent symmetry of these statements, they also agree that A is the aged less than B at their return.

    To understand how this works, read through the Twin Paradox FAQ, and pay particular attention to the relativistic Doppler effect explanation.
     
  16. Jan 27, 2015 #15

    Nugatory

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    That doesn't work here. The problem is that the difference between the time of emission and time of reception is different in different frames, so if you assume that it can always be set to zero you end up assuming that there are two numbers that are not equal to one another but are both equal to zero.

    (Actually, you can save yourself by assuming that the speed of light is infinite - but that's not true in the world that we live in, so the conclusions you draw after making that assumption tell us nothing about our world).
     
  17. Jan 28, 2015 #16
    I'm aware of the limitations. i think we may be looking at my question in different ways. I'm trying to look at it in a video feed of both participants being shown at the same time. if we wanted to we could wait till the trip is complete where both observers have had their time during the trip recorded in real time then after the trip both recordings are played back side by side twin A who went away at speed would have a shorter recording than twin B who stayed behind in actual length of recording. to make both recordings play in sync you'd need to speed up or drop frames from twin B's recording. correct?
     
  18. Jan 28, 2015 #17

    PeterDonis

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    This is correct, yes; if the two videos were compared after the trip was complete, twin A's recording would be shorter, if played at standard frame rate, than twin B's, because twin A experienced less proper time than twin B did.

    Note, btw, that you could also make the recordings play in sync by slowing down twin B's recording, so this comparison does not pick out either one's "speed of video" as the "correct" one.
     
  19. Jan 28, 2015 #18
    I'm thinking you meant twin (A's) video could be slowed.
     
  20. Jan 28, 2015 #19

    PeterDonis

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    Oops, yes, got them mixed up. Twin A's video has less proper time elapsed, so it would need to be run at a slower frame rate to take up the same amount of playback time as Twin B's video.
     
  21. Jan 29, 2015 #20

    pervect

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    If you transmit a video signal, the carrier frequency and decoded video signal will be frequency shifted by the same amount. This amount is the relativistic doppler factor. If ##\beta = v / c## the relativistic doppler factor is ##k = \sqrt{\frac{1+\beta}{1-\beta}}## when the motion is directly toward or directly away from the receiver.

    For a reference, see the wiki at http://en.wikipedia.org/wiki/Relativistic_Doppler_effect

    The relativistic doppler shift takes into account both time dilation and the non-relativistic doppler effect.

    You will also have the usual lightspeed delays. If we pick two inertial frames, a "receiver frame" and a "transmitter frame" then we can state that in the receiver frame, a signal emitted at a time ##t_0## in that frame will be received at time ##t_0 + d_0/c## where ##d_0## is the distance that the emitter is away in the receiver frame at the time of transmission of the signal, and that the received frequency of the signal will be the transmitted frequency (as measured in the transmitter frame) multiplied by a factor of ##k = \sqrt{\frac{1+\beta}{1-\beta}}## Here ##\beta## is taken to be positive if the transmitter is approaching the receiver, and negative if it is moving away.

    If you draw a space time diagram, you'll get a consistent account of who receives what signal when. See for instance
    https://www.physicsforums.com/threads/spacetime-diagram-twin-paradox.671398/page-2#post-4270805

    It's hard to give more detail without a specific scenario - you can use the one that the above space-time diagram was drawn for, or create your own.
     
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