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Change of frames in relativity

  1. Feb 10, 2014 #1
    Hi guys, this is my first post, I'm a newbie regarding physics and relativity so I hope you can help me here.

    I've just recently started learning about relativity, and my question will be about the relative simultaneity, or precisely Andromeda paradox.

    So in the original setup, if an observer is at rest with respect to Andromeda, he will consider some event, let's call it X, as his present. If another observer moves away from Andromeda, he will consider an earlier event of Andromeda's history, let's call it Y as his present. That's at least what should be the setup.

    But if two observers are at rest mutually and of course with respect to Andromeda, and one starts moving away from it, accelerating to get in the position that I've previously mentioned, how will he be able to get from the moment that they both agree about to consider their present (since they are mutually at rest) to a moment when he considers his present to be 'earlier' than the present of the stationary observer. Basically, what happens during acceleration or the change of frames, so that one observer can get from one position of simultaneity defining to another?

    Thanks in advance. Johnny
     
  2. jcsd
  3. Feb 10, 2014 #2
    Is there a question here? Seems like you do understand the paradox but are uncomfortable with is implications to the meaning of present, past and future. When you accelerate and changes frames, you are using different coordinates to locate events in the past-prensent-future. An example might help. Imagine you come to an intersection and that there is a building ahead on the left side of the street. Than you take a left turn (change of frames) and the building is now on the right side of the street. The building didn't move, but you see if from a different perspective. Same thing is true about the Andromeda paradox. Nothing changes to Andromeda but you see it from a different perspective and an event that seemed to be in the future, now is in the past (or vice-verse, depending on how you accelerate).
     
  4. Feb 10, 2014 #3
    Hey there, I understand what you mean, but can you answer me more concretely regarding the scenario I mentioned, when two observers agree about simultaneity and then one accelerates and they disagree. It will make it much easier, but your answer is helpful, nonetheless. :)
     
  5. Feb 10, 2014 #4

    JesseM

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    It may help to understand that frames aren't something that are "naturally attached" to a given observer, any observer is free to use any coordinate system they wish (with any definition of simultaneity they wish) to assign position and time coordinates to events. It is a matter of convention that purely inertial observers should consider "their own frame" to be the inertial frame in which they happen to be at rest, but there isn't really any standard convention for the coordinate system of observers who accelerate. You can if you wish design a non-inertial coordinate system for an observer who accelerates that has the property that its definition of simultaneity at every point on their worldline agrees with the definition of simultaneity in the inertial frame where they are instantaneously at rest at that point, and in this case the simultaneity surface will rapidly "swing forward" during the accelerated phase of the Andromeda example, but this is purely a matter of choice, nothing in physics forces a given observer to use any particular coordinate system.
     
  6. Feb 10, 2014 #5

    ghwellsjr

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    Welcome to PF. We're glad to help.

    I'm going to draw some spacetime diagrams for you using the Lorentz Transformation process to get from one to the other. In my diagrams, I will assume that the observer at rest with Andromeda is on the Earth and show him as the thick blue line, Andromeda as the thick red line 2.5 million miles away and the observer moving away from Andromeda at 0.4c in a rocket shown as the thick black line. Dots on each of the thick lines represent 1 million year increments of time. I show your event X as a red star exploding at Andromeda and your event Y as a yellow star exploding 1 million years earlier according to Andromeda time. The thin lines represent signals traveling at the speed of light between an observer and Andromeda. I will be showing both how a spacetime diagram depicts simultaneity and how each observer can measure it using radar techniques. In both cases, the key to understanding is Einstein's second postulate, that it takes the same amount of time for a signal to get from an observer to a target as it takes for the signal to get from the target back to the observer, and his first postulate that each observer uses his own clock to measure time and that each observer will measure the round-trip speed of light to be the same value (called c).

    Here's the mutual Earth/Andromeda rest frame showing how the red star exploding at time 0 is simultaneous with the Earth clock being at 0. The yellow star exploded at -1 million years at both Andromeda and at Earth in this Inertial Reference Frame (IRF). You can see the black rocket observer's worldline but we are not yet concerned about his observations or measurements:

    attachment.php?attachmentid=66471&stc=1&d=1392061624.png

    To see how the Earth observer uses radar to measure the times, consider the signals that he sent earlier and received later. He has been sending signals continually millions of years earlier but I'm only showing the ones that are significant for our purposes. The one that he sent at his time of -3.5 My (million years) hits the yellow star when it explodes and the return echo arrives at the Earth observer at his time of 1.5 My so he takes the average of those two numbers and gets -1 My as his measurement of the time when the yellow star exploded, the same as his rest frame depicts it. Similarly, he can measure the time of the red star exploding as 0 My according to his clock.

    Here's the rest frame of the black rocket observer. Except for the signals, this diagram was derived by taking the coordinates from the first diagram and Lorentz Transforming them at a speed of 0.4c. Note that the event of the yellow star exploding is simultaneous with his time of zero and the red star explodes at his time of about 1.05 million years (My):

    attachment.php?attachmentid=66472&stc=1&d=1392061624.png

    Now we can see how he also can make these measurements. He also has been sending out radar signals continually but again, I'm only showing the ones that hit the exploding stars. The one that he sent at his time of -2.25 My hit the exploding yellow star and echoed back to him at his time +2.25 My and the average is 0 My. The radar signal that he sent at -1.6 My gets averaged with the one he received at +3.7 My for a measurement of 1.05 My.

    Do you see how this works?

    The observers can also measure the distance to the events by taking the difference between the received and sent times, in other words, the interval of time between sending and receiving the radar signals and dividing by two and calculating how far light travels in that amount of time.

    Let's see how that works for the black rocket observer. For the yellow star, half of 4.5 My is 2.25 My which corresponds to 2.25 Mly distance, exactly as the diagram depicts. For the red star, half of 5.3 My yields 2.65 Mly distance.

    For this situation, let's consider the Earth observer to hop on the rocket just as it is passing by Earth at time zero (for everybody, since it is the origin of all the IRF's). Here is a diagram similar to the first one but showing the radar signals starting prior to time zero with the blue Earth observer and ending with the echoes returning to the black rocket observer after time zero:

    attachment.php?attachmentid=66473&stc=1&d=1392061624.png

    For the yellow star, the observer sent the radar signal at -3.5 My and received the echo at 2.3 My and calculates a time of -0.6 My at a distance of 2.9 Mly. For the red star, he sent the signal at -2.5 My and got the echo at 3.8 My for a time of 0.65 My at a distance of 3.15 Mly.

    But note that these measurements are not depicted on the diagram because it is inertial and the observer is not.

    Now I want to show you that he gets the same thing if we look at it from the rest IRF of the rocket:

    attachment.php?attachmentid=66474&stc=1&d=1392061624.png

    Again, this diagram does not depict the observer's measurements for the same reason as before (he is not inertial and the diagram is).

    Next I want to show you how he can construct a non-inertial rest frame by making radar measurements throughout his entire trip. We have to show this starting earlier and ending later and I'm going to show it in another IRF in which the observer's speed is the same both before and after he transfers from the Earth to the rocket. His speed is 0.208712c, first towards Andromeda and then away:

    attachment.php?attachmentid=66475&stc=1&d=1392061624.jpg

    The observer keeps track of all the sent and received times and does the calculations as before and then constructs this non-inertial diagram:

    attachment.php?attachmentid=66476&stc=1&d=1392061624.png

    Note that the times and distances that he calculated for the two stars are now depicted accurately. In fact, if you want, you can copy this diagram and drawn in all the same radar signals that were in the previous diagram to validate the process.

    You're welcome. I only hope that you still feel like thanking me.
     

    Attached Files:

  7. Feb 10, 2014 #6

    Dale

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    I just want to second what JesseM said. Reference frames are not an inherent part of physics (they are part of the analysis), any observer is free to use any frame that they like. So if they disagree it is only because they have chosen to disagree. The details depend on that choice. ghwellsjr described some of the acceptable ways of disagreeing.
     
  8. Feb 10, 2014 #7

    ghwellsjr

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    And if they agree it is only because they have chosen to agree as I pointed out in another recent Andromeda thread:

     
  9. Feb 10, 2014 #8

    Dale

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    I have chosen to agree :smile:
     
  10. Feb 11, 2014 #9
    Brian Greene gave a similar example in his Nova TV series on the "Fabric of the Cosmos" (and also on his book of the same title). I thought his explanation was very good.
     
  11. Mar 14, 2014 #10
    Ghwellsjr, I looked through your diagrams and I apologize, but they seem very complicated to me at this point. I don't really understand them and the relation to my question. Maybe you or somebody else can explan this if I give more details to my questions. Or if I specify the example.

    So we have two observers on Earth mutually at rest and at rest wrt Andromeda. Their clocks are in sync so they agree on simultaneity.

    One of them starts to move away from Andromeda, and from the other observer that stays in the same inertial frame. He accelerates and the starts to move away inertially, and therefore he should consider the present of Andromeda the past of what the stationary observer considers to be present. So my question is how? If we consider the time to flow 'normally' in the reference frame of the stationary observer, and by that I mean that the clock on Andromeda ticks at the same rate as his, how is it possible that the 'moving-away' observer saves a day of time, or more, during the short acceleration period? How do the two simultaneity 'surfaces' compare?
     
  12. Mar 14, 2014 #11

    pervect

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    Why don't you draw them for yourself? If you draw your own, perhaps GH's will make more sense to you when you compare them to yours.

    Hints:

    Let the stationary observers coordinates be t,x,y,z. Let the moving observer's coordinates be t', x', y', z'

    Let the motion of the moving observer be in the x direction according to the stationary observer

    Then y' = y, and z' = z, so you don't need to draw them, meaning you can draw the simultaneity surfaces on a 2 d diagram of x and t.

    Set c=1 for simplicity, by choosing units of years and light years

    Surfaces of simultaneity will be t=constant for the stationary obserer and t'=constant for the moving observer

    Use the Lorentz transform to transform between (t,x,y,z) and (t', x', y', z')
     
  13. Mar 14, 2014 #12

    ghwellsjr

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    Maybe my diagrams are too complicated or maybe they don't have enough appropriate detail or explanation.

    I'll redraw my first two diagrams with an attempt to remedy these problems. Here is the first one showing the observer on Earth in blue and the observer on Andromeda in red 2.5 million miles away with two events labeled X and Y:

    attachment.php?attachmentid=67618&stc=1&d=1394811434.png

    I have labeled the dots that represent one-million year increments of time for both observers with the times on their previously synchronized clocks and aligned them with the Coordinate Time of the diagram. They both consider the explosion of the red star, event X, to be in their present because it is at their synchronized time of 0 years. The correct way of saying this is that the events of their clocks being at 0 years and event X have the same Coordinate Time of 0 years. They both consider the explosion of the yellow star, event Y, to be in their past because it is at their synchronized time of -1 million years. The correct way of saying this is that the events of their clocks being at -1 million years and event Y have the same Coordinate Time of -1 million years. I hope I have made this simple enough to clear up any confusion. Does it make perfect sense to you?

    Now you originally stated that there was another observer moving away from Andromeda. So here is a diagram showing just that observer in black and the observer on Andromeda in red along with the same two events of the stars exploding:

    attachment.php?attachmentid=67619&stc=1&d=1394811434.png

    Since the black observer is moving, he can't synchronize his clock to the Andromeda clock because his own clock is Time Dilated in this diagram meaning that the dots marking the one-million intervals of time are stretched further apart than the Coordinate Time. However, I have set the time on the black clock so that it is at 0 years when the Coordinate Time is 0 years. That means that in this diagram, it is event X that is simultaneous with the moving observer's present. The correct way of saying this is that the event of the moving observer's clock being at 0 years and event X have the same Coordinate Time of 0 years.

    Does this diagram and explanation make perfect sense to you? Any questions?

    Now to see how the black observer establishes simultaneity according to his own rest frame, we have to use the Lorentz Transformation process to get from the coordinates of the previous diagram to this diagram:

    attachment.php?attachmentid=67620&stc=1&d=1394811434.png

    Do you understand how the Lorentz Transformation process works? If not, then, of course this diagram won't make any sense to you.

    But assuming that the Lorentz Transformation is not a problem, then we can see how in the black observer's rest frame, it is event Y that is simultaneous with his time of 0 years or in his present. The correct way of saying this is that the event of the black observer's clock being at 0 years and event Y have the same Coordinate Time of 0 years.

    So before we can go on, I need to know if everything so far is perfectly understandable to you? Please post any points of confusion.
     

    Attached Files:

  14. Mar 14, 2014 #13
    @Ghwellsjr, thank you very much for the diagrams, I've understood them quite well, I think. The only problem with this is that they don't quite follow the scenario I was talking about. In the scenario that I'm seeking the answer for, the blue observer in some kind of way becomes the black observer.

    I'll again describe it. We have two observers on Earth which are at rest wrt Andromeda and they agree on simultaneity. Suddenly, one starts accelerating to move away from Andromeda and this should shift his simultaneity hyperplane in some kind of way so that after he finishes acceleration and starts to move inertially, his 'now' of the Andromeda should be the event of Andromeda which happens before the event that the stationary observer considers to be present. So how does this happen? If the observer didn't start to accelerate he would keep on agreeing with the second one about what's happening on Andromeda. The one that didn't accelerate continues to have his definition of simultaneity, while the first one changes frames, and what does he get as a result?
     
  15. Mar 14, 2014 #14

    Dale

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    Not necessarily. There is no standard definition of simultaneity for a non-inertial observer's frame, like this one. So you have to define the convention you want to use for determining his coordinates.

    That depends on the convention he arbitrarily chooses to adopt. Here is a paper on the topic which outlines a convention that I personally like. Note, it is merely a convention, and not required by nature in any way.

    http://arxiv.org/abs/gr-qc/0104077
     
  16. Mar 14, 2014 #15
    Hey Dale,

    I understand that there is no standard convention, but what's certain is that the observer comes from one meaning of simultaneity (before acceleration) to another (after acceleration). So he shifts his lines of simultaneity, while the lines of the other observer remain constant, right? For me, the biggest question is what both of them consider to be their present of Andromeda before, during, and after acceleration of the 'moving one'.
    Btw thanks for the link, I'll certainly check it out. I hope it will help.
     
  17. Mar 14, 2014 #16
    @Dale, I've red the convention you gave me, I understand it but the part that confuses me is the fact that the lines of simultaneity during of the travelling twin described here are different than those in the 'textbook' example, which is the one that is most mentioned. Is this true or am I missin something? And if it is, why so?
     
  18. Mar 14, 2014 #17

    JesseM

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    Why is that certain? Do you understand that the laws of nature don't force any particular simultaneity definition on anyone, that any observer is free to use any simultaneity convention he pleases? Even for purely inertial observers, it is just a matter of human convention that physicists often describe the inertial frame where they are at rest as "their own" frame, it's not as if there would be anything physically "wrong" with an inertial observer adopting some different definition of simultaneity.
     
  19. Mar 14, 2014 #18
    So basically there is no paradox in assigning more than one time of happening (relativized to different inertial frames) to an event, or to assign the same event to different equivalence classes under simultaneity?
     
  20. Mar 14, 2014 #19

    PAllen

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    Correct. Certain rules for the assignment have desirable properties, especially for inertial frames, but that's about all you can say.
     
  21. Mar 14, 2014 #20

    JesseM

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    Not sure what you mean by "different equivalence classes under simultaneity"--under any specific definition of simultaneity there is of course a unique truth about which groups of events are simultaneous with each other according to that definition, but there's no paradox in the fact that you there are multiple equally valid definitions that define simultaneity differently. It's similar to how, if you have a 2D plane with a pair of dots on it, and you want to draw some Cartesian x-y axes in to assign coordinates to each event, you have a choice of different angles that you can orient your axes, and depending on how you orient them you can get different answers to the question "do both dots have the same y-coordinate?"
     
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