Graphical Derivation of the CADO Equation

Discussion in 'Physics & Math' started by Mike_Fontenot, Sep 5, 2018.

  1. Mike_Fontenot Registered Senior Member

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    622
    The CADO equation can handle any kind of accelerations by the traveling twin. An example is given in Section 7 (entitled "Some CADO Equation Results for Finite Accelerations") of my webpage:

    https://sites.google.com/site/cadoequation/cado-reference-frame

    for a sequence of +/- 1g accelerations by the traveler, and an "age-correspondence" diagram is described which shows the current age of the home twin, for each instant t of the traveler's (his) life, according to him. Specifically, the diagram gives CADO_T(t) during a -1g acceleration that lasts two years of his life, followed immediately by a +1g acceleration that lasts for another 2 years of his life. (The negative acceleration means that he points the nose of his rocket toward her, and the positive acceleration means that he points the nose of the rocket away from her.)

    At the beginning of the -1g acceleration, they are separated by about 40 lightyears (according to her), which took 26 years of his life, at a constant speed of 0.774 ly/y (constant except for a short initial 1g acceleration to get him up to the 0.774 ly/y speed) . She is 17 years old then (according to him). At the end of the -1g acceleration, he is 28 years old, and she is 81.24 years old (according to him). During that -1g acceleration, she gets 64.24 years older (according to him), whereas he only gets 2 years older. Her rapid age increase is smooth and continuous during that acceleration.

    Then he does the two year +1g acceleration. At the end of the +1g acceleration, he is 30 years old, she is 21.75 years old (according to him). During that +1g acceleration, she gets 59.49 years younger, whereas he gets only two years older. Her rapid age decrease is smooth and continuous during that acceleration.

    In my paper:

    "Accelerated Observers in Special Relativity", PHYSICS ESSAYS, December 1999, p629,

    a plot of the her age (according to him) vs his age, during those back-to-back accelerations, is given. The webpage gives a qualitative description of the shape of that plot, but the plot itself isn't given. (The scenario in the paper is somewhat different from the scenario in the webpage, but they are qualitatively similar).
     
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  3. NotEinstein Valued Senior Member

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    Sorry, I should have been more clear. I wrote: "I'm pretty sure you'll find that even Charlie disagrees with you." What I meant with that is that if Charlie calculates what's going on *after* he switches back to the train frame, he'll get an answer that conflicts with his previous calculation.

    That's why I asked you to draw the travelling light into your Minkowski diagram as well. You'd have noticed that all the information that Charlie receives (i.e. everything that's physically relevant) gives a single, coherent picture about what's going on. In other words, Charlie will end up observing a reality that disagrees with this first calculation.

    It's not that his first calculation is wrong or incorrect. It just doesn't apply to his new situation anymore.

    No, you got that the wrong way around. It's not me that must convince you about my "overwrites" idea, it's that you must resolve the conflict. If my "overwrites" idea turns out to be a (the?) solution, great! And if it turns out there's a better answer than my "overwrites" idea, I'm happy to be corrected/improved upon.
     
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  5. Neddy Bate Valued Senior Member

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    Yes, *before* Charlie jumps from the ground to the train he calculates t=40, x=0. And *after* he switches to the train frame, he calculates t=10, x=0. I'm not sure those two calculations "conflict" but they certainly are different. Is the reason you think they conflict just because they are different, or because 40 is greater than 10?

    I'm not sure what difference you think it makes for us to worry about when the information actually reaches Charlie. For example, the light from the t=10, x=0 event reaches Charlie as he is riding on the train at his own time t'=37.32 x'=0. From there, Charlie would have to consider how far the light traveled, (which would be 17.32 light years because the light originated from x=-17.32), and then calculate how long ago that event happened, (17.32 years ago), and then subtract that from his current time to get t'=37.32-17.32=20.

    But back when his time was t'=20, x'=0 he could have calculated t=10, x=0 and known from that SR calculation that those events are both simultaneous in the train frame. He doesn't get a different result for that time calculation whether he waits for the information or not. But of course it is always good to verify that the results are the same, using both methods.

    If you are concerned with what is actually seen by eye, a good question and answer on the matter was given in posts #37 and #38 in the other thread:

    Confused2's Question:
    http://sciforums.com/threads/instan...n-the-twin-paradox.161164/page-2#post-3543720

    Janus58's Answer:
    http://sciforums.com/threads/instan...n-the-twin-paradox.161164/page-2#post-3543727

    I don't really see it as a conflict. Different frames have different planes of simultaneity. Both before and after Charlie jumps from the ground to the train, he would say that the local ground time is t=40, x=34.64. To me, an example of a "conflict" would be if Charlie had to explain how a local clock in his immediate vicinity changed times, but fortunately that doesn't happen. SR is not supposed to have conflicts, I don't think.
     
    Last edited: Oct 5, 2018
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  7. phyti Registered Senior Member

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    The CADO Frame, for the Standard Twin Paradox Scenario


    In the drawing two clocks c1 and c2 are static. At the origin, c2 sends a light signal (event A) which reflects (B) and returns to c2 (C). c2 assigns B to 1/2 of transit time

    A-C, with simultaneity axis (sa) indicated by the green line.

    If c2 accelerates toward c1, the signal returns earlier (D), with sa indicated by the red line. If c2 accelerates away from c1, the signal returns later (F), with sa indicated by the magenta line. The speed profiles that show acceleration require finite intervals of time and therefore cannot provide instantaneous position data.
    The SR clock synch convention requires constant linear (inertial) motion, and is only practical on a local level. A universal system of synchronized clocks is an ideal and imaginary concept.
     

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  8. Mike_Fontenot Registered Senior Member

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    622
    I don't understand your statements above (that I have highlighted in red). Could you please elaborate?
     
  9. phyti Registered Senior Member

    Messages:
    732
    To attempt to synchronize a clock even 1 ly distant is not possible since the earth would continuously change position, while the light is in transit. Consider the clocks of the GPS system are updated in very short intervals because they drift, and are accelerating in direction. yet these are local.
     
  10. Mike_Fontenot Registered Senior Member

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    Special relativity theory assumes that no gravitational fields are present anywhere. Therefore there is no earth in the twin "paradox" scenario.
     
  11. NotEinstein Valued Senior Member

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    1,986
    Apparently, you've forgotten what we were talking about again, so I'll repeat myself: the conflict is that t=40 comes after t=10 according to what Charlie observes, while he calculates t=40 happening before t=10. (In fact, Charlie calculates t=40 happening twice.)

    Please do try to keep up.

    I know you don't; you are having huge trouble figuring out what's "physically meaningful" and "real", and thus information means nothing to you.

    You are skipping the most important part: does t=40 happen before t=10, even according to Charlie's measurements? That was what we were talking about, remember?

    Please do try to keep up.

    I'm obviously not; see my distinction between observation and measurement.

    Please do try to keep up.

    I know you don't; you don't understand the "arrow of time".

    Which was never in question.

    Which isn't in the slightest relevant here, so I don't know why you'd bring that up?

    So t=40 happening twice isn't a problem to you? t=40 happening before t=10 isn't a problem to you? Breaking causality isn't a problem to you?

    Sure, if you ignore or refuse to acknowledge the problems, you don't see them. Doesn't mean they're not there.

    And I know you at least understand *that*, because you're going through great efforts to avoid addressing these problems directly.

    Exactly, which is why the conflict has to be resolved, as I explained before.

    Please do try to keep up.
     
  12. Neddy Bate Valued Senior Member

    Messages:
    2,548
    Well "Charlie's measurements" may be what you are talking about. I have been talking about the two DVRs which Charlie set to record after pre-calculated time delays, and the two video broadcasts which end up being recorded on those two DVRs.

    There is no disputing that the ground-DVR was set correctly and recorded the coordinates t=40, x=0 being the case at the moment when Charlie is on the ground. There is also no disputing that the train-DVR was set correctly and recorded the coordinates t=10, x=0 being the case at the moment when Charlie jumped onto the train.

    We know Charlie was on the ground before he jumped on the train, so there cannot be any disputing that t=40, x=0 came before t=10, x=0 in the case of Charlie doing his jumping routine. It is a valid result of SR, so it cannot be a "conflict," and it cannot need to be "resolved" in some other way.

    Sorry, I'm still not seeing the 'conflict'. And if there was one, I don't see how you could just 'resolve' it with your words. Does it really change anything if you say t=10, x=0 "overwrites" t=40, x=0? Does it really change anything if you say Charlie never "measures" a clock going backwards? Not really, I don't think so. It doesn't change the DVR recordings.

    Why don't you try just accepting SR at face value? It works for me.
     
    Last edited: Oct 12, 2018
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  13. Neddy Bate Valued Senior Member

    Messages:
    2,548
    I think you mean Charlie calculates t=10, x=0 twice. Just before he jumps off the train he calculates t=10, x=0, then just after he jumps off the train he calculates t=40, x=0, then just after he jumps back on the train he calculates t=10, x=0 again.

    You seem to have no problem with the first two calculations, because 10<40 but then you suddenly seem to become concerned with the last two calculations because 40>10. This just shows how arbitrary your concerns are, because there is no physical difference between 'jumping off' vs. 'jumping on' a train.

    Also, when Charlie jumps off the train, he leaves a hypothetical assistant on the train in his place. That assistant calculates t=10, x=0 the whole time Charlie is gone, and when he returns. Naturally Charlie should agree with that assistant once he returns after his jumping routine.

    Of course if we make the jumps of non-zero duration, then we might have something more like t=10.000, x=0 followed by t=40.000, x=0 followed by t=10.001, x=0 if that makes you feel any better.
     
    Last edited: Oct 12, 2018
  14. phyti Registered Senior Member

    Messages:
    732
    Mike
    It's not about Earth or gravity. It's about curved motion in space. Two anauts in space are tracking each other. If one sends signals to synch their clocks, while changing his velocity, when the signal returns his velocity will be different from when it was sent. I.e. the clock synch method requires constant linear speed and a finite amount of time. This eliminates the possibility of instantaneous data.
     
  15. Mike_Fontenot Registered Senior Member

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    622
    The only synchronization of clocks being done is within a given perpetually inertial reference frame. The
    clocks being synchronized are all perpetually motionless wrt each other.
     
  16. NotEinstein Valued Senior Member

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    1,986
    In the ground frame, yes.

    In the moving frame, yes.

    But what does Charlie say about his moment on the ground, when the information about Alice's clock reaches him? You seem to fail to understand what the word "overwrite" means: when Charlie jumped back on the train, his calculation no longer applies. It's still correct, but no longer applicable to his situation. It's clear as day in the Minkowski diagram, that by now, after all these weeks, you must have drawn. The answer is literally in front of your face.

    Hint: Charlie's plane of equivalence sweeping backwards in time doesn't make Alice travel back in time, as I've pointed out before.

    The calculations, yes. The travelling back in time, no. Which poses the conflict.

    You continue to deny the conflict, because you are ignoring half of it. Your continued purposeful selective ignoring is quite pathetic, to be honest. A great pseudo-scientific tactic, though, and a wonderful display of intellectual dishonesty. Really not something reasonable people would do, but hey, you do you.

    So what about the violation of the arrow of time? The violation of causality? The non-violation of T-symmetry?

    Oh right, you choose to ignore all of that, because otherwise you would have to claim they are valid results of SR.

    If t=40 precedes t=10, then you have entropy in a closed system decreasing. That's a violation of the second law of thermodynamics. I would consider that a bit of a problem, wouldn't you?

    I've already mentioned multiple issues multiple times, so it's clear there is one.

    You do know that words can form arguments, and that arguments can lead to logical conclusions?

    Oh, I definitely agree that the *choice* of words may be poor. Feel free to suggest a better one.

    So measured reality doesn't matter to you? Now I understand why you can't give a definition of "real" or "physically meaningful"! That's for clear that up for me.

    True, but what's Charlie's conclusion about what the DVR's have been doing after he recieved all the relevant information from Alice's clock? You keep intentionally forgetting about the second half of the story.

    Why don't you? Look at your Minkowski diagram; point me to where Alice is travelling back in time.

    Clearly not, or you wouldn't feel the need to continually ignore half of everything said.

    He calculates Alice to be at t=40 twice too: when he jumped off, and way later after he jumped back on. See, you're ignoring that second half again.

    Yep; time travelling forwards happens all the time, so that isn't an immediate red flag. Time travelling backwards, though...

    You really should read that "arrow of time" article I linked earlier.

    I fully agree that if one is a problem, the other one is too. However, one doesn't stand out like a sour thumb, while the other one does. Thus my concerns only seem arbitrary to people that fail to understand the underlying basis of those concerns. Basic things like the arrow of time and causality.

    Alright, and what does this assistant answer when Charlie asks him/her whether he/she calculates/measures/observers Alice travel back in time? What does the assistant answer when Charlie asks him/her whether Alice's clock showed t=40 at any point during his jumping routine? What does the Minkowski diagram look like as drawn by the assistant?

    Hint for that last one: it's the same one as Charlie and Alice draw. Now point me to the place where the assistant has Alice travelling back in time.

    Sure, and I never claimed otherwise.

    As I said earlier: that's irrelevant. Please do try to keep up.
     
  17. Neddy Bate Valued Senior Member

    Messages:
    2,548
    I'm not sure what you mean. The whole purpose of the ground DVR recording was to show that letting the information travel a known distance at the speed of light does not change the end result. The time information received by the ground DVR (t=40 x=0) matches Charlie's calculation.

    So that's why I am not sure what you mean. Perhaps you mean that after Charlie jumps on the train, and he waits for the time information to arrive to himself as he rides on the train, then the information would be t=10, x=0 for the moment he landed on the train? Yes, that is true, and that is what the train DVR records. But that pertains to the moment he landed on the train, not the moment he landed on the ground.

    Yes, I agree. The t=40 x=0 calculation is specifically for the moment he is on the ground. Once he has jumped back on the train, his calculation produces t=10, x=0 which is the only calculation that applies to him then. That is the whole point isn't it? First it was t=40 then it was t=10.

    1. If someone asks Charlie what time (t) it was at x=0 just before he jumped off the train, he would have to answer t=10.

    2. If someone asks Charlie what time (t) it was at x=0 just after he jumped off the train, he would have to answer t=40.

    3. If someone asks Charlie what time (t) it was at x=0 just after he jumped back on the train, he would have to answer t=10.

    Even if the answer to #3 "overwrites" the answer to #2, those are still the answers he has to give to those questions.

    I honestly don't see the conflict. Sorry you find that so offensive. If I could see it, I would say so. I've been doing SR calculations a long time, so maybe I'm just used to it being this way. I guess it does seem strange sometimes, but I just accept that's how it is.

    But you say none of that happens, even though Charlie's answers to the 3 questions are t=10, t=40, t=10 respectively, right? So you have already explained to yourself that none of that happens. So why is it still a problem?

    Well if the word "overwrites" prevents all of those problems, then I guess it is a good word to use. I didn't really realise there were all those problem to begin with, so I didn't realise how important it was to use the right words.

    Measured reality is what is on the two DVR recordings. They corroborate Charlie's answers to the 3 questions t=10, t=40, and t=10 respectively. That's all there is to this, right?

    It's not intentional. I honestly don't understand what else there is to this besides Charlie's answers to the 3 questions t=10, t=40, and t=10 respectively.

    I thought I already did that. On the Minkowski diagram, there are two different planes of simultaneity for the two frames. One intersects the t axis at t=40 x=0 and the other intersects the t axis at t=10 x=0. When Charlie jumps from the ground to the train, his plane of simultaneity changes from the former to the latter.

    I honestly don't know what else you think there is to this.

    Yes that is true! But that is because I thought the only moments we were concerned with were the moment before he jumped off the train, the moment after he jumped off the train, and the moment after he jumped back on the train. It would be 60 years after that (in Charlie's own time) when he would calculate t=40 x=0 again.

    Don't worry, the sore thumb is there whether Charlie jumps off the train or on the train. You just have to look at the right location on the x axis to find the sore thumb in both cases. There is no difference jumping off or on a train, it is just changing frames either way.

    The train assistant never leaves the train frame, so it is t=10 x=0 during the whole jumping routine. But Charlie and the assistant are not in the same frame when Charlie is on the ground. He could have an assistant on the ground tell him it is t=40 in the moment he is on the ground, and he would have to trust the ground assistant, not the train assistant.

    The train assistant doesn't have Alice traveling back in time. But Charlie's answers to the 3 questions are still t=10, t=40, and t=10 respectively, because of his jumping routine.
     
    Last edited: Oct 19, 2018
  18. Neddy Bate Valued Senior Member

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    2,548
    I thought it would be a good idea to post this, for reference:

    Please Register or Log in to view the hidden image!

     
  19. Neddy Bate Valued Senior Member

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    Here is something interesting I came up with over the weekend.

    This spreadsheet shows the traveling twin's perspective. The top half is the outbound leg of the trip, and the bottom half is the inbound leg. In the middle you can see where his t=10 calculation suddenly becomes t=70 after the turnaround point.

    Please Register or Log in to view the hidden image!



    The far right hand column shows what the traveling twin actually sees with his eyes, as he looks at the stay-home twin's clock. For example, on the outbound leg, he sees 5.09 on the distant clock at the same time that his own clock displays t'=19. Also, on the inbound leg, he sees 72.53 on the distant clock at the same time that his own clock displays t'=38.

    I chose to use those examples because they happen to be close to some values for t that the traveling twin had calculated earlier, (please refer to the blue boxes & arrows). What is interesting is that this provides visual confirmation to the traveling twin that his earlier calculations were correct, as follows:

    Back when the traveling twin's own clock displayed t'=10, he calculated t=5 for the stay-home twin's clock. And at that time, the distance to the stay-home twins clock was -x'=8.66 so it follows that he should actually see the t=5 after 8.66 more time has elapsed, which would be at t'=10+8.66=18.66 but the closest thing on the chart is t'=19 and he sees 5.09 at that time (see the red boxes & arrows on the top half of the chart). That is a confirmation that his calculations for t are correct on the outbound half of the trip.

    And back when the traveling twin's own clock displayed t'=25, he calculated t=72.5 for the stay-home twin's clock. And at that time, the distance to the stay-home twins clock was -x'=12.99 so it follows that he should actually see the t=72.5 after 12.99 more time has elapsed, which would be at t'=25+12.99=37.99 but the closest thing on the chart is t'=38 and he sees 72.53 at that time (see the red boxes & arrows on the bottom half of the chart). That is a confirmation that his calculations for t are correct on the inbound half of the trip.

    If all the inbound and outbound calculations are correct, then it follows that the t=10 calculation just before the turnaround, and the t=70 calculation just after the turnaround are both correct. The traveling twin could even visually confirm the t=70 in this same manner at his own time t'=20+17.32=37.32, but it just does not happen to be on the chart. The closest is t'=37 and he sees 68.8 at that time.
     
    Last edited: Oct 22, 2018
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  20. phyti Registered Senior Member

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    732
    There is no such thing. the discovery of finite, independent light speed prevents any type of instantaneous universal time.
     
  21. Mike_Fontenot Registered Senior Member

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    622
    It's not at all clear to me what you mean by "instantaneous universal time". Care to elaborate?
     
  22. phyti Registered Senior Member

    Messages:
    732
    Prior to the 1600's, light was thought to propagate instantaneously, so human observation was considered to be in 'real time'. An astronomer Ole Romer making observations of Jupiters moons, discovered a delay when Jupiter was on the far side of the sun vs the near side. This led to a finite speed of light. The end result, all observation is historical, you observe things as they were, not as they are.
     
  23. Mike_Fontenot Registered Senior Member

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    622
    I still don't see what you're objecting to, in my definition of the CADO reference frame.
     

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