Why Didn't Einstein FULLY Address Simultaneity-at-a-Distance?

Discussion in 'Alternative Theories' started by Mike_Fontenot, Apr 11, 2021.

  1. Beaconator Valued Senior Member

    Messages:
    1,486
    They share the same destination
     
  2. Google AdSense Guest Advertisement



    to hide all adverts.
  3. phyti Registered Senior Member

    Messages:
    732
    mike;

    You are thinking in terms of an ideal perfect world, where nothing goes wrong or deviates from a plan. In the real world, things vary, and there is no such thing as a perpetually inertial observer since gravity will alter the paths of objects over time, and why Einstein titled the theory 'Special Relativity'. Short segments of curved paths may be considered inertial for short time intervals.
    Assuming distant synchronized clocks:
    In the left example the TDE predicts he will be 8 arriving at x=6, triggering a clock signal t=10, which is received by her at t=16. The experience agrees with prediction.
    She concludes he WAS 8 when she WAS 10.
    In the right example, he has an unanticipated problem requiring a reduction in speed. He is 10.3 arriving at x=6, triggering a clock signal t=12, which is received by her at t=18. The experience does not agree with prediction. Even if she receives an intermediate signal from when he is 7.2, she will only be aware of an increase in his clock rate, but will not know IF it remains constant or changes. The final signal from the destination allows a conclusion.

    Please Register or Log in to view the hidden image!

     
  4. Google AdSense Guest Advertisement



    to hide all adverts.
  5. Neddy Bate Valued Senior Member

    Messages:
    2,548
    Okay, but in your prior post you wrote, "she can set up a grid of clocks, stationary wrt her, and synchronized via light signals..." so there is clearly a chosen simultaneity convention already in place throughout her reference frame. That is, unless you want to allow some people who are stationary wrt to her to disagree as to whether or not her grid of clocks is synchronised.

    If someone who is stationary wrt to her claims that grid of clocks is not synchronised, then that person is effectively claiming that the speed of light was not a constant in her reference frame during the time when the synch process was performed. Everyone is entitled to their own opinions, but they are not entitled to their own facts.

    Yes I know about your simultaneity method. But do you see now that your method is essentially allowing him, a guy who is now stationary wrt to her, to legitimately (according to you) claim her grid of clocks is (temporarily!) not synchronised? Do you see now that he would be claiming the speed of light was not a constant in her reference frame during the time when the synch process was performed, which would have been long before he ever accelerated?

    And then later, after the DI, your method tells him to change his tune and say, "Well, now the grid of clocks is synchronised, but it wasn't before!" even though nothing was done to the clocks in the meantime.

    Once again, everyone is entitled to their own opinions, but they are not entitled to their own facts.
     
    Last edited: May 26, 2021
  6. Google AdSense Guest Advertisement



    to hide all adverts.
  7. Mike_Fontenot Registered Senior Member

    Messages:
    622
    Let T- denote the instant (in his life) infinitesimally before his speed change, and let T+ denote the instant immediately after his speed change. There is only an infinitesimal time difference between those two instants.

    Let PIO_A and PIO_B be co-located with him at both T- and T+. At T-, he is stationary wrt PIO_A. At T+, he is stationary wrt PIO_B.

    With the CMIF method, at T-, he tells PIO_A that A's array of clocks are all perfectly synchronized, and he tells PIO_B that B's array of clocks are terribly synchronized. At T+, he tells PIO_A that A's clocks are terribly synchronized. and he tells PIO_B that B's clocks are all perfectly synchronized. A VERY abrupt change.

    With my method, at T-, he tells PIO_A that his clocks are all perfectly synchronized. And he tells PIO_B that his clocks are terribly synchronized. At T+, he tells PIO_A that his clocks are still perfectly synchronized. And he tells PIO_B that his clocks are still terribly synchronized. No change from what he told them before, but this is the beginning of the disagreement interval, because he is now stationary wrt PIO_B, and he disagrees with him. Later, about 10% of the way through the DI, he tells PIO_A (via a message) that his clocks aren't perfectly synchronized any more, but they are not very badly synchronized. And he tells PIO_B that his clocks are still badly synchronized, but a little better than before. Later still, he tells PIO_A that his clock synchronization has gotten worse, but is still not terrible. And he tells PIO_B that his clocks are still not synchronized, but not as bad as at the last report. Finally, at the end of the DI and also thereafter, he tells PIO_A that his synchronization is terrible. And he tells PIO_B that his synchronization is perfect.

    It seems to me that my method comes out of that comparison looking pretty good. In both methods, clocks that were perfectly synchronized become terribly synchronized, and vice versa. But in my method, the changes in the degree of synchronization are gradual, not abrupt as in the CMIF method. That could be considered to be a good thing.
     
  8. phyti Registered Senior Member

    Messages:
    732
    Mike;

    [Nothing happens in zero time.
    The motion shown in the 3 sided drawing is discontinuous, an unreal motion not described with physical laws, and therefore can't be analyzed as such.
    Clocks are either synchronized or not synchronized.
    The results you claim are due to faulty application and interpretation of simultaneity.
    There is no instantaneous knowledge.]
     
  9. Neddy Bate Valued Senior Member

    Messages:
    2,548
    No, it's not a good thing, because there is no physics behinds it.

    There is a physical explanation why the grid of clocks that are stationary with respect to the stay "home" twin (she) can be safely considered to be syncrhonised in the "home" reference frame. It is because the speed of light is a constant in that reference frame, and the velocity of all of those clocks was constantly zero with respect to that reference frame throughout the entire synchronisation process. The synchronisation process relies on those physical facts in order to legitimately claim that those clocks are syncrhonised in the home reference frame.

    There is also a physical explanation why the traveling twin (he) can legitimately claim that her grid of clocks is not syncrhonised, while he is traveling at non-zero velocity with respect to her. It is because even though the speed of light is still a constant, he can legitimately say that the velocity of all of those clocks was non-zero throughout the entire synchronisation process. Remember, he considers the whole "home" frame to be moving at constant velocity in one direction, and so the traveling reference frame concludes correctly that her clock grid was moving during the synch process.

    Thus, the traveling twin, while he is traveling at non-zero velocity with respect to her, can legitimately claim that the light signals used to synch the clocks must have taken different amounts of time to travel between her clocks, depending on whether they were sent in the same direction her clocks were moving, or the opposite direction her clocks were moving.

    While he is traveling at non-zero velocity with respect to her, he can legitimately say that, for those light signals which were sent in the same direction the clocks were moving, to him it is as if the receiving clock would have been running away from the light, and thus the light would have had to travel for more time while it catches up to the clock that is moving away from the light. Likewise, he can legitimately say that, for those light signals which were sent in the opposite direction the clocks were moving, to him it is as if the receiving clock would be running toward the light, and thus the light only needs to travel for less time to reach the clock that is moving toward the light.

    You should be able to prove this to yourself with some diagrams. You can show that for a clock-grid moving to the right, the Einstein synch process would make a clock at the left end of the grid be ahead of a clock at the right end of the grid. And vice-versa for a clock-grid moving to the left -- the Einstein synch process would make a clock at the left end of the grid be behind a clock at the right end of the grid.

    So before the traveler stops, he can legitimately claim that her clocks were not synchronised, (for the above physical reasons). But after he stops, he must acknowledge that all of his valid reasons only held true in the traveling frame, not the home frame. So, once he becomes stationary with respect to the home frame, and for as long as he is stationary with respect to the home frame, he must agree that those clocks are properly synched in the home reference frame, for purely physical reasons.

    So, in short, simultaneity in SR is based on the actual physics of the Einstein synch process.
     
  10. Mike_Fontenot Registered Senior Member

    Messages:
    622
    I don't find your argument compelling. Deciding that a line of clocks, which were perfectly synchronized an instant ago, are now suddenly very unsynchronized seems more bizarre to me than a line of clocks which gradually become unsynchronized.

    I am, at the moment, much more interested in your reaction to my possible proof, in the other thread, that negative ageing doesn't occur in the CMIF simultaneity method.
     
  11. Neddy Bate Valued Senior Member

    Messages:
    2,548
    Well, you have that backwards for the specific case that we were discussing. The traveler should find the home grid of clocks to be unsynchronised before he comes to a stop with respect to them, and then he should find the home grid of clocks to be synchronised after he comes to a stop with respect to them. You would not have been backwards if the traveler was considering the synchornised grid of clocks mounted inside his spaceship, (that grid of clocks would have gone from synchronised before he jumped out of the spaceship, to unsynchronised after he jumped out of the spaceship and landed on the ground, now stationary with respect to the home twin). However, the spaceship grid of clocks has not been discussed, as far as I know.

    But let's explore what is and what is not compelling to you. Have you ever heard of the "Crackpot Doppler Theory"? It is basically identical to Doppler in that when an observer is traveling away from a light source, the light appears red shifted to him. The difference in Crackpot Doppler Theory is that if the observer decelerates and comes to a stop with respect to the light source, the observer can still claim the light is red shifted for a certain period of time.

    In Crackpot Doppler Theory, the observer can make this unsubstantiated claim for a period called the "Disagreement Interval". Perpetually inertial observers stationary with respect to the light source will see the light as the true color of the light source, naturally. However, during the Disagreement Interval, the recently accelerated observer simply "disagrees" with them and says the light is red shifted, although the amount of red shift is becoming less and less, gradually fading away on its own, even though there does not appear to be any underlying mechanism at all for this change. Then, at the end of the Disagreement Interval, the formerly accelerated observer is finally in agreement with the other inertial observers, and says the light appears as the true color of the light source.

    There is only one small problem with Crackpot Doppler Theory, which is that there is no physics behind it at all. There is a physical reason behind the idea that moving away from a light source causes the light to appear red shifted, just as there is a physical reason behind the idea that being stationary with respect to a light source causes the light to appear as the true color of the light source. But there is no physical reason behind the idea that the red shift can persist after the observer has come to a stop, as the "Disagreement Interval" allows. That is just a made-up idea that has no place in physics.

    But I understand from your statements that you are not going to find the "no physics behind it at all" argument to be "compelling." So let me give you the crackpot reasoning instead, and see if it sounds familiar to you:

    "Deciding that light, which was red-shifted an instant ago, is now suddenly not-red-shifted seems more bizarre to me than light which gradually becomes not-red-shifted."

    Do you find that compelling? If not, then why would you expect anyone to find your "seems more bizarre to me" argument to be compelling?

    The cause seems hopeless to me, as exemplified above. You are not interested in any physics that might contradict something that you "invented" and thus have pride of ownership in. If you would like to see how the threads always go, then you should try arguing against the Crackpot Doppler Theory, and I will play devil's advocate and support it, no matter what valid objections you raise.
     
    Last edited: Jun 15, 2021
  12. Mike_Fontenot Registered Senior Member

    Messages:
    622
    I can't remember what your position is on negative ageing (with the CMIF simultaneity method). In the standard twin paradox scenario, I think you DO believe that the distant person (she) ages instantaneously when the traveler (he) instantaneously reverses course. But if instead he instantaneously increases his speed away from her, do you believe he'll say that she instantaneously gets younger? Since you are a committed CMIF believer, I would think you would answer "yes" to that question, but I seem to recall that you don't believe that negative ageing occurs. Can you set me straight on where you stand on that question?

    Because of my argument that two back-to-back instantaneous velocity changes (with no time elapsing between them), the first one from +v to -v (where + means "away from her", and - means "toward her"), followed by the second one from -v to +v, must be equivalent to no velocity change at all, it follows that if an instantaneous age increase can occur, an instantaneous age decrease must also be able to occur. That's why my "proof" (IF it's correct) is potentially important: if sudden negative ageing doesn't occur, then sudden positive ageing can't either, which would mean that CMIF can't be true.

    Do you see a flaw in my proof? I haven't been able to spot one yet. There certainly could be an error in there somewhere, and if so, I want to know that.

    I don't really "have a dog in this race" (between CMIF and my new method). I like the simplicity of CMIF, and I have no problem with negative ageing. What bothers me is to not KNOW what the correct simultaneity-at-a-distance for an accelerating observer IS. I'm convinced that there IS a correct simultaneity-at-a-distance for an accelerating observer, and I want to know what it is.
     
    Last edited: Jun 15, 2021
  13. Neddy Bate Valued Senior Member

    Messages:
    2,548
    That is a bit of a cop-out, because you are not only questioning the correct simultaneity-at-a-distance for an accelerating observer, you are questioning the correct simultaneity-at-a-distance for an inertial observer who has accelerated in the past. Please consider that you are deliberately ignoring the possibility that the concept of simultaneity in SR might depend solely on the reference frame, and not on any particular observer's acceleration history.

    Your concept is absurd. You are willing to accept that all of the perpetually inertial observers in the home-frame know for a fact that she is 40 years old at the time when the clocks in their home-grid display 40. But then for the case of the traveler being 20 years old and jumping off his spaceship and landing stationary next to a home-grid clock that displays 40, you suddenly can't figure out how old he should say she is? You want to throw in the possibility that she could be both 10 (for him) and 40 (for everyone else) at the same time, with all of them stationary with respect to each other? That would mean the "famous time dilation equation" that she uses to calculate their respective ages at the end would have to be updated so that it leads to multiple ambiguous answers -- she is 40, he is 20, and she is also 10 ?!?!? Come on, you think that might be valid SR, and it was just not clarified enough in all these years?

    I call bullshite. You are trying to make it look like there is an unexplained question in SR, when there isn't. Nowhere in SR, or anyplace else in physics for that matter, would the acceleration history of a person have to be known so that they could be set aside as having different physical interpretations than everyone else stationary with respect to them.

    Once you accept that he should say she is currently 40 (not 10) in this scenario, then you should be able to figure out the answers to the questions you asked me. Of course if he rapidly jumps back on the spaceship, he would say she is 10 again, and if he rapidly jumps off again, he would say she is 40 again. There is no problem with his simultaneity going in either direction, because his simultaneity does not affect the way she ages in her own reference frame.
     
    Last edited: Jun 16, 2021
  14. Mike_Fontenot Registered Senior Member

    Messages:
    622
    You are still ignoring the really important question: "Is my proof, that negative ageing doesn't occur, valid?" If my proof IS valid, then CMIF simultaneity ISN'T valid.
     
  15. Neddy Bate Valued Senior Member

    Messages:
    2,548
    You don't think it is invalid that all of the perpetually inertial people who are stationary wrt the home frame say that "she" is 40 years old when the home-grid synchronised clocks display 40 years. So to PROVE that it would be invalid for "him" to use CMIF simultaneity after he has stopped moving and become stationary with respect to her, you would have to prove the following:

    1. All of the perpetually inertial people who are stationary wrt the home frame are CORRECT when they say that "she" is 40 years old when the home-grid synchronised clocks display 40 years, but...
    2. The one person "he" who is stationary wrt the home frame who has not been perpetually inertial is INCORRECT if he says that "she" is 40 years old when the home-grid synchronised clocks display 40 years.

    Go ahead, I'll wait.
     
    Last edited: Jun 16, 2021
  16. Mike_Fontenot Registered Senior Member

    Messages:
    622
    To show that my proof is invalid, you need to identify at least one of my statements in the proof, and show how and why that statement is invalid. You haven't done that.
     
  17. Neddy Bate Valued Senior Member

    Messages:
    2,548
    Let's leave that topic to the other thread.

    In this thread, you are arguing that she is both 40 years old and 10 years old, at the same time, in the same inertial reference frame. And you are wondering why Albert Einstein never fully addressed why that would never be possible in SR. Maybe because it is total nonsense, and he never thought anyone would be so confused about SR to even consider it?

    And, in this thread, you have also declared that her being both 40 years old and 10 years old, at the same time, in the same inertial reference frame, seems "less bizarre" to you that if she were just plain old 40 years old.

    That's what this thread is about. Please explain, or forever hold your peace.
     
    Last edited: Jun 17, 2021
  18. Bells Staff Member

    Messages:
    24,270
    Mod Note

    Moving this to a more appropriate sub-forum.

    Mike_Fontenot, you seem somewhat obsessed with this (given you have been pushing this across multiple platforms and sub-forums) and you appear to push your pet theory in multiple threads. Please stop doing this. I've already had to move one to the alternative theories sub-forum.
     

Share This Page