Interesting. Do you mean how to replicate the condition before time of departure?
What I meant by the question was, "are my assumptions up through this point in the post correct?"
Time travel theories and temporal anomalies
(68 posts) (4 voices)-
Wed Jun 30 2010 8:55 pm #
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Max
Memberif the traveler is somehow told a replicator pattern
Sorry for the confusion, the question below asks about the replicator pattern that is quoted above.
Do you mean how to replicate the condition before time of departure?
If it is not used to replicate the condition, what does the replicator pattern used for?
Other than that, it seems alright to me.
Wed Jun 30 2010 9:07 pm # -
If it is not used to replicate the condition, what does the replicator pattern used for?
It doesn't matter yet. It doesn't even have to be used at all until after the anomaly is resolved. Use of the pattern is immaterial at this point; so far I'm just trying to establish equivalency between a written note and a replicator pattern. If the scenario I inexpertly described could potentially result in an N-jump, then I might go on to discuss something more complex.
(Of course, the ultimate goal is to refine my understanding of what is sufficient to result in an infinity loop.)
Wed Jun 30 2010 9:26 pm # -
So, then, if the traveler is somehow told a replicator pattern, travels to the past, and goes on to tell his CD self that same pattern, and the CD self travels back with this knowledge, this has the potential to be an N-jump; the CD-traveler exactly repeats the actions of the original traveler (including telling "himself" the pattern), and the timeline resolves when his younger version goes back from point D. (Notice that no cloning is involved here, only the knowledge of a replicator pattern.)
I'm good with that. The only thing that is traveling is information, and the information has an origin in this world.
For a quick analogy, the watch in Somewhere In Time keeps making the same loop, and eventually decays to dust. However, the formula for transparent aluminum in Star Trek IV: The Voyage Home makes exactly the same kind of loop, but it does not decay. This is because the original source of the information has become irrelevant. Scotty will learn the information in school regardless of who originally discovered it, and in delivering it to Dr. Peters guarantees that he will again learn it in school. In the same way, if we assumed it were possible for a replicator pattern to be memorized and transmitted from one person to another, the traveler could bring back the pattern and deliver it to his alternate self, who would then bring it back and deliver it to himself, and that would establish the loop.
Note at this point that if the traveler were to bring back the pattern and use it himself, but not teach it to his alternate self, then when his alternate self comes back he will not have the pattern and will not be able to use it.
Here it is assumed that Y chromosome does not recombine (or even if it recombine it takes a similar genetic information) so actually Traveler is 50% father or stranger.
Ah, but Father gets more than just a Y chromosome from Stranger; he gets twenty-three chromosomes from Stranger and twenty-three from Grandmother. If we ignore the Y chromosome, the rest of the DNA is still changing. Traveler might have Stranger's Y chromosome completely unaffected, but he has Grandmother's X chromosome and forty-four other chromosomes with mixed genes from each of them. When Traveler replaces Stranger, he brings twenty-three of Mother's chromosomes with him as well as half of twenty-three of Grandmother's chromosomes. Apart from the Y chromosome, he also has half of twenty-three of Stranger's chromosomes. But then with the next pass his share of Stranger's chromosomes decreases to a quarter of that, replaced by genes drawn from Grandmother and Mother (because Father gets some of Mother's genes from Traveler); and with each iteration that reduces again.
Traveler's identity cannot stabilize.
--M. J. Young
Wed Jun 30 2010 11:09 pm # -
Okay. Try this.
--AB--
1) Xander scans himself, memorizes the pattern, and then goes back in time.--CD--
2) Upon arrival, Xander creates a person using that pattern. We'll call him Yohan.
3) Xander communicates the pattern to Yohan, who memorizes it.
4) Yohan puts Xander into stasis.*
5) Yohan carries out the mission. He knows the entire plan (including these steps) because he remembers coming up with it (as Xander).
6) At point D, Yohan goes back in time, arriving at point E.**--EF--
7) Upon arrival, Yohan creates a person using the pattern he memorized shortly after his own creation. We'll call him Yohan Prime.
8) Yohan communicates the pattern to Yohan Prime, who memorizes it.
9) Yohan Prime puts Yohan into stasis.***
10) Yohan Prime carries out the mission. He knows the entire plan (including these steps) because he remembers coming up with it (as Xander).
11) At point F, Yohan Prime goes back in time, arriving at some point.****My question is, at what temporal point does the traveler arrive after step 11 - point G, or point E? I'm inclined to think point E, but I've been wrong before.
_____
* (This is to get him out of the way; it's less complicated without Xander hanging around.)** (Xander Prime - that is, the local Xander, the Xander who has never time traveled, the Xander who is not currently in stasis - does not make the return trip. Yohan has gone back instead.)
*** See note *, replacing "Xander" with "Yohan".
**** See note **, replacing "Xander Prime" with "Xander Double Prime" and "Yohan" with "Yohan Prime." References to "Xander" remain unchanged, with the caveat that he does not exist at all in the EF timeline, his presence and actions from the CD timeline having been replaced by Yohan's presence and actions.
Thu Jul 1 2010 2:45 am # -
Max
MemberTraveler might have Stranger's Y chromosome completely unaffected, but he has Grandmother's X chromosome and forty-four other chromosomes with mixed genes from each of them. When Traveler replaces Stranger, he brings twenty-three of Mother's chromosomes with him as well as half of twenty-three of Grandmother's chromosomes.
First correction is the Traveler has Mother's (recombined) X chromosome and Stranger's Y chromosome.
Second correction is when the Traveler replaces Stranger, he brings twenty-three of Mother's chromosomes, half (recombination of) twenty-two of Stranger and Grandmother chromosomes with him as well as Y chromosome from Stranger.Lets me simulate the situation for autosomes with the formula:
T=aS+bG+0.5M where 0=<a<=0.5 and b=0.5-a1) Here the original father uses 50% of Stranger genetics and 50% of Grandmother genetics.
F=0.5S+0.5G2) Let us assume that the autonomes are recombined equally between stranger and grandmother.
T=0.25S+0.25G+0.5M3) Here let us assume that Father becomes a new person (unless this 0.25M is similar to the other 0.25S which would make them be the same person) because he is born using all of Stranger genetic and half genetics of Mother that still exist within Traveler and half of genetics of Grandmother. This would make the 0.25G+0.25M portion from Traveler genetics to be discarded.
F=0.25S+0.25M+0.5G4) Here let us assume that Traveler is born using all of Stranger genetic and half genetics of Grandmother that still exist within Father which would make the 0.25M+0.25G portion from Father genetics to be discarded.
T=0.25S+0.25G+0.5MBecause (4)=(2), time stabilizes. As you can see, time could stabilize if the recombination process is consistent in selecting which genetic should be pass on.
*****
For a quick analogy, the watch in Somewhere In Time keeps making the same loop, and eventually decays to dust. However, the formula for transparent aluminum in Star Trek IV: The Voyage Home makes exactly the same kind of loop, but it does not decay.
So the only thing that prevents a self causing event is actually the decay right?
If the mass could repair the decay exactly the same as before the time travel event happen, could this help the mass to exist indefinitely?
This makes me remember about the ship that replaces a part of it at a time and when all of the parts have been replaced it actually becomes a new ship that is indistinguishable from the old one.*****
If Yohan is actually indistinguishable from Xander at the point of stasis, I think CD time line will go into N-jump instead of going to EF time line but if Xander is actually distinguishable from Yohan then we probably go through EF time line before it terminates into N-jump.
Thu Jul 1 2010 4:54 pm # -
If Yohan is actually indistinguishable from Xander at the point of stasis, I think CD time line will go into N-jump instead of going to EF time line but if Xander is actually distinguishable from Yohan then we probably go through EF time line before it terminates into N-jump.
Yohan is most definitely not indistinguishable from Xander at the time of stasis, the obvious differences being that Yohan remembers nothing from the AB timeline that occurred after the pattern was made, never traveled through time, and had to be told the pattern by Xander (whereas Xander studied the pattern himself).
Actually, you can ignore all of that because it doesn't matter. For the CD timeline to stabalize, Yohan at point D would need to be indistinguishable from Xander at point B, which is obviously not the case. (Remember, the "usual" method for creating an N-Jump would involve Xander Prime making the same trip that Xander did for the same reasons, repeating the same actions. We're deviating from that by having Yohan travel back instead.)
I am quite certain that this is a sawtooth snap. My only question concerns whether the timeline stabilizes on the third leg, continuing forward from point F.
Thu Jul 1 2010 8:18 pm # -
Max
MemberYohan at point D would need to be indistinguishable from Xander at point B, which is obviously not the case.
That raises the question of stability because what Xander do to Yohan might be a little different from what Yohan do to Yohan Prime. This in turn might affect what Yohan Prime might do to the next Yohan clone but I think it probably have a high chance of stabilizing if Yohan is indistinguishable from Yohan Prime at point F. The act of putting Yohan in stasis is probably to limit further changes by him to history.
I still have a problem with understanding the causality of sawtooth snap and infinity loop because to me if Yohan goes to EF time line, the existence of Xander to cause the existence of Yohan would be erased. Thus, Yohan could not cause the existence of Yohan Prime which makes EF time line to return to its original state. This is just my opinion of course.
Similarly, if the note is changed by one letter from the original note but this new note is send to the next version of Abe, I wonder if time would stabilize if the same note is send from then on or would it return to the original state because the new note is different from the original note which would make it an uncaused cause when the new note overwrites the original note.
Thu Jul 1 2010 9:45 pm # -
O.K., I'm ignoring Maxx for the moment so I can focus on Scott; I'll try to get to Maxx next, if I manage to resolve things with Scott.
I must congratulate you on creating an incredibly difficult scenario. I think I'm starting to wrap my head around it, but as I start writing here I have not yet determined how it ends. However, let me start by distinguishing the original original history as AB1, in which no one arrives from the past, and Xander departs from point B1.
That puts us in CD1, in which AB1-Xander creates from memory CD1-Yohan. CD1-Yohan puts AB1-Xander in stasis so that there won't be three identicals here, and CD1-Yohan then leaves from point D1.
We have a minor problem. If we assume that CD1-Yohan's pattern was exactly identical to CD1-Xander at the moment CD-1 Yohan was created, then either CD1-Yohan does not know the plan or CD1-Xander has already conceived it. The other problem is why CD1-Xander either does not conceive the same plan or does not execute it. However, we'll assume that among other things, CD1-Yohan does something that will prevent CD1-Xander from doing what AB1-Xander did.
That means that CD1-Xander does not make the trip, and Anomaly 1 ends in an infinity loop. However, at the instant that would happen, CD1-Yohan departs from point D1 to attempt to prevent it by substituting himself for CD1-Xander. This means that we've got a second anomaly interacting with the first, preventing the first from falling into an infinity loop. AB2 is exactly congruent with CD1, and point C2 is the same moment as point C1. At point D1 which is also point B2, AB1-Xander and CD1-Xander both cease ever to have existed, along with everyone else, as time resets to C2, the second timeline of the second anomaly.
I am inclined to think that because this is an overlapping anomaly, the AB1-Xander who left from B1 and arrived at what is A2 must also arrive at C2; therefore at C2 you have both CD1-Yohan and AB1-Xander, both of whom are targeting the same place and time with the intention of creating duplicates of themselves whom they will send to the past. I would assume that our time machine has some sort of matter displacement system to prevent travelers from materializing inside solid objects (including moving ones), and therefore that they arrive pressed against each other.
At that point, the fact that AB1-Xander and AB2-Yohan can't fail to notice each other is going to impact their actions in the CD2 timeline. However, you've now got interlocking anomalies, and if either of them sticks to the plan, you'll have more interlocking anomalies--CD3 will be that history in which two distinct AB3 Yohans, one created by AB2 Yohan and one by AB1 Xander, arrive to find AB1 Xander and AB2 Yohan also arriving with plans to do what they are planning to do.
I think that's how it resolves, and why it has been so difficult to resolve as one anomaly.
--M. J. Young
Fri Jul 2 2010 12:31 am # -
As you can see, time could stabilize if the recombination process is consistent in selecting which genetic should be pass on.
The problem here is that it is impossible for Traveler's DNA to be identical to that of Stranger; it is not even plausible that all the genes Stranger gave to Father would be transmitted to Traveler. Father has only half of Stranger's DNA, and Traveler has only half of Father's DNA.
If I'm understanding you aright, you're arguing that it's possible that that every single gene passed by Stranger to Father is then passed by Father to Traveler, and that when Traveler replaces Stranger he in turn passes every single gene he got from Father back to Father. You are thus attempting to knock Grandmother and Mother out of the essential side of the equation, noting that they are both constants (their DNA is not impacted by the transition) and that despite the staggering improbability of it, it is within the realm of the possible that the exact same DNA would be transmitted in each case.
I am a man who believes in miracles, and I find that hard to accept.
I will note that if I had set this up such that the connection in the middle generation was mother (not father), your solution would crash and burn--in that case, the Y chromosome would come from the outsider in the intervening generation, and traveler could not have stranger's Y chromosome unless father and stranger were close relatives. However, that's not the scenario as you painted it.
If there were some miraculous intervention that caused stranger's DNA to be completely preserved from Stranger to Father to Traveler to Father, your scenario would work. That's not a natural event, though. The odds against it are astronomical, so intervention would be essential to its success.
--M. J. Young
Fri Jul 2 2010 12:42 am # -
Max
Memberdespite the staggering improbability of it, it is within the realm of the possible that the exact same DNA would be transmitted in each case.
That would depend on how the genetic recombine. If it recombine exactly the same way as it have before then it is not impossible for the same thing to happen again.
If somehow, the recombination process are changed then time would not stabilize.
This is similar to an act of trying to choose different things than what have happened before. If this happens then what is the cause that make the recombinant process to choose this specific way?I will note that if I had set this up such that the connection in the middle generation was mother (not father), your solution would crash and burn--in that case, the Y chromosome would come from the outsider in the intervening generation, and traveler could not have stranger's Y chromosome unless father and stranger were close relatives.
If that is the case then the Mother X chromosome is treated as autosomes so that all of Stranger/Traveler X chromosome will be pass on (because male only have one X chromosome) while half of Grandmother (recombined) X chromosome will be pass on.
There is an example which the stranger genetic is not needed at all.
Lets me simulate the situation for autosomes with the formula:
T=aS+bG+0.5F where 0=<a<=0.5 and b=0.5-a1) Here the original mother uses 50% of Stranger genetics and 50% of Grandmother genetics.
M=0.5S+0.5G2) Let us assume that the traveler uses the grandmother chromosomes only from his mother side.
T=0.5G+0.5F
(As you can see, there is no stranger chromosome here.)3) Here let us assume that Mother becomes a new person (unless this 0.5F is similar to the other 0.5S which would make them be the same person) because she is born using all of Father genetic that still exist within Traveler and half of genetics of Grandmother. This would make the 0.5G portion from Traveler genetics to be discarded.
M=0.5F+0.5G4) Here let us assume that Traveler is born using all of Grandmother genetics that still exist within Father which would make the 0.5F portion from Father genetics to be discarded.
T=0.5G+0.5FBecause (4)=(2), time stabilizes. As you can see, time could stabilize if the recombination process is consistent in selecting which genetic should be pass on.
*****
I am inclined to think that because this is an overlapping anomaly, the AB1-Xander who left from B1 and arrived at what is A2 must also arrive at C2; therefore at C2 you have both CD1-Yohan and AB1-Xander, both of whom are targeting the same place and time with the intention of creating duplicates of themselves whom they will send to the past.
If AB1-Xander is prevented to go back to the past but instead AB2-Yohan targets the same place and time to replace the cause of himself, would time still stabilize after that?
Fri Jul 2 2010 3:47 pm # -
Max
MemberI forget to put the simulation for X-Y chromosomes. In below examples, it is assumed that similar genetic information in X chromosome could produce genetic anomaly (which could be similar to Turner Syndrome) but that might not be the case.
The example where there is no stranger genetics from the original traveler.
1) Stranger = 100%S (XsY); Grandmother = 100%G (XaXb). Mother = 50%S+50%G (XsXa)
2) Mother = 50%S+50%G (XsXa); Father = 100%F (XfY). Male Traveler = 50%G+50%F (XaY)
3) Male Traveler = 50%G+50%F (XaY); Grandmother = 100%G (XaXb). Mother = 100%G (XaXa)
4) Mother = 100%G (XaXa); Father = 100%M (XfY). Male Traveler = 50%G+50%F (XaY)
The new mother will probably have some genetic anomaly because she will have two similar X chromosome from grandmother. This anomaly could be avoided if in step (3) Xb is chosen instead of Xa but this will make the process to choose the genetic information to be inconsistent which could disrupt time. The above assumes that no recombination takes place between Xa and Xb.The example where there is 25% stranger genetics from the original traveler.
1) Stranger = 100%S (XY); Grandmother = 100%G (XX). Mother = 50%S+50%G (XX)
2) Mother = 50%S+50%G (XX); Father = 100%F (XY). Male Traveler = 25%S+25%G+50%F (XY)
3) Male Traveler = 25%S+25%G+50%F (XY); Grandmother = 100%G (XX). Mother = 25%S+75%G (XX)
4) Mother = 25%S+75%G (XX); Father = 100%M (XY). Male Traveler = 25%S+25%G+50%F (XY)
The new mother will probably have some genetic anomaly (with lower rate than previous example) because about half of her X chromosome is similar to each other. The above assumes that no recombination takes place between grandmother's X chromosomes. If the recombination takes place then the genetic anomaly could be avoided if in step (3) the non-similar genetic information is chosen instead of the same one but this will make the process to choose the genetic information to be inconsistent which could disrupt time.The example where there is 50% stranger genetics from the original traveler.
1) Stranger = 100%S (XY); Grandmother = 100%G (XX). Mother = 50%S+50%G (XX)
2) Mother = 50%S+50%G (XX); Father = 100%F (XY). Male Traveler = 50%S+50%F (XY)
3) Male Traveler = 50%S+50%F (XY); Grandmother = 100%G (XX). Mother = 50%S+50%G (XX)
4) Mother = 50%S+50%G (XX); Father = 100%M (XY). Male Traveler = 50%S+50%F (XY)
This new mother will be similar to the initial example that uses Y chromosome but in this example, the X chromosome that will be pass on to traveler is self-causing instead of the Y chromosome. In this example, genetic anomaly due to similar chromosome could be avoided.Fri Jul 2 2010 8:06 pm # -
Either I can't get you to understand the genetic problem, or you can't get me to understand the solution to it. Either way, I've spent a lot of time on it and am going to leave it alone now.
If AB1-Xander is prevented to go back to the past but instead AB2-Yohan targets the same place and time to replace the cause of himself, would time still stabilize after that?
If AB-1 Xander is prevented from going back in time at point B1, there is no Yohan and no time travel at all. The initial premise is that Xander leaves from point B1 and arrives at C1 where he creates Yohan. The only way AB-1 Xander can be prevented from making that trip is if in the original history in which no time traveler arrives he fails to make it. In that case, there are no other histories, as no anomaly ever happened here.
--M. J. Young
Sat Jul 3 2010 12:08 am # -
Max
MemberI can't get you to understand the genetic problem
If I understand correctly, you are actually saying that the recombination process chooses different genetic each time the traveler goes back in time. If that happens then either sawtooth snap or infinity loop happens.
you can't get me to understand the solution to it.
The solution that I proposed is how the recombination process could lead to N-jump termination by choosing consistently. If no significant changes will be done to alter the act of choosing the genetic combination, then probably the same genetic will be chosen.
*****
If AB-1 Xander is prevented from going back in time at point B1, there is no Yohan and no time travel at all.
Even if Yohan Prime will replace what Yohan did earlier so the time could stabilize?
Will this result in infinity loop?
That is one of my problem with sawtooth snap.
If the cause for change that have time traveler in it have been erased, could time still stabilize even if this new time line is self causing?
If it does not stabilize, would it go to infinity loop or something?Sat Jul 3 2010 2:17 pm # -
The solution that I proposed is how the recombination process could lead to N-jump termination by choosing consistently. If no significant changes will be done to alter the act of choosing the genetic combination, then probably the same genetic will be chosen.
Ah, the light appears. I get it.
I agree that if when history repeats the same two people couple at exactly the same time under exactly the same circumstances, a child conceived by them will be exactly the same child with exactly the same genetic makeup.
That is not the situation we have here.
What you want is for Stranger to give twenty-three chromosomes to Father, and for those same twenty-three chromosomes to pass intact from Father to Traveler, and then to do so again in passing from Traveler back to Father and in such a fashion that Father also gets exactly the same twenty-three chromosomes from Grandmother.
I concur that if that happened to happen the first time, it would happen again. I also note that the odds are considerably higher that Grandmother would contribute the same genes given that it is the same time, since ovulation gives her one contribution per month; the same would be so for Mother, again assuming that Traveler is conceived in the same month (which is somewhat less likely but not impossible).
The problem is, as you noted, that twenty-two of twenty-three chromosome pairs are busily swapping genes in the creation of gametes. The roughly twenty-three thousand genes are unevenly distributed across the forty-six chromosomes, but we can estimate roughly a thousand per pair, five hundred per gene, on average, which gives us eleven thousand genes on the twenty-two chromosomes from Stranger, all of which are subject to swapping with the comparable eleven thousand genes from Grandmother. That means on the passing from Father to Traveler, the chance of the exact same set of genes being passed is roughly one in eleven thousand, less than one one-hundredth of one percent.
The same gene swapping and gamete creation occurs in Traveler, such that the probability of the same genes he received from Father being passed again to Father is also the same, less than one one-hundredth of one percent. That means for both of those events to happen correctly the first time, the odds are less than one one-hundredth of one one-hundredth of one percent, or less than one ten-thousandth of one percent, or less than one in ten million.
If it happened the first time it would happen every time thereafter, assuming nothing else changed (that is, assuming that the fact that Traveler has replaced Stranger has not changed Father's life such that his relationship with Mother is altered in any way). That's a separate but not insignificant factor, though: whatever Father knows about his father is going to impact his life, his identity, and his interaction with mother and ultimately with his own son, and Traveler is different from Stranger, so he has that impact. Moving the conception date of Traveler (or indeed of Father) one month changes half his DNA, and in doing so means we have to play the DNA matchup game again, with the same probabilities against getting the same genes at the same time.
The problem is not whether the same genes will be passed between the same people; it's whether the same genes will be passed by different people in new situations, and the odds are very much against it.
*****
If AB-1 Xander is prevented from going back in time at point B1, there is no Yohan and no time travel at all.
Even if Yohan Prime will replace what Yohan did earlier so the time could stabilize?
Will this result in infinity loop?You're not understanding the concept of AB-1 Xander here. Let's take it a step back.
The original scenario said that Xander, whom we're labeling AB-1 Xander, left from point B1 in the first ever (relevant) trip to the past, and arrived at C1.
The thought was that the CD-1 Xander would be prevented from making that same trip, but that AB-1 Xander was going to create a clone of himself, CD-1 Yohan, and at point D CD-1 Yohan was going to travel, aiming for point C.
My statement was that because CD-1 Xander does not depart from point D, no Xander would arrive at point C, and thus we would have an infinity loop. The argument is that Yohan replaces Xander and does what Xander would do.
My long-considered conclusion concerning that is that when CD-1 Yohan departs from point D, he is not confirming the history of the CD timeline but creating not merely a new history but also a new anomaly, and by leaving at the instant that is point D he is preventing the first anomaly from resolving. That means that CD-1 Yohan is also AB-2 Yohan, the same person in the same history that is identified by two different labels according to which anomaly we are discussing. He thus creates point C2, the CD-2 timeline.
However, the second anomaly is contained within and overlaps the first; that means that when Yohan arrives at that time which was point C1, he arrives in the world as it existed at C1. That world included that AB-1 Xander had just arrived from B1, and so it still includes that. AB-1 Xander must also arrive at that exact place and time.
Now you're trying to find a way to say that AB-1 Xander is prevented from leaving from point B1, so that he does not arrive at point C2 (which being a rewrite of point C1 has the same facts present as C1, plus the arrival of AB-2 Yohan). However, by definition no one can travel to the AB-1 timeline a prevent AB-1 Xander from departing (because that is the history of the world in which no time traveler has interfered), and therefore no one can prevent AB-1 Xander from arriving at point C1, no matter how many other anomalies attempt to change C-1, no matter whether time traveler arrive before or after this. The only way it can be resolved is based on whatever action CD-1 Xander takes at point D, which has not been determined because AB-2 Yohan prevented that action from being completed. CD-1 Xander did not travel to the past, but he also did not not travel to the past; time stopped before his inaction could alter the past.
This must be so, because if it is determined that CD-1 Xander did not depart from point D1, then AB-1 Xander never arrived at point C1, CD-1 (equals AB-2) Yohan was never created, and we revert to point A1 and the original infinity loop.
So your choice is either Yohan prevented Xander from making the trip, in which case Yohan does not exist, or Yohan did not prevent Xander from making the trip, in which case both Yohan and Xander arrive at the same moment in the past, because they both made their respective trips.
Yeah, it took me quite a while to work that out, but that has to be the answer.
--M. J. Young
Sun Jul 4 2010 10:19 pm # -
Max
MemberThe problem is not whether the same genes will be passed between the same people; it's whether the same genes will be passed by different people in new situations, and the odds are very much against it.
The odd could be raised if the traveler study his own genetics and could choose the exact genetics to be passed on to Father. The traveler must also learn his father exact time of conception so that the right genetic will be pass on. Not knowing those variables or the way to manipulate it would probably make the traveler depends on luck or fate for his own existence.
Another factor that would help is the similarity of stranger at point A to the traveler at point C. If they are 100% similar then time would probably stabilize.*****
This must be so, because if it is determined that CD-1 Xander did not depart from point D1, then AB-1 Xander never arrived at point C1, CD-1 (equals AB-2) Yohan was never created, and we revert to point A1 and the original infinity loop.
That is exactly my problem with sawtooth snap. A sawtooth snap would not stabilize because the cause of change is changed which will make the future change to be not viable in the first place. Any sawtooth snap will probably act like an infinity loop because of this reason.
Yeah, it took me quite a while to work that out, but that has to be the answer.
That raises the question for other sawtooth snaps.
Could you give one example of sawtooth snap that will stabilize?Mon Jul 5 2010 12:38 am # -
JohnA1nut
MemberI haven't been following this, and I know I of all people should be. I just wanted to say though that I'm watching Star Trek Enterprise on DVD, and it looks like they go by the "Time is always running" theory. Changes in the past don't affect the future until they happen in the past. If the Terminator took 48 hours to kill Sarah Connor, John Connor would have 48 hours to send Reese back, before the timeline changed. Actually, 47 hours, 59 minutes and 59 seconds, plus or minus a microsecond. I think Enterprise ran by that theory. The timelines are running side by side. Contact between the present and the past is possible, because they're running together, but on a different plane of existence. Imagine a deck of cards with an infinite number of cards in the stack. Now imagine a band saw cutting through the cards. That band saw represents time, and time travel. When the traveler leaves, the future isn't changed until the band saw "catches up" to his point of departure. It's a cool theory, I think.
(Doing Therapy)
Mon Jul 5 2010 7:18 pm # -
The "time is always running" theory does not make any sense to me at all. It's just silly.
Could you give one example of sawtooth snap that will stabilize?
No, but I can give you one example of a sawtooth snap that might stabilize: Back to the Future, part one.
We see in the film that Marty McFly travels from 1985 to 1955 where he interferes with his parent's meeting and then gets them back together on different ground. It happens that when he returns to the future, he sees himself leaving for the past, but that when he returns home it is a much nicer home in the same house. His father is a confident successful author.
That means that the Marty he saw leave grew up in this home, and never heard the story of how his mother's father hit his father with the car and his mother nursed his father back to health and that's how they connected. Instead, he heard the story of this guy named Marty who kept insisting that they were perfect for each other, and of how George saved Lorraine from Bif Tannen at the dance, and the kissed and eventually got married. This "affluent" Marty will probably try to connect to someone, find and follow his father, wind up hit by the car, and realize at some point that he is the Marty that brought his parents together. He will do everything he can to do that.
At this point we have a sawtooth snap--the AB history in which no one comes from the future, the CD history in which the Marty we see in the movie comes from the future, and the EF timeline in which the affluent Marty comes from the future. At point F there are three possibilities. If Affluent Marty failed to bring his parents together as Original Marty did, then there is no Marty in this history and we revert to the original AB history in an infinity loop termination, a cycling causality. If Affluent Marty manages to bring his parents together somehow, but the story is different and his life will be changed, then we continue with the GH history in which a yet different Marty makes the trip. If, however, Affluent Marty manages to bring his parents together in essentially the same way that Original Marty did, then he'll create the same history that leads to his own birth and life and to his identical self, Affluent Marty, traveling to the past from point F, confirming that history and creating an N-jump termination.
So it is possible.
I consider N-jumps highly unlikely, and believe that they become increasingly less likely if a sawtooth snap begins; but they are possible, and when a movie presents a story in which they happen my job is to figure out whether in that story it could have happened.
--M. J. Young
Tue Jul 6 2010 1:39 am # -
Max
MemberIf, however, Affluent Marty manages to bring his parents together in essentially the same way that Original Marty did, then he'll create the same history that leads to his own birth and life and to his identical self, Affluent Marty, traveling to the past from point F, confirming that history and creating an N-jump termination.
Does Affluent Marty meet up with Original Marty because both of them target the same time?
If the answer is no, then why does time traveling Xander need to exist at point E when Yohan try to stabilize time by departing at point D?*****
The below is what I think about the "time is always running" theory. If you think it is a silly theory then you could skip reading the below to save time.
Contact between the present and the past is possible, because they're running together, but on a different plane of existence.
That sounds like parallel dimension theory.
When the traveler leaves, the future isn't changed until the band saw "catches up" to his point of departure. It's a cool theory, I think.
That sounds like the original dimension splits into two at the time of arrival then converge later at two times the time difference of departure and arrival points.
When the converge happens, the time line with the time traveler will be the dominant one so any attempt to go to the (parallel) past will probably make that new time line dominant.
Below is the formula for the convergence point:
Convergence point = Point of arrival + 2(Point of departure - Point of arrival)If the people in the future send something to the future compared to the time traveler, it will arrive late in the time traveler time line. In the case of more than one time traveler, the item sent from the future can arrive (without splitting the time line) to any future time of time traveler as long as that parallel dimension is targeted.
If the people in the future try to send something before the current point of time traveler, the time line will further split and this new time line will probably be the dominant time line. Time traveling when the time line does not converge yet will complicate things so before the time converge, time travel event should be minimized.For example, the time traveler departs at 10:00 and arrives at 05:00.
This will make the time to converge to be 15:00.
If two hours later at 12:00 future time and 07:00 time traveler time, people from future send another time traveler to 06:00, this second time traveler will arrive in the world that the first time traveler have changed but this second traveler will further split this time line. The future time line will converge with the first time traveler time line at 15:00 and after that this time line will converge with the second time traveler time line at 18:00.The people from the future could still contact the first traveler when the second traveler is sent but it will arrive late if the same configuration is used to target the dominant time line so the communication with the first traveler in the dominant time line will have to be resynchronized. The communication with the first time traveler in the time line where there are no second time traveler could be resumed with the same configuration.
The first time traveler could probably travel to parallel world future. If this happens, there would probably be a duplicate of first time traveler in the targeted time line.I wonder what would happen if people in current future and current time traveler sends an item to the past of both time line but at different places (because if it is in the same place then it would probably crash with each other). I think targeting the same place and the same time is one of the way to deny someone from time traveling if one party knew the target of the other.
Tue Jul 6 2010 5:47 pm # -
Does Affluent Marty meet up with Original Marty because both of them target the same time?
If the answer is no, then why does time traveling Xander need to exist at point E when Yohan try to stabilize time by departing at point D?In 1955, Original Marty and Affluent Marty were the same (as yet unborn) person. Thus Affluent Marty is replacing Original Marty.
Yohan was never Xander; he is a copy of Xander. Xander exists separately. Thus if Yohan travels back in time, he is creating a new anomaly.
I agree that the "time is always running" theory feels like some variant of parallel dimension theory. It does not seem to be a workable one.
--M. J. Young
Wed Jul 7 2010 1:46 am # -
Max
MemberThus if Yohan travels back in time, he is creating a new anomaly.
What if Yohan kidnaps the new Xander to travel with him by freezing the new Xander.
Would time continue then?*****
It does not seem to be a workable one.
Do you mean it is not workable because it is a parallel dimension theory or because of some other matters?
Wed Jul 14 2010 9:45 pm # -
What if Yohan kidnaps the new Xander to travel with him by freezing the new Xander.
Would time continue then?My overall feeling is that the continued existence of Yohan creates a temporal problem that cannot be resolved.
Apart from that, we'll work with the notion that Yohan 1, copied in the CD1 history from Xander 1, takes Xander 2 (of the CD1 history) back with him. That is functionally the same as Xander 2 making the trip, but taking Yohan 1 with him. At that point, the history is EF because Xander 2 and Yohan 1 have both arrived, and if it will resolve at all (of which I am still not persuaded), it will require that Yohan 1 create Yohan 2 from Xander 2, and then Yohan 2 take Xander 3 back with the intent of doing exactly the same thing. There's still the problem that Yohan 2 is not identical to Yohan 1 because of his interaction with Yohan 1 and his failure to interact with (the frozen) Xander 2, but we might be able to manage s short sawtooth snap into an N-jump. That's a really improbable outcome from this point, but it is not impossible.
Max quoted part of this from me:
I agree that the "time is always running" theory feels like some variant of parallel dimension theory. It does not seem to be a workable one.
He then asked
Do you mean it is not workable because it is a parallel dimension theory or because of some other matters?
Someone once painted an image for me of a multiverse in which all the universes lay side by side, each temporally out of phase with its adjacent ones by plus and minus the smallest possible increment of time. That is, if we give the universes sequential Greek letters starting somewhere in the middle and decide that there is no smaller unit of time than a picosecond, universe lambda is one picosecond behind universe kappa and one picosecond ahead of universe mu. In this conception, if you were in universe kappa and you traveled back in time one picosecond, you would be in universe lambda. Return forward, and you are again in universe kappa. Travel back another picosecond and you are in universe from lambda and you are in mu. The thought then was that if you made a change in lambda, you also changed mu, but you did not change kappa.
I saw no reason why in this situation you would change mu but not kappa; it seemed to me that if you traveled from kappa to lambda you changed lambda but had no impact on mu. At the same time, it also seemed to me that if you traveled from kappa to lambda, it must also be that the you in alpha, unaffected by any time traveler coming to alpha from the future, must have traveled to beta, and changed beta. If the change to beta meant that the you in beta did not travel to gamma, then the you in gamma would be default travel to delta and cause the same change preventing the delta you from traveling much as the alpha you stopped the beta you. That means that the epsilon you does interfere with the zeta you, who doesn't interfere with the eta you, who does interfere with the theta you, who doesn't interfere with the kappa you, who interferes with the lambda you, who doesn't interfere with the mu you, who does interfere with the nu you, who doesn't interfere with the xi you, who does--well, I've shone off my familiarity with the Greek alphabet sufficiently.
What I don't get from this, though, is how a cause in lambda would ever have an effect in kappa. That is, I can see how in parallel dimension theory you could be in 2000 in a world you just changed, and when 2010 comes in that world it will be different from what it was, but I can't see how what you just did in 2000 would ever change the history of 2010 in the world you left behind. The only way a change in 2000 can change 2010 is both of those dates are in the same universe; yet if they are in the same universe, then in 2010 anything that existed in 2000 has been that way since 2000, and the "change" happened then, not twenty minutes from now when we are the same temporal distance from the change in the past as we were from the arrival of the time travel at his point of departure.
There's also a non sequitur in it. Supposing that I did believe that the change made in 2001 by a traveler who left from 2010 who arrived in 2000 would not change my history until 2011, how would I know before 2011 that the change was coming? Suddenly I have a time traveler who can change the world in 2001 and leap forward to 2005 to find that it has not changed, make changes in 2005 based on the world that isn't going to be the way he found it, and then have those changes compounded--none of it makes any sense. It's like, I can leap back in time and steal the Mona Lisa from the Louvre in 2005, and then go back to 2000 and steal it again, and I have two, and then I can travel back to 1995 and arrange for the Mona Lisa to be on a world tour (and thus not in the Louvre) from 1999 through 2009, so that I never stole it, and still have the ones I took--and what's worse, I can do that in the reverse order. That is, I can travel to 1995, arrange the tour that will remove the Mona Lisa from the Louvre for a decade, leap ahead and get there in 2000 "after" it is going to leave for its tour but "before" my changes to history have gotten that far, and steal a painting from a place it's not going to be, then leap ahead another five years and steal the painting I've already stolen, because the fact that it went on tour in 1999 and the fact that it was stolen in 2000 are both changes made to history by a time traveler which won't get here for a few hours yet, so the painting that won't be here is still here for me to take.
I see this kind of thinking in <i>Frequency</i> and I think maybe in <i>Butterfly Effect</i>, and something like it is probably the only thing that really spoils <i>Millennium</i>, and it just does not work in any rational way: if the past was changed, it has now always been what it became, and there is no rational sense to saying that the affect in the past is going to take time to reach the future.
--M. J. Young
Thu Jul 15 2010 11:29 pm # -
Max
Memberit will require that Yohan 1 create Yohan 2 from Xander 2, and then Yohan 2 take Xander 3 back with the intent of doing exactly the same thing.
Yohan 1 could create Yohan 2 by using the pattern that is told to him by Xander 1.
So when time stabilize, Xander 2 and Yohan 1 will be in frozen state.There's still the problem that Yohan 2 is not identical to Yohan 1 because of his interaction with Yohan 1 and his failure to interact with (the frozen) Xander 2
I think N-jump would be achieved easier if the interaction is done using a note, computer terminal or something so information could be pass on consistently.
*****
That is, I can travel to 1995, arrange the tour that will remove the Mona Lisa from the Louvre for a decade, leap ahead and get there in 2000 "after" it is going to leave for its tour but "before" my changes to history have gotten that far, and steal a painting from a place it's not going to be
Parallel dimension in the theory that I said only have now or the future. Any travel to the past will create a new parallel dimension. If you wanted to change the history of original time line, you will create a new parallel dimension. So it is possible to steal something multiple times but when you go to the (parallel) past, the same amount of mass will be exchanged. So for example if you steal a planet or something, the same amount of mass will have to travel to the (parallel) future. This is to prevent the point of convergence to have more energy than the original time line. The point of divergence (where the time traveler arrive in the past) will have to create negative energy to divide the time line into two. If this new parallel dimension takes all of the energy from point of departure then I think it will converge at the point of departure rather than at a future point unless at the future point of convergence, the negative energy will stop everything from moving for as long as the point of convergence have pass the point of departure. So I think this variant of parallel dimension theory probably needs the existence of negative energy to make it work.
Thank you for your time.
Fri Jul 16 2010 9:26 pm #
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