We previously addressed the question of whether the film Primer might work under divergent dimension theory, and found difficulties. Might it be possible, however, that the film works under pure parallel dimension theory? Are Aaron and Abe traveling to different universes which already exist?
Again we have the problem that if this is the case, there will be universes from which they vanish, never to be seen again. Each of them makes a solo trip near the beginning of their travels, and that puts them out of synch with each other. This is not insignificant. It means that from universe 1, Abe goes to universe 2, and Aaron in universe 1 never knows what happened to him; but then after Abe explains time travel to Aaron in universe 2, Aaron makes his secret trip back to the beginning, and so he leaves universe 2 for universe 3 and the next day Aaron cannot find him ever again.
However, the problem created by the shotgun party is insurmountable for this theory.
It was previously mentioned about parallel dimension theory that making the number of universes infinite does not solve problems, it only makes them more difficult to discuss. Thus let us suppose that there are only ten universes for illustration; it should be apparent that the number of universes becomes irrelevant. Skipping to the moment that Abe decides to be the hero, he and Aaron leave from universe one to visit the party in universe two; they also prevent their counterparts here from doing the same thing, so no one appears in universe three, and that pair go to universe four and stop their counterparts--in any case, there are two pairs in all even numbered universes and none in the odd ones.
However, Aaron says they repeated this multiple times. That means that the spare pair in universe two went to universe three, where the boyfriend was waving a shotgun; and then they went to universe four--but wait, the pair from universe three are already at that party, and they are now at universe six where the pair from universe five are trying to intervene, who are now in universe eight confronted by the pair from seven, and so on. Then if the boys make another trip to universe five, the pair from universe three is also there.
What parallel dimension theory is supposed to do is make it possible to change the past without changing anything that happened; but once the universes are so linked, travelers from different universes begin to interfere with each other, and so history changes anyway. In this case, it is not possible for Abe and Aaron to have gone to that party more than twice and not encountered themselves as time travelers; yet they do not do so. Thus parallel dimension theory does not resolve the problem at the party, and cannot make a coherent story of the movie.
If you are using some form of multiple dimension theory, either the universes already exist or your actions are creating them. If the latter, you face the problem described last time of duplicating yourself with each trip; if the former, the actions of your doppelgangers in all other dimensions must be included in considering the outcomes.
We should consider whether the film works under fixed time theory, but that is a question for a future article.
Comments
Can't buy your theory. Obviously not what the writer had in mind. Besides, when one Aaron is at the party, the previous Aaron is in a hotel room. There are never two Aarons at the party. You should read the book if you want to know more, although, I'm not sure if it directly answers how time operates. It seems to exclude multiple universes and parellel universes. It seems like Primer uses replacement theory or more likely the rewind the tape theory, for lack of a better name.
Thanks for your comment, Randy. If you look at the previous articles in this series (follow one of the "Primer" links to get a full listing), you'll see that I attempt to analyze the film using replacement theory, and it fails; this article is an attempt to determine whether parallel dimension theory solves the problems better.
The primary problem is that Aaron states they went back to the party multiple times. Supposing Abe to be 30 years and 30 days old the first time he does this, then the second time he is 30 years and 31 days old, and the third time he is "and 32 days" old. When the "and 32 days" old Abe gets to the party, where are the "and 30" and "and 31" days old versions? If he erased them, he has undone his own past and no longer exists as having made those trips. If he does does not, they must also be at the party.
I hope this helps. I think the authors intended some form of replacement theory, but I don't think it works.
--M. J. Young
When 32 day Aaron arrives, 31 day Aaron , his previous version, is at the hotel. The 30 day version who was at the party, did not travel back in thie previous timeline, only 31 Aaron did. Therefore only 31 and 32 Aaron exist in this timeline. 30 Aaron has been replaced by 32 Aaron. This is so simple, I can not understand how you fail to grasp it, even from just one viewing of Primer. If you don't have a clear grasp on this, why do you write as if you were some kind of expert? I'm sure I am not the first to tell you that Primer has conquered your intellect. Victory for the author, yet you blame him for your lack of understanding. Sad, just sad.
Randy, it's inconsistent.
At one point, the time travelers are downtown but their duplicate selves are at the hotel; both exist.
At another point, the time travelers attend a party which their duplicate selves (not the originals, but a previous set) also attended, but those duplicates are not there.
The film wants it both ways. Either every trip creates duplicates, or no trip does.
Remember, Aaron says that they went to the party multiple times to get it right. So where are the versions of their selves who attended the party in previous timelines? If they're not there, then their alternate selves are not at the hotel, either.
In fact, if their alternate selves ceased to exist, then the travelers can't fight with their past selves.
Either they're duplicated or they aren't; either those are their past selves or they aren't.
The film is very smart, but it is rather confused on this critical point.
--M. J. Young
M.J., I think maybe a better way to look at Primer is to get away from the idea of the characters creating a new alternate universe upon their emergence from the box, and more toward the box creating a conduit to an alternate universe, with one terminus in the new alternate universe where/when the box is turned on, and one in the current universe where/when the box is turned back off. The number of available conduits (activated boxes) at any one time is the limiting factor that prevents previous versions of Abe and Aaron from showing up at the party also.
Thanks for your suggestion, Barry. I admit I'm a bit like a post-game analyst using the benefit of hindsight to suggest what the coach did wrong, but parallel dimension theory is always confusing and rarely solves anything--as I just posted relative to http://www.examiner.com/time-travel-movies-in-national/timeline-part-12-... Timeline part 12: multiple dimensions. It doesn't work in Primer, either.
Abe and Aaron (A1) leave from universe 1 (U1) to universe 2 (U2). If they prevent their doppelgangers (A2) from leaving for universe 3 (U3) there are two sets of them in U2 thereafter, but A3 will leave U3 for U4, so all even-numbered universes have two sets of Abe and Aaron, and all odd numbered universes have none. If A1 does not prevent A2 from leaving, then there is an A(#) in every universe. If A2 never left and A1 now leaves U2 (trying to do the party visit right) they get one shot in U3 because there is no A there, but if they then move to A4 we hit the problem. We hit the problem sooner if A2 did leave for U3. The problem is that whether it is in U3 or U4, either A2 or A3 is already at the party.
Your solution is that only one "A" can emerge from the box, and therefore only one "A" will be at the party. Which one, and where is the other? If it is A1, somehow they have destroyed the other A which never emerges from the box and so ceases to exist; if not A1, all memories of A1's first trip must be lost as they cease to exist.
More in a moment.
Certainly your solution prevents multiple time travelers from arriving together, but it creates the issue of where they went. You haven't prevented them from leaving their points of origin, but you have prevented them from reaching their points of arrival. Therefore they entered the box and never emerged.
Either that, or the problem is that when A1 tries to leave from U2, A2 is already there trying to use the boxes, and they have to argue about who gets to travel to U3.
One way or another, your dimension travelers are going to interfere with each other.
Thanks again for your comment. I hope this helps.
--M. J. Young
Here's a thought:
Going back to the weeble-wobble example, the wobble goes from A (Box on) to B (Box off) and back to A, round and round. They are separate paths, or the weeble would collide with itself in the middle. When the guys are using the boxes, the box is only occupied on the trip from B to A, because at A, they exit. Nothing in the box is going from A to B. For that trip, it's empty.
They can exit, look back in the box, expecting to see their past selves still sleeping, but see nothing. Which means an exited box is eligible for another trip by someone else to go on their own ride from a box off point to that same A. This is what Aaron means by the boxes are "recyclable" rather than "one time use".
And because a Box in U1 will only go to Box On in U2, there isn't a worry that the A1 from U1 will emerge at Box On point in U3 as well.
Thanks for your note, Carl; I agree that there is some sense in which the travelers neither meet nor see those going the opposite direction, as if caught in a loop in a second time dimension. I think, though, that you can't get the result you want from multiple dimension theory, and the party is the problem. I addressed this in http://www.examiner.com/examiner/x-15701-Time-Travel-Movies-Examiner~y20... Primer question 3: multiple dimension theory; here's a summary.
The assumption in divergent dimension theory is that when A1 travels to the past he creates U2, the universe exactly as it was in the past of U1, and that events progress from there exactly as they did but for his interference. Thus when A1 leaves from U1, he goes to the party in U2; but when he leaves U2 and goes to the party, he creates U3 based on U2, and thus finds himself already at the party.
It will be easier to understand if we postulate that there is a second time traveler, B. If A travels to U2 and goes to the party, and B2 decides that he will go to that same party, when he creates U3 and arrives at the party he will find A1 already there. It does not matter whether A1 has already arrived or will arrive; his arrival is part of the history of U2 that is being copied.
It should not be different if A is making that second trip.
If the same box is being (re-)used, the mechanics of the box prevent that, I think. Because the mechanics of the box initiate a hard-coded time of arrival, each universe would diverge from the same point, a point without any travelers from the future in it yet, just their past selves. Once a traveler arrives, that's a new branch, containing only himself and that past self.
The divergence looks like a chicken foot, with a growing number of toes.
Now if B stops A from taking the trip in the box in one of those branches, and uses the box for himself, he'd arrive at that same branch point, in a universe containing only the past versions of A and B.
At least, that's how I see it.
That was me... forgot to put my name. (Sometimes I see replies automatically use my Facebook account.)
Greetings, Carl. The long answer will have to be in three parts. As to the Facebook connection, sometime in the past year The Examiner revamped the articles section such that comments on new articles are Facebook-linked, but the old articles are still on the old system. That's why sometimes you get that.
Your view still leaves problems in my mind.
The trip to the party was not the first trip the boys made; they had already made several trips and so built up cash playing stocks. Presumably when they travel to the party, they arrive in a universe in which that cash exists, because someone who is in some sense they already arrived in that universe and did what they did. Otherwise their efforts are wasted, because if they arrive in a universe in which no time traveler has ever arrived, then they can never build their financial resources beyond a single transaction. Further, since each transaction alters the financial market some, it becomes increasing probable that the world to which they travel will be different, and thus that their expected windfall will not occur. We assume that those previous arrivals have arrived.
--M. J. Young
The problem occurs if we are dealing with either simultaneous or subsequent arrivals. Using subsequent arrivals first, if Aaron travels to eight o'clock and does something stupid, and then Abe manages to travel to six o'clock, will Aaron, who left before him but arrived after him, arrive at eight, or not? If he does not, is that because time travelers travel to different universes (in which case all the other trips must go to different universes) or because history is anticipating that Abe will prevent Aaron from departing? But no, if Abe prevents Aaron from departing, Aaron will arrive in this universe because Abe did not prevent him from departing from the other, but will not arrive in the next universe, where Abe will not arrive either, not having any reason to leave this universe to interfere with the next. So it appears necessary that if someone leaves after you but arrives before you, you still arrive; and if someone leaves before you and arrives after you, they, too, must arrive. So if someone leaves when you leave and arrives when you leave--as happens whenever they use parallel boxes (which we must admit are not going to be perfectly in synch with each other) both must arrive in the same history; and that must apply even if the person who arrives at the same time is a duplicate version of you.
--M. J. Young
The only thing that really keeps it from seeming wrong that there are not multiple versions of the time travelers there is that we can't figure out how they would come out of the same box at the same time; and that's a flaw in the time travel device, something the movie fails to address. It accepts that multiple time travelers can interact with each other (Abe deals with his duplicate selves), but fails to deal with the problem of when they happen and when they do not happen.
Thanks again for your note. It's really a remarkable movie, but ultimately it become inconsistent and incoherent. It does so subtly enough that it's hard to spot, but it really does not work under any coherent theory of time.
--M. J. Young
It is a great movie. And one I think I understand better and better through discussing it.
And I do understand the idea that if two travelers can travel together and end up in the same universe, then so must a traveler and his duplicate be able to. And that certainly does happen in the movie. But the duplicate is traveling in a different Box (fail safe).
It's pretty clear that you can't use the same Box to travel to a point earlier than when the box was turned on ... it's not on. And that you can't get out of it later than it's turn on point, due to some bad disorientation. This fixes the arrival time.
Nor could you get into a Box after it was turned off. Turning it off sends you on your way to the "on" point AND prevents someone from getting into it after you did. They'll find the Box is turned off and can only start a new arrival time by turning it on.
So given a single Box, there's no way to have more than one person using it at a time.
I think the the playing the stocks trips happen on day 1 and the party trips on day 2, so I think that's how they're still holding onto some stock market fortune from day 1 as they get to day 2 and the party trips.
It is a very good movie, but it still has problems.
Hypothetically, what would happen if Abe1 turned the box on at noon and off at nine, traveling from nine until noon, then while he was there for some reason Abe2 turned the box off at six intending to travel to noon?
The party poses a particular problem. Suppose Abe1 travels back to fix things. If Abe2 does not do so, what happens to Abe1, whose existence is dependent on that trip? If Abe2 makes the same trip, though, he will encounter Abe1 trying to make his second trip to the party. That is, they will both be attempting to depart at the same time using the same box, but if Abe2 does not make the trip Abe1's history is undone--unless we accept a divergent dimension theory. But in that case, it does not matter whether Abe1 came from anywhere, because when the universe of Abe3 diverges from that of Abe2, Abe1 is already arriving in it, and if Abe1 arrives in it again he has to find himself there.
So it's a very clever movie, but it is not entirely consistent.
Thanks again for your note.
--M. J. Young
I realized the timeline "splays" (the chickfoot) I see when a Box is recycled, is topologically like the Sawtooth Snap diagram, with all the start points (C, E, etc.) pulled to A. This is what Aaron does by redoing the party, just like a time traveler repeating a Sawtooth Snap until it continues on with an N-Jump.
Based on Aaron's recycled trips, I would assume Abe2 would find Box2 empty at 6pm. Abe1 had already exited Box2 at noon and is already in T2. He's not in Box2 in the "forward arc", unless he misses The Exit (disaster!). Abe2 can safely Turn Off Box2 at 6pm in T2. Abe1 won't be effected, since Box1 was still On at 9pm in T1. Abe2 can't turn Box1 off in T1!
Next step: Abe2 climbs into Box2 at 6pm in T2. He exits Box3 at noon in T3, with a blissfully ignorant Abe3. Where's Abe1? He's still back in T2. He can't travel to T3 and join Abe2 so they both Exit Box3 at noon together because Abe2 Turned off Box2 in T2. Abe1 could turn it On again, but that's a new start point, to a new set of timelines, for him.
Abe1 doesn't show up in T3 anymore than multiple Terminators do in the later Sawtooth Snaps. Skynet1 uses TimeMachine1 in Timeline1 to send Terminator1 to kill Sarah2 in Timeline2, but if another round is needed, it's Skynet2 in Timeline2 sending Terminator2 using TimeMachine2. Why isn't Terminator1 showing up in Timeline3 along with Terminator2? I say it's the use of different Time Machines. Primer's timelines have their own Boxes, going to the next one over.
Thanks again, Carl. I understand the concept, but find it flawed.
We don't have multiple Terminators because each one is an original replacing the prior one. That differs from this situation.
Abe1 goes to Time1, creating a new history, but what happens to the old one? Under replacement theory, Abe1 has undoes his own history, and if Abe2 does not do exactly what Abe1 did, Abe1 will cease ever to have existed--and so will not arrive at Time1. We are attempting to avoid that via divergent dimension theory: Abe1 created the new universe, U2. He left U1, and if Abe2 does not leave U2 there will be two Abes in U2.
Abe1 repeats his trip, traveling again to Time1. U3 is created--but from what? To be consistent, U3 must diverge from U2, but U2 contains Abe1(a) already, and so in U3 Abe1(b) duplicates not only Abe2 but Abe1(a). However, you want U3 to diverge from U1. That means you are "unmaking" U2, and if you unmake U2, Abe1 can't be in it. But if Abe1 is not in U2, he can't leave it for U3.
In short, for Abe1 to travel from U2 and create U3, he must create U3 as diverging from U2, and his slightly younger sequentially earlier self must already be in the U2 he copies. If he is not there, then he undoes his own existence by undoing his history; if he is there, then he encounters himself.
Continued...
…continuing
You want the new universe to diverge sometimes from the last universe created and sometimes from an original universe. It is an inconsistency the film obscures by being vague about how things work, but some things cannot work under the same rules as other things in the same film. It is an interesting exploration of some of the problems, but ultimately it fails.
Thanks again.
--M. J. Young
No, the original universe only gets diverged from once, but the history of the two timelines after Time1 are the same. Just like in Replacement theory, the history before A is the same as the history before C, and as E, etc. But C-D comes from A-B, E-F comes from C-D, so, even there, you have timelines diverging from a previous timeline, yet keeping the same original timeline's history up to A.
So... up to Time1 (noon) in U1, there's Abe1 turning on Box1. Which means in U2 at noon, there's just Abe2 turning on Box2. And in U3, just Abe3. And so on. Up to noon, all the timelines look identical.
At noon, Abe1 turns on Box1 in U1. This is the diverging point, NOT when an Abe Exits the Box. The backward traveling Abe Exits the Box just "before" he gets to the Turn On point. He doesn't want to miss it and end up on the forward arc by accident. And trying to time his Exit *exactly* at noon would be problematic. Let's say he always gets out of the box 5 minutes "early", which is would be 12:05pm.
So, in U2, there's just Abe2 up until 12:05pm, when Abe1 exits. Abe2 turned on Box2 at noon, though, so U3 up to noon, has just Abe3. When Abe2 makes his trip in Box2, he'll exit in U3 at 12:05pm. There's no copy of Abe1 from U2, as Abe1 was not in U2 at noon, when Box2 was turned on.
, which, from his "backward arc", ends up being that he Exits shortly after the Box turns on. *After Time1.*
When Abe2 makes his trip to U3 to Time1, he doesn't find Abe1 there.
Hmmmm, I left a couple lines in that post I forgot to remove.
Anyhow, I think I'm not exactly addressing the party scenario. If Abe1 travels to U2 and goes to the party, Abe2 has no reason to travel to U3, so, yes, you have two Abes. We see this in the movie and causes no issues of Abe undoing himself.
Still Abe2 will probably travel to U3 anyway, to play the stock market or whatever. But he could just as well reset the Box. He has to reset it occasionally, or else the journey back becomes less and less convenient, like the 3-Day Fail Safe trip requires food, extra oxygen, a place to potty?
Anyhow, let's say Abe1 doesn't get the result he likes, so wants to travel again. Recycle the Box. He prevents Abe2 from using the Box, and instead makes the trip again himself. Now Abe1 is in U3, and tweaks the party. If it's still not what he wants, he'll have to similarly disable Abe3 as he did Abe2, and travel again.
Is Abe2 in U3? No. He got stuck in U2. Is Abe1 at the party with the Abe1 that went to the U2 party? No. The timelines diverged at noon. Abe1 arrived in U2 at 12:05pm.
I apologize in advance; my sleep was disrupted last night, and my mind is not completely clear, but I know if I delay answering it will be too long before I return to it.
I understand the concept; I'm hung up on the divergent dimension concept, though. If U3 is a copy of U2, then everything which happened in U2 must happen in U3 except as altered by the traveler who comes from U2. That means the traveler who comes from U1 must also arrive, because his arrival is part of the history of U2 not altered by the arrival of the traveler from U2 to U3. How can there be another such person? The same way as everyone else in U3 exists, as copies of the people and events of U2.
Certainly the departure of A1 from U2 prevents A2 from doing so; but it does not prevent A1 from leaving U1. You will object that A1 arrives in U2, not U3, but since U3 is a perfect copy of U2, it must also copy the arrival of A1 from U1.
I understand why you think not, but I'm stuck with the notion that a perfect copy of U2 must include the arrival of A1 from U1, even if it also includes the separate arrival of A1 from U2. It is the flaw in divergent dimension theory on this point: http://www.examiner.com/time-travel-movies-in-national/temporal-theory-1...
I hope this helps.
--M. J. Young
I suppose I have a different take on Divergent theory.
At the divergence point, the timelines are allowed to be different from there on. The only thing "copied" is the history. The future is not. It's like branches on a tree. When the traveler goes to a new timeline, a new branch is sprouted and grows from the same trunk.
The future is not "copied" over, any more than a new branch on a tree needs to first copy an existing branch in order to start growing anew.
You say the arrival of A1 from U1 is part of the "history" of U2, but since it happens after the divergence point, I'd say it's part of the future of U2, so isn't copied over.
We may just have different ideas on how Divergence works.
And your earlier comment about terminators "replacing" earlier iterations has be wondering if I quite get Replacement theory either. I'd been viewing it as a form of Divergence theory (my interpretation) with some added contraints. Now I think I need to sit and look at each of these differently to understand your position better. I brought my own biases to the table. :)
Concerning replacement theory, that happens a lot. The critical point to remember is that when someone travels to the past, they are, moment by moment, erasing the original history and replacing it with a new one, such that ultimately the original history will be as if it had never existed, including that the traveler never left from it. That means that the traveler from that original history cannot have arrived in the new history, and he will only be there if an exactly identical traveler left from the new history to the same point in the past for the same reason. In essence, it allows you to prevent the traveler from changing the past by preventing him from leaving the future, but there are serious consequences to doing so.
Continuing...
...continuing:
I see your position on divergence, but am going to pose a problem. Aaron (hereafter "A") 1 leaves from Wednesday (U1) to Tuesday (U2). Abe (hereafter "B") 2 becomes aware of this only after A2 leaves for U3, but he has a box he started on Monday. B2 leaves U2 on Thursday, reaches Monday of U3.
By your model, A2 will never arrive in U3, but will cease to exist once he leaves U2, or else A2 lands in U3 and B2 creates U4 in which Aaron never arrives. Assuming B2 does not prevent Aaron's (3? or 4?) departure, Aaron creates another universe in which A2 has been duplicated--and at this point I'm going to leave the problem to you, because at least one Aaron seems to have passed out of existence and at least one Abe appears to have landed in two different universes. But it's your theory, so perhaps it makes sense to you.
Thanks again for the comments.
--M. J. Young
You're correct that mt interpretation of Divergence would mean that B2's trip creates U4, not that he would slide into U3. (The numbering system is arbitrary.)
Let's look at the universes created so far, and, for sanity's sake, I'll assume A3 and A4 don't make their trips (as it will create more universes). The trips made by A2 and B2 are the last trips.
Monday - Tuesday - Wednesday - Thursday
U1 - A1B1 - A1B1 - B1 - B1 (A1 left on Wednesday... appears on U2 Tuesday)
U2 - A2B2 - A1A2B2 - A1B2 - A1 (A2 left on Wednesday... appears on U3 Tuesday)
U3 - A3B3 - A2A3B3 - A2A3B3 - A2A3B3 (B2 left on Thursday... appears on U4 Monday)
U4 - B2A4B4 - B2A4B4 - B2A4B4 - B2A4B4
When I look at the 4 Thursday's, I see all participants. No drop-offs, no duplicates.
U1 is stuck with B1 wondering what happened to A1.
U2 is stuck with A1, knowing A2 left like he did, but wondering where B2 went.
U3 has A2 joining A3 and B3.
U4 has B2 joining A4 and B4.
That's all 8.
Thanks again, Carl. I'll admit that it's going to take me more time to wrap my head around your post than I have right at the moment (particularly since the chart formatting did not come through at all), but hopefully I'll get it figured out.
I do object inherently to the idea of divergent dimension theory being considered time travel; my objections are published elsewhere, and certainly if we are going to work from a divergent dimension theory foundation they are irrelevant to the question of what happens. (My analysis of Hot Tub Time Machine ended http://www.examiner.com/time-travel-movies-in-national/hot-tub-time-mach... with the suggestion that with multiple dimension theory there were a lot of really disastrous universes created by all that, but we got to see the one good one.)
I'll try to get back to this.
--M. J. Young
I thought about drawing something and posting it somewhere, but I googled instead.
This guy has a TT page with some pictures, and essentially adheres to my interpretation of Divergence also: http://thisischris.com/feature/2011/timetravel.html
He discusses midway down about what happens if you travel back before your arrival point in an attempt to get back to your original time, but fails, as it just creates a new divergent reality. It's similar to the "fail safe" box. The image he has is this: http://thisischris.com/feature/2011/img/timetravel03.gif
He also discusses 3/4's of the way down about two travelers (yourself and Roger) arriving at the same point, and how they each just go into their own new branch, never running into each other (well, except for the local versions). The image has has is this: http://thisischris.com/feature/2011/img/timetravel05.gif
I'll admit his thoughts on a divergence happening when you travel forward in time is not something I've thought about much. He doesn't really explain why he thinks it's so, nor can I come up with one myself.
But we're on the same page with backward TT.
I'm getting to understand why a trip to the future might also cause a Divergence. It depends on whether the future already exists or not. There could already be a future where the Traveler didn't appear from the past. His arrival would cause a separate divergent timeline.
It gets into Sideways Time. We're traveling through time at one second per second. But that's a perceptional value. All of time may be churned out in an instant (Speed of Time = Infinite), but we only experience it @ 1 sec./sec. Or, time may be built piece-by-piece by hard-working Time Gnomes who take 1 Year to construct just 1 Second of Time (Speed of Time = 1 yr./sec.), but our perception would remain at 1 sec./1sec.
If you're going to a future that hasn't been built yet, your arrival will be "built-in"' to the single emerging Timeline. But if you're arriving in a Future that's already been built, you create a divergence.
IF you believe in the Divergence Theory.
Another thought goes along the lines of your Video Tape analogy, where a Traveler records over the previous program. How fast does the re-recording go? They can rip 2 hours movies in minutes, but the movie still plays at 1 sec./sec. If you modernize the analogy a bit, and make it a hard drive, you can have the old recording keep recording, while it's being overwritten at the same "time".
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