# using a paradox to multiply time machine

I've asked myself many times if it's possible to multiply a time machine, but never really found any statement on how.

Let's say we're in a field with one time-machine (the First one and still only built and functional), and a hundred (soon to be) time-travelers. It's an official presentation of the machine. Every participant has been here since the early morning. At the end of the day, the experiment begins.

So the first traveler gets into the machine, and his destination is 5 minutes earlier. What happens? If we repeat this step 7 times do we get 128 time-machines? Or does all of this fall apart from the first try as the machine disappears from its reality when the journey starts?

• Only one man can use the machine at a time
• The machine can choose any location on earth as a destination
• It really depends on the rules you have for time travel, and how paradoxes are resolved. But to multiply, your machine has to "move with the traveller" as well as stay in its original position. That sound like a strange machine. – clem steredenn Aug 27 '15 at 11:40
• @bilbo_pingouin It's not moving with the traveller and staying at the same position. It's moving with the traveller and staying at the destination. It doesn't multiply, but each travel introduces a single new copy, so it adds. – Taemyr Aug 27 '15 at 11:47
• The first step of my thought was if you enter a time-machine and select a date of arrival 5 minutes earlier than now, you'll be in a room with two time-machines. (without thinking of any location problem) – MickMRCX Aug 27 '15 at 11:48
• Ok, so you have then 5 min before the "old" disappears. Fair enough. I still think that it depends on many different rules of time travel. – clem steredenn Aug 27 '15 at 12:00
• and after as this are time-machine each traveler could go when they want, but with the only obligation to come back a this point of time. but as @Taemyr said the travellers multiply with the machines – MickMRCX Aug 27 '15 at 12:06

With no paradox you only get 7 travellers in 7 "steps" - Each traveler only introduces one copy of the machine. However the steps are part of the same timeline so it's wrong to call them steps. If you allow paradoxes the situation becomes complicated and the answer depends on your model for time travel.

6 minutes before the demonstration starts there are one machine and 100 soon-to-be time travelers.

5 minutes before the demonstration starts 100 machines appear, and the 100 travellers exit. There are now 101 copies of the time machine and two copies of each of the 100 travelers.

5 minutes later traveler #1 enters the machine that no one exited. Traveler #2 enters the machine traveler #1 exited, and so on. After this there will be 1 machine and 100 timetravelers present.

If each travel wrote in a log in the machine before he exited the machine then traveler #100 would see that each of the 99 other travelers had used his machine "before" him. - And this machine would have traveled a total of 8 hours 20 minutes. (In case there was any doubts: This is the one machine that is present after the demonstration is through)

Exponetial growth would require that traveler #2 enters the machine that #1 arived in while traveler #3 enters the machine that #1 is going to use before #1 gets into it. However this creates a paradox because there will be no machine available for #1 to enter, and hence he will not exit any machine.

• For the multiplying, i've forget that even if you come back in a room with one time-machine ("static", waiting to be used) and the ones that comes from the future. I've been thinking ( i admit the error) that you'll arrived in a room with as many (static) machine as the room you come from has. – MickMRCX Aug 27 '15 at 11:55

While Taemyr's answer is correct, it covers the version of time travel where there is only one universe, and any time travellers must travel into their own past. While this is a common model to use for time travelling stories, I'm more of a fan of the system used in the movie 'Primer', wherein every time you travel to the past, the universe splits off into a new future.

In this case, let's say you're in the field and no new time machines show up. You are the 'prime' version of yourself; the universe has not yet split up, no other universe has sent a time traveller to you. Either that, or the time machine doesn't work.

Anyway, the prime you, let's call it Y1, steps into the time machine and goes back five minutes. When Y1 arrives, they get a chance to meet a new version of themself, let's call them Y2. Y2 was the exact same person as Y1 up until Y1 arrived in the past, but now they are two separate people.

There are two options at this point: either Y2 decides to go back in time like Y1 did, or he doesn't. Y1 could convince him not to; somewhat paradoxically, the fact that Y1 convinces Y2 not to go back in time may lead to the very conditions that led Y1 to go back in time, leading to an infinite loop of Yx going back, then Yx+1 not going back. But let's assume these time travelers are willing to test this out, and they both go back in time. Assuming the time machines end up in the same tangent universe, after the journey there will now be three time machines, and three copies of yourself, Y1, Y2, and Y3. This can continue for as long as Y1 is willing to keep going; it's possible the effects of time travel will cause him to stop eventually.

The interesting part about this model of time travel is that you are not guaranteed to enter the loop at the beginning. You could show up in this field and be met with a thousand copies of yourself, or none at all. They could have been travelling for days, or years, or maybe it was the first jump. That's part of the reason I prefer this model, because virtually anything could step out of that time machine.

Anyway, I'd like to point out that in this scenario, the number of time machines increases linearily; that is, you have one, then two, then three, and so on. However, if you intend to mass-produce time machines this way, there is a very good chance you'll get in on it partway through, so your initial gains could potentially make up for this slow growth. Also, no one but the time travellers and the residents of the last universe are going to benefit from this; for every other universe, every iteration of Y has been erased permanently, along with all the time machines. So it's probably a safer bet to gamble on getting just the second machine, as it's 50/50 odds on double or nothing.

• The problem with this model is that it's not time travel. Instead you are traveling to a different reality. Also I am not sure why you can assume that two time travelers will end up in the same line. – Taemyr Aug 27 '15 at 14:21
• @Taemyr It's not so much a different reality as it is a branch off of the same reality. The past is still the same, it's just the future that will be different. But as to your second point, I agree, the behavior of parallel time travel is undefined. – DaaaahWhoosh Aug 27 '15 at 14:28

I'm gonna make this simple, because it can be.

The time traveller creates a list of who needs a time machine. They goes back in time five minutes and tell themselves what number is next. "No 75 needs a time machine! Go back and give him one!" They wait for the original departure time and the younger version goes back. Then he increases the number for the next one and so on.

Of course, this does leave 100 versions of themselves, since they've duplicated themselves along with the time machine. A better idea could be to use a simple tally on a piece of paper, increasing the increment each time.

This is just it's simplest form. There are all kinds of complications and stuff, but this is a common way that time travel works in fiction.

This answer is to offer something the others haven't yet. The issues with the math in this question is the word "multiply" in the question title and the condition

Only one man can use the machine at a time

If this means that the machine must be the outer limit of what is moved through time, then the above answers cover it nicely. If you can tie one machine to another, and get them both to move, it creates a huge paradox when that other machine is itself the first machine, but you can actually multiply them.

The above answers go over paradoxes and splitting universes, and the other parts of time travel that make it very frustrating and confusing. If you just want more machines from one, and you have 16 people to do it, here is how. The 16 people and the 1 machine stand in a field for n number of minutes (to soon be calculated). This creates the window of certainty for the machine multiplication. Traveler1 takes the machine back 1 minute, ties it to itself, takes it forward 2 minutes. T1 unties the machines and goes back 3 minutes while T2 gets in the "copy" and goes back 4 minutes. T1 and T2 bring back new copies. T1, T2, T3, and T4 now all go back and get copies and come back 1 minute after their departure. You now have 8 machines in 10 minutes, where initial wait n is only 7 minutes and initial traveler T1 has only made 6 jumps (counting return trips).

Now T1 through T8 all jump back with 1 minute spacing, get copies, and return to 1 minute after their jump. Getting 16 machines requires 15 minutes of waiting in the field before the first jump, and 19 minutes total (4 return jumps).

Now your 16 people can all go where they want.

• Assuming the universe does not collapse when the machine touches its past self.
• Four rounds of trips makes 16 machines. 2^4 = 16
• Let us say the 1 minute spacing is to tie and untie.
• Initial prep time in the field is 2^4 - 1 minutes with 4 minutes for the return trips.
• 7 round trips gets you 128 machines in 2 hours and 14 minutes.
• You are creating a paradox. "Traveler1 takes the machine back 1 minute, ties it to itself, takes it forward 2 minutes" This means that, when Traveler1 is supposed to enter the machine, it's no longer there. Since you took the machine that Traveler1 was supposed to enter and moved it into the future past the point when he entered it. – Taemyr Aug 28 '15 at 7:54