The Problem

I like time travel as a concept, but I’ve always found something to nitpick about the time-travel model in every science fiction story that I’ve read or watched. It seems like they all have at least one of two problems:

• Unpredictability: the universe is not predictable in advance by the audience.
• Anthropocentrism: the universe seems to respect human ideas too much.

Unpredictability

The unpredictability problem is typical of any time-travel model that allows causal loops. If time travel allows a chain of events that ends in someone going back in time and setting off that same chain of events, then any such loop of events could occur at any moment. In other words, even if I know everything about the universe today, I cannot predict what it will look like tomorrow, because tomorrow morning something might happen for no reason other than that it eventually will cause itself to happen.

Note that universes that allow this sort of causal loop cannot be predicted even probabilistically: if two or more futures could cause themselves through time travel, then I have no way of saying in advance which of them is more likely.

One way of solving this problem is to just disallow causal loops. “Branching timeline” models do this: when you go back in time, you don’t go to your past, but to an exact copy of your past except that you’re in it. Then things develop from there according to the mechanistic nature of the universe, resulting in an alternate present/future to the one you left.

The branching timeline model can work well, but it is—in a sense—cheating. One of the most compelling things about time travel as a concept is that it messes with our understanding of causality. Time travel that doesn’t allow things to touch their own past at all denies the audience the pleasure of being confused in this way.

This leads us to the “potential futures” model of time travel. In this model, things can touch their own past, but their ability to do so depends on how probable they are (where that probability is, implicitly, calculated without the possibility of more time travel). Films like Back to the Future and Looper work like this. Marty McFly goes back in time and, when his actions start making his own future conception less likely, he begins to fade out of existence.

This mostly solves the problem of unpredictability. Unpredictable things can’t arrive from the future just because they will eventually cause themselves to arrive from the future. Rather, anything that comes from the future will be something whose existence (and propensity to gain access to a time machine) is predictable from the present.

Anthropocentrism

Unfortunately, stories using the Back to the Future / Looper model tend to fall prey to the second nitpick, anthropocentrism. The basic issue is that the real world is deterministic. Not totally deterministic, because of quantum uncertainty, but very close to deterministic at the scale at which humans operate. The uncertainty about the future exists mostly in the characters’ and the audience’s head.

For example, at one point in Looper, someone goes back in time (not far, to a period that includes a past version of them) and then goes on the lam. The people chasing them kidnap their past self and start mutilating them. As they cut fingers off the past self, the fingers of the self-from-the-future begin to disappear—because, once the finger has been cut off, there’s no longer any possible future in which the self-with-the-finger exists and comes back in time.

But this is standard operating procedure for the baddies in this universe! As soon as this person came back in time, this outcome was going to happen. A human observer might not be sure, in advance, whether the past self would be successfully captured, or whether the people doing the mutilating would have a change of heart before going through with it—but the idea that physics itself is unsure of the outcome only makes sense if you buy into very fluffy notions of free will, and sometimes not even then.

(Those fluffy notions of free will, by the way, can be very interesting. By all means feel free to explore them in a story, and even cross them with time travel if you like. But a time-travel story that accepted such a physics-defying notion of free will would look very different to any time travel story I’ve seen, because there’d be no guarantee that, on going back in time, you’d find people making the same decisions as in the “original” timeline—even if you did nothing to change their decision.)

Another kind of anthropocentrism that comes up is in ignoring the butterfly effect. Marty McFly brings a family photo from the future back with him. When he messes with the timeline, he and his family start to disappear from the photo—but the rest of the photo remains the same. So even once his parents are unlikely to get together, there’s apparently still a likely future where somebody took a photograph at exactly the same angle and time of day, but without anyone in it. Why?

This anthropocentrism is annoying for two reasons. The first is philosophical: to me, part of the appeal of storytelling—particularly science fiction—is the ability to find (or invent) meaningful narrative in an uncaring universe. Constructing a universe which does “care” about human things feels like cheating.

The second brings us full circle: anthropocentric mechanics tend to introduce unpredictability, because each person is different. The question of how “unlikely” it is for Marty McFly’s parents to get together given some initial conditions is a subjective question, with the answer depending on how much the observer knows about the situation. Predicting the story requires you to guess at the level of ignorance the writers want you to pretend to have.

The Solution

This leads me to propose a model of time travel that is:

• Predictable
• Non-anthropocentric
• Headache-inducing, but in the fun way

I call it information-only weighted-branching-paths steady-state time travel.

Intuitive Explanation

The key idea is to extend the notion of “possible futures” to get “possible presents”. This is not actually a novel concept: under the many-worlds interpretation of quantum mechanics, this is already how the world works. Whenever something truly uncertain happens (e.g., a radioactive atom decays, or doesn’t), the universe splits in two. Someone standing before the split should expect to “find themselves” in one of those two universes, with probability depending on the event.

But back to time travel. In this model, only information can be sent back in time. You set up a time machine to receive some information from the future. As soon as you turn it on, the present splits into multiple timelines—one for each possible message you could receive. The probability of each of these presents depends on the probability of the futures that lead to them, and vice versa (yay circular causality!).

For example, let’s say we accept only a single bit from one minute into the future. Before starting the experiment, we decide as follows: if we receive a 0, we will send back a 1; if we receive a 1, we will send back a 0.

Then we have the following set of equations:

• Because we will send back a 0 if (and only if) we receive a 1: \(P(\text{send }0) = P(\text{get }1)\)
• Because we will send back a 1 if (and only if) we receive a 0: \(P(\text{send }1) = P(\text{get }0)\)
• Because we will receive a 0 only if we send a 0: \(P(\text{get }0) = P(\text{send }0)\)
• Because we will receive a 1 only if we send a 1: \(P(\text{get }1) = P(\text{send }1)\)
• Because 0 and 1 are the only possible messages we can receive, their probabilities must sum to a certainty: \(P(\text{get }0) + P(\text{get }1) = 1\)

Chaining together equations 4 and 2, we get \(P(\text{get }1) = P(\text{send }1) = P(\text{get }0)\). In other words, the probability of getting a 0 and a 1 are equal. Since they must sum to 1, the probability of each one is 50%.

One way to think of this is to imagine that we’re using an alternate-pasts model, where each time you go back in time you create another timeline, and we’re the infinity-th timeline. The loop may not yet have settled on a single timeline—in the example above, there is no timeline that it could settle on, as each timeline leads to the other one—but their relative frequencies have settled, which gives us our steady-state probability distribution.

Formal Model

The more general mathematical model follows. If the machine is configured to get one of \(m\) possible messages, then when we turn it on we split into \(m\) possible worlds, one for each message. Each of those worlds (say, world number \(i\), where \(1 \leq i \leq m\) has a probability which we will call \(\pi_i\). Note that since these are mutually exhaustive probabilities, we have:

• For each \(i\), \(0 \leq \pi_i \leq 1\)
• \(\sum_{i=1}^m \pi_i = 1\)

Each of those messages, if received, will lead to a certain message being sent back. Actually, that’s not quite true, since the universe isn’t really deterministic (though again I stress that real physical randomness isn’t the same thing as unpredictability-by-humans). It would be more accurate to say that for each received message \(i\) and each message that might be sent \(j\), we have a certain probability that receiving \(i\) will lead to sending \(j\). Call this \(P_{ij}\).

Now we close the loop: the total probability of receiving a message must equal the total probability of sending it. So we have:

For each \(i\): \(\pi_i = \sum_{j=1}^m \pi_j P_{ji}\).

(Or, for those more comfortable with matrix notation: \(\vec \pi^T = \vec \pi^T P\).)

Clearing Up Some Technicalities

Multiple Steady-States

The astute reader (or the reader who has studied Markov chains) may argue that I have not entirely solved the unpredictability problem. To see why, consider a slightly different scenario: we still only have one bit of bandwidth, but we decide that we will always send back whatever message we receive.

This gives us the following set of equations:

• \(\pi_0 = \pi_0\)
• \(\pi_1 = \pi_1\)
• \(\pi_0 + \pi_1 = 1\)

The problem is that this set of equations has multiple solutions. In particular, \(\pi_0=0, \pi_1=1\) is a solution, so we’ll definitely get a 1. But \(\pi_0=1, \pi_1=0\) also is, so we’ll definitely get a 0. Oops!

To solve this problem, we go back to quantum uncertainty. Even if the operator of the time machine has decided they will send back exactly the message they receive, there is nevertheless some probability that a different message will be sent. Real subatomic particles jump around sometimes—at any moment, they may turn out to be elsewhere than where you thought they would be. So there exists a tiny probability that, for example, the stuff inside the time machine user’s brain will jump to a different state, causing them to change their mind about what to send. Or that the stuff in the time machine itself will jump to a different state, causing the wrong thing to be sent.

To be clear, the probability of this actually happening is, for most practical purposes, nil. I would not write a science fiction story that hinges on any human-noticeable thing spontaneously moving around due to quantum randomness. If you’re allowing things that unlikely to just happen, then you’ve made your story ludicrously unpredictable, and also divorced it from any reality the reader is familiar with.

But this tiny probability of something crazy happening is, ironically, enough to make our time machine predictable. Call the tiny probability \(\epsilon\). Now there are two possible ways for a 0 to be sent back in time: in the timeline where we got a 0, we will send a 0 with probability \((1-\epsilon)\); and in the timeline where we got a 1, we will send a 0 with probability \(\epsilon\). The ways for 1 to be sent are symmetrical. So we get:

• \(\pi_0 = \pi_0 (1 – \epsilon) + \pi_1 \epsilon\)
• \(\pi_1 = \pi_1 (1 – \epsilon) + \pi_0 \epsilon\)
• \(\pi_0 + \pi_1 = 1\)

Which, when you work it out, has only one possible solution: \(\pi_0 = \pi_1 = \frac{1}{2}\). Which is as we’d expect: given that nothing makes either timeline more likely, they have equal probabilities.

Applications

Well that’s all well and good, but what kind of story possibilities does this model present? What can the characters actually use it for?

Getting Information From The Future

The simplest use-case is when there’s a piece of information that would be useful in the present, but which we will only learn in the future. Assume for the sake of the argument that the information doesn’t change based on our actions, and that we’ll learn it at the same point in the future regardless of what we do in the present. Lottery numbers, for example, satisfy all these criteria.

Assuming our time machine has enough bandwidth, we can scheme as follows:

• Tell our time machine to receive a message from after the lottery numbers are announced.
• Buy a ticket matching the numbers received from the future.
• Wait until the real winning numbers are announced.
• If we did in fact win the lottery, send the same winning number back in time and go party.
• Otherwise, send the winning numbers back in time and go feel disappointed.

Following this plan, every possible message received results in the true winning numbers being sent. Thus, the probability of the true winning numbers being received is basically 100%.

Information Hidden By Risk

Now for a slightly more complex scenario. What if we’re facing a risky course of action that might or might not work out, and—unlike the lottery setup—the only way to find out whether it would have had a good outcome (other than time travel) is to do it? Examples would be agreeing to a date, launching an attack, or performing a difficult surgery. We can’t copy our lottery-number scheme because, in the world where we decide not to take the risk, we don’t find out what the correct decision would have been, and so can’t send it back. But instead, we can do as follows:

• Number all possible messages from \(1\) to \(m\).
• Tell the time machine to receive a message from some point in the future when the outcome of the risky action, if we do it, will be clear.
• If we receive any message other than \(m\), do not take the risk. Call the message we received \(k\). When the appropriate point in the future comes, send message \(k+1\).
• If we receive message \(m\), take the risk. If it works out well, send message \(m\). Otherwise, send message \(1\).

To see how this works, consider that the only future from which we will send message \(1\) is the future in which we received \(m\), and then only if the risky action turns out to be a bad idea. Therefore, if it turns out to be a good idea, there is no future from which we will send message \(1\), and therefore the probability of receiving message \(1\) is approximately \(0\). But receiving message \(1\) is the only way we could ever send message \(2\)—so, if the risky action is a good idea, we will never receive message \(2\) either. Following the same logic, we will never receive messages \(3\), \(4\), etc… all the way up to \(m-1\).

Message \(m\), on the other hand, is sent from two possible futures—the future where we received \(m-1\), and the future where we received \(m\). The future where we received \(m-1\) happens with probability \(0\). That leaves the future where we received \(m\). What probability does that have? Well, that’s precisely what we’re trying to figure out, but there’s nothing forcing it to be zero. In fact, since message \(m\) is the only message we can receive with nonzero probability, and probabilities must sum to \(1\), it must happen with probability \(1\). So we will certainly receive message \(m\), and we’ll go ahead and take the risk, which will pay off.

Conversely, if the risky action turns out to be a bad idea, then we will send back message \(1\). Since this is the only future which sends back message \(1\), the probability of message \(1\) must be equal to the probability of message \(m\). Also, since \(m\) no longer sends itself back, now the only future which sends \(m\) is the one where we received \(m-1\), so the probability of \(m\) must equal that of \(m-1\). As before, all the other messages must also have equal probability to each other. Therefore, all the messages now have equal probability, so the probability of message \(m\) is \(1/m\). Therefore our probability of taking the risky action (which is a bad idea) is \(1/m\). As our time machine gets more bits (that is, more possible messages), this probability becomes close to zero.

No Binary Outcome

In the previous setup, we assumed that the risky action, if we took it, had obvious “good” and “bad” outcomes: the operation succeeds or fails, the battle is won or lost. In reality, most choices don’t work like that—the decision we made has a certain outcome, but we could only say it’s “good” or “bad” relative to the outcome of the road not taken, which is unknown to us.

Time travel, of course, can fix this petty trouble of the human condition. Assume for simplicity you’re choosing between two options. Here’s what you do:

• Decide arbitrarily which choice will count as “choice \(0\)” and which as “choice \(1\)”

• Decide on a metric for the “goodness” of the outcome. This might be:
  • Life satisfaction from 1 to 10 (for an individual)
  • Quarterly profits (for a corporation)
  • A citizen welfare index (for a government)
• Decide in advance when you’re going to evaluate your metric and send the message back.
• Code the time machine to receive a message from that point in the future. When it arrives, do whatever the time machine says to do.
• Let things develop naturally until it’s time to send the message back.
• Compute your goodness metric.
• Use a physically random process (like radioactive decay) to generate a random number on the same scale as your metric.
• If the random number is higher than your observed “goodness”, send back the number of the choice you didn’t make. If the random number is lower than your observed “goodness”, send back the number of the choice you did make.

Here, for the first time, we deliberately construct a scenario where the message sent is not a deterministic function of the message received. Why?

To do the maths, let’s normalize the goodness metric to get a number between 0 and 1. Assume the goodness of choice 0 is a and the goodness of choice 1 is b. Then we have:

• \(P_{00} = a\)
• \(P_{01} = 1-a\)
• \(P_{10} = 1-b\)
• \(P_{11} = b\)

Putting these together, we get:

• \(\pi_0 = \pi_0 a + \pi_1 (1-b)\)
• \(\pi_1 = \pi_0 (1-a) +\pi_1 b\)
• \(\pi_0 + \pi_1 = 1\)

From the third equation we get \(\pi_1 = 1-\pi_0\). Substituting this into the first equation, we get:

\(\pi_0 = \pi_0 a +(1-\pi_{0})(1-b) = a \pi_0 + 1-b-\pi_0 + b \pi_0 = (a+b-1)\pi_0 + 1-b\)

Which gives us:

\(\pi_0 = \frac{1 – b}{2 -a – b} = \frac{1 – b}{(1 – b) + (1 – a)}\)

Obviously, this gives us \(\pi_0 > \frac{1}{2}\) if and only if \(1-b > 1-a\).

Remember that \(a\) and \(b\) are the “goodness” of choices \(0\) and \(1\) respectively, on a scale from \(0\) to \(1\). Therefore \(1-a\) and \(1-b\) are the “badness” of \(0\) and \(1\) respectively. Therefore, we are more likely to receive message \(0\) if and only if choice \(1\) is the worse choice.

Counterplay

Let’s say you’re in direct conflict or competition with someone who has a time machine. The time machine will give them a significant informational advantage over you. For example, if you are at war with a nation with a time machine, they will only meet you in battle when the time machine tells them in advance that they’ll win. How can you neutralize this advantage?

The first option is to come up with an absolutely airtight plan, such that no matter what your opponent does, they will lose. Thus knowing the future will not help them. This isn’t really a useful suggestion, as you probably can’t think of such a plan—and if you could, you’d probably use it whether or not the opponent had a time machine.

The second is to make your behavior unpredictable even to the time machine. You can do this by conditioning your decisions on something truly random, like the decay of radioactive particles.

The third is to mess with the enemy’s information flow by destroying or seizing control of the time machine in any timeline where you are victorious. If you are successful, you can send the enemy’s past a false message implying that the current timeline (the one where you defeated them and seized control of their time machine) actually worked out well for them. This will make their past self do whatever it was that led to you winning.

Multiple Time Machines

This is left as an exercise for the reader.