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I get email: Do FTL signals let you communicate back in time?

by dave on September 29th, 2011

First and foremost, let me go on the record, once again, and say that I think that the faster than light neutrino result is almost certainly flawed. I’m not going to debate the point, since I’ve already made a number of objections, here and elsewhere, and at the end of the day, the results will either be experimentally confirmed or they won’t. Probably on the “won’t” side.

That said, a number of you emailed me asking about the implications of what would happen if we could send a signal back in time. I got an email from a reader named Jason (who previously asked me about Hawking radiation), who wanted to know:

Even if the communication took place instantaneously, I don’t see how one would use that to communicate with one’s own past. Can you explainit in plain English?

Specifically, Alice and Bob want to chat, Bob sends a message and N seconds later, Alice sends a reply. Even if the transit time for the message were zero, N seconds have still elapsed for both individuals. How does a superluminal transit speed equate to negative elapsed time? Wouldn’t, at best, a message have a zero delay?

Well, I’m not going to answer Jason’s question entirely intuitively (that’s why there’s a “technical” flag), but hopefully I can shed a little light (no pun intended) on the situation.

Also, in order to make some really kickass spacetime diagrams, I wrote a nice little package for pylab. Download my code, as well as a sample program showing how to use it if you’d like to make your own.

So first things first: Jason is absolutely correct in one very important respect. Say we’re standing about 1 light-second away from one another, and I send him a beam which (for convenience) travels at 4c. I realize that this is much faster than the purported neutrino effect, but the difference is only one of degree. In that case, it’ll take 1/4 of a second to send a signal. If Jason then responds immediately, I get his response 1/4 of a second after that.

The entire series of events can be summed up in a spacetime diagram:

Jason and I are both represented by vertical lines in this diagram. We’re not moving through space — just through time.

The situation is exactly as Jason said it would be. I receive a response to my signal exactly half a second after I send my original message. Or, to put it another way event “3″ (receiving the response) occurs after event “1″ (sending my original signal) exactly as I expected it to.

But things get kind of strange once we consider frames moving compared to one another. For instance, if Jason is flying to the right at speed v (I’ll be using so-called “natural units” throughout, so that c=1, light-seconds and seconds can be used interchangeably, unit-wise, and all speeds are in fractions of the speed of light), then the time and space interval as seen by each of us will be different.

For instance, if the time between the two events (as seen by me) is \Delta t and the space interval is \Delta x, then Jason will see:

  \Delta t'=\gamma \Delta t-v\gamma \Delta x

and
  \Delta x'=\gamma \Delta x-v\gamma \Delta t

where
  \gamma\equiv \frac{1}{\sqrt{1-v^2}}

If Lorentz transforms, and the idea that two different observers can measure the amount of distance or time between two different events bother you, there’s little I can do — short of doing a complete course on special relativity.

If you want a quick explanation, the basic idea is that those transformations are basically unique in that they produce the same speed of light for all inertial observers.

What this means is that suppose I measure two events occurring to Jason and me simultaneously (\Delta t=0), Jason won’t necessarily measure it that way. Supposing the event in question was Jason and me both popping a balloon, and that at that instant, I measure Jason’s ship to be about 1 light-second away. In that case, \Delta x=1\ ls. If Jason is traveling at 0.6 (60% the speed of light) then \gamma=1.25, and thus, using the relation above, we’ll find that:

  \Delta t'=-0.75\ s

and
  \Delta x'=1.25\ ls

In other words, he’ll claim that my balloon popped first.

Even if I sent him a message at the speed of light (warning him about the balloon poppage, presumably), my message won’t reach him until after his own balloon has popped. In other words, because of finite signal speed, this difference in frame normally doesn’t matter.

But what happens if you can send a signal faster than light? Then things get very interesting.

We’ll call v_s our signal speed, and rather than go through all of the math, I’ll show you the result of a communication graphically. The only thing you need to verify is that in my diagram, photons always make 45 degree angles with the horizontal (because they cover 1 ls/s), and that from the perspective of the person sending it, neutrinos always have a slope corresponding to 4ls/s (much closer to the horizontal).

What I’m plotting here is exactly what we described before. I send a signal to Jason (event “1″), and Jason receives it (event “2″). I also shot off a photon at the same time, just for reference (the squiggly line). Hold off on worrying about event “3″ for the moment.

Let’s look at things from Jason’s perspective. That’s the whole point of relativity. According to Einstein, there’s nothing preventing Jason from claiming that he’s at rest, and that I’m the one who’s moving, and hence he’s represented by a vertical line in his own space-time diagram:

The first thing you’ll notice is that, according to Jason, event “2″ occurs before event “1.” In other words, he receives my signal before I sent it. In fact, there’s nothing terrible about this, since these two events occur in the same place. The craziness occurs when Jason sends his signal back. He sends his signal to me at very high speed, and what looks to be his future. However, seen from my perspective (look at the “unprimed” figure above again), event “3″ (me receiving his message) occurs before the signal was sent out.

In other words, if you can send a signal faster than light, then it really is possible to communicate with the past, provided the observers are moving away from each other fast enough. How fast? If two observers are moving away from one another at speed, v, then the signal needs to propagate at at least:

  v_s> \frac{1+\frac{1}{\gamma}}{v}

which for our example where Jason is traveling at v=0.6, corresponds to v_s=3.

In other words, if you want a succinct answer to “Why would it matter if we could send particles faster than light?” You have one: You’d be able to send messages into the freakin’ past!

-Dave

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13 Comments
  1. Yet it is cool :P the freaking past !

  2. “This is not a dream..not a dream. We are using your brain’s electrical system as a receiver. We are unable to transmit through conscious neural interference. You are receiving this broadcast as a dream. We are transmitting from the year one, nine, nine, nine. You are receiving this broadcast in order to alter the events you are seeing. Our technology has not developed a transmitter strong enough to reach your conscious state of awareness but this is not a dream. You are seeing what is actually occurring for the purpose of causality violation.”

  3. Ben Grimm permalink

    You cannot send messages into the past. You can only send messages faster than the receiver is able to SEE you send the message. The receiver’s reply, even if it’s traveling faster than light, will still reach you *after* you sent the message.

    • dave permalink

      If 1) special relativity is correct and 2) if we can send signals faster than light (of which I most sincerely believe the first and disbelieve the second) then you are incorrect, I’m afraid. You really can signals to the past. It’s not just a misperception.

  4. It’s not that I don’t believe Special Relativity it’s that I can’t justify it in my mind. Using the Lorentz transform to show that FTL is impossible seems circular to me since it includes the speed of light. Also using diagrams like above couldn’t you show that either twin is younger and thus have a contradiction.
    Dave K

    • dave permalink

      Ah, but the thing with the twin paradox is that it isn’t resolved through special relativity alone. The thing that makes the stay-at-home twin older than the traveling twin is that the traveling twin accelerated a number of times during her trip.

      In this case, we could easily imagine the two observers flying apart with equal and opposite speeds from their initial rest position. Of course, by inspection it should be clear that no matter what the combination of speeds or signal speeds, they still can’t send a message back before they started moving (or before they “built the antitelephone.”)

      • Isn’t the stay at home twin older? Let’s keep the accelerating parts the same. Two twins take off and accelerate to 0.9c. One immediately decelerates and returns to earth. The other continues on for a couple of years (pick your frame of reference) and then returns. What are their age differences? Their accelerating parts of the trip were the same so it should have the same effect.

        What about my circular reasoning argument about using the Lorentz transform?

        Sorry to be so obstinate, I’m just trying to understand.

        • dave permalink

          Pardon me. That’s what happens when I don’t read what I write. Fixed now. Of course, the stay at home twin is older and the traveling twin is younger.

          You’re correct in your implication that the twin who’s traveling for longer will be time-dilated more even though the accelerations are the same. The issue, however, is more that you can’t imply a simple Lorentz transform between frames at all if there has been an acceleration. It’s more than acceleration voids the assumptions.

          • What about my circular reasoning argument about using the Lorentz transform?

          • dave permalink

            We call ‘c’ the speed of light, but really, it’s the speed of any massless particle. The assumptions underlying SR aren’t that nothing can exceed the speed of light. Rather, they are 1) that c is invariant for any observer and 2) that experiments will be the same in any inertial frame. The fact that FTL signals produce acausality may be messed up, but it doesn’t actually invalidate SR.

  5. come and go permalink

    If I understand correctly, detection is defined as interaction with *incoming* particles, and emission as creation of *outgoing* particles. Neutrinos are relatively easy to create but hard to detect, and even harder to reflect. Superluminal mirror should create asymmetrical pairs of particles/antiparticles or change one to another? How about entropy? There should be events like cosmic background neutrino incoming bursts from all directions simultaneously, something like retarded EM waves of “anti-photons”.

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