4 ms·

But is it necessarily impossible? Yes, because of time dilation. In particular, suppose that, in some reference frame, you can send a message from location A t

But is it necessarily impossible? Yes, because of time dilation. In particular, suppose that, in some reference frame, you can send a message from location A to location B faster than light. In that reference frame, it arrives at B after it leaves A. Then, in some other reference frame (moving at less than the speed of light relative to the first), it arrives at B before it leaves A. So, we can reverse that: Let's say I (in Boston) want to send you (in Toronto) a message that you'll receive before I send it. Then I simply figure out a reference frame where the message would travel forward in time but faster than light. Then I use the faster-than-light communication in that reference frame to send it to you. If you now do the same -- choose a 3rd reference frame, where the message from Toronto to Boston travels forward in time but faster than light, but where on the Earth's surface it travels backward in time -- you can now send your reply before I send my message. Then I can arrange a simple grandmother paradox. e.g. I plan to send a message to you at 2pm, but if I get your reply before 2pm, I don't actually send it. EDIT: In particular, if one observer measures an event as happening at position-time (x, y, z, t), and another observer is moving at speed u along the x axis, that observer will see: x' = (x - ut) / sqrt(1 - u^2/c^2) y' = y z' = z t' = (t - ux/c^2) / sqrt(1 - u^2/c^2) (This is the Lorentz transformation.) Suppose the first observer can send info faster than light. Suppose she sends it along the x axis at a speed of 2c. You can repeat the derivation with any kc where k > 1. If the sending event is at position-time (0, 0, 0, 0), and it takes one time unit to get there, then the receiving event in her frame is (2c, 0, 0, 1). What does the other observer see? Let's suppose they're moving more than half the speed of light, say 3/4 the speed of light. The sending event is the origin, and the origin maps to the origin, so that's easy. For the receiving event: x' = (x - ut) / sqrt(1 - u^2/c^2) = (2c - (3/4 c) * 1) / sqrt(1 - (3/4 c)^2/c^2) y' = y = 0 z' = z = 0 t' = (t - ux/c^2) / sqrt(1 - u^2/c^2) = (1 - (3/4 c) * 2 * c/c^2) / sqrt(1 - (3/4 c)^2 / c^2) = (1 - 3/4 * 2) / sqrt(1 - (3/4)^2). In the formula for t', note that the first term, 1 - 3/4 * 2, is negative. So, if I can send messages at a speed of k * c, then an observer moving with a speed greater than c/k in the same direction sees the message arrive before it leaves. That description is very easy to reverse.

raganwald 15y ago
Time dilation: http://en.wikipedia.org/wiki/Time_dilation http://en.wikipedia.org/wiki/Time_dilation Ok, so let's take two twins, Raj and Ragan. Raj is orbiting a massive cosmological object, Ragan is on the surface. Ragan's time appears to travel slower than Raj's time. Does FTL communication between Ragan and Raj ever create a situation where messages can travel backwards in time? Does bringing an observer into it help?
- martincmartin 15y ago
  As you were posting this, I edited my post to fill in more details. You've got me re-reading my special relativity texts, which I haven't touched in 20 years, and for which I thank you. :)
Groxx 15y ago
I've seen that explanation over and over, but I don't see why there's a problem with something arriving at B before B sees it leave A. We have the same situation with faster-than-sound travel, but it doesn't appear to be impossible. Granted, it's not impossible by that proof because we have faster means of communication (you can detect something before its sound reaches you) (or maybe because it's slower than light), but that's not a problem with faster-than-light communication either. By hypothetically accepting that there's a FTL communication method, didn't you just accept that there's something faster than light? Wouldn't that imply there's something we don't know about time dilation (if FTL comm is achieved), given that it defines such a thing as impossible? It's a circular argument, it's impossible by this proof because this proof relies on it being impossible. So arguing a hypothesis which rejects that requirement with this proof is ignoring the hypothetical universe that was constructed, and it's not a proof at all.
- bermanoid 15y ago
  ...I don't see why there's a problem with something arriving at B before B sees it leave A. The problem is that if something arrives at B before B sees it leave A, then there is a way to arrange for a message to be sent from B to A, and a response back from A to B, such that the response arrives before the message was sent. This is a consequence of special relativity, which allows any spacelike vector to be transformed into any spacelike vector (and any timelike into any other timelike) by a subluminal (less than light speed) change in observer velocity. If you can do any spacelike (FTL) transmission of data, then you can do every spacelike transmission of data if you care to go through the trouble (which is not to say it would be easy, just that physics doesn't prohibit it). Combining spacelike vectors lets you move back in time, so causality is screwed beyond belief.

raganwald 15y ago

Time dilation: http://en.wikipedia.org/wiki/Time_dilation http://en.wikipedia.org/wiki/Time_dilation Ok, so let's take two twins, Raj and Ragan. Raj is orbiting a massive cosmological object, Ragan is on the surface. Ragan's time appears to travel slower than Raj's time. Does FTL communication between Ragan and Raj ever create a situation where messages can travel backwards in time? Does bringing an observer into it help?

martincmartin 15y ago
As you were posting this, I edited my post to fill in more details. You've got me re-reading my special relativity texts, which I haven't touched in 20 years, and for which I thank you. :)

Groxx 15y ago

I've seen that explanation over and over, but I don't see why there's a problem with something arriving at B before B sees it leave A. We have the same situation with faster-than-sound travel, but it doesn't appear to be impossible. Granted, it's not impossible by that proof because we have faster means of communication (you can detect something before its sound reaches you) (or maybe because it's slower than light), but that's not a problem with faster-than-light communication either. By hypothetically accepting that there's a FTL communication method, didn't you just accept that there's something faster than light? Wouldn't that imply there's something we don't know about time dilation (if FTL comm is achieved), given that it defines such a thing as impossible? It's a circular argument, it's impossible by this proof because this proof relies on it being impossible. So arguing a hypothesis which rejects that requirement with this proof is ignoring the hypothetical universe that was constructed, and it's not a proof at all.

bermanoid 15y ago
...I don't see why there's a problem with something arriving at B before B sees it leave A. The problem is that if something arrives at B before B sees it leave A, then there is a way to arrange for a message to be sent from B to A, and a response back from A to B, such that the response arrives before the message was sent. This is a consequence of special relativity, which allows any spacelike vector to be transformed into any spacelike vector (and any timelike into any other timelike) by a subluminal (less than light speed) change in observer velocity. If you can do any spacelike (FTL) transmission of data, then you can do every spacelike transmission of data if you care to go through the trouble (which is not to say it would be easy, just that physics doesn't prohibit it). Combining spacelike vectors lets you move back in time, so causality is screwed beyond belief.