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by raganwald
5481 days ago
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I am constantly reading that information cannot travel faster than light, and I accept each of the explanations as to why the various methods for attempting to send information faster than light do not work. What I do not understand is whether sending information faster than light would cause a paradox of any sort. It may be that it happens to be impossible. But is it necessarily impossible? |
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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.