# actual FTL travel question

http://www.nytimes.com/library/national/science/053000sci-physics-light.html

Describes a way to make pulses of light travel up to 300% FTL. It also says that there is no way to make use of this information and send information back to “change the past” in anyway.

I am having trouble with that.

Lets say you have a two tubes of X length. Light pulse of 1 second enters tube one and you accelerate it to only 200% of the speed of light. A pulse of light for one second triggers a switch to send a pulse back down the second tube, this time at 300% the speed of light, telling another switch there NOT to send the first pulse of light, or to at least stop it short of the 1 second length needed to send the second pulse.

Now, I know the mechanical or optical or whatever switches would be the slowest links in the chain. But you make the length of the tubes (X) long enough to compensate for the mechanical delays, since even at 200% the speed of light it takes time to get there.

I realize I am missing something someplace, I just need it put into some simple terms. Well, as simple as they can be I guess.

CandyMan

I can’t see the link you posted (because I refuse to register with the NYT) but I’m guessing it’s about the tunnelling effects they’ve been testing to speed light up.
{WAG}
I’m pretty sure the reason you can’t send FTL information with them is that these are quantum effects, and thus are based on wave-functions and probabilities. When they send the light pulses through the apparatus, the leading edge of the wave tunnels through the barrier almost instantaneously, and it’s the leading edge measurements that give those FTL travel times. The average place the photon will be detected, though, is not at the leading edge of the wave-function (which tends to look like a bell curve, high probability in the middle dropping off sharply to the edges) but at the center of the wave packet, a tightly packed series of waves of similar frequency travelling together. So while they’ve changed the wave velocity of the light (moving the wafefront forward) they haven’t changed the packet velocity (the average travel time for the photon) – they’re just spreading the wavefunction out over more area, smooshing the bell curve flatter. So while there’s a better chance of detecting the photon AHEAD of the expected arrival time, the average should remain unchanged, with the photon sometimes arriving slower than light. So you can see, it’s all about quantum uncertainty – in a sense, there’s ALWAYS a chance of detecting a photon sooner than one would expect given the speed of light. These scientists have just found a way to encourage that kind of thing.
{/WAG}

Corrections welcome =)

Actually the post says you will see the trailing edge of the pulse leave the tube before the trailing edge enters the other side.

That part was what made me post here.

CandyMan

Remember, though, that the location of the ‘trailing edge’ of the pulse is determined, not as it actualy leaves the apparatus, but is inferred from a distribution of photons hitting the detector. So again, there is the possibility of parts of the pulse exceeding light speed, while the pulse on average travels as expected.