Physically, you math guys aren't allowed to cross near the boundary $c$ (speed of light). Special Relativity does that. SR says that it would be impossible for a particle to be accelerated to $c$ because the speed of light (maximum possible measured velocity) is constant in vacuum for all inertial observers (i.e.) Observers in all inertial frames would measure the same value for $c$. Not only the fact that infinite energies are required to accelerate objects to speed of light, (but) an observer would see things going crazy around the guy (or an object) traveling at $c$ such as length contraction (length would be contracted to zero), time dilation (time would freeze around him) & infinite mass. You can't enjoy anything when you travel at $c$. But, the stationary observer who's measuring your speed (relative to his frame) would definitely suffer..!
Note: But, there are some quantum mechanical solutions that allow negative masses like the expression for relativistic energy-momentum. Let's try not to make the subject more complicated.
$$E^2=p^2c^2+m^2c^4$$
There are hypothetical particles (having negative mass squared (or) imaginary mass) always traveling faster than the speed of light called Tachyon. This was assumed by Physicists in order to investigate the faster than light case. So When $v>c$, the denominator becomes a imaginary. But, Energy is an observable. It should be some integer. A consistent theory could be made if their mass is made to be imaginary and Energy to be negative. Using these data in the E-p relation, we would arrive at a point $p^2-E^2=m^2$, where $m$ is real. This makes Tachyons behave a kind of opposite to that of ordinary particles. When they gain energy, their momentum decreases (which strongly disproves all our assumptions).
The first reason that this investigation blown off is Cherenkov radiation where particles traveling faster than light emit this kind of radiation. As far as now, No such radiation has been observed in vacuum proving the existence of these..! It's like making a pencil to stand at its graphite tip. If it would stand, physicists would've to blow up their heads :-)
There are tougher stories on the topic when you Google it out...
(There's a couple of these questions kicking around, but I didn't see anyone give the "two boosted copies" answer. Generically, I'd say that's the right answer, since it gives an actual causality violation.)
In your scenario, the two planets remain a hundred thousand light years apart. The fact is, you won't get any actual causality violations with FTL that way. The trouble comes if the two planets are moving away from each other. So, let's say that your warp drive travels at ten times the speed of light. Except if the two endpoints of the trip are moving, then what does that mean? Ten times the speed of light relative to which end?
Let's say Tralfamadore is moving at a steady 20% of $c$ (the speed of light), away from Earth. (So, Earth is moving at a steady 20% of $c$ away from Tralfamadore.)
If I leave Tralfamadore (in the direction of Earth) and I am travelling at anything less than 20% of $c$ relative to Tralfamadore, then I am still moving away from Earth. I'll never get home.
Let's say instead I am travelling at 60% of $c$ relative to Tralfamadore. I will catch up to Earth. Relative to Earth, how fast am I approaching? You might guess the answer is 40% of $c$, but it's 45.45%.
Generally, the velocity subtraction formula of relativity is: $$w = (u-v)/(1-uv/c^2)$$
Let's say instead I am travelling at 100% of $c$ relative to Tralfamadore. Plug $u=c, v=0.2c$ into the formula and get $w=c$. Relative to Earth, I am approaching at 100% of $c$! The speed of light is the same for everyone.
So finally, let's say instead I am using your warp drive to travel at 1000% of $c$ relative to Tralfamadore. Relative to Earth, I am approaching at -980% of $c$. In Earth's reference frame, I will arrive on Earth before I leave Tralfamadore. Now you may say this in itself isn't a causality violation, because we've applied Earth's calendar to Tralfamadore. And that's true, but I'll make a round trip:
- In the futuristic Earth year of 3000, Tralfamadore is 98,000 light years away, and receding at 20% of $c$. I leave Earth at 1000% of $c$, relative to Earth.
- In Earth year 13000 Tralfamadore is 100,000 light years away, and I catch up to it. I turn around and leave Tralfamadore at 1000% of $c$, relative to Tralfamadore.
- In Earth year 2796, I arrive home.
Earth's calendar certainly applies to Earth, and I arrived home two centuries before I left. No two ways about it, I'm a time traveller!
There is nothing special about ten times the speed of light. Given a warp drive that moves a certain amount faster than light, you can make the above time machine using two endpoints that are moving apart a certain amount slower than light, provided that the warp drive can move faster than light relative to either end. This time machine works for any form of FTL: tachyons, warp drives, wormholes, what have you.
Best Answer
The conclusion derives from looking at how tachyon signals would behave as seen in slower-than-light inertial frames, not from trying to consider a tachyon's own "time" (if you can call it that, since a tachyon's worldline would have to be space-like, not time-like)--basically, it's a consequence of the relativity of simultaneity. It can be shown that any signal that moves even slightly faster than light in one frame would move instantaneously in some other inertial frame, and if the first postulate of SR applies to tachyons, then if it's possible to send a message instantaneously in one frame, this must be possible in all inertial frames. Likewise, if a signal travels instantaneously in one frame, there must be some other inertial frame where it actually travels backwards in time (i.e. the event of the signal being received occurs before the event of it being sent), and if that's possible in any frame, it must be possible in all frames too.
If you're familiar with the basics of SR spacetime diagrams and how a surface of simultaneity in one frame looks tilted in other frames, then you could take a look at the helpful explanation with diagrams on this page, which shows how the ability of two different slower-than-light observers to send signals that move instantaneously as seen in their own frames implies that they can bounce a message back and forth, and in each one's frame the other one's signal is going backwards in time, so that the message gets returned to the original sender at an earlier point on his worldline than the point where he sent the message, a clear causality violation in all frames. You could also take a look at the tachyonic antitelephone article which goes into more detail, with equations and a numerical example.
Note that this answer is really about why the ability to send FTL signals would violate causality--as other answers have noted, in quantum field theories there could be tachyons that would be impossible to use for transmitting information, in which case no causality violation would occur (the last section of this article has a helpful discussion about why tachyons in QFT wouldn't be usable for information transmission).