The discussion about two observers misses the point of the OP I believe. I interpreted the question as one observer as in
#1 I send light to a reflector 30 light-seconds away. It is impossible for me to see it before 1 minute has elapsed.
#2 I send light to a reflector 30 light-seconds away. It is impossible for me to see it before I sent it.
Are these impossible for the same reason?
And my question
#3 I send out a tachyon burst. Will that result in #1 or #2?
It’s also not something that grade-school students can measure. Even laser beams spread out some, enough that over those distances, you need to start with a really big laser and then use a very good telescope to see the reflection.
But these are irrelevant details to this thread.
Aiming laser beams at the moon has a long and storied history of going badly awry: Laser Pointer - What If - xkcd.
For the tachyon case, every observer traveling slower than light will see 1) or 2).
In special relativity, the combined distance (in the Minkowski metric) makes it so that any two events (an event is “something that happens at a particular time and place”), like "X emits a particle at point a and time a’ " or "Y detects the particle at point b and time b’ " can stand in three possible relationships: 1) the two events can have a light-like separation (light emitted at point a at time a’ would get to point b at time b’ (or vice versa)), 2) the two events can have a time-like separation (light emitted at point a at time a’ arrives at at point b before time b’, or 3) the two events can have a space-like separation (light emitted at point a at time a’ arrives at point b after time b’). And according to special relativity, all observers traveling slower than light will classify the relationship between the two events the same way. So 1) and 2) are really the same - in either case, a signal is getting between two space-like separated events - which is highly unusual
Yes, I don’t remember much from my Special Relativity class in college, but I do remember one thing: simultaneity is relative.
Again, not talking about any observer. Talking about the observer at rest relative to the tachyons at the source. Would that be #1 or #2?
Sorry about not answering before. I think #1 is the answer. You see the particles come back remarkably quickly, but not before you sent them. But it’s possible I’m missing something.
Speaking as the OP, I’m definitely missing something: wouldn’t it be a tachyon if the ‘observer at the source’ hypothetically beams something at the moon that’s a little faster than light, and so then sees the reflection a mere two seconds later or one second later instead of roughly two-and-a-half seconds later?
And wouldn’t it also be a tachyon — which is to say, wouldn’t it equally be a tachyon — in a wacky hypothetical where it’s moving so fast that said observer sees the reflection one or two seconds before he beams it at the moon?
I think (but I’m not sure) that for this case, even an infinite speed only gets a zero delay, not a negative one, because the two points in space in question are the same spot in the observer’s frame.
Mostly. If two things happen in the same place and at the same time, then all observers will agree about this. Though that can really only happen if it’s actually the same event.
If two events have timelike separation–that is, they’re close enough that one can influence the other (B happens one second after A, but is less than 1 light-second away), then all observers will agree about which event happened first, though they won’t agree on the duration or distance.
This was exactly what I thought of when I read the OP.