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14 November 2007 @ 07:55 pm
 
Okay, jon_leonard gave me an answer to the first relativity question, but i'm still a little confused, so here's a simpler question.

Thought experiment 2:

Same basic setup as before. Ships A and B pass each other each in their own inertial frame but at a relative velocity of 0.866 c.

Both ships send out an encoded time signal by laser burst every second towards the other ship.

Each ship will see via this signal the other ships's clock advancing one second for every two seconds of their own time, correct? At least they will when they account for the distance the signals had to travel.

After traveling time.. let's say 1.866 t, they'll each be receiving signals that were sent time t ago, correct? (Sent at time/distance t, and in the time it took those signals to travel distance t at c the ships have moved another 0.866t, right?)

The time signal they are getting will claim it's from time t/2 though, right? Or am i thinking about this wrong?

Now, because these are magic thought experiment ships they have more reaction mass than god and inertial compensators to match. Ship A accelerates up to 0.866c to bring themselves to rest in relation to Ship B in the space of about half a second.

So now... the distance between them is cut in half so the signals start coming twice as often?

But the important issue is that just before the course change they received a signal saying "it is now time x:y:z." They are now going to be getting a signal saying it is now time x:y:z + 1" Being the ones that made the course change should have put them in the "we're younger than Ship B" frame, but nothing seems to have changed. Earlier evidence would have told both ships that they other one was aging half as fast, but now they both see the other as aging at the same rate and although the rate at which they're getting signals has changed they haven't suddenly switched to a new time so what's the discrepancy?

(Where did i put my Helliwell book??? =P)
 
 
 
jon_leonardjon_leonard on November 15th, 2007 04:40 am (UTC)
After traveling time.. let's say 1.866 t, they'll each be receiving signals that were sent time t ago, correct? (Sent at time/distance t, and in the time it took those signals to travel distance t at c the ships have moved another 0.866t, right?)

No; You need to use the Lorentz transform (see, for example, Wikipedia). Also, you need to be careful to talk about events from some reference frame or another. In Newtonian physics it doesn't matter so much, but if you omit that part in a relativistic question, you'll get nonsense. When you talk about "sent time t ago", you must specify which frame, and you get different lengths of time depending on which one you pick. Furthermore, when you accelerate, your answer for 'how long ago was that?' changes. (And changes in a way that's rather at odds with ordinary intuition.)

(Where did i put my Helliwell book??? =P)

That's probably the most relevant question, yes.

DonAithnendonaithnen on November 15th, 2007 07:06 am (UTC)
Sent at time t/2 in ship A's reference frame.

As long as neither one of them is accelerating we shouldn't need anything as complicated as the lorentz transform should we? It's just "straightforward" time dilation. Everything moving at 0.866 is at double or half the length, and time is progressing at half the normal rate (from the point of view of a "stationary" observer of course.)

You're certainly right about what happens at the point of acceleration being contrary to ordinary intuition though!

Let's see, if you have an observer in the middle where they both pass each other, and it sees them each heading off at 0.433 c in opposite directions the time dilation factor will be... 0.9?

So after 5 years of travel in ship c's time they'll be 4.33 LY apart in its time, and according to it their clocks will read 4.5 years.

So according to Ship A ship B will be 4.5 * 0.866 LY away, so 3.897. That matches the 0.9 space dilation from ship C's point of view.

Ship A will think Ship's B clock is at 2.25 Y.

From Ship C's perspective the signal reach Ship A right at that point will have been fired from Ship B at... damn, i hate these catching up type problems, i always forget the formula =P

if it was fired at time t, and hits at 5Y by Ship C's clock, it's total traveled distance t*0.433 + 5 * 0.433, so d = 0.433t + 2.165, since it's a light beam, the total distance will also equal 5-t (in ship C's frame) so 5-t = 0.433t + 2.165, 1.433 t = 2.835, t = 1.978.

Again, from C's perspective, B fired it's signal at 1.978 years, at distance 0.856 LY, it passed C at 2.834 LY, it will then cross another 2.166 LY in 2.166 Y to hit ship A at 5 Y.

C thinks that B fired the beam at B's time 1.780, actually everyone will think that, cause that better be what the timestamp on the beam says. C thinks the beam hit A at A's time 4.5.

So A sees the 1.780 timestamp, and thinks the beam was fired at 3.560 their time. At 3.560 their time ship B would have traveled a distance 3.083... wait, that's not working.

If A gets the signal at 4.5 their time, they're going to think it left B at t = 4.5 - d in A's time, and the total distance traveled will be 0.866t, so t = 4.5 - (0.866t), 1.866t = 4.5, t = 2.41. At time 2.41 Ship B will have traveled 2.09 LY, so it will take 4.5 Y for the light signal to arrive, check.

Damn, two answers that are internally consistent, but which i can't get to relate to each other, clearly i'm doing something very wrong. I think i'm gonna get back to this later, perhaps when i'm more rested.
Andrewneonelephant on November 17th, 2007 06:06 am (UTC)
Let's see, if you have an observer in the middle where they both pass each other, and it sees them each heading off at 0.433 c in opposite directions the time dilation factor will be... 0.9?

I'm relatively certain that a stationary observer would see no such thing. Recall that if they're both heading towards the stationary observer, from opposite directions, at .9c, then from ship A's perspective ship B is not approaching at 1.8c....
DonAithnendonaithnen on November 17th, 2007 07:18 pm (UTC)
Doh! Good point! =P
Vespera: Umm...ooh-kaysol_rei on November 16th, 2007 03:01 am (UTC)
I wish my brain functioned that way. @_@