0
$\begingroup$

This question is an exact duplicate of:

I recently posted a version of the twin paradox with a twist but here I'd like to propose a new thought experiment all together. Take two clocks, A & B, and place A 1 AU away from B. Once the clocks are separated and stationary relative to each other sync them up again; to do so send a light pulse from A to B, and when the pulse reaches B the clock will start ticking while A waits the exact amount of time it takes light to reach B to start ticking. The idea is to have two clocks that are in sync but separated by a large distance, and I imagine there are multiple ways to do this. Now take a third clock, clock C, which is also synced with A and B and have it travel the distance between A and B at .866c so that time runs at 1/2 the rate for clock C. However to Clock C, A & B would be running at half speed. When it arrived at B though we'd be able to compare the elapsed times and say for sure whether C or A & B had in fact been running slower, even though special relativity dictates both viewed each other as running slower and both are correct. The only decent explanation I can think of is that for C, simultaneity between A & B is broken rendering B's readings inaccurate, but that still leaves us in a very bizarre place as ultimately to C, B should read half the time and to B, C should read half the time.

*Note: Please don't try and use acceleration to explain the disparity, it is a common misconception that acceleration resolves the twin paradox when its actually that acceleration is used to confirm the change in reference frames. To avoid that conundrum all together you can imagine that clock C was already up to speed when it passed A and synced itself up with A as it passed C. Furthermore we can imagine C never actually slowed down when it passed B, rather it just took the reading from B to compare. This way C is in one inertial frame for the entire duration of the experiment.

$\endgroup$

marked as duplicate by Ali, Kyle Kanos, Brandon Enright, John Rennie, Danu Jul 4 '14 at 18:37

This question was marked as an exact duplicate of an existing question.

  • 2
    $\begingroup$ i.stack.imgur.com/qzMK3.png $\endgroup$ – pfnuesel Jul 4 '14 at 15:59
  • $\begingroup$ possible duplicate of Twin Paradox Without Acceleration $\endgroup$ – pfnuesel Jul 4 '14 at 16:01
  • $\begingroup$ Haha I've seen that before $\endgroup$ – Krel Jul 4 '14 at 16:01
  • $\begingroup$ It is a duplicate I'm pretty confused by both circumstances. Ultimately I'm trying to correct my intuition about how time dilates $\endgroup$ – Krel Jul 4 '14 at 16:03
2
$\begingroup$

You are still trying to use intuition about space-time that is simply wrong. Experimentally wrong. The world doesn't work the way you (and I) think it does. Spatially separated clocks can not be synchronized in all frames. No, really. They can't.

If you synchronize clocks A & B in their common rest frame (which can be done) then they are unavoidable unsynchronized in C's frame. This is not just an effect of the delayed signal, it is a real change in the time after correcting for the transmission delay.

You are, in fact still testing the same old thing, but now you should be tracking the proper time of two "message"s. One is just the message that $A$ writes to himself and keeps handy. The other I'll define that as

  1. The synchronization signal from A to B (proper time is zero if you used a light)
  2. Possibly a delay while B sits ready for C to pass. (proper time positive)
  3. The physical return journey of C from B to A (proper time positive)

You them compare this to the time experienced at A between sending the synchronization message in step 1 and C arriving. This time will always be longer than the time experienced by the message for the same reason that the traveling twin arrives home younger in the classic version of the "paradox".

Because it is the same "paradox".

$\endgroup$
1
$\begingroup$

Now take a third clock, clock C, which is also synced with A and B and have it travel the distance between A and B at .866c

That's impossible.

Let clock C, traveling at $0.866c$, pass by clock A just when both clocks read $t_C = t_A = 0$.

Now, according to clock A, clock B also reads $t_B = 0$ at this instant.

But, according to clock C, clock B does not read $t_B = 0$ at this instant. That is to say, according to clock C, clocks A & B are not synchronized.

Indeed, according to clock C, clock B is ahead of clock A. According to clock C, when $t_C = t_A = 0$,

$$t_B = \frac{0.866 \cdot 1 AU}{c}$$

This would all be quite plain to see if you would draw a spacetime diagram.

$\endgroup$
0
$\begingroup$

After reading this and your other question I'm going to recommend you try to find a copy of this book, Modern Physics for Scientists and Engineers by John R. Taylor. It has great explanations on special relativity especially these weird ones.

The geometry of your problem is critical. When you say C trave;s the distance between A & B do you mean it runs the line directly between A & B? Let's assume so...

If C runs from A to B at 0.866c then relative to A & B, clock C runs at half time. At the instant C passes B, it's clock would have read half the time that A & B did. Because A & B are stationary and synced (inertial frame of reference), their clocks can be considered the "real" time of the problem.

If you want to look at clock C then the problem changes. Now you have Clock C stationary with A & B moving past C (at the same speed that C had been moving before). A&B's clock will have half the time that C measured for the time it took from A to leave C and B to arrive.

For these problems you HAVE to pick a single inertial frame. If you try to compare them from both sides the problem makes to sense.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.