Time Dilation in SR: Exploring Clocks & Perception

[Biker]

I don't really understand how time dilation works in SR.
I just know this that "Now" in one reference frame is not the same in another so for example when Observer A says his time is 10s then at the same moment he will say Observer B's time who is moving relative to him is 5s for example.
The same applied to B if he says his time is 5 then A's time is 2.5 s.

But this just doesn't "Tick" in my head. Suppose we have A and B who synchronize there clocks when they pass each other. A sees himself stationary and puts a flag some distance away from him. He measures the time it takes for B to reach that and says it is for example 10s , He then says it has to be 5 secs on B's clock.
From B's perspective, Distance is shorter so he does take 5 secs to complete the journey (?) however he says that A's clock should measure 2.5 secs. Doesn't this contradicts itself? Say both of them have a very large telescope what would they see on each other's clock ?

 

[Ibix]

As well as length contraction and time dilation, there is a third and much more important effect called the relativity of simultaneity. What the two frames mean by "at the same time as B passes the flag" is different. So there isn't a contradiction - it's just that natural language hasn't the concepts to express that clearly.

The phenomenon is closely related to a simple Euclidean analogy. Put a toy car on a table facing North. Put another due East of it, facing North East. According to the first car, the second one is beside it. According to the second car, the other one is way behind it.

In relativity it turns out that "at the same time as me" is similar to "beside me" in space - it means different things to different people.

 

[Nugatory]

From B's perspective, Distance is shorter so he does take 5 secs to complete the journey (?) however he says that A's clock should measure 2.5 secs. Doesn't this contradicts itself? Say both of them have a very large telescope what would they see on each other's clock ?

Using coordinates in which A is at rest, A's clock reads 10 seconds at the same time that B's clock reads 5 seconds: the events "A's clock reads 10" and "B's clock reads 5" are simultaneous in the frame in which A is at rest. Thus in that frame B's clock is running slow.

However, we have to remember the relativity of simultaneity. Using the frame in which B is at rest, these two events are not simultaneous. Instead, the event that happens at the same time as "B's clock reads 5" is "A's clock reads 2.5" so B concludes that A's clock is running slow.

In both cases, "at the same time" is allowing for light travel time. If something happens ten light-seconds away from me, and light from that event reaches my eyes and I see it when my clock reads 15 seconds, I know that it actually happened when my clock read 5. (This is no different than knowing that an airliner that spends an hour in flight and lands at 4:00 must have taken off at 3:00, and it's the only sensible definition of when a distant event happens). Thus, they aren't seeing the clocks running slow, they are calculating the clock rates from their observations and what they know of the light travel time.

As for what they see if they watch each other's clocks through telescopes? If they're approaching each other, they both see the other clock running fast, and if they're moving apart they both see the other clock running slow (Google for "Doppler effect", and remember that the distance and hence the light travel time is continually changing). Only when they allow for the fact that whatever they see in their telescopes happened sometime in the past because of the light travel delay does the time dilation become apparent.

 

[Biker]

I have been thinking about what you both said. I would ask you please to check if this is correct.

Assume lorentz factor is 2

Say Observer A knows he will unfortunately explode when his clock reads 5 secs and he is moving relative to observer B. When both of them were at the same point they agreed that Observer B will wave to observer A when observer B's clock reads 1 sec and that B will wave again at 10 secs in B's clock

So if we look from A's perspective, When B's clock reads 1, A's clock reads 2. It will take some time for the light to travel and reach A to see the wave. A explodes when his clock reads 5, He sees B's clock read 2.5 so from his perspective, He will die before B does his 2nd wave.

B will wave at A when his clock reads 1 and he will calculate that at this time A's clock reads 0.5. Then he waits to 10 secs and wave again but he knows that he did his wave as A exploded.Another example:

Say observer B decides to do his 2nd wave in such time t so that it will reach A at time 10 according to his clock. So from B's perspective he thinks A will see the two waves. A does his calculation and agrees that he will see the two waves the 2nd wave will reach him as he explodes however B's clock wasnt at 10 when he did but it was 2.5

Is all of this correct?

 

[Nugatory]

Assume lorentz factor is 2

...meaning that their relative speed is about .87c

When both of them were at the same point they agreed that Observer B will wave to observer A when observer B's clock reads 1 sec and that B will wave again at 10 secs in B's clock

...and also zero their clocks, right? Subsequent discussion assumes they did.

So if we look from A's perspective, When B's clock reads 1, A's clock reads 2. It will take some time for the light to travel and reach A to see the wave.

about 1.74 seconds because B has been moving away at .87c for two second so is 1.74 light-seconds away at that time (all using A's frame of course). Thus A sees the first wave when his clock reads 3.74 seconds, while he is still alive.

A explodes when his clock reads 5, He sees B's clock read 2.5 so from his perspective, He will die before B does his 2nd wave.

Yes, A explodes when his clock reads 5. No, he does not see B's clock read 2.5, because the light from that event leaves B at the same time (still using A's frame) that A's clock reads 5 and he explodes, so he he's dead before the light can get to him. Yes, he certainly doesn't see the second wave.

B will wave at A when his clock reads 1 and he will calculate that at this time A's clock reads 0.5.

Yes, and at that time A is .87 light-seconds away. However, A is moving away from him, so it will take more than .87 seconds for the light from the wave to get to A... in fact, the light will get to A when A's clock reads 3.74 seconds (same result as in A's frame, because "light reached A when his clock read 3.74" is one event).

Another example:
Say observer B decides to do his 2nd wave in such time t so that it will reach A at time 10 according to his clock. So from B's perspective he thinks A will see the two waves. A does his calculation and agrees that he will see the two waves the 2nd wave will reach him as he explodes however B's clock wasnt at 10 when he did but it was 2.5

Using A's definition of "at the same time", B's clock does read 2.5 at the same that A explodes. Using B's definition of "at the same time" the explosion happens at the same time that B's clock reads 10. B will have to wave well before his clock reads 2.5 if he want A to see the wave.

An aside: When you're constructing thought experiments like these, you can often make the arithmetic easier by choosing $v=3/5(.6)$ or $v=4/5(.8)$. The gamma factors then come out to 4/5 or 3/5 and you can do the arithmetic exactly in your head without rounding.

[wise people will approach my calculations above with some caution - I've already corrected several arithmetic errors]

 

[Biker]

Brilliant, That cleared things so much for me. Thank you Nugatory.