Relative Speed Impact on Synchronization: Clock Comparison

[cronnin]

If two clocks start from the same point, and one travels double the speed of another, will they be showing the same time when they get together at the destination point.

Obviously, faster clock will arrive first, and will have to wait for the slower one to arrive.

 

[Ibix]

I don't believe that this is correct. Please show your working.

 

[russ_watters]

If two clocks start from the same point, and one travels double the speed of another, will they be showing the same time when they get together at the destination point.

They will not.

 

[phinds]

If two clocks start from the same point, and one travels double the speed of another, will they be showing the same time when they get together at the destination point.

Obviously, faster clock will arrive first, and will have to wait for the slower one to arrive.

What you seem to not be taking into consideration is the acceleration one of them will have to undergo in order for them to meet up again. This makes your statement simply a variation on the twin paradox which you should study if you have not already

 

[MarekKuzmicki]

When you KNOW that one travels double the speed of another that means there is no symetry - there will be different times on clocks.

 

[cronnin]

OK, twins start from Earth, and go to the asteroid 10 LY away. Brother is traveling at 80% speed of light and sister is traveling at 40% speed of light.

gamma(brother) = 5/3
gamma(sister) = 1.0911

So, they will be arriving on asteroid after:

t(brother) = 6 years
t(sister) = 9.165

of their own time.

So, after 10 years on Earth, brother will arrive on asteroid, with his clock showing 6.0.
And, after 10 years on Earth, sister will arrive on asteroid, with her clock showing 9.165.

Now, although the same time has passed on Earth, that can’t be the same moment. After all brother was going twice faster and they can't both arrive at same moment. And that is when I get confused :)

I'm assuming Earth-asteroid system is not moving and is an IRF?

 

[cronnin]

What you seem to not be taking into consideration is the acceleration one of them will have to undergo in order for them to meet up again. This makes your statement simply a variation on the twin paradox which you should study if you have not already

I'm assuming they are both accelerating when they start from Earth, and decelerating when they arrive on asteroid.

It's more like a triplets problem, where one stays on Earth, and second and third are traveling but with different speeds. I'm assuming that the (Earth - first triplet - asteroid) constitute an inertial reference frame, and it would be another kind of problem if they both turned back to Earth. This one, I don't know how to digest.

 

[phinds]

So, after 10 years on Earth, brother will arrive on asteroid, with his clock showing 6.0.
And, after 10 years on Earth, sister will arrive on asteroid, with her clock showing 9.165.

I'm assuming they are both accelerating when they start from Earth, and decelerating when they arrive on asteroid.

Irrelevant. They have to accelerate/decelerate by different amounts

 

[cronnin]

Yes I know that, but where does it get them time wise.

At first all triplets are stationary and synchronized.
Then the two start accelerating compared to Earthling, but not to one another.
Then the second one stops accelerating, which leaves only the third one accelerating for some time.
Then two of them do the opposite.

 

[phinds]

So, after 10 years on Earth, brother will arrive on asteroid, with his clock showing 6.0.
And, after 10 years on Earth, sister will arrive on asteroid, with her clock showing 9.165.

This does not fit with the rest of your statements. If the target is 10LY away and you travel there at 40% of c, Earth will see your arrival after 35 years and at 80% of c, Earth will see the arrival after 22.5 years. I realize you are stating what will happen after 10 years on Earth but your statement is incorrect. Where are they really, after 10 years of Earth time? This does not require SR, just simple d=rt

 

[PeterDonis]

twins start from Earth, and go to the asteroid 10 LY away. Brother is traveling at 80% speed of light and sister is traveling at 40% speed of light.

Which means that, in the Earth's rest frame, brother will arrive at the asteroid in 10/0.8 = 12.5 years, while sister will arive in 10/0.4 = 25 years.

So, after 10 years on Earth, brother will arrive on asteroid, with his clock showing 6.0.
And, after 10 years on Earth, sister will arrive on asteroid, with her clock showing 9.165.

The "after 10 years on Earth" doesn't match the above for either brother or sister. Go back and recalculate your own scenario. (Your numbers for how much time elapses on brother's and sister's clocks are wrong too, because you calculated them from the incorrect "10 years on Earth".)

 

[cronnin]

Oh, sorry, my mistake. So, they should cover the distance of 10 light years. It was pretty much stupid of me to say that they travel slower than light and are there years before the light is :)

Brother is traveling at 80% speed of light, and he'll do it in 12.5 Earth's years.
Sister is traveling at 40% speed of light, and he'll do it in 25 Earth's years.

Now, brother arrives after 12.5 our years, and is left waiting for sister for the next 12.5 years. So, at the moment sister arrives, he should be seeing his 6 years of traveling plus 12.5 years of waiting, which is 18.5 years total.

At that moment sister arrives, and her clock is stating 9.165 years.
Just to note, she is younger, although she traveled slower. During the time he waited, he got older faster than her.
Am I correct now?

 

[PeterDonis]

at the moment sister arrives, he should be seeing his 6 years of travelling

No, he doesn't travel for 6 years by his clock. Go back and recalculate that since you now know he spends 12.5 Earth years traveling, not 10.

At that moment sister arrives, and her clock is stating 9.165 years.

No, it doesn't. Go back and recalculate sister's elapsed time on her clock during the trip, since you now know she spends 25 Earth years traveling, not 10.

she is younger, although she traveled slower

Not when you correct your numbers. Although you should note that "she traveled slower" (relative to Earth) is not true for the whole trip, because brother only spends part of his time traveling faster than her (relative to Earth); the rest of the time he spends at rest (relative to Earth) waiting for her to arrive.

 

[cronnin]

Oh, this is getting cumbersome, writing one thing on paper, different in Word, and third in here :)

Brother's time dilation factor is 5/3, which means that he'll see our 12.5 years as 7.5 years.
Sister's time dilation factor is 1.0911, which means that he'll see our 25 years as 22.91 years.

So at the moment sister arrives, brother will be 20 years older, and sister 22.91 years older than when they went in space.
At the same time Earthling will be 25 years older, but it will take him another 10 years to get the info about their age. If they decide to meet all three together, the difference will change again.

 

[PeterDonis]

at the moment sister arrives, brother will be 20 years older, and sister 22.91 years older than when they went in space.

Yes.

 

[cronnin]

Ok, now we're getting somewhere :)

But does this mean that you can never get synchronized clocks, because they go "desynced" as soon as you move some of them to different location? Or at least, if you want them to be in sync, you should calculate the dilation, and calculate it in before syncing.

So, if I want a clock in Japan to be in perfect sync with mine, I should set mine to noon, and the other one to 7 seconds later (number speculated), and then send that other in a high speed space vehicle to Japan?

 

[PeterDonis]

does this mean that you can never get synchronized clocks, because they go "desynced" as soon as you move some of them to different location?

No. Einstein clock synchronization works for spatially separated clocks. But the clocks do have to remain at rest relative to each other.

if I want a clock in Japan to be in perfect sync with mine, I should set mine to noon, and the other one to 7 seconds later (number speculated), and then send that other in a high speed space vehicle to Japan?

No, you should put the clocks where you want them to be, and then Einstein synchronize them.

 

[PAllen]

If two clocks start from the same point, and one travels double the speed of another, will they be showing the same time when they get together at the destination point.

Obviously, faster clock will arrive first, and will have to wait for the slower one to arrive.

I don’t know why all the hoopla about many answers to this. The problem seems perfectly well posed. The answer is, no they won’t show the same time, and the one that has to wait will read less time. I assume you mean accealerations and deceleration are instant, a common idealization in such problems, which ensure the acceleration phases make no contribution to the answer.

The result that the one that moves faster and waits is younger, trivially follows, without math, from the geodesic principle. The slower one is following an inertial path except for start and stop. The other is a bent path in spacetime (two legs of a triangle, with the slow path being the third side). In Minkowski spacetime, a non geodesic path always elapses less time than a geodesic between the same events.

Despite the title, I see nothing about synchronization in the question. It seems just a trivial variation on standard twin differential aging scenario.

 

[Nugatory]

What you seem to not be taking into consideration is the acceleration one of them will have to undergo in order for them to meet up again. This makes your statement simply a variation on the twin paradox which you should study if you have not already

The acceleration is a red herring, both in this problem and in the standard twin paradox problem. The easiest way of seeing this is to follow PAllen's lead:

accealerations and deceleration are instant, a common idealization in such problems, which ensure the acceleration phases make no contribution to the answer.

 

[cronnin]

Despite the title, I see nothing about synchronization in the question. It seems just a trivial variation on standard twin differential aging scenario.

My problem was/is with keeping synchronized clocks in sync when you're moving them.

For example (just one of the many), you and I synchronize clocks in Florida, and then go same distance but different speed. You go Concord to France, and I go same distance to Africa but in an ordinary airplane. Our clocks are no longer in sync, as is comes out.

Anyways, I've just learned from PeterDonis' last post that there is some form of convention about clock syncing, and I'm reading it.