Comparing the clocks via synchronization

[Mike_bb]

Hello, PF!

I encountered with problem when I tried to understand following explanation about comparing the clocks. (See below)

Given:

We have two synchronized clocks at rest relative to each other, and one moving clock flying past them.

Let us consider two reference frames: Earth (observer A) and the rocket (observer B).

Scenario 1: Measurements are made by Earth (Observer A)

1. Two clocks—Clock 1 and Clock 2—are positioned on Earth along the rocket's trajectory. They have been synchronized with each other
beforehand.
2. The rocket flies past Clock 1. At that moment, Observer A records the readings of the rocket's clock and Clock 1.
3. The rocket continues its flight and passes Clock 2. Observer A again records the readings of the rocket's clock and Clock 2.
4. Result: Observer A compares the difference between the readings of Clock 1 and Clock 2 with the time elapsed on the rocket's clock. He observes that less time has elapsed on the rocket's clock.

Scenario 2: Measurements taken by the rocket (Observer B)

To the observer in the rocket, things look different. The rocket is stationary, and it is the Earth that is flying past it.

1. From B’s perspective, Earth’s Clock 1 first approaches his rocket. He records the time.
2. Then Clock 1 moves away, and Earth’s Clock 2 approaches in its place. He records the time again.
3. The result: Observer B compares the readings of his single clock—taken at two different moments—with two different Earth clocks. And from his perspective, Earth’s Clock 1 and Clock 2 were not synchronized! Due to the relativity of simultaneity, Clock 2 was initially "ahead." Therefore, when B subtracts the Earth clock readings, it appears to him that more time has passed on Earth, and that his own clock is lagging behind the Earth clocks.

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From Scenario 1 we have $\Delta t_{A1} = Clock 2-Clock 1$. This difference between Clock 2 and Clock 1 is more than difference on rocket clock. Rocket clock is slowed ($\Delta t_{B1} < \Delta t_{A1}$).

From Scenario 2 we have $\Delta t_{A2} = Clock 2-Clock 1$. This difference between Clock 2 and Clock 1 is more than difference on rocket clock. Rocket clock is slowed ($\Delta t_{B2} < \Delta t_{A2}$).

In accordance with the Scenario 1 clock of Observer B is slowed relative to Observer A ($\Delta t_{B1} < \Delta t_{A1}$) but otherwise clock of Observer A is slowed relative to Observer B.
Nevertheless, in accordance with the Scenario 2 clock of Observer B is slowed relative to Observer A ($\Delta t_{B2} < \Delta t_{A2}$).

But I can't understand how is it possible if in accordance with the Scenario 1 Observer B is slowed relative to Observer A ($\Delta t_{B1} < \Delta t_{A1}$) and otherwise (Observer A is slowed relative to Observer B) then we have that from Scenario 2 Observer B read $\Delta t_{B2}$ instead of $\Delta t_{A1}$?

Thanks!

 

[Sagittarius A-Star]

But I can't understand how is it possible if in accordance with the Scenario 1 Observer B is slowed relative to Observer A ($\Delta t_{B1} < \Delta t_{A1}$) and otherwise (Observer A is slowed relative to Observer B)

The explanation is:

And from his perspective, Earth’s Clock 1 and Clock 2 were not synchronized! Due to the relativity of simultaneity, Clock 2 was initially "ahead."

 

[Mike_bb]

Sagittarius A-Star,

In accordance with the principle of relativity: if Observer B is slowed relative to Observer A then otherwise Observer A is slowed relative to Observer B.

 

[Sagittarius A-Star]

Sagittarius A-Star,

In accordance with the principle of relativity: if Observer B is slowed relative to Observer A then otherwise Observer A is slowed relative to Observer B.

Yes. But in the scenario is an asymmetry: 1 clock of observer B is compared to 2 clocks of observer A.

 

[FactChecker]

To the observer in the rocket, things look different. The rocket is stationary, and it is the Earth that is flying past it.

1. From B’s perspective, Earth’s Clock 1 first approaches his rocket. He records the time.
2. Then Clock 1 moves away, and Earth’s Clock 2 approaches in its place. He records the time again.
3. The result: Observer B compares the readings of his single clock—taken at two different moments—with two different Earth clocks. And from his perspective, Earth’s Clock 1 and Clock 2 were not synchronized! Due to the relativity of simultaneity, Clock 2 was initially "ahead."

Correct.

Therefore, when B subtracts the Earth clock readings, it appears to him that more time has passed on Earth,

But he thinks the Earth clocks were not properly synchronized, so why would he use them to judge the passage of time on Earth? In the IRF of the rocket, a good way to judge the passage of time on the Earth, from the rocket perspective, would be to mimic Scenario 1 with two of his own synchronized clocks, separated along the path of an Earth clock.

Where did this explanation come from?

 

[Mike_bb]

FactChecker, Google AI

Why can we conclude that " it appears to him that more time has passed on Earth," from "due to the relativity of simultaneity, Clock 2 was initially "ahead.""?

 

[Histspec]

FactChecker, Google AI

Why can we conclude that " it appears to him that more time has passed on Earth," from "due to the relativity of simultaneity, Clock 2 was initially "ahead.""?

You may have a look at the images in the German Wikipedia that precisely describe, and solve, your scenarios:

In scenario 1 (rest frame of Earth), there is a rocket C traveling between synchronized clocks A and B. When C reaches B the clocks were advanced by $\Delta C=2$ and $\Delta B=3$, thus $$\Delta C < \Delta B$$.

In scenario 2 (rest frame of rocket), A and B are not synchronized because A=0 and B=1.7 at the start due to relativity of simultaneity. When B reaches C they again indicate C=2 and B=3. Yet we know that B=1.7 at the start, so B only advanced by $\Delta B=3-1.7=1.3$ during motion while C advanced by $\Delta C=2$, thus $$\Delta B < \Delta C$$.

 

[Mike_bb]

Histspec, Thanks! In your example $\Delta C < \Delta B$ and $\Delta B < \Delta C$ but in explanation it was written that $\Delta C < \Delta B$ in both scenario.

 

[Histspec]

Histspec, Thanks! In your example $\Delta C < \Delta B$ and $\Delta B < \Delta C$ but in explanation it was written that $\Delta C < \Delta B$ in both scenario.

As you can see on the images, in both scenarios the clock indications of B and C when they meet are B=3 and C=2, thus $$C=2<B=3$$
But time dilation is about intervals, which gives (as explained above) $$\Delta C=2 < \Delta B=3$$ in scenario 1, and $$\Delta B=1.3 < \Delta C=2$$ in scenario 2 .

 

[Mike_bb]

As you can see on the images, in both scenarios the clock indications of B and C when they meet are B=3 and C=2, thus $$C=2<B=3$$
But time dilation is about intervals, which gives (as explained above) $$\Delta C=2 < \Delta B=3$$ in scenario 1, and $$\Delta B=1.3 < \Delta C=2$$ in scenario 2 .

It's ok if it's so!

 

[Mike_bb]

Histspec, Why "When B reaches C they again indicate C=2 and B=3"?

 

[Histspec]

Histspec, Why "When B reaches C they again indicate C=2 and B=3"?

In Scenario 2, clock C indicates 0 at the start, advances by interval 2 during motion and therefore indicates time C=2 when it meets B; and clock B indicates 1.7 at the start and advances by interval 1.3 during motion, thus B indicates B=1.7+1.3=3 when it meets C. What's the problem?

 

[Sagittarius A-Star]

Why "When B reaches C they again indicate C=2 and B=3"?

Both "scenarios" describe the same events:

1. Meeting $(C=0, A=0)$
2. Meeting $(C=2, B=3)$

 

[berkeman]

@FactChecker, Google AI

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[PeterDonis]

Google AI

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