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I have some doubts about getting the time-dilation concept right, so I would like to suggest a scenario and see if my conclusions are right or wrong.

Person A is on Earth, Person B is on a spaceship. Both are on Earth at $t=0$. Person B leaves Earth and heads for planet X. When Person B reaches planet X his clock reads $t_B$, and person A's clock reads $t_A$.

Now, simplified time dilation formula is $$t=\frac{t_0}{\sqrt{1-\frac{v^2}{c^2}}} $$ where $t_0$ is the proper time.

My thoughts on this:

Person B's time is the only proper time because only in his frame the two events (leaving Earth/arriving at planet X) happened in the same place - they have the same $x$-coordinate. So, by this formula, Person A measures longer time compared to Person B's clock.

Now, if my thoughts are right (?) I am troubled with this statement:

Moving clocks run slower.

From Person B's perspective, Person A is moving. But, he does not see this clock run slower. What am I concluding wrong? Should all clocks moving relative to us run slower? Is there a situation in which Person B (moving away from Earth) would see Person A's clock run slower? Maybe when looking at an event that happened and ended on Earth?

Sorry about many questions, but this all bugs me a lot.

I kind of think that all problems with conceptually understanding dilation arise when comparing one single event across two frames, if the event happened and ended only in one of these frames.

Maybe this statement has sense only when comparing two events: one in frame A, and the same event (like a heartbeat) in frame B?

Help me wrap this up!

Jakov
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2 Answers2

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If person A on Earth looks at his own clock A', he sees his proper time, because he is at rest with respect to his own clock. Similarly, if person B on the rocket looks at his own clock B', he sees his proper time as well. But if person A looks at the other guy's clock B', he will see it being slower, because it is moving with respect to him. Similarly, if person B on the rocket looks at the clock A' on Earth, he will see it being slower as well.

Photon
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So $A$'s clock is going to tick off a time of

$$t_A=t_0$$

which is the proper time between the two relevant events because $A$ is at rest.

As $B$ approaches the planet, he has been moving at $\beta$ in the unprimed frame, so his clock approaches:

$$ t'_B = \frac{t_A}{\gamma} $$

where $\gamma$ is the Lorentz factor.

Since $B$ is at rest in the primed frame, $B$ sees $A$'s clock running slow, and thus says:

$$ t'_A=\frac{t'_B}{\gamma}=\frac{t_A}{\gamma^2} $$

Note that $B$ is still moving w.r.t. to $A$.

At the planet, $B$ stops and is now reenters the unprimed frame of $A$. Their clocks now both run at the same speed, and they both agree on the simultaneity of events. $B$'s clock reads $t_0/\gamma$, while $A's$ clock reads $t_0$.

JEB
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