Beginner question special relativity: How many clocks does each observer use when measuring simultaneity?

Question

I am reading a bit about special relativity and saw this picture in a book:

If I understand correctly, the author is using it to demonstrate that when we consider observer Alice at rest, she will measure a moving observer's (Bob) clocks to run slowly, but likewise Bob will observe Alice's clocks to run slowly, because we can just as well consider Bob to be at rest. And this has to do with different definitions of simultaneity.

So it is not the case that when Alice measures Bob's clocks to run slowly, Bob will measure Alice's to tick faster - they both measure each other's clocks to run slowly.

But I don't understand the explanation. The author says that Alice uses two clocks and compares them with only one of Bobs's clocks.

So there is one clock for Bob, moving from A to B (his time axis is $ \bar{t}$). but at event A, Alice has one clock and then another clock at event F. Alice's clock at A coincides with Bob's clock at A, and Alice's clock at F coincides with Bob's clock at B - from her perspective.

So Alice says that at A, both clocks read the same time, but at event B, her clock F will read a later time than Bob's clock at B.

And I don't understand that: Why is there only one clock for Bob moving from A to B, but two for Alice? Aren't there supposed to be clocks at every location in space in a spacetime-diagram? So aren't here a whole bunch of clocks on the $ \bar{t} $-axis?

Or from Bob's perspective: Alice's first clock ticks at event E - and that is what Bob sees as simultaneous to B, while he notices that F ticks later than A.

So again: Why does Alice have to use two of her clocks to compare to one of Bob's clocks and why doesn't Bob use two clocks in his analysis of the situation?

I think it would help me to have a real-world example of what type of event this diagram could represent.

Answer 1

I think clock counting is semantic confusion. What the author is trying to say is that Bob at $B$ is simultaneous with Alice at $E$, while Alice at $C$ considers herself simultaneous with Bob at $B$. (So we mentioned only Bob at $B$, while Alice is considered at $C$ and $E$).

In Alice's frame, she moves from $A$ to $C$ in time $t_{AC}$, while Bob moves to $x=\beta t_{AC}$. The length of $AB$ is then $(t_{AC}^2 + (\beta t_{AC})^2)^{\frac 1 2}$. However, in Minkowski space, transformed axes' tic marks are stretched by:

$$ \sqrt{\frac{1+\beta^2}{1-\beta^2}} $$

so Alice sees Bob's elapsed time as (Bob is on a primed frame, no bars):

$$ t'_{AB} = \frac{\sqrt{t_{AC}^2 + (\beta t_{AC})^2}}{\sqrt{\frac{1+\beta^2}{1-\beta^2}}} = t_{AC}/\gamma$$

Summary: Alice at $C$ sees Bob at $B$, and his clock is dilated by $\gamma$.

Bob at $B$ sees Alice at $E$. Since

$${\angle CEB} = \tan^{-1}\beta $$

you can show that:

$$ t_{AE} = t_{AC} / \gamma^2 = t'_{AB}/\gamma$$

That is, Bob sees Alice's clock dilated by $\gamma$.

Of course, boosting the diagram by $-\beta$ makes a symmetric situation where Bob's time axis is vertical and Alice is moving to the left.

Answer 2

Bob has to pass two clocks in two separate places in Alice's frame in order for the time dilation effect to be apparent. That's because the effect is caused by the fact that clocks in Alice's frame appear increasingly out of synch to Bob along his direction of travel- it is not caused by clocks in one frame running slower than clocks in another.

To see how that works, imagine you are at the start of a corridor, on the wall of which are clocks set 10m apart. Now let's suppose the clocks are out of synch with each other, so that each clock down the corridor is set one second ahead of the previous one. If you set your watch by the first clock, then start to walk down the corridor at a metre per second, what happens? When you come to the next clock along, ten seconds will have passed on your watch, but when you compare it with the clock it will seem like eleven seconds have passed, because that clock is running a second ahead of the previous one. If you continue walking, the next clock you pass will show that 22 seconds have passed, while your watch will tell you only 20 seconds have passed. By the time you get to the tenth clock, say, it will tell you that 110 seconds have passed, while your watch will tell you that only 100 seconds have passed. Your watch, therefore, will seem to be time-dilated compared to the clocks you are passing. However, your watch and the clocks are all ticking at the same rate- it is the fact that the clocks are out of synch, according to you, that results in the time dilation effect.