How Does Time Differ for Observers in Relative Motion?

[yuiop]

Firstly about those clocks. I see them as recording clocks. The observer's two clocks are synchronised at the start.

OK, let's say the observer's two clocks (A and B) are 1 light second apart in the the rest frame of the clocks and the observer. You need to be clear that they are synchronised only in this reference frame and in any other reference frame they will not appear to be synchronised.

Every clock, when it observes an event, prominently displays its recorded time so that it can be read by the observer, the astronaut, (and by us!). (Is that the "proper time"?)

The proper time is the normally the time indicated by a singe clock such as the clock of the astronaut. The astronaut's clock requires no synchronisation and does not require any calculations to obtain the elapsed time between the two events. The time measured b the observer is a calculated time using two clocks and this type of time interval is known as a coordinate time.

I used the term "fly-past" to indicate that when the event occurred, the astronaut's clock and the observer's clock were both present in the same position in space.

This is good practice. When the clocks are present at the same position we do not need to worry about how light travel times affect what is seen.

Secondly, I am puzzled by the statement made by Ghwellsjr that:-
Quote: "The first thing to note here is that as the rocket ship is approaching observer O, observer O sees the astronaut's clock running at twice the rate of his own and then as it goes past observer O, he sees it running at one-half the rate of his own clock."

Ghwellsjr is talking about what is actually "seen" when the clocks are in transit and this is affected by light travel times. When a clock approaches an observer, the gap between consecutive pulses appears to be shorter than the ticking rate of the clock because the clock is closer to the observer with each tick. This is known as the Doppler affect and occurs in Newtonian physics where all clocks are considered to run at the same rate. You can think of it as an optical illusion because the classic Doppler shift does not mean the clocks are actually running slower. In relativity, time dilation and Lorentz transformations allow for light travel time and remove any classic Doppler effects so that only the true time dilation is calculated.

I thought that a clock in a moving frame will always be seen to be running slow, or "dilated".

The word "seen" is used carelessly in relativity and can cause confusion. It is best to imagine collocated clocks that are next to each other when comparing times (as you did in your post). To continue with the example above, imagine the astronaut is traveling from A to B with a velocity of 0.8c. When the astronauts clock is next to clock A let's say both the astronaut's clock and the A clock read zero. On arrival at B, the B clock reads 1.25 seconds and the astronauts clock reads 0.75 seconds. According to the observer, the difference between his two clocks is greater than the time recorded on the astronaut's single clocks so he says the astronauts clock is ticking slower.

Now we look at things from the astronauts point of view. In his reference frame his clock and the A clock are both initially reading zero but the B clock reads L*v = 0.8 seconds. This is because the two clocks of the observer do not appear to be synchronised in the astronaut's reference frame. On arrival at B the astronauts that his clock reads 0.75 seconds and the B clock reads 1.25 seconds just as in the other reference frame, but according to the astronaut the B clock already has 0.8 seconds indicated at the start so the B clock has only advanced by (1.25-0.8 = 0.45) seconds. The astronaut concludes that the B clock is ticking slower than his own clock. The same is true when he compares notes with readings on the A clock.

Note that in either reference frame, the ticking rate of any clocks moving relative to the reference frame is slower than clocks at rest in the reference frame. Thus the observer and the astronaut calculate the other person's clocks to be running slower than their own. This is always true in Special Relativity but it also true that the elapsed proper time recorded by a single clock moving inertially between two events (such as the clock of the astronaut), will always be less than the coordinate time interval calculated from spatially separated clocks. The difference is subtle and you should do some calculations of your own or draw some spacetime diagrams to get your head around it. Closely studying the excellent spacetime drawings presented by Ghwellsjr would be a good place to start.

 

[ghwellsjr]

Thanks to everyone trying to sort out my confusion. Apologies if I have not used the accepted notation for the times.
Firstly about those clocks. I see them as recording clocks.

Yes, you made that very clear in your first post.

The observer's two clocks are synchronised at the start.

Even though you didn't make this clear, I assumed that you meant them to be synchronized.

Every clock, when it observes an event, prominently displays its recorded time so that it can be read by the observer, the astronaut, (and by us!). (Is that the "proper time"?)

Yes, every clock always displays its own Proper Time.

I used the term "fly-past" to indicate that when the event occurred, the astronaut's clock and the observer's clock were both present in the same position in space.

Yes, you also made that very clear.

Secondly, I am puzzled by the statement made by Ghwellsjr that:-
Quote: "The first thing to note here is that as the rocket ship is approaching observer O, observer O sees the astronaut's clock running at twice the rate of his own and then as it goes past observer O, he sees it running at one-half the rate of his own clock."
I thought that a clock in a moving frame will always be seen to be running slow, or "dilated". A classic example is the muon particle created in the Earth's upper atmosphere, which travels very fast but has an extremely short half-life. The muons seem to age more slowly according to an observer on the ground, and so that many more muons actually reach the ground than would be the case in a non-reletivistic environment.

You can't see a clock running at the determined Time Dilation rate, you see it at the Doppler rate. Even in the case of muons, in the Earth's rest frame, we don't see their creation at the time it happens, we see it ever so slightly before we see the muon's arrival. So from creation to arrival is a very small fraction of a microsecond, even if the muon lasted for 2 microseconds in its rest frame and almost 45 microseconds in Earth's rest frame, assuming a speed is 0.999c. Here's a spacetime diagram for the blue Earth's rest frame showing the muon in red:

It's impossible to see in this drawing but there are two red lines showing how Earth views the creation of the muon and how it views it at 1 microsecond later, so I have zoomed in on the end of the life of the muon where we on the Earth can see its creation and half-way through its life:

In effect, we see the muon's "clock" ticking at 44.7 times faster (the Doppler factor) than the rate of our own clock, not 22.4 times slower (the Time Dilation factor). Note that the Time Dilation factor is clearly visible on the first spacetime diagram since the 1-microsecond tick marks for the muon are spaced out to 22.4 times according to the Coordinate Time. If you want, I can show you how the Time Dilation can be determined by Earth observers for their rest frame in the muon scenario or in your scenario. And keep in mind that in other frames, the propagation time for the signals can change but what the observers actually see remains the same.

Aside from the Doppler puzzle, were the diagrams clear or did you have any more questions regarding them?