Which Clock Shows Less Time When They Collide?

[matheinste]

Hello all.

Question for Janus.

---When the clocks meet they will read the same time according to all frames. ----

Just to clear up a semantic point does this mean that what an observer in any/all frames will see the clocks reading the same times, not necessarily the same as each other. That is they can show different times but all will agree what these times are.

Thanks Matheinste.

 

[Dale]

I Do agree.

OK, now, note carefully that the worldline of B from B0 to AB is about two ct' units while the worldline of A form A0 to AB is about 5 ct units. Note also that the x-axis is parallel to the line from A0 to B0 (they are simultaneous in A's frame) while the x' axis is not parallel to that line (they are not simultaneous in B's frame). Finally, note that the worldline from Flash to B0 is about 3 (ct+x) or 20 (ct'+x') and the worldline from Flash to A0 is about 3 (ct-x) or 0.5 (ct'-x')

Does that give you enough information to draw the other graph?

 

[Janus]

I had read and reread all the posts (Janus's posts more carefully), but can't see how is my question answered. Can you point out precisely?

Again situation restated, from the frame of the origin of light flash both clocks started at zero time "simultaneously". The observer in this frame obviously knows that A is stationary and B is moving wrt him. Now, after collision, he brings the stopped clocks and compares times. According to Janus's post and Mentz114's spacetime diagram, he finds that both stopped clocks have accumulated same time.

Now, as the observer in light-flash frame knows that B was moving while A was stationary, thus B should have accumulated less time, for the same reason we say time dilation is real and not apparent. clocks actually slow when moving which is experimentally tested as has been pointed out earlier. Then how come we say that the light-flash frame observer (or any other observer because the clocks had been stopped earlier) would see the same accumulated time? where is the time dilation?

I think I see a problem here when referring to the frame of the origin of the light flash. In Mentz114's diagrams in my posts, it is treated as a frame in which the origin of the flash remains equidistant from both A and B at all times. In this frame, neither A or B is stationary but both have a equal velocity towards the origin of the flash. OTOH, you seem to be treating it as a frame in which A is stationary. I didn't catch this earlier because this is just the same as A's frame, when it comes to dealing with the problem and so I wouldn't treat it as a separate frame of its own.

So when you treat this like we did, then A and B start together, tick at the same rate and stop together, accumulating the same time.

When you treat it like you did, then things behave just as they would in A's frame where B starts ticking before A.

 

[Dale]

Hi Janus,

Yes, I noticed that also after I posted my diagram and then looked back at yours and Mentz's in detail. You and I interpereted the original scenario significantly differently from each other. The original phrasing is ambiguous since the word "equidistant" is frame-variant and no reference frame is specified. It is also ambiguous if the distance is equal at the time of emission or reception of the flash (again, frame variant).

IMO, this kind of ambiguity is yet another reason to use diagrams rather than words. A diagram can be wrong, but it cannot be ambiguous.

 

[Janus]

Hello all.

Question for Janus.

---When the clocks meet they will read the same time according to all frames. ----

Just to clear up a semantic point does this mean that what an observer in any/all frames will see the clocks reading the same times, not necessarily the same as each other. That is they can show different times but all will agree what these times are.

Thanks Matheinste.

In this particular situation as posed by the OP, the clocks real read identical times when they meet. In a different situation they could read different times (such as in the standard twin paradox), but all observers in all frame would agree as to what each clock read.

 

[matheinste]

Hello Janus.

Thanks.

Matheinste.

 

[AntigenX]

Hello all,

Thanks. Especially the senior members of the forum for their patience and efforts to clarify the things for me.

There are variety of stand points proposed by all of you, and I am confused as to which one am I supposed to endorse. I think we are analyzing and reanalyzing and re-re-analyzing the simple situation. I have tried to restate the original statements but without any results. Here is my fresh attempt to refine my situation. But before that let me ask this...

For two clocks A (stationary) and B (moving wrt A), is there any mechanism possible so that both A and B will agree that their clocks were started simultaneously with initial reading zero?

If not, there is no situation, but previous posts suggest that it is not impossible. If at all it is possible, we start our situation by setting the two clocks at zero by some mechanism, and, as I have said, the method to do that is irrelevant. I will accept any logical method anybody comes up with.

Now, there are two frames A and B, agreed upon their clock matching. Eventually, their clocks collide and stop. Now, the clocks are no more dynamic, but static, like time written on a piece of paper, and can be matched by anyone. Their matching is real. We can not say that according to A, B's clock will be slow, and according to B, A's clock will be slow. The result is unique, and can be verified. There are no extra clocks or frames of references, and no "ambiguity" as I see it. If at all anyone sees one, please point-out and I will rectify it.

Two more things restated...

As it is impossible in SR to say which clock is moving, It also implies that it is impossible to say which clock is slow. I understand this, and that is my situation. But there is a solution. The solution is time dilation, which is a real (and not apparent) effect in SR. We can match the readings and decide that the clock which accumulated less time was moving. But here also is one catch. If at all we can decide which clock is slow experimentally, we can decide the absolute velocity of that clock, which is impossible in SR (due to isotropy of light speed).

The situation is comparable with that of two identical local earth-clocks, one kept at equator and the other at pole (any, south or north). This is example given for SR (and not GR) by einstein himself, which states, that the clock on equator will actually be slower then the pole clock for all frames of reference, because, the equator clock is moving. And similar (not same) tests have experimentally been performed, and certified to be consistent with SR. I would further clarify that, As it was example given for explanation of SR (and not GR) there was no mention of gravitational time dilation or time dilation due to acceleration. Though I wouldn't stress much upon this example, It is significantly similar to our situation, though our situation is much more simplified.

As for "ambiguity" of the situation, it is all because we are stressing upon the method of synchronization, which I have clarified and reclarified, that, is not important, unless at all it is impossible.

And the spacetime diagrams are not doing anything but confusing the situation with the additional math. I will honestly give it a full fledge effort though, but the situation is so symmetric after matching the clocks, that spacetime diagrams will only tell us the point of view of A and B. And again we will be left to decide, which clock is moving? Needless to say, it is impossible, except arbitrary assumption.

 

[AntigenX]

To determine how much time each clock accumulates, you have to take into account:
1) When did it start running?
2) At what rate did it tick while running.
3) When did it stop running?

Since we are comparing two clocks, we can figure that above as seen by each clock.

Thus for A and B considered from their own reference they start at a certain time reading 0, run at a certain rate and stop at a certain time reading time "t"

But what of B as seen from A, and A as seen from B.

A sees B start at some time before A itself starts (IOW, by the time A itself starts ticking clock B has already built up a "head start"), then tick at a slower rate than A, and then stop at the same time as A (when they collide), at which time it will read the same (t) as Clock A(by ticking faster, A closes the gap of B's head start). So even though clock B ticked slower than A while running, it started ticking before A, and thus accumulates the same time as A between starting and stopping.

B sees the same thing with A and B switched.

There is some problem. As we are not ready to ignore the time delay for clock matching light flash to reach to both the clocks (or observers), we may not ignore light travel time delay when A sees B's clock started. Well, the best possible way for A and B to know that each other's clocks have started is, if their clocks send a light signal as soon as they start to the other clock. In which case, a will see that he is receiving the light signal after his own clock has started. So there is no head start for B but for A, who thinks that he is ahead of B, and B will slow down later again (due to time dilation) up to the collision. Thus, A will always think he is ahead of B. Ditto for B. There is no compensation of time dilation by head start, instead they add to the final effect. Further, how come we know that the head-start, if there is any, will exactly compensate the time dilation. Secondly, head start would be only apparent, while time dilation is real.

Edit: I am pleading again people, If it's not impossible to start two clocks simultaneously so that both clocks agree, tell me, else, put aside the method of matching the clocks and start considering that the matching has done "by some mechanism".

 

[granpa]

For two clocks A (stationary) and B (moving wrt A), is there any mechanism possible so that both A and B will agree that their clocks were started simultaneously with initial reading zero?

The only way i know of is for both clocks to be at the same place at the same time.

 

[AntigenX]

For two clocks A (stationary) and B (moving wrt A), is there any mechanism possible so that both A and B will agree that their clocks were started simultaneously with initial reading zero?

The only way i know of is for both clocks to be at the same place at the same time.

That is prerequisite for synchronization, but we are talking about matching only one reading for future reference.

Edit: This is done all the time in GPS (though they actually do synchronization), matching time for both clocks for any arbitrary instant would be even trivial, No?

 

[Dale]

Hi Antigen

Two spatially (spacelike) separated events will only be simultaneous in one unique standard-configuration reference frame. That is called the "relativity of simultaneity". That is why the details of your starting procedure are so important to the scenario and cannot be ignored. If the clocks start simultaneously in one clock's rest frame (my diagram) then that clock will record the most time. If the clocks do not start simultaneously in either rest frame then it is possible for them to record the same time (Janus and Mentz).

The spacetime diagrams are not confusing the situation at all. In fact, they have clarified exactly the source of your confusion: the relativity of simultaneity. You did not describe the situation ambiguously on purpose. You did it because you didn't realize that what you wanted to do was impossible due to relativity of simultaneity.

 

[AntigenX]

Hi Antigen

Two spatially (spacelike) separated events will only be simultaneous in one unique standard-configuration reference frame. That is called the "relativity of simultaneity". That is why the details of your starting procedure are so important to the scenario and cannot be ignored. If the clocks start simultaneously in one clock's rest frame (my diagram) then that clock will record the most time. If the clocks do not start simultaneously in either rest frame then it is possible for them to record the same time (Janus and Mentz).

The spacetime diagrams are not confusing the situation at all. In fact, they have clarified exactly the source of your confusion: the relativity of simultaneity. You did not describe the situation ambiguously on purpose. You did it because you didn't realize that what you wanted to do was impossible due to relativity of simultaneity.

Well, I'm getting your point, but nobody is getting my point.

Suppose I am experimenter and am doing the same experiment as has been described before viz, when I am equidistant to both the clocks, I am sending a signal to both the clocks. Here, It does not make any difference wether I am in moving frame or rest frame, because the speed of light does not depend on the speed of the source. It will also be wrong to say that the light will reach B faster than A, because the speed of light does not depend on observer either. The observer will notice the speed of light c, and not c-v. So light will reach both the observers at same time, not different time. Now, irrelevant of my motion, I know that I have matched both clocks. The clocks do not need to know anything. When they collide, I will collect them and notice the time of collision.

Now if I explicitly say that A was stationary and B was moving, Even thought, After I collect them, I am bound to get same readings.
The relativity of simultaneity comes into play when I am either in A or B. But I am in neither of them, nor do I state my motion to A and B. Nobody is constantly observing them. Taking advantage of isotropy of speed of light I match two clocks, I wait till collision and I compare the stopped clocks.
If I am asked, what time will I observe for both clocks when they collide, I am supposed to reveal my motion, but when the clocks have stopped, my motion is irrelevant.

 

[Doc Al]

Well, I'm getting your point, but nobody is getting my point.

Suppose I am experimenter and am doing the same experiment as has been described before viz, when I am equidistant to both the clocks, I am sending a signal to both the clocks. Here, It does not make any difference wether I am in moving frame or rest frame, because the speed of light does not depend on the speed of the source. It will also be wrong to say that the light will reach B faster than A, because the speed of light does not depend on observer either. The observer will notice the speed of light c, and not c-v. So light will reach both the observers at same time, not different time. Now, irrelevant of my motion, I know that I have matched both clocks. The clocks do not need to know anything. When they collide, I will collect them and notice the time of collision.

We get your point--but it's wrong!

You seem to think that just because you are equidistant between two clocks (according to you), the signal you send must reach each clock "at the same time". It still hasn't sunk in that simultaneity is relative.

The time it takes for a signal to reach a clock doesn't just depend on the speed of the signal--which everyone agrees is just c--it also depends on the speed of the clock, which depends on the reference frame.

Of course, if the clocks are at rest with respect to each other and are synchronized, then the clocks will read the same when the signals reach them. (Of course, in other frames those clocks are not synchronized.)

If the clocks are moving with respect to each other, they won't even agree that you were equidistant from both at the moment you sent out your signal.

 

[Dale]

Suppose I am experimenter and am doing the same experiment as has been described before viz, when I am equidistant to both the clocks, I am sending a signal to both the clocks. Here, It does not make any difference wether I am in moving frame or rest frame, because the speed of light does not depend on the speed of the source.

The speed of light does not depend on the speed of the source, but "when" and "equidistant" do. So you must specify the reference frame used to determine "when I am equidistant".

Also, because the speed of light is finite if you emit the light when they are equidistant to you they will not receive the light when they are equidistant to you except in the reference frame shown by Mentz.

 

[Dale]

AntigenX,

Here is a spacetime diagram I constructed years ago that really helped me learn relativity. There is no observer or object on it, it is just the Lorentz transform of an unprimed "rest" frame (black lines) and a primed frame moving at 0.6 c (white lines). If you study this diagram carefully you can get a geometric understanding of all of the important basic features of relativity: length contraction, time dilation, relativity of simultaneity, and symmetry. I would encourage you to sit down with the Lorentz transform and derive something similar yourself, perhaps even doing a pair of diagrams, one in each frame.

With the relativity of simultaneity the point is the following: if you look at the line t=2, all of those events are simultaneous in the unprimed (black) frame, but none of them are simultaneous in the primed (white) frame. The reverse is true from the line t'=2. In general, you can pick any straight line between ±45º and this line will represent a set of simultaneous events in exactly one reference frame. The events will not be simultaneous in any other reference frame. Does that help?

 

[AntigenX]

We get your point--but it's wrong!

You seem to think that just because you are equidistant between two clocks (according to you), the signal you send must reach each clock "at the same time". It still hasn't sunk in that simultaneity is relative.

The time it takes for a signal to reach a clock doesn't just depend on the speed of the signal--which everyone agrees is just c--it also depends on the speed of the clock, which depends on the reference frame.

Of course, if the clocks are at rest with respect to each other and are synchronized, then the clocks will read the same when the signals reach them. (Of course, in other frames those clocks are not synchronized.)

If the clocks are moving with respect to each other, they won't even agree that you were equidistant from both at the moment you sent out your signal.

Thank you for the bold face wrong!

Though, I must say, you are still not getting my point. From post#1 of this thread, I have been asking "is it at all possible for any observer to make two clocks (in relative motion) read zero reading at some arbitrary instant?". I haven't got any plain answer yet, though plenty I got.
Janus's posts have some very close indications that it is possible for some observer who is in relative motion wrt both A and B.
Also you will notice that I have stressed several times throught the thread to suggest some method of starting the clocks simultaneously for at least one observer. The plea has gone unheard.
Through this thread I wish to understand "how time dilation is real and not apparent (physical and not mere mathamatical)?". This has not been attempted at all.
The word "equidistant" has been the word of choice to falsify the reasoning behind the post, but as I have said several times, there is no reasoning after the method of matching the clocks. I am using this method due to lack of a better method, not suggested by anyone.
Let's replace the word "equidistant" with "equidistant in spacetime", does it make any sense now? Can an observer match readings of two clocks (in relative motion with each other) simultaneously (according to himself) when he is equidistant from both wrt spacetime (not just spatial) co-ordinates? If this doesn't work, suggest me some better method from your point of view, or tell me it's just not going to happen.
Well that sounds simple, No?

 

[AntigenX]

... Does that help?

That does in fact, but I'm afraid not much. Can you suggest where should I start, to get 'real' understanding of space-time diagrams? from word go?

 

[Doc Al]

Though, I must say, you are still not getting my point. From post#1 of this thread, I have been asking "is it at all possible for any observer to make two clocks (in relative motion) read zero reading at some arbitrary instant?". I haven't got any plain answer yet, though plenty I got.

Here's a plain answer to that question (if I understand it): Of course! (You might want to reread your post #1 if you think you asked that question.)

As to how to do it, that will take some prearrangement. One way is to send signals to each clock that will reach them simultaneously according to you. You may have to send the signals at different times. Arrange to have both clocks reset to zero at the instant they receive their signals. Is that what you want?

 

[Dale]

I have been asking "is it at all possible for any observer to make two clocks (in relative motion) read zero reading at some arbitrary instant?". I haven't got any plain answer yet

I am sorry about the communication difficulties here, but I did answer this when I said:

The speed of light does not depend on the speed of the source, but "when" and "equidistant" do. So you must specify the reference frame used to determine "when I am equidistant".

In other words, yes, you can do it but you must specify the reference frame in which the clocks read 0 at the same "arbitrary instant" and are "equidistant". There is nothing inherently wrong with the idea of an instant (simultaneous) nor is there anything wrong with the idea of equidistant. But simultaneity and distance are relative (frame variant), so the reference frame must be specified.

 

[Mentz114]

AntigenX,

Also you will notice that I have stressed several times throught the thread to suggest some method of starting the clocks simultaneously for at least one observer.

This diagram illustrates a method for M to send 2 light pulses so that A and B's clocks agree momentarily at the orange line. Red lines are light beams.

So when A sees his clock reset, he knows that if he had instantaneous comms ( horizontal line ) he'd see B's clock reseting as his does.

I have not tried to work out the times involved.

M

 

[Dale]

so that A and B's clocks agree momentarily

In A's rest frame.

 

[Phrak]

This problem is simply the one-clock vs. two-clock problem presented in a somewhat obscure manner.

 

[Mentz114]

DaleSpam,
yes, in A's rest frame. I didn't say it, but it's part of the (obscure) requirement.

The geometry in that diagram is unbelievably tricky. I had a go at working out the transmission times in M's frame but failed.

M

 

[robphy]

DaleSpam,
yes, in A's rest frame. I didn't say it, but it's part of the (obscure) requirement.

The geometry in that diagram is unbelievably tricky. I had a go at working out the transmission times in M's frame but failed.

M

It shouldn't be too difficult... trigonometrically.
What are the relationships among A,B, and M?

 

[Dale]

That does in fact, but I'm afraid not much. Can you suggest where should I start, to get 'real' understanding of space-time diagrams? from word go?

Hi AntigenX,

The wikipedia http://en.wikipedia.org/wiki/Minkowski_diagram" page is a decent intro, but IMO the real thing is to actually sit down and construct some on your own. Get some feel for how to build one that is numerically accurate. For a good start do a standard "twin paradox" diagram. Then have one twin send light flashes to the other at regular intervals and determine what the other twin receives. It may not teach you anything new about the physics, but it will help you get used to drawing the diagrams in a scenario that is fully understood. You might also try a light clock.

 

[Janus]

Thank you for the bold face wrong!

Though, I must say, you are still not getting my point. From post#1 of this thread, I have been asking "is it at all possible for any observer to make two clocks (in relative motion) read zero reading at some arbitrary instant?". I haven't got any plain answer yet, though plenty I got.
Janus's posts have some very close indications that it is possible for some observer who is in relative motion wrt both A and B.
Also you will notice that I have stressed several times throught the thread to suggest some method of starting the clocks simultaneously for at least one observer. The plea has gone unheard.
Through this thread I wish to understand "how time dilation is real and not apparent (physical and not mere mathamatical)?". This has not been attempted at all.

Yes, it is perfectly possible to arrange things such that both clocks start simultaneously for any given observer. For instance, by placing the origin of the flash closer to A than to B, (as shown in the attachment)you can have both clocks start simultaneously as far as A is concerned.

With such an arrangement, according to A: Both clocks start simultaneously. B runs slower during the period before collision and then after both clocks have stopped will have accumulated less time.

So according to A, B "really" runs slower.

However, according to B: Clock A starts significantly before clock B. B runs slower than A until the collision. After A and B stop, B will have accumulated less time than A even though A ran slower than B. This is because clock A had accumulated a great deal of time before clock B even starts.

So according to B, A "really" runs slower, but it also "really" started ticking sooner.

You can't just look at the accumulated time on both clocks at the end, and say which clock "really" ran slower.

I think the part of the problem you are having is that you are equating "real" with "absolute", that unless you can say which clock absolutely ran slower, can dilation isn't "real".

But time itself is relative and not absolute, and relative time measurement is as "real" as time gets.

 

[AntigenX]

Sorry people for very very late reply.

I was trying to learn some space time diagrams, but I think it will take some time to learn their proper representation and more importantly, their correct interpretation. Meanwhile, thank you all, especially, Janus, DaleSpam, Mentz114 and others for their concern.

Here are my replies to the responses...

Here's a plain answer to that question (if I understand it): Of course! (You might want to reread your post #1 if you think you asked that question.)

I didn't mean literally from post#1, instead, from the very beginning was what I meant. Thanks for reading post#1 though.

As to how to do it, that will take some prearrangement. One way is to send signals to each clock that will reach them simultaneously according to you. You may have to send the signals at different times. Arrange to have both clocks reset to zero at the instant they receive their signals. Is that what you want?

I think that is what I want, provided, the source of signals is stationary w.r.t. at least one clock A or B. It has been suggested that, in such a situation, there is no need to consider three frames but only two, but I would like to retain three frames (as a personal preference), if at all it is possible.

I am sorry about the communication difficulties here, but I did answer this when I said:In other words, yes, you can do it but you must specify the reference frame in which the clocks read 0 at the same "arbitrary instant" and are "equidistant". There is nothing inherently wrong with the idea of an instant (simultaneous) nor is there anything wrong with the idea of equidistant. But simultaneity and distance are relative (frame variant), so the reference frame must be specified.

Don't feel sorry, you have been very cooperative. I should have made it clear that I do not understand anything except plain things.

AntigenX,

This diagram illustrates a method for M to send 2 light pulses so that A and B's clocks agree momentarily at the orange line. Red lines are light beams.

So when A sees his clock reset, he knows that if he had instantaneous comms (horizontal line) he'd see B's clock reseting as his does.

I have not tried to work out the times involved.

M

This would work, but, As I mentioned earlier, the person matching the clocks is in relative motion with both the clocks, which is the case I wish to avoid.

This problem is simply the one-clock vs. two-clock problem presented in a somewhat obscure manner.

DaleSpam,
yes, in A's rest frame. I didn't say it, but it's part of the (obscure) requirement.

The geometry in that diagram is unbelievably tricky. I had a go at working out the transmission times in M's frame but failed.

M

I would like to know what is so obscure in this problem? And the extent of obscurity as well.

 

[AntigenX]

Yes, it is perfectly possible to arrange things such that both clocks start simultaneously for any given observer. For instance, by placing the origin of the flash closer to A than to B, (as shown in the attachment)you can have both clocks start simultaneously as far as A is concerned.

I want both the clocks started simultaneously by somebody stationary w.r.t. clock A or B, and suggested method can do that, If the light flesh is emitted from closer to A, such that the light beams have to cover same distance in both direction in the source's rest frame. In such a case, As is (obscure?) requirement of my problem, the source will be at rest wrt A. Now, the source is convinced that he started both clocks simultaneously according to himself. He also knows that A is at rest and B is moving wrt him. The question is, what will he find when, after collision, he will match the clocks? To the case, following discussion from you is also applicable...

With such an arrangement, according to A: Both clocks start simultaneously. B runs slower during the period before collision and then after both clocks have stopped will have accumulated less time.

So according to A, B "really" runs slower.

However, according to B: Clock A starts significantly before clock B. B runs slower than A until the collision. After A and B stop, B will have accumulated less time than A even though A ran slower than B. This is because clock A had accumulated a great deal of time before clock B even starts.

So according to B, A "really" runs slower, but it also "really" started ticking sooner.

It is explained that, According to A, B is slower, and According to B, A is slower. But the clocks have stopped now. Who will be wrong?

You can't just look at the accumulated time on both clocks at the end, and say which clock "really" ran slower.

Exactly why not?

I think the part of the problem you are having is that you are equating "real" with "absolute", that unless you can say which clock absolutely ran slower, can dilation isn't "real".

But time itself is relative and not absolute, and relative time measurement is as "real" as time gets.

Do I have any way to convince you guys that this is not the case? For me the strongest reason for not believing in absolute time is I haven't seen any clock showing any yet, else I would synchronize my own clock with it to be on time everywhere!

 

[Doc Al]

I think that is what I want, provided, the source of signals is stationary w.r.t. at least one clock A or B. It has been suggested that, in such a situation, there is no need to consider three frames but only two, but I would like to retain three frames (as a personal preference), if at all it is possible.

As far as I can see, there are only two reference frames: Clock A and the observer are in one; Clock B is in the other.

I want both the clocks started simultaneously by somebody stationary w.r.t. clock A or B, and suggested method can do that, If the light flesh is emitted from closer to A, such that the light beams have to cover same distance in both direction in the source's rest frame. In such a case, As is (obscure?) requirement of my problem, the source will be at rest wrt A. Now, the source is convinced that he started both clocks simultaneously according to himself. He also knows that A is at rest and B is moving wrt him. The question is, what will he find when, after collision, he will match the clocks? To the case, following discussion from you is also applicable...

Is the following an accurate summary of the set up?
(1) There's a clock A at some position along the x-axis.
(2) An observer (call him O) is at rest with respect to A and is at some position to the right of A along the x-axis.
(3) There's a clock B, which is moving towards A at some speed v, at some point to the right of O along the x-axis.

We all agree that it's perfectly possible to reset clock A and clock B to both read t = 0 at the same instant according to frame O (which is also frame A). Let's say we do that. Is your question: When clock B collides with Clock A, what will they both read?

Assuming that's your question, the answer is: Clock B will read less time than Clock A.

It's easy to understand from A's point of view: Clock B runs slow, so less time accumulates as it travels from its original location to point A.

It's a bit trickier to understand from B's point of view (it requires understanding of the relativity of simultaneity): According to B, Clock A runs slow. When observer O sent B the signal he thought it reached B at the same time it reached A (and it did--according to frame O-A). But according to B, the signal reached A long before it reached B. Even after accounting for time dilation of the moving clock A, more time accumulates on clock A than on on clock B.

Everyone agrees that clock A will read more time than clock B when they finally meet.

 

[AntigenX]

Is the following an accurate summary of the set up?
(1) There's a clock A at some position along the x-axis.
(2) An observer (call him O) is at rest with respect to A and is at some position to the right of A along the x-axis.
(3) There's a clock B, which is moving towards A at some speed v, at some point to the right of O along the x-axis.

Precisely! I should have done it this way, poor me...

We all agree that it's perfectly possible to reset clock A and clock B to both read t = 0 at the same instant according to frame O (which is also frame A). Let's say we do that. Is your question: When clock B collides with Clock A, what will they both read?

Yes.

Assuming that's your question, the answer is: Clock B will read less time than Clock A.

It's easy to understand from A's point of view: Clock B runs slow, so less time accumulates as it travels from its original location to point A.

got it.

It's a bit trickier to understand from B's point of view (it requires understanding of the relativity of simultaneity): According to B, Clock A runs slow.

got it.

When observer O sent B the signal he thought it reached B at the same time it reached A (and it did--according to frame O-A).

got it.
Edit: Just for clarification, Though O & A are in same rest frame, O will agree that the light signal reached to both A & B simultaneously. I don't see any reason for A to think so.

But according to B, the signal reached A long before it reached B. Even after accounting for time dilation of the moving clock A, more time accumulates on clock A than on on clock B.

How will B decide that the signal reached A long before it reached B? I hope not from the space-time diagram.

Everyone agrees that clock A will read more time than clock B when they finally meet.

Not clear yet.