Dingle's Dilemma: Solve the Puzzle

[mitesh9]

Dingle's Dilemma again!

Dear all,
Following is a problem, I suppose, is related to some controversy called "Dingle's Dilemma", as I read somewhere on some relativity forum, and I wish to understand where is the catch.

Consider two inertial frames A and B, stationary with respect to each other, with two identical and synchronized clocks (one in each). Now, another inertial frame C (say a rocket) passes by A with a uniform velocity (say v). When the co-ordinates of A and C coincide while the process of passing by, C synchronizes its clock with A (and as A and B are still synchronized).

Now, by the time C reaches B (assuming that C's motion is in the line connecting A and B), for an observer in A, C's clock has been slowed down by some time (time dilation, say 1 sec), while, as there is no mechanism to establish wether A is in absolute motion or C, for an observer in C, the clock in A has been slowed down, and as clocks in A and B are already synchronized B has been slowed down as well.

What will happen, when the C reaches B? Will their clocks be still synchronized (I don't think so)! And if not, which one will be slower, B's clock or C's clock? (At the moment when B and C coincides, If the need arise, there should be some kind of accident, where, C doesn't move at all, abruptly stops, and merges with B.)

In case you wish to respond to this one, please note that, I'm not against SR (or Einstein for that matter, nor am I gifted enough to ever expect this), instead, being a chemist, the best fit for my role in PF can be as a hobbyist relativist.

 

[paw]

Since A and B are not moving wrt to each other they can be considered to be the same frame. It seems to me you can eliminate the second frame (A) and ask if the clocks at C and B were synchronised at some time t0 what is the state at some other time t1. This gives the standard answer C sees B's clock ticking slow while B sees C's clock ticking slow.

 

[mitesh9]

Do we do such generalizations? I mean, though A and B are stationary, their clocks are at different places (co-ordinates may be). I don't know if this has anything to do with the situation, however. By the way, is it the dingle's dilemma?

 

[paw]

Do we do such generalizations? I mean, though A and B are stationary, their clocks are at different places (co-ordinates may be).

Assuming a flat spacetime I don't see why not. A different position in a flat spacetime shouldn't affect the clocks rate in any way. It amounts to the same question as a spaceship travels from alpha centauri to procyon at constant speed 0.9c, what do observers on Earth and the ship 'see'. Obviously the ship observer sees Earths clock running slow while the Earth observer sees the ship clock running slow.

 

[mitesh9]

So is it the same thing that happens in muon lifetime case? Muons see our clocks slower, while we see muons' clock slower?

 

[paw]

So is it the same thing that happens in muon lifetime case? Muons see our clocks slower, while we see muons' clock slower?

Yes.

BTW Dingles Dilema seems to have been his inablity to accept experimental evidence. :grin

Look up Herbert Dingle on wiki and you'll see he had a number of objections to SR and ended up suggesting a conspiracy, more or less. I didn't see any reference to a specific 'dilema'.

 

[mitesh9]

Thanks, I had read about that somewhere, and I only could remember this problem about him.
Now, as we have concluded here, both observers on B and C frames think they are running faster than other in time (i.e. their clocks are faster then the other one), but when C coincides with B, he can actually check, wether what he thinks is true or not. What I wish to know is that, when comparing clocks from B and C, at the time of coinciding the two frames, keeping both clocks in two hands, what will one come to know?
That is to say, the rocket-man crashes (meets with an abrupt accident, though I don't think that's required) on B and being in B now, copares the clock in B with his own. What will he find then?

Edit: Or alternatively, when C coincides with B, someone from B throws the clock into C (which is rocket), now, the rocket-man can compare the clocks, keeping them in both hands. Or more simply, when the frames C and B coincides, the rocket-man can take a look at both the clocks (considering both of them to be big and observable )

 

[paw]

Now, as we have concluded here, both observers on B and C frames think they are running faster than other in time (i.e. their clocks are faster then the other one

Ok so far...

but when C coincides with B, he can actually check, wether what he thinks is true or not.

Now we need to start being more careful. How and what is C checking?

What I wish to know is that, when comparing clocks from B and C, at the time of coinciding the two frames, keeping both clocks in two hands, what will one come to know?

Suppose C passes B close enough to actually see B's clock (remember it was previously synchronised at t0), without light travel delays, then C will observe B's clock has accumulated less time than C's. B, of course, can say the same about C's clock.

If you have C stop at (or crash into) B you introduce an acceleration and that's going to complicate the situation. Acceleration will break the symmetry. This would best be solved with a spacetime diagram. I'm reasonably sure that C will have the longer world line and thus the shortest 'proper' time in this case. Or to put it another way, when the situation is no longer symmetric C will have aged less than B (or A for that matter since A and B are equivalent).

 

[mitesh9]

Suppose C passes B close enough to actually see B's clock (remember it was previously synchronised at t0), without light travel delays, then C will observe B's clock has accumulated less time than C's. B, of course, can say the same about C's clock.

Let's remove crash, or anything related to acceleration. And let's go a little slow as well. What I gather is that, when C and B coincide (that is they are sharing the same coordinates of space), and removing light travel delays, both will see each other's clocks slow? When they are face to face (for a fraction of time may be) each having his clock in his hand so that he and other one, both can see the clock? This seems improbable, because, they are not calculating time, but actually seeing the clocks, and clocks can give a single time to all observers sharing same spatial coordinates!

 

[paw]

This seems improbable, because, they are not calculating time, but actually seeing the clocks, and clocks can give a single time to all observers sharing same spatial coordinates!

That's the problem right there. They may share the same spatial coordinates momentarily but it's the spacetime coordinates that matter. B and C can only share the same spacetime coordinates if C stops (or crashes) and that has to involve an acceleration.

 

[mitesh9]

Well, I get that now, but that means that they are at same place, but different time, because, they are sharing same spatial coordinates (x,y z), but their time is different. I think that's what you wish to convey (or is there any other definition of space time coordinates, that I am not aware of).

So in such a case, will they collide (being at same place but different times)? This is like, we are both in same school and same home, and you were rushing from school to home following same path in the evening, which I followed in opposite direction to reach to school in the same morning. Should we collide?

 

[paw]

Well, I get that now, but that means that they are at same place, but different time, because, they are sharing same spatial coordinates (x,y z), but their time is different. I think that's what you wish to convey (or is there any other definition of space time coordinates, that I am not aware of).

So in such a case, will they collide (being at same place but different times)? This is like, we are both in same school and same home, and you were rushing from school to home following same path in the evening, which I followed in opposite direction to reach to school in the same morning. Should we collide?

Try to think of a body that's in relative motion to an observer as getting further and further away from that observer on the time axis. Try not to think of the time coordinate as an actual time like a clock on the wall. Remember, there is no absolute time.

This way you can see that, even though C might miss B by the smallest distance in space, C will be at a completely different point in spacetime. The only way C and B can share the same point in space time is for one of them to accelerate to match speeds as well as position.

They can collide of course but they won't agree on what time the collision occurred at. They can only work with their clocks that were synchronised at some earlier time t0.

 

[mitesh9]

Sure, but my point is, as there is no absolute time (irrelative to any other time, but truly absolute), there is no absolute place as well (we arbitrarily select them as per our convenience).
Ideally, they must not collide. Two observers can only collide if their all spacetime coordinates (that is x, y, z & t) are same for both of them (as I tried to explain in my school-home example), other than that they can not collide.
And If they collide, they must have "observed" same space time coordinates. That is to say, even if they don't try to observe the other's time (or clock), the collision itself is the proof for both of them, that their spacetime coordinates were same at the time of collision (i.e. before acceleration/de-acceleration, as that will happen after collision). If they remember their own time of collision, as the collision it self is evidence of them being at sametime-sameplace, they may confirm afterwords, without looking their clocks again, that they were same.

Edit: To make it more clear, though there is no absolute time, they defined their own relative time by synchronization of their clocks, so at the time of synchronization, they were both at the origin of the time axis, yet having different spatial coordinates (x, y, z) and thus they did not collide. Similarly, if one of the observer is moving away from the other observer on time axis, even if their spatial coordinates coincide, they must not collide. This further clarifies why should their all four spacetime coordinates be same for a collision to occur.

 

[paw]

the collision itself is the proof for both of them, that their spacetime coordinates were same at the time of collision (i.e. before acceleration/de-acceleration, as that will happen after collision). If they remember their own time of collision, as the collision it self is evidence of them being at sametime-sameplace, they may confirm afterwords, without looking their clocks again, that they were same.

Sure the collision is proof they share the same spacetime coords. But my point is you can't have a collision without an acceleration. If in the split instant before C hits B they compare clocks the situation will be as before, C sees B's clock as ticking slow and vise versa. When the collision is complete, and assuming both observers and their clocks aren't just puffs of vapor, both will agree that C has accumulated less proper time.

I still think you are looking at the time axis as a clock rather than a coordinate axis. In order to share the same spacetime coords both observers must be at rest wrt to each other and in the same place. As long as one is moving wrt to the other they can't share the same spacetime coords without an acceleration.

 

[mitesh9]

I exactly see what you want to show me (see the Edit, sorry for the delay), however, the acceleration (or de-acceleration) requires collision. And without (or before) collision, there won't be any acceleration (or de-acceleration). In turn collision requires same spacetime coordinates, which as you correctly pointed out, can not happen without acceleration (or de-acceleration) within the domain of SR.

I think, we both know now, that there is some catch, which we are not able to figure-out.

 

[mitesh9]

Dear pav,

Thank you for your extensive support and help. See you tomorrow again...
In the meantime, is not there anybody willing to help or join the discussion?

 

[phyti]

The time interval from event AC to BC is the same for all, (in terms of the invariant interval).
The clocks do not indicate time, only the rate of time. Time dilation is a function of v/c. The clock that moves faster has a slower rate. When C is coincident with B (event BC), they could transmit a signal containing the time to each other. Coincident meaning close enough to observe without colliding. Objects cannot truly be 'in the same place', therefore there is always a distance involved, but not significant enough to affect measurements. The clock with the smaller value would be the faster.

 

[MeJennifer]

Do we do such generalizations? I mean, though A and B are stationary, their clocks are at different places (co-ordinates may be). I don't know if this has anything to do with the situation, however. By the way, is it the dingle's dilemma?

Clocks at different places do generally not matter except in the case of accelerating frames.

For instance a frame that is accelerated cannot maintain a constant time for each coordinate value in the direction of acceleration.

 

[Ich]

When C meets B, C will read less time than B.
B and C could still claim that the other clock experienced less time during the experiment. Because, from C's perspective, A and B are not synchronized, there is no problem with the above result.

 

[mitesh9]

The time interval from event AC to BC is the same for all, (in terms of the invariant interval).

Which of the observer will observe invariant interval? If none, than we must not bring in any invariant time interval into discussion.

The clocks do not indicate time, only the rate of time. Time dilation is a function of v/c. The clock that moves faster has a slower rate. When C is coincident with B (event BC), they could transmit a signal containing the time to each other.

Well, yes, clocks do not indicate the time, but the rate of time, however, our problem also involves rate of time only, and not absolute time. And surely, While coincident, B and C can both transmit the signal, but what we are interested in is, will their signals show same passage of time interval or different intervals for both clocks?

Coincident meaning close enough to observe without colliding. Objects cannot truly be 'in the same place', therefore there is always a distance involved, but not significant enough to affect measurements.

Sure, but instead of observers, if the entities traveling in the frames C and B (as there is not absolute motion, but only relative one, both are traveling with respect to each other) are two particles, and if they collide, is not the point of impact where both particles are at same place same time (may be their centers are separated, yet at the point where their surfaces meet for impact, they are there same place same time), yet as you pointed out, this are not significant details.

The clock with the smaller value would be the faster.

But which clock will that be?

I think I should repeat my question.
Three inertial frames A, B and C. A and B stationary while C moving with a relative velocity v with respect to A and B, and traveling in the line connecting A and B, so that it will first coincide with A and eventually B (coincide means their origins share same spatial co-ordinates x, y and z). A's clock is slightly (say 1m) away from A's spatial Origin, while, B and C's clocks are perfectly at their spatial origins. Further assume that, clocks have an intuitive mechanism, by virtue of which, they stop as soon as they are touched by anything. Now when C coincides with A, all three clocks are synchronized. Considering that A's clock was off-origin 1m, and clocks were smaller than that, the clocks of A and C won't collide, even when their origins coincide.
Eventually, C reaches B and their origins coincide, However, there are clocks at their origins, so they will collide. At the same time of collision (when the clocks have touched each other) the clocks will stop. If the clocks are intact, will there be any time difference in the time value (or rate or interval) shown by the clocks?
We should note that, As has already been posted earlier in this thread, The clocks should not collide, if their spacetime coordinates (of at least one point on their surface, where the point of impact is) are same. But, if we apply Lorentz transform on any of the clock, we will come to know that it is off-sync with the other, and hence, as it's spacetime co-ordinates are different then the other clock (time coordinate to be precise), it must not collide, yet we know it will collide.

To make it more clear, though there is no absolute time, they defined their own relative time by synchronization of their clocks, so at the time of synchronization, they were both at the origin of the time axis, yet having different spatial coordinates (x, y, z) and thus they did not collide. Similarly, if one of the observer is moving away from the other observer on time axis, even if their spatial coordinates coincide, they must not collide. This further clarifies why should their all four spacetime coordinates be same for a collision to occur. (this has also been pointed out in earlier posts)

I'm pretty sure, there is some catch in the situation, but can't see what it is, may you people help me out.

Regards,

Mitesh

 

[mitesh9]

Clocks at different places do generally not matter except in the case of accelerating frames.
For instance a frame that is accelerated cannot maintain a constant time for each coordinate value in the direction of acceleration.

Well, paw told me similar thing. Though I'm not sure will I be able to translate the situation in two frame situation, because there may be some complications.

When C meets B, C will read less time than B.
B and C could still claim that the other clock experienced less time during the experiment. Because, from C's perspective, A and B are not synchronized, there is no problem with the above result.

Dear Sir, as MeJennifer and pawhas pointed out, A and B being stationary with respect to each other, and no acceleration being involved, how can C see them not synchronized? And yet, A's purpose was to synchronize the clocks of C and B, so that there can be a fair competition between them. Now when B and C are synchronized, C moves towards B, their clocks collides and stops, which would have accumulated more time? In fact, the first question is should they collide at all?

 

[Doc Al]

Dear Sir, as MeJennifer and pawhas pointed out, A and B being stationary with respect to each other, and no acceleration being involved, how can C see them not synchronized?

Dear Sir, the moving frame in which C is at rest will not see clocks A and B as synchronized. This is the relativity of simultaneity: Look it up!

And yet, A's purpose was to synchronize the clocks of C and B, so that there can be a fair competition between them.

Since the clocks of C and B are in relative motion, they cannot be synchronized. Of course, you could arrange that when the two clocks pass each other that they read the same at that instant. But that's hardly the same as being synchronized.

 

[Ich]

Dear Sir, as MeJennifer and pawhas pointed out, A and B being stationary with respect to each other, and no acceleration being involved, how can C see them not synchronized?

Dear Sir, I am in the fortunate position that I don't have to rely on others to work this kind of problem.
C and B meet.
C reads less time than B when they meet.
A and B are not synchronized in C's frame.

 

[mitesh9]

I think I should stop this discussion from my side.

It was good to learn that this is how "spirit of the game" is defined on PF. In last three days of my participation in PF discussions on relativity, I have collected more infraction points (rather than information) than anybody else who keeps on insulting others. I would suggest to review my posts, if you do not believe me. And hang me till death, If you find any insulting word in any of my posts. PF is not a learning forum, but a rigid community I must say, who do not want to answer questions, but instead prefers insulting people for their ignorance.

Anyways, learning physics may be less difficult task, than learning it from any man, I suppose.

Bye bye, PF.

Regards,

Mitesh Patel

 

[Ich]

If you find any insulting word in any of my posts. PF is not a learning forum, but a rigid community I must say, who do not want to answer questions, but instead prefers insulting people for their ignorance.

Maybe you want to read my statement in the appropriate context: "Dear Sir, as MeJennifer and pawhas pointed out, A and B being stationary with respect to each other, and no acceleration being involved, how can C see them not synchronized?".
I don't have to rely on other user's statements to work the problem, meaning that I'm confident enough to contradict them if necessary.
I never meant to ridicule you for asking here. Asking and giving answers is the intention of this forum.

 

[paw]

Dear Sir, as MeJennifer and pawhas pointed out, A and B being stationary with respect to each other, and no acceleration being involved, how can C see them not synchronized?

I'm sorry but you have misunderstood. I (and MeJennifer) said A and B are equivalent frames. As such you don't need A at all. In your scenario A serves only to define the time t0 that B and C's clocks have the same reading. That could be done just as easily with a light flash equidistant between B and C. B and C set their clocks when the light pulse arrives.

Any time after t0, C will observe B's clock as running slow and B will observe C's clock as running slow. This will remain true for all time unless B or C undergo an acceleration.

As for whether B and C will collide if they are at some time coincident in space, the answer is yes they will. However B and C will not agree on the time the collision occured. When the collision has occurred C will have accumulated less proper time than B because C has undergone an (rather severe) acceleration.

 

[paw]

I don't have to rely on other user's statements to work the problem, meaning that I'm confident enough to contradict them if necessary.

No need to contradict either me (or MeJennifer). : ) Mitesh9 misunderstood the statements. See above.

 

[paw]

Dear Sir, I am in the fortunate position that I don't have to rely on others to work this kind of problem.
C and B meet.
C reads less time than B when they meet.
A and B are not synchronized in C's frame.

Now there's an interesting point. You agree A and B are equivalent frames right? That is, there's no acceleration and no relative motion between them.

So, while C will always say both A's clock and B's clock are running slow, A's clock and B's clock will always read the same time according to C. Right?

Isn't this the same as saying A and B's clocks are synchronized according to C? Or is the word 'synchronized' reserved for more specialised use?

 

[Doc Al]

simultaneity

Now there's an interesting point. You agree A and B are equivalent frames right? That is, there's no acceleration and no relative motion between them.

So, while C will always say both A's clock and B's clock are running slow, A's clock and B's clock will always read the same time according to C. Right?

Wrong.

Isn't this the same as saying A and B's clocks are synchronized according to C? Or is the word 'synchronized' reserved for more specialised use?

A and B are two clocks in the same frame and are presumeably synchronized within their own frame. They are not synchronized according to frame C, since they are separated along the direction of motion. According to frame C, clock B is ahead of clock A.

When clock C passes clock A let's assume that they both read the same time (by special arrangement). According to frame A-B, at the moment that C passes A all three clocks read the same time. Frame C agrees that at the moment he passes A, clocks C and A both read the same (again, by arrangement), but does not agree that clock B reads the same time.

 

[paw]

A and B are two clocks in the same frame and are presumeably synchronized within their own frame.

Right. That's what I meant.

They are not synchronized according to frame C, since they are separated along the direction of motion. According to frame C, clock B is ahead of clock A.

Arrrrr, I missed that. They do tick at the same rate though, correct?

When clock C passes clock A let's assume that they both read the same time (by special arrangement). According to frame A-B, at the moment that C passes A all three clocks read the same time. Frame C agrees that at the moment he passes A, clocks C and A both read the same (again, by arrangement), but does not agree that clock B reads the same time.

Thanks. Luckily I don't think any of that changes the results I predicted though.

1. Any time after C passes A, C will see both A and B ticking slower while A and B will see C ticking slower.

2. A collision will occur.

3. B and C will not agree on the time the collision occured.

4. After collision C will have accumulated more proper time than B.

5. A was not required for any of this analysis.

Does this all seem right to you?

 

[Doc Al]

Arrrrr, I missed that. They do tick at the same rate though, correct?

Sure.

Thanks. Luckily I don't think any of that changes the results I predicted though.

1. Any time after C passes A, C will see both A and B ticking slower while A and B will see C ticking slower.

Sounds good.

2. A collision will occur.

3. B and C will not agree on the time the collision occured.

Sounds good. Assume that clocks A & B are a distance L apart (along the direction of motion) and are synchronized and that C passes A when both read t = 0. Then when C collides with B, B will read t = L/v and C will read L/(\gamma v).

(If I misunderstood the scenario, please correct me.)

4. After collision C will have accumulated more proper time than B.

Not sure what you mean. Proper time between what two events?

 

[paw]

Sounds good. Assume that clocks A & B are a distance L apart (along the direction of motion) and are synchronized and that C passes A when both read t = 0. Then when C collides with B, B will read t = L/v and C will read L/(\gamma v).

(If I misunderstood the scenario, please correct me.)

No, you didn't misunderstand. That's exactly how I saw the scenario.

Not sure what you mean. Proper time between what two events?

Between t=0, that is when C passed A, and when C collides with B. But I see you answered in the affirmative anyway in the previous quote.

 

[AntigenX]

the acceleration (or de-acceleration) requires collision. And without (or before) collision, there won't be any acceleration (or de-acceleration). In turn collision requires same spacetime coordinates, which as you correctly pointed out, can not happen without acceleration (or de-acceleration) within the domain of SR.

This is like running in circles...

Three inertial frames A, B and C...
...though there is no absolute time, they defined their own relative time by synchronization of their clocks, so at the time of synchronization, they were both at the origin of the time axis, yet having different spatial coordinates (x, y, z) and thus they did not collide. Similarly, if one of the observer is moving away from the other observer on time axis, even if their spatial coordinates coincide, they must not collide. This further clarifies why should their all four spacetime coordinates be same for a collision to occur. (this has also been pointed out in earlier posts)...

Is this correct? Do we require all four spacetime coordinates equal for a collision?

 

[Doc Al]

Do we require all four spacetime coordinates equal for a collision?

By definition, a collision means that two things are at the same place at the same time.

 

[AntigenX]

By definition, a collision means that two things are at the same place at the same time.

But if collision occurs, whose coordinates we should consider? For event collision, the coordinates of space will naturally be same for two frames colliding, but the time coordinates of both will be different, because one is moving and other is stationary. And if both are moving, then also their time coordinates will be different. This means that two things can never collide!

 

[Doc Al]

But if collision occurs, whose coordinates we should consider? For event collision, the coordinates of space will naturally be same for two frames colliding, but the time coordinates of both will be different, because one is moving and other is stationary. And if both are moving, then also their time coordinates will be different. This means that two things can never collide!

You can use anyone's coordinates.

For any two things to collide they must have the same spacetime coordinates at the time of collision, according to anyone's frame of reference. Of course, different observers will use different coordinates to represent the same event. So what?

 

[AntigenX]

I can't say I understand much of what you just wrote.

For any two things to collide they must have the same spacetime coordinates at the time of collision, according to anyone's frame of reference. Of course, different observers will use different coordinates to represent the same event. So what?

What I want to ask is, is it when a collision occur when all four coordinates of observer in one frame match those of an observer in another frame? Or is it that they may match their coordinates and they are different according to them, and even though the colision will occur?

 

[Doc Al]

When two things collide everyone will agree that they are at the same time and place. Of course, the particular spacetime coordinates depend on what frame you are measuring with respect to. Make sense?

 

[AntigenX]

When two things collide everyone will agree that they are at the same time and place. Of course, the particular spacetime coordinates depend on what frame you are measuring with respect to. Make sense?

Ok, I think the question is not "LOUD AND CLEAR", as it should be...

The question is, after the collision has happened, obviously, everybody will be sharing the same frame of reference and so will agree about the same time and same place, however, will observers of colliding frames agree upon the spacetime coordinates of the event "collision"?

Because, being in relative motion, even their clocks may have been synchronized earlier, they will be observing different time coordinates of the other frame, and will be unable to extrapolate their own spacetime coordinates leading to collision. Simply saying, they will not be able to predict the collision, which is bound to happen, by virtue of their different readings of the other frame of reference.

 

[Doc Al]

The question is, after the collision has happened, obviously, everybody will be sharing the same frame of reference and so will agree about the same time and same place, however, will observers of colliding frames agree upon the spacetime coordinates of the event "collision"?

(1) The fact that a collision happens has nothing to do with sharing the same frame of reference. The same collision can be viewed from any frame.
(2) What do you mean by "colliding frames"? I assume you mean a case where say person A (who measures things from his own frame, moving with him) is about to collide with person B (who has his own frame). Right?
(3) If two different frames view the same event (the collision) they will not necessarily agree upon the spacetime coordinates of the event.

Because, being in relative motion, even their clocks may have been synchronized earlier, they will be observing different time coordinates of the other frame, and will be unable to extrapolate their own spacetime coordinates leading to collision. Simply saying, they will not be able to predict the collision, which is bound to happen, by virtue of their different readings of the other frame of reference.

The different frames have nothing to do with each other. (Their clocks are not synchronized.) Assuming that observers in each frame have full knowledge of what's going on, they are equally capable of predicting a collision.

Using my example from (2) above: Person A and B are on a collision course with some relative speed. A sees B moving towards him; B sees A moving towards him. They both can predict the exact time and place that they will collide according to their respective distance and time measurements. (If A and B are both located at the origin of their frames, then they will agree that the collision takes place at location x = 0! )

 

[AntigenX]

(1) The fact that ... at location x = 0! )

I think you are right about everything.
Now one question, As there is no absolute time, as the OP has insistently said, does synchronizing clocks means that obsrevers are setting themselves (both) at origin of the time axis? and one more, can two clocks, moving with relative velocity be synchronized?

 

[Doc Al]

Now one question, As there is no absolute time, as the OP has insistently said, does synchronizing clocks means that obsrevers are setting themselves (both) at origin of the time axis?

Synchronizing clocks that are in the same frame (at rest with respect to each other) is certainly meaningful and doable. The "origin" of the time axis can be arbitrarily set to be when the clocks in a frame all read t = 0. (Of course, from another frame's perspective those clocks are not synchronized.)

and one more, can two clocks, moving with relative velocity be synchronized?

No. While you can arrange for two clocks moving with relative velocity to read the same at a given instant--for example, when they pass each other--that does not mean that they are synchronized.

 

[AntigenX]

No. While you can arrange for two clocks moving with relative velocity to read the same at a given instant--for example, when they pass each other--that does not mean that they are synchronized.

What does it mean to be synchronized? I mean, Synchronization means that "at the instance of synchronization the clocks should read t=0 (or any arbitrary but same value).", Then and then we will be able to calculate the time dilation in case of relative motion, no?

 

[Doc Al]

Clocks are synchronized if they always read the same time, not just at one instant.

 

[Antenna Guy]

What does it mean to be synchronized?

Sorry to interrupt, but I'd be interested in hearing if/how "proper time" might fit into this discussion.

Regards,

Bill

 

[AntigenX]

Clocks are synchronized if they always read the same time, not just at one instant.

That seems quite useless. If the synchronized clocks are always going to show the same time, why would we synchronize clocks? Let's say we do it so that, when later anyone of them is in motion, we can compare their time and deduce time dilation. In such a case, only the instance, when one of the two clocks is set moving is of importance, and no relevance is there of their history. In other words, while comparing their rates, and calculating the accumulated time by any clock, we will only be concerned with the last time when they both read same time. Does it make sense?

 

[DocZaius]

Here is a diagram I made of this scenario. I'd like a more experienced forum-goer to take a look at it and tell me if he/she agrees with it. If there is an error in it I'll take it down and/or will advertise what is wrong with it so that we might all come to a better understanding of this supposed dilemma.

 

[Doc Al]

That seems quite useless. If the synchronized clocks are always going to show the same time, why would we synchronize clocks? Let's say we do it so that, when later anyone of them is in motion, we can compare their time and deduce time dilation. In such a case, only the instance, when one of the two clocks is set moving is of importance, and no relevance is there of their history. In other words, while comparing their rates, and calculating the accumulated time by any clock, we will only be concerned with the last time when they both read same time. Does it make sense?

Sorry, I don't really know what point you are trying to make. If you'd like to continue this discussion, I suggest using a very specific example so that we don't go around in circles.

 

[Doc Al]

Here is a diagram I made of this scenario. I'd like a more experienced forum-goer to take a look at it and tell me if he/she agrees with it. If there is an error in it I'll take it down and/or will advertise what is wrong with it so that we might all come to a better understanding of this supposed dilemma.

Very good. I like it.

Perhaps AntigenX can use this specific example to make his point.

 

[AntigenX]

Sorry, I don't really know what point you are trying to make. If you'd like to continue this discussion, I suggest using a very specific example so that we don't go around in circles.

You are right.

Let me try again.
I assert, the only purpose of synchronizing two clocks is to compare their rates when they are in relative motion. To compare rates, only two points on their time axis are required. Hence, If we make to read two clocks in relative motion the same arbitrary reading, the synchronization is achieved. (we have set both clocks to origin of time axis, let their space co-ordinates be different), now while comparing them, we only need their respective times at the instance of comparison.
The moral of (my ) story is, two clocks in relative motion can be synchronized, even when their space coordinates are different.

Very good. I like it.

Perhaps AntigenX can use this specific example to make his point.

Example is quite carefully made, but involves lots of details. And I think I understand it as explained by the poster. May be I will put a more correct situation later, when I can think of it correctly.